Physics 212
Honors topics in electrodynamics
Introduction and syllabus
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003.
Introduction and syllabus
.1
2
Honors topics in electrodynamics
Physics 212
(...though it's officially PHYCS 199&HO, and it will be renumbered "PHYCS 222" next fall)
Instructor: George Gollin; Loomis 437D. 333-4451; g-gollin@uiuc.edu
Teaching assistant: To be announced
Course secretary: Donna Guzy; Loomis 441. 333-4452; d-guzy@uiuc.edu
Prerequisites: Physics 111 and associated math courses.
Corequisites: Concurrent registration in Physics 112.
Description: In Physics 212 (called PHYCS 199&HO by the registrar for the time being),
students will investigate some of the most interesting topics in electricity and magnetism at a
more sophisticated level than would be appropriate for Physics 112. We welcome all interested
students who have done well in Physics 111; it is likely that about half of Physics 212 students
will be physics majors. Enrollment is limited (by fire safety considerations) to 30 students.
Electrodynamics is a beautiful subject which has, at its heart, the fever-dream alienness of
Special Relativity. The subject offers students their first opportunity to see how profoundly
different physical reality is from our familiar (and grossly inaccurate) classical world. We will
cover a number of subjects related to those presented in Physics 112, but with a different slant:
our intention is to delve more deeply into a limited number of topics in order to supplement the
material of Physics 112. We will call upon the superior mathematical skills of our Physics 212
students to allow the course to deal with electrodynamics in a more sophisticated manner which
complements (but does not replace) Physics 112.
The format of the course comprises weekly meetings in which students will work together in
small groups (closely supervised by the instructor and teaching assistant) to solve a number of
simple problems, then to discuss the conclusions one can draw from the results. In this way,
students will derive for themselves some of the surprising features of our post-classical physical
world. For example, the simple fact that the speed of light is constant gives rise to the existence
of magnetic fields, the nature of special relativity, and the origins of electromagnetic radiation.
We will review mathematical techniques as required. Students will be expected to become
comfortable using calculus to solve problems as necessary.
Completion of each of the weekly homework sets will require an hour or two. Grades will be
based most strongly on participation in the weekly meetings and performance on problem sets.
Physics 212, Spring 2003.
Introduction and syllabus
.2
3
We want the course to be interesting, engaging, challenging, and lots of fun for both the students
and the staff. Since Physics 212 is a "topics" course, we'll get to focus on the cool stuff, while
relying on Physics 112 to provide broad coverage of the entire subject. It is likely that most
students who are capable of working at the required level for this honors course will receive high
course grades.
In more detail, we will cover:
1. Special relativity: time dilation and Lorentz contraction
2. Special relativity: non-simultaneity; the Lorentz transformations
3. A microscopic version of Gauss’ law
4. Why superposition works
5. Pinwheels (curl and the vector potential)
6. The origin of the magnetic field as a consequence of special relativity
7. Operational amplifier circuits (meet in Loomis 336)
8. Electronic models of mechanical systems (meet in Loomis 336)
9. Analog computers (meet in Loomis 336)
10. Calculating magnetic fields using the vector potential
11. Maxwell's equations
12. Visualizing fields and potentials with Mathematica (meet in Loomis 257)
13. Studying electromagnetic radiation with Mathematica (meet in Loomis 257)
14. Oops! It doesn't really work like that! (Quantum mechanics enters the picture)
15. The electrodynamics of barbecuing
We will review techniques involving differential equations and vector calculus as necessary. Do
not be alarmed by the amount of calculus in some of the units! It’ll sink in as you work with it.
Office hours: Professor Gollin: just call, or walk in. Email is also usually a reasonably quick way
to reach me.
Teaching assistant: TBA
Physics 212, Spring 2003.
Introduction and syllabus
.3
4
Physics 212, Spring 2003.
Introduction and syllabus
.4
5
Physics 212
Honors topics in electrodynamics
Unit 1
Special relativity: time dilation and
Lorentz contraction
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.1
6
Unit 1: Special relativity-- time dilation and Lorentz contraction
Introduction: the speed of light is finite, and constant.
You're going to love this: its consequences are the weirdest things you'll see in physics before
you learn about quantum mechanics.
All the strange features of special relativity come from the fact that the speed of light is exactly
2.9972458 × 108 meters per second. It doesn't matter whether the source of light moves with
respect to the measuring apparatus: any device measuring c (the speed of light) will obtain,
within experimental error, this value. Nothing else works this way: sound travels at fixed speed
with respect to the air, for example. The value 2.9972458 × 108 meters per second really is
exact: it serves as a definition, along with some time standard, of the length of one meter. The
speed of light is very close to 1 foot per nanosecond, so we'll use that as a convenient
approximation this week and next week.
Here’s what you’ll discover today:
• Events which are simultaneous in one frame of reference are not necessarily simultaneous in
another frame.
• The rate at which time passes in a moving frame of reference is slowed down.
• Moving objects become shorter along their direction of motion.
Exercise 1.1: constant c; simultaneity.
(a) Mr. Urkin, sociopathic co-proprietor of Urkin's Deli and Hardware Emporium, is capable of
throwing a snowball at 50 miles per hour. While riding as a passenger in a delivery truck
traveling 40 miles per hour, he hurls snowballs at stationary pedestrians ahead of, and behind,
Urkin's Deli
and
Hardware Emporium
40 mph
rearward victim
foward victim
the moving truck.
Calculate the speeds of the snowballs from the perspectives of the two pedestrians shown in the
diagram.
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.2
7
(b) The starship Nostromo, drifting in deep space, passes a large interstellar recreational facility
which is also adrift, as shown in the figure. Observers on the recreational facility see the
Nostromo as moving to the right, with speed 0.8c. While the ship is between the ends of the rec
facility, it fires its communications lasers to send signal pulses in the forward and aft directions
to receivers built into the ends of the rec facility.
receiver
burst of
laser light
N
Nostromo
v=0.8c
burst of
laser light
receiver
Six Flags Over τ Ceti iv
According to technicians on the recreation facility, how fast are the two light pulses traveling as
they pass through the receiver hardware built into each end of the rec facility?
Note that (slowly moving) snowballs do not behave the same way as bursts of light!
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.3
8
(c) The Sulaco, another fast ship, glides past the same drifting interstellar recreational facility.
As before, observers on the recreational facility see the ship moving to the right with speed 0.8c.
Unfortunately, Sulaco's X-band antenna strikes the rec facility's TV parabola, interrupting the
final episode of Pride and Prejudice. Sulaco's wrecked antenna is midway along the ship's
length; the rec facility's ruined parabola is also midway along the facility's length. Observers on
the recreation facility see the Sulaco's length as 600 feet. The rec facility's length, according to
the manufacturer, is 2000 feet.
S
receiver
Sulaco
boom!
v=0.8c
receiver
Six Flags Over τ Ceti iv
600 feet
2000 feet
According to technicians on the recreation facility, when does the flash of light from the
collision reach the rec facility's right- and left-receivers (assume the flash occurs at t = 0)?
According to technicians on the recreation facility, when does light from the collision reach
Sulaco's aft periscope (located at the rear of the ship)? When does light reach the forward
periscope?
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.4
9
(d) From the perspective of the Sulaco, the collision happens because the incompetently-piloted
recreational facility smashes into Sulaco’s antenna as it lumbers past at 0.8c, moving to the left,
as shown in the figure.
S
Sulaco
boom!
v=0.8c
Six Flags Over τ Ceti iv
According to cosmonauts on the Sulaco, does the flash of light from the collision reach the
Sulaco's forward and aft periscopes simultaneously?
According to cosmonauts on the Sulaco, does the flash of light from the collision reach the rec
facility's right- and left-receivers simultaneously?
Discussion
Imagine we use the receipt of the light flash as a way to synchronize clocks on the rec facility:
we set a pair of stopped clocks, one at each end of the rec facility, to t = +1000 nsec. When the
light flash arrives at the left end, we start the left clock ticking. When the light flash arrives at the
right end, we start the right clock ticking. How do the clocks look from the two frames of
reference (recreation facility's and Sulaco's)?
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.5
10
Exercise 1.2: a light clock; time dilation
(a) Mrs. Urkin, the irritable, but surprisingly fit co-proprietress of Urkin's D&HE, drifts in deep
space aboard the gigantic starship Vulcan Flashrocket. Since she realizes that the speed of light
is approximately one foot per nanosecond, Mrs. Urkin constructs a clock using two mirrors,
separated by a distance of 500 million feet, and a bouncing light pulse. The pulse is of brief
duration-- the laser that produced it fired for only a nanosecond-- and it bounces between the
mirrors without attenuation.
500 million feet
Urkin's Deli
and
Hardware Emporium
p ath of
bouncing light
p ulse
Vulcan Flashrocket
Bombs
R Us
Free delivery
As shown in the figure, she holds the lower mirror in one hand, moving it briefly into place to
reflect the descending light beam. After each successful reflection she removes the mirror.
Because Mrs. Urkin knows her resting pulse rate, she has constructed the device so that she will
need to move the mirror into position immediately after each heart beat. It is fortunate that she
does not suffer from cardiac arrhythmia since the bomb stored below the light clock will
detonate if its optical sensor is struck by the light beam.
What is Mrs. Urkin's pulse rate?
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.6
11
(b) The Flashrocket is passed by that hazardous interstellar recreational facility while Mrs. Urkin
is occupied with her light clock. Fortunately, no collision takes place. From the perspective (the
"frame of reference") of technicians on the rec facility, the Flashrocket is seen to glide past the
facility at speed v, as shown in the figure below.
Urkin 's Deli
a nd
Hard ware Emporiu m
Vulca n
Flashro cke
v (=0.8c)
Free delivery
Six Flags Over τ Ceti iv
As seen from the rec facility's perspective, does Mrs. Urkin's bomb detonate?
500 million feet
(c) As seen from the recreational facility, Mrs. Urkin's light pulse travels along a diagonal path,
as shown below. Keep in mind that the Flashrocket is seen to move with speed v, and that the
light pulse is seen to travel one foot per nanosecond!
mirror assembly
moves with speed v
c
path of
light pulse
How long is the path followed by the light pulse during one lower-mirror-to-upper-mirror-tolower-mirror trip? (Express your answer in terms of v, c, and so forth.) If v = 0.8c, how long
does it take for the light pulse to make the bottom-to-top-to-bottom trip?
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.7
12
(d) If we define ∆t to be the time required for one bottom-to-top-to-bottom light pulse cycle in
Mrs. Urkin's frame of reference, and ∆t' to be the time required for one cycle according to
observers on the recreational facility, what is the ratio ∆t' / ∆t? Express your result in terms of v
and c.
(e) What is Mrs. Urkin's pulse rate, according to observers on the recreational facility?
Discussion
• To Mrs. Urkin, everything seems fine. To observers in the other frame, Mrs. Urkin is doing
everything slowly.
• This MUST be true if something (light) exists which has the same velocity in all frames
regardless of any source-observer relative motion.
• All systems used to measure time are affected the same way (heart beat, light clocks,
mechanical clocks,...)
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.8
13
Exercise 1.3: measuring the length of a moving object using a single clock.
(a) Floating comfortably aboard the Vulcan Flashrocket, Mrs. Urkin measures the length of the
Nostromo by timing how long it takes for the Nostromo to coast past her position, moving to the
right at speed v.
Here's how Mrs. Urkin's length measurement looks to the crew of the Nostromo: they see the
Flashrocket gliding past with speed v and, naturally, see that Mrs. Urkin's clock is ticking
slowly. The Nostromo crew can measure the length of their own ship (let's call it L) by seeing
how long it takes the Flashrocket to fly the full length of the Nostromo. (Let's call the time
required ∆t).
L
N
∆t
v = 0.8c
Free delivery
Flashrocket
Nostromo
Nostromo
∆t′
Flashrocket clock stops
when its clock lever strikes
Nostromo's aft probe
Urkin's Deli
and
Hardware Emporium
Flashrocket
∆t
Flashrocket clock starts
when its clock lever strikes
Nostromo's forward probe
(As seen from the Nostromo's frame of reference)
What is L in terms of ∆t and so on? (L is the Nostromo's rest length. This is a ridiculously easy
problem!)
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.9
14
(b) In Mrs. Urkin’s rest frame, the measurement looks as shown in the figure. Her clock is
started and stopped by the probes mounted near Nostromo’s nose and tail. Naturally, she sees
that Nostromo’s clocks (not shown) are ticking slowly, and are not synchronized. (Note that
Nostromo’s aft clock must still read ∆t, however, as it passes Mrs. Urkin.)
L´
N
∆t
Nostromo
Flashrocket clock stops
when Nostromo's aft probe
strikes clock lever
Free delivery
Flashrocket
v = 0.8c
∆t´
Urkin's Deli
and
Hardware Emporium
?
Flashrocket clock starts
when Nostromo's forward
probe strikes clock lever
(As seen from the Flashrocket's frame of reference)
Define L′ to be the Nostromo length measured by Mrs. Urkin; the time interval which passed on
the Flashrocket clock is ∆t′. In terms of v and ∆t′, what is the value she obtains for L′? (Clearly,
we have another difficult problem here!)
(c) What is the ratio L' / L, expressed only in terms of v and c? (Lorentz contraction!) Keep in
mind that ∆t′ < ∆t.
Discussion
All methods of performing a length measurement (tape measure, timing how long it takes a
moving object to coast past a fixed point,...) will yield consistent results in one frame of
reference.
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.10
15
Exercise 1.4: is there any “transverse” contraction?
Lengths parallel to the direction of motion become shorter as
v → c. What happens to lengths perpendicular to the
direction of motion?
Consider a long stick which carries a pair of paint balls,
spaced along the edge of the stick which is perpendicular to
its velocity. The stick crashes into a stationary ruler, as
shown in the figure.
1
2
How far apart are the paint marks which are left on the ruler?
3
paint balls on moving stick
Imagine that a second observer views the same process (the
very same ruler-stick collision shown in the above diagram),
but from a frame in which the stick is at rest and the ruler is
in motion.
1
How far apart will this observer think the paint marks are?
2
What conclusion can you draw concerning contraction
perpendicular to the direction of motion from this system?
3
paint balls on stationary stick
Summary and wrap-up discussion
•
Time is messed up:
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.11
16
1. moving clocks take longer per tick by a factor of 1
•
•
•
•
(
)
1 − v2 c2 .
2. simultaneity doesn't work: clocks "at the front" read earlier times than clocks "at the
back"
All time references are affected identically: light clocks, heartbeat,...
Conclusions are valid if anything (light, neutrinos, gravitons,...) is found to move with
constant velocity, independent of source-observer motion.
Some things don't change when one switches frames of reference (e.g. bomb did/didn't
explode).
(
)
Length is messed up: moving objects are shorter by a factor of 1 − v 2 c 2 , but only along
their direction of motion.
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.12
17
Reading and homework should be completed by this Friday, 4pm. Please leave your
finished problem set in the TA’s mailbox by that time. We are available for help and
questions as they arise; please call (let the phone ring until a secretary picks it up if I am not in
my office), drop in, or send us email.
Required reading:
1. Introduction and syllabus handout.
2. Notes on Special Relativity, pp. 1 – 23 as necessary to clarify points that remain obscure after
class.
Optional reading, just for fun:
Einstein’s Dreams, Alan Lightman. Warner Books (paperback), 179 pages, 1994. Comments
from Amazon.com: "The book takes flight when Einstein takes to his bed and we share his
dreams, 30 little fables about places where time behaves quite differently. In one world, time is
circular; in another a man is occasionally plucked from the present and deposited in the past: "He
is agonized. For if he makes the slightest alteration in anything, he may destroy the future ... he
is forced to witness events without being part of them ... an inert gas, a ghost ... an exile of time."
Einstein's Bridge, John Cramer. Avon Books (paperback), 1998. A high energy physics
experiment at the SSC lab in Texas opens a channel into another universe. There are good things
and bad things lurking out there, and both find their way through the portal. I suppose this really
is about a quantum version of General Relativity, rather than Special Relativity, but it was really
fun to read.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises and write a
(true!) statement on your problem set asserting that you have successfully printed your personal
copy.
Problem 1 (10 points): KS0 meson decay
Neutral K mesons (“kaons”) are unstable particles composed of a quark and an antiquark. They
can be produced copiously in energetic collisions between stable particles. The neutral kaon
mass is roughly 1/2 the proton mass; short-lived neutral kaons usually decay into pairs of
(lighter) pi-mesons through the decays KS0 → π+π- and KS0 → π0π0. The average KS0 lifetime
for a kaon at rest is 0.8935 × 10-10 seconds.
A beam of fast KS0 is produced at a national laboratory. The average distance traveled by a kaon
before decaying in flight is found to be approximately 9.2 cm. How fast are the kaons moving?
(They are not traveling faster than c!!) You should assume all kaons are moving at the same
speed v and use the value c = 3 × 1010 cm/sec in your calculations.
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.13
18
Problem 2 (10 points): The space telescope sees clearly
Two photons from a pair of faint, distant objects simultaneously enter the open end of the
Hubble Space Telescope’s barrel. One of the objects is moving at 0.5 c directly towards the
telescope, while the other is moving at 0.5 c directly away from the telescope. Assuming that the
distance from the telescope’s open end to its primary mirror is 20 feet, calculate how long it
takes for each of the two photons to arrive at the primary mirror after entering the open end of
the telescope barrel (as determined by observers in the rest frame of the telescope).
Problem 3 (10 points): The Minoans synchronize their watches
The identical sister ships Linear A and Linear B pass each other in deep space, as shown from
the perspective of the Linear A’s crew in the figure below. In its rest frame, each ship is 2000
feet long; the crew of Linear A sees the other ship glide by with speed 0.8 c. As the midpoint of
Linear B passes the midpoint of Linear A, parabolic antennas on the two ships collide. The brief
flash of light from the collision travels outwards from the point of contact.
v=0.8c
?
?
Linear B
L-B
boom!
Linear A
L-A
2000 feet
(a) According to the crew of Linear A, how long is the ship Linear B?
(b) According to the crew of Linear B, how long is the ship Linear A?
(c) Both ships carry (initially stopped) clocks at their forward and aft ends. Crew members
stationed next to each of the clocks start them ticking as soon as they see the light from the
explosion. After all clocks have finally begun ticking, how do the clocks on Linear B look to the
crew on Linear A? If you conclude that the Linear A crew believes them to be out-of-synch,
which Linear B clock reads the later time according to Linear A observers?
Physics 212, Spring 2003. Unit 1
Special relativity: time dilation and Lorentz contraction
1.14
19
Physics 212
Honors topics in electrodynamics
Unit 2
Special relativity: Non-simultaneity; the
Lorentz transformations
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.1
20
Unit 2: Special relativity-- Non-simultaneity; the Lorentz transformations
Recap of last week
Introduction: reference frames and measurements.
The only quantitative features of special relativity we've seen so far are the rates at which
moving clocks slow down and moving objects shrink. If we're clever, we can use this to
bootstrap our way into drawing other quantitative conclusions concerning measurements such as
the non-simultaneity of events which are simultaneous in a different reference frame.
What do we mean by "reference frame?" The idea is simple: it's just a coordinate system in
which all our clocks and rulers are at rest, and all our clocks are synchronized. A measurement
of a spatial interval between two events amounts to determining the distance between the two
clocks placed next to the events. The time interval is just the difference in the readings on the
faces of the two clocks. For example, the space-time interval between waking up and entering
my office involves a distance of about 2 miles and a time interval that depends on how sleepy I
am.
Synchronizing clocks in a reference frame is easy: we preset all of them so that a clock which is
R feet from the origin reads +R nanoseconds. Each clock (initially paused) starts running when it
senses the light from a strobe, placed at the origin, which fires at time zero.
Keep in mind that most measurements involve both position and time: measuring the length of a
moving object as it glides past our (stationary) ruler requires us to see where the two ends of the
object are, in comparison to the scale printed on our ruler, at the same time. Timing how long a
moving object needs to coast past our stopwatch requires that the stopwatch remain in the same
place during the timed interval.
We'll continue to use the approximation c = 1 foot per nanosecond this week.
Here’s what you’ll find/work on today:
• A quantitative expression for the non-simultaneity of events viewed from one frame when
the events occur simultaneously in another frame
• Putting it all together: deriving the Lorentz transformations
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.2
21
Exercise 2.1: quantitative description of non-simultaneity.
We know from last week that clocks at the front of a moving object read earlier times than
clocks at the back of the object. Let's work this up, quantitatively, now.
Here's the setup: a flash is produced at the tail of the Nostromo as its rear clock reads zero. The
light travels the length of the ship, arriving at a sensor at its nose. We can describe the
production and reception of the light in terms of a pair of events (an event is something which
happens at one point in space and lasts only an instant). The first "event" is the production of the
spark near the Nostromo's aft clock. The second event is the arrival at the forward sensor of light
from the spark.
event 1
N
t=0
Nostromo
t=0
Nostromo
t = L/c
flash of light
rest length L
event 2
N
t = L/c
flash of light
Since Nostromo's rest length is L, and all its clocks are synchronized in its rest frame, we know
that the light will arrive at the forward clock when that clock reads L/c. Perhaps the light from
the arriving flash is used to expose some photographic film to record permanently the reading on
the forward clock when the light arrives. There is a well-defined spatial and temporal interval
between the two events in this reference frame (∆x = L and ∆t = L/c).
From the perspective of observers who see Nostromo moving to the right, the ship's length is
Lorentz contracted to L 1 − ( v c ) . Nostromo's forward clock is seen to show an earlier time
2
than its aft clock. (Let's call this difference δ.) The two events look like this:
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.3
22
L 1 − ( v c)
event 1
2
N
Nostromo
t=0
t = -δ
v
flash of light
0
Six Flags Over τ Ceti iv
L 1 − ( v c)
event 2
0
2
N
t = L/c + δ
vt'
Nostromo
t = L/c
flash of light
ct'
t'
Six Flags Over τ Ceti iv
t'
Event 1: spark is produced at the aft end of Nostromo. The Nostromo's tail clock reads 0; the
nose clock -δ. (We already know that δ is a positive number, since the forward clock reads an
earlier time than the aft clock.) Since we are viewing events from the rest frame of the observers
on the rec facility, all clocks on the recreation facility read zero.
Event 2: light from spark reaches forward end of Nostromo. From the photograph taken in the
Nostromo rest frame, we know that Nostromo's forward clock will read L/c when the light
arrives. The tail clock always reads δ later than the nose clock, according to rec facility
observers. The rec facility clocks all read t'. In time t', the light from the spark has traveled a
distance ct' to catch up with the nose of Nostromo, while the Nostromo itself has moved a
distance vt'.
There is a well-defined spatial and temporal interval between the two events in this reference
frame. (∆x' = ct' and ∆t' = t' -- note that I'm using coordinates with primes to describe variables
in the rec facility frame.).
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.4
23
(a) Note that (in the rec facility rest frame) the light beam travels a distance ∆x' = c∆t' which is
the same as the sum of the Lorentz-contracted Nostromo length and the distance traveled by the
Nostromo.
Write an equation which says the same thing as the previous sentence, and then use this to solve
for ∆t' in terms of L, v, and c.
(b) When event 2 occurs in the rec facility rest frame, Nostromo’s forward clock must read L/c
and its aft clock must still be ahead of the forward clock by an amount δ. As a result, the reading
on Nostromo’s slowly-ticking aft clock has advanced from 0 to L/c + δ during the time it took
for a rec facility clock to advance (in the rec facilty rest frame) from 0 to ∆t'.
Since the moving Nostromo clocks are ticking slowly in comparison to the rec facility’s clocks,
we must have L c + δ = ∆t ′ 1 − v 2 c 2 . Use this fact to derive an expression for δ in terms of L,
v, and c. Please simplify your expression as much as possible.
Discussion
How do the rec facility clocks look to the Nostromo crew?
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.5
24
Exercise 2.2: Deriving the Lorentz transformations
Observers in different frames of reference which are in relative motion will tend to disagree
about lengths, distances, time intervals, and the synchronization of clocks.
Here’s a summary of what we know so far:
1.
A single, moving clock ticks slowly so that a time interval
∆t between two events measured by this one clock will be shorter than the time interval
∆t' measured by observers who see that this clock is in motion: ∆t = ∆t ′ 1 − v 2 c 2 .
2.
The length L' of a moving object is shorter than the length
L of the object as measured in the object’s rest frame: L 1 − v 2 c 2 = L′ .
3.
Clocks which are separated by a distance ∆x in their rest
frame and are synchronized in their rest frame are seen to be out-of-synch in a frame in
which the clocks are moving with speed v by an amount −v∆x c 2 . (The negative sign
means that the forward clock reads an earlier time than the aft clock.)
Imagine that a pair of events is seen by observers in the two different coordinate systems O and
O'. Observers in O (the "unprimed" frame) measure the space-ime intervals between the events
to be ∆x, ∆y, ∆z, ∆t while observers in O' measure the interval to be ∆x′, ∆y′, ∆z′, ∆t′.
(I’m defining ∆x = x2 – x1, ∆t = t2 – t1, and so forth.)
Let’s assume that the origins of O and O' coincide when clocks at the origins of both frames
read zero. Frame O' (according to observers in O) is moving with velocity v along the x axis of
O as shown in the figure. (Note that the sign of v matters.)
O
O'
y
y'
v
x
z
x'
event 2
(x2, y2, z2, t2) or (x2', y2', z2', t2')
event 1
(x1, y1, z1, t1) or (x1', y1', z1', t1')
z'
The Lorentz transformations allow us to calculate the space-time interval between the two events
measured by observers in one frame if we know the interval between the events measured by
observers in the other frame and also know the relative velocities of the two frames. They allow
us to write ∆x′ as a function of ∆x, ∆t, v, and c and also ∆t′ as a (different) function of ∆x, ∆t, v,
and c.
(a) Let’s redraw the figure used in the last exercise in order to make the choice of primed and
unprimed coordinates agree with the discussion in this exercise. Here’s the redrawn figure:
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.6
25
event 1
N
Nostromo
t′ = 0
v
flash of light
t=0
Six Flags Over τ Ceti iv
t=0
∆x′ = rest length
event 2
N
Nostromo
v∆t
t′ = ∆t′
flash of light
∆x = c∆t
t = ∆t
Six Flags Over τ Ceti iv
t = ∆t
In the rec facility frame the light beam travels a distance which is the same as the sum of the
Lorentz-contracted Nostromo length and the distance traveled by the Nostromo. The equation
which says the same thing as this is ∆x = ∆x′ 1 − v 2 c 2 + v∆t .
Solve this equation for ∆x' as a function of ∆x and ∆t.
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.7
26
(b) Since the two events are separated by ∆x′ and ∆t′ in the Nostromo rest frame, we know that
∆t ′= ∆x′ c . We also know that ∆t = ∆x c and that ∆t c = ∆x c 2 .
Use these facts to rewrite your answer to the previous exercise as an equation for ∆t', and
simplify it as much as possible to express ∆t' as a function of ∆t and ∆x.
Discussion
You have derived the Lorentz transformations! The frames O and O' correspond to the rest
frames of the rec facility and the Nostromo respectively; the space-time separation of any pair of
events measured by observers in the unprimed frame can be used to derived the space-time
interval between the events that would be measured by observer in the primed frame. Just to
make sure we have them written down compactly: define: γ = 1 1 − v 2 c 2 so that
∆x′ = γ ( ∆x − v∆t )
∆x = γ ( ∆x′ + v∆t ′ )
∆y′ = ∆y
∆z′ = ∆z
(
∆t ′ = γ ∆t − v∆x c 2
)
(
)
∆t = γ ∆t ′ + v∆x′ c 2 .
If we know the intervals in one frame, we can calculate what they'll be in the other frame. Keep
in mind that positive v means that O' is moving to the right, as viewed from O.
Note that we can extract all of the quantities we’ve calculated so far using the Lorentz
transformations as long as we choose the right pair of events. Keep in mind that almost all
measurements require you to know something about space and time. For example, a time
interval measured using a single clock is done so that the spatial separation between the
“starting” event and “stopping” event used to define the time interval is zero.
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.8
27
Summary and wrap-up discussion
•
Time AND space are messed up.
1.
rate of passage of time is slowed in a moving frame:
(
∆t = ∆t ′ 1 − v 2 c 2
•
•
)
2.
simultaneity doesn't work: clocks "at the front" read earlier
by v∆x / c than clocks "at the back," where ∆x is the separation between the clocks as
measured in their own rest frame.
3.
moving objects are shorter: L ′ = L 1 − v 2 c 2 where L is
2
(
)
the rest length of the object.
The Lorentz transformations can be used to calculate the intervals ∆x, ∆y, ∆z, ∆t between
events observed from one reference frame using the intervals ∆x', ∆y', ∆z', ∆t' between the
same events observed from another reference frame.
v → c.
Comments about homework assigned for next week
Please do not ever ever ever search for solutions to assigned problems in texts, lecture notes or
problem set solutions from other courses, or solutions distributed in previous offerings of this
course! Scout’s honor!
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.9
28
Reading and homework should be completed by 4pm Friday.
Required reading:
Notes on Special Relativity, pages as appropriate to our work.
Optional reading, just for fun:
The Feynman Lectures on Physics (volume I), chapter 15.
The Forever War, Joe Haldeman. Avon books (paperback), 1974. This science fiction novel won
both the Hugo and Nebula awards. Time dilation plays a role in the messes the characters make
of their lives.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): Can anything go faster than c?
In this problem you'll derive the relativistic velocity addition formula. To replenish the
Nostromo's perilously low supply of jell-o, Mrs. Urkin's fast shuttle is launched on a resupply
mission. As seen from Nostromo, the shuttle streaks away with constant velocity u in the
positive x' direction. At position x1' and time t1' the shuttle strikes a dust mote, vaporizing it.
Somewhat later, the shuttle hits a micrometeoroid.
N
Nostromo
U
event 1: Mrs. Urkin
clobbers a dust mote
u
∆x', ∆t'
(Nostromo rest frame)
N
Nostromo
event 2: Mrs. Urkin strikes
a micrometeoroid
U
u
The light flashes from both events are seen by Nostromo's cook, who calculates (after correcting
for the light's propagation time to the ship) that the space-time interval between the events was
(∆x', ∆t') = (x2' - x1', t2' - t1'). Not surprisingly, the cook notices that u = ∆x' / ∆t'.
The same events are viewed by Mrs. Urkin's personal trainer on board the interstellar recreation
facility. He sees the Nostromo travelling with speed v in the positive x direction, and determines
that the space-time interval between the two events was (∆x, ∆t) = (x2 - x1, t2 - t1). He concludes
that the shuttle craft was moving in the positive x direction with speed ∆x / ∆t.
(a) By making use of the Lorentz transformations, calculate ∆x / ∆t in terms of u (the shuttle's
velocity as seen from Nostromo), v (Nostromo's velocity, as seen from the rec facility), and c.
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.10
29
(b) How does your expression for ∆x / ∆t behave in the limit that both u and v are close to c?
Problem 2 (10 points): The train-tunnel "paradox"
Here's how this one works. We have a tunnel, 800 feet long in its rest frame, with doors on each
end which can be used to seal the tunnel. The train is 1,000 feet long in its own rest frame, as
shown in the illustration:
Tunnel rest frame:
The train travels at speed 0.8 c so that its Lorentz-contracted length allows it to fit entirely inside
the tunnel. When (synchronized in their rest frame) tunnel clocks by both doors read zero, just as
the train is neatly centered inside the tunnel, the doors slam shut, trapping the entire train inside
the tunnel. The front of the train crashes through the right-side tunnel door 125 nanoseconds
later, but the (closed) left-side door was able to close without interfering with the train.
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.11
30
Train rest frame:
From the train's rest frame, things look rather different: the Lorentz-contracted tunnel rushes
towards it as shown in the following figure.
The tunnel is much too short to be able to trap the train entirely inside itself, yet both doors will
drop, and neither will touch the train (although the front of the train will crash through the
already-closed right-side door).
How can this be possible?
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.12
31
Problem 3 (10 points): An introduction to partial derivatives
We will begin using calculus to address problems next unit. Here’s a reminder about the
definition of an ordinary derivative, and an introduction to partial derivatives.
(a) The Urkins are hiking in eastern France’s Jura mountains. They have been told by residents
of a nearby village that the altitude above sea level is well-described by the function
h(x,y) = 10-7 x3, where x and y are the distances due east and due north (in meters) of the center
of town. A plot of the elevation near the town would therefore look something like this:
altitude is a function only of x
village
100
50
h(x,y) = 10-7 x3
1000
0
500
-50
-100
0
y
-500
-500
0
x
500
-1000
1000.
(i) The Urkins hike due east from the point (x, y) one morning, stopping for lunch at the point
(x + a, y). Write an algebraic expression for their change in altitude, ∆h = h ( x + a, y ) − h ( x, y )
and for the average slope of their path, m = ∆h a . Please simplify your result as much as
possible, expressing it as a function of x and a.
(ii) Evaluate the average slope in the limit that a → 0 , expressing your answer as a function of x.
(Since the definition of the derivative of a function of one variable is
df ( x ) dx = lim  f ( x + ∆x ) − f ( x )  ∆x , your answer shouldn’t come as a surprise to you!)
∆x → 0
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.13
32
(b) The next day the Urkins are hiking in a region where the altitude is well-described by the
function h(x,y) = 10-10 x3y, where x and y are the distances due east and due north of the center of
a different town. An elevation plot would therefore look something like the following:
altitude as a function of x,y is 10-10 x3y
100
50
altitude
0
500
-50
-100
0
y
-500
-500
0
x
500
(i) The Urkins hike due east from the point (x, y) to the point (x + a, y). Write an algebraic
expression for their change in altitude, ∆h = h ( x + a, y ) − h ( x, y ) and for the average slope of
their path, m = ∆h a . Simplify your result, expressing it as a function of x, y, and a.
(ii) Evaluate the average slope in the limit that a → 0 as a function of x and y.
You have just evaluated a partial derivative! The definition of a partial derivative of a function of
several variables is quite similar to the definition of the derivative of a function of one variable:
∂f ( x, y )
f ( x + ∆x , y ) − f ( x , y )
∂f ( x, y )
f ( x , y + ∆y ) − f ( x , y )
= lim
and
= lim
.
∆x → 0
∆y →0
∂x
∆x
∂y
∆y
y constant
x constant
The machinery is easy to use: just treat all “the other” variables as if they were constants and
take the derivative of the function in the usual way.
Physics 212, Spring 2003. Unit 2
Special relativity: Non-simultaneity; the Lorentz transformations
2.14
33
Physics 212
Honors topics in electrodynamics
Unit 3
A microscopic version of Gauss’ law
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.1
34
Unit 3: A microscopic version of Gauss’ law
Recap of last week
Introduction: the connection between Coulomb’s and Gauss’ laws
The Physics 112 discussion of the electric field and electrostatic potential describes them as
quantities which depend on the global distribution of electric charge:
V (r ) =
E (r ) =
Qi
⇒
all charges 4πε 0 r − ri
∑
1
∑
1
Qi
all charges 4πε 0 r − ri
2
ρ ( r ′)
1
∫
4πε 0 r − r ′
all space
r − ri
r − ri
⇒
∫
1
d3r ′
ρ ( r ′) r − r ′
4πε 0 r − r ′ 2 r − r ′
all space
d3r ′ .
By “global” I mean that you’ll need to know where all the charge is in order to calculate the
sums or integrals. Don’t be alarmed by the volume integrals—an integral is just a gigantic sum,
which is probably why an integral sign is meant to resemble a large letter “S.”
It is possible to craft equations which relate characteristics of the fields (or potentials) to the
charge distribution using only on local properties of the charge density. (You’ll do that today.) In
particular, knowing the charge density at a single point in space lets us know about the spatial
derivatives of the electric field at that same point in space.
The electric field falls off like 1/r2 while the area of a surface surrounding a finite charge
distribution grows like r2. One consequence of this is Gauss’ law: the net flux of electric field
through any closed surface depends only on the amount of charge inside the surface. This is
another global statement since knowledge of the field over an extended surface is necessary to
calculate the flux in order to establish the connection between the electric field and the
(extended) charge distribution.
Again, it is possible to concoct a local-properties-only formulation of Gauss’ law.
Naturally, the electric field is the force per Coulomb felt by a point test charge placed
somewhere, while the electric potential is the test charge’s potential energy per Coulomb.
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.2
35
Here’s what you’ll find/work on today:
• Verifying that Gauss’ law is true even when the electric field is not perpendicular to the
Gaussian surface enclosing a point charge
• A microscopic (differential) version of Gauss’ law which depends only on local properties of
the electric field and charge distribution.
• Another differential equation (Poisson’s equation) which relates the potential to the charge
distribution.
Exercise 3.1: Coulomb ⇒ Gauss in cylindrical geometry
Gauss’ law describes the connection between the flux of electric field lines passing through a
closed surface and the net charge contained inside the surface. The idea is relatively simple:
electric field lines can only begin on positive charges and can only end on negative charges. If
there are no charges inside a closed surface, any field line entering the surface must also exit. If
the surface does enclose a net positive charge, more field lines will leave it than will enter.
Naturally, the situation is reversed if the surface encloses a net negative charge.
The flux of electric field lines passing through a small area patch ∆A depends on how strong the
field is, how large the area of the patch is, and how close the electric field lines are to being
perpendicular to the surface of the patch. In symbols, ∆Φ = E i ∆A = E ∆A cosθ , where the
direction associated with the area patch is along the (outward) normal to the surface of the patch.
Summing the flux of electric field over a closed surface involves a sum (an integral) over all area
patches which comprise the surface:
Φ=
∑
surface
∆Φ =
∑
surface
E i ∆A 
→
∆A→dA
∫
surface
dΦ =
∫
E idA .
surface
The formal statement of Gauss’ law is
∫
surface
E idA =
Qenclosed
ε0
.
It’s easy to evaluate this integral when the surface is a sphere of radius R which is centered on a
point charge Q: the integral is just the product of the strength of the electric field at a distance R
and the surface area of a sphere, since all the cosines in the dot product are unity:
Q
Q
E idA = E × 4π R 2 =
× 4π R 2 = .
2
∫
4πε 0 R
ε0
surface
The dot product inside the integral is important: it is what makes Gauss’ law true regardless of
the shape of the bounding surface used in the integral.
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.3
36
To convince yourself (or at least to make it seem more plausible) that the shape doesn’t matter,
consider an infinitely long cylindrical surface of radius a surrounding a point charge Q as in the
following figure.
φ
dA
x
a
E
θ
z
Q
y
The charge is at the origin, so the distance to an area patch on the surface of the cylinder at
(r, z, φ) = (a, z, φ) is (a2 + z2)1/2. In addition, the cosine of the angle between the electric field and
the normal to the area patch is cos(θ) = a / (a2 + z2)1/2. An area patch of extent dφ, dz has area
dA = a dφ dz.
For this surface, evaluate the electric field flux through the cylinder, namely
∫
E idA .
surface
Don’t get spooked by this: a surface integral involves taking a double integral, which can be
written this way:


a
E idA = ∫ E 
dA
∫
2
2 
 a +z 
surface
surface
=



Q
a


∫ 4πε a 2 + z 2  a 2 + z 2  dA

surface 
0
(
=∫
+∞
−∞
.
)
 2π 



Q
a
∫ 
 dz.
φ
a
d

 0  4πε 0 a 2 + z 2   a 2 + z 2 



(
)
E
cos θ
dA
(A correct calculation will yield the answer Q/ε0. You will probably find the trig substitution
z = a⋅tan(α), in combination with the identity 1 + tan2(α) = sec2(α) = 1/cos2(α), to be helpful.)
Discussion
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.4
37
Exercise 3.2: A microscopic version of Gauss’ law
If a surface encloses a distribution of charge with density ρ ( r ) C/m3, Gauss’ law can be written
Q
1
∫ E idA = enclosed = ∫ ρ ( r ) dV
ε0
surface
ε 0 volume
since the amount of charge contained inside a small volume element dV is ρdV and the integral
just sums up the charge contained by all volume elements inside the bounding surface.
Let’s develop a microscopic version of this. Consider a small volume which is a parallelepiped
centered at (x,y,z) with side lengths dx, dy, dz as shown in the figure. Assume that ρ and E vary
slowly over the size of the volume so that the charge inside the volume is .
y
dz
dy
x
dx
z
If the volume is small enough, the surface integral can be rewritten as a sum over the faces of the
box so that Gauss’ law becomes
i =6
ρ ( x, y, z ) dxdydz
.
∑ Ei idAi =
i =1
ε0
(a) Write an explicit expression for each of the six area patch vectors dAi in terms of the unit
vectors x̂ , ŷ , ẑ and the volume’s side lengths dx, dy, dz.
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.5
38
(b) The electric field at the center of one particular face of the box can be written as
dx
dx
dx
dx








E  x − , y, z  = Ex  x − , y, z  xˆ + E y  x − , y, z  yˆ + Ez  x − , y, z  zˆ








2
2
2
2
while the field at the center of the opposite face is
dx
dx
dx
dx








E  x + , y, z  = Ex  x + , y, z  xˆ + E y  x + , y, z  yˆ + Ez  x + , y, z  zˆ .








2
2
2
2
There are similar expressions for the electric field at the other four faces of the box.
Use this fact, and the definition of a partial derivative, to rewrite Gauss’ law for a vanishingly
small volume entirely in terms of ρ and partial derivatives of the components of the electric field.
(You should obtain
∂E x ( x, y , z ) ∂E y ( x , y , z ) ∂E z ( x, y , z ) ρ ( x , y , z )
+
+
=
.)
ε0
∂x
∂y
∂z
In Cartesian coordinates, the definition of the operator ∇ (“del”) is ∇ ≡ xˆ
∂
∂
∂
+ yˆ
+ zˆ
so
∂x
∂y
∂z
that the divergence of any vector field G can be written
∇iG =
∂G x ( x , y , z ) ∂G y ( x, y , z ) ∂G z ( x , y , z )
+
+
.
∂x
∂y
∂z
As a result, the microscopic (differential) version of Gauss’ law can be written compactly as
∇i E =
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
ρ
.
ε0
3.6
39
Discussion: divergence (and the divergence theorem)
• A net “outflow” of electric field lines automatically means that the electric field has positive
divergence. A net “inflow” means the field has negative divergence.
• The divergence theorem (true for all vector fields): use ∇i E =
∫
E idA =
surface
ρ
to rewrite Gauss’ law as
ε0
ρ (r )
∫ ε 0 dV = volume
∫ ∇i E dV :
volume
(
)
The flux of field through a closed surface is the same as the integral of the field’s divergence
over the volume enclosed by the surface. In general, for any vector field G it is true that
∫
G idA =
surface
∫
volume
( ∇iG) dV .
Exercise 3.3: Poisson’s equation
The equation ∇i E =
ρ
depends only on the local properties of the charge density and the
ε0
derivatives of the electric field.
We can work up a similar equation connecting the potential with the charge density with the help
of the work-energy theorem. (This only applies to static systems, where all free charge stays put,
and all currents are constant!) Recall that the (small) change in an object’s potential energy is
∆U = F i ∆r when it is pushed through a displacement ∆r by a force F . If you are applying the
force to overcome the effects of an electric field opposing your efforts, your force will satisfy
F = −QE . Imagine that you move the test charge through a displacement
∆r = xˆ ∆x + yˆ ∆y + zˆ ∆z in three steps, as illustrated below.
y
3. dz
2. dy
Q starts here
Q ends here
1. dx
x
z
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.7
40
The work you do will be ∆U = ∆U1 + ∆U 2 + ∆U 3 with
∆U1 = F i xˆ ∆x = Fx ∆x, ∆U 2 = F i yˆ ∆y = Fy ∆y, ∆U 3 = F i zˆ∆z = Fz ∆z .
∆U1
∂U
∂U
∂U
= Fx ,
= Fy ,
= Fz so
= Fx , with some thought it should be apparent that
∂x
∂y
∂z
∆x
∂U
∂U
∂U
+ yˆ
+ zˆ
= ∇U .
that F = xˆ
∂x
∂y
∂z
Since
Because F = −QE and U = QV , we can conclude that
E = −∇V (static systems only) .
Use this to construct Poisson’s equation, which relates various second derivatives of the
potential to the charge distribution, from the differential form of Gauss’ law.
ρ ( x, y , z )
∂ 2V ∂ 2V ∂ 2V
. The operator ∇ 2 is called the
(You should obtain ∇ V ≡ 2 + 2 + 2 = −
ε0
∂x
∂y
∂z
“Laplacian.”)
2
Discussion
Summary and wrap-up discussion
•
•
The 1/r2 nature of the electric field leads to Gauss’ law which relates the flux of electric field
through a closed surface to the net charge enclosed by the surface.
In combination with the connection between the electric field and the electrostatic potential,
the microscopic (differential) version of Gauss’ law can be used to produce Poisson’s
equation, which relates various second derivatives of the potential with the local charge
density.
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.8
41
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.9
42
Reading and homework should be completed by 4pm Friday.
Required reading:
Purcell, chapter 1, sections 7 through 10. Chapter 2, sections 1 through 5 and 7 through 12. Note
that Purcell uses CGS, instead of MKS (SI), units so there won’t be any 4πε0’s anywhere.
Read over (but don't worry about understanding) the Unit 4 material we'll tackle in class next
week.
Optional reading:
The Feynman Lectures on Physics (volume II), chapter 2, sections 1-5 and 7-8; chapter 3,
sections 1-3.
Comment on notation:
In general, I’ll use xˆ , yˆ , zˆ to represent unit vectors parallel to the x, y, and z axes.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): This is not why the universe is still expanding
Theoretical food chemists in the Urkin's D&HE Research Division are completing an analysis of
a hypothetical universe filled with electrically charged mayonnaise. In this alternative reality the
electrostatic potential created by the mayo is found to be
V ( x, y, z ) = V0 ( sin kx + cos ky )
where k is a constant. (Note that the potential depends only on x and y.)
(a) Calculate the electric field E ( x, y, z ) produced by mayonnaise.
(b) Calculate the net charge density in Coulombs per cubic meter ρ ( x, y, z ) responsible for
creating this potential.
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.10
43
Problem 2 (10 points): At the surface of the FlashOven
The electric field inside, and at the external surface of Urkin’s Vulcan FlashOven is accurately
described as E ( x, y, z ) = zˆ E0 x 2 yz. The oven is a rectangular box whose sides are bounded by
the planes x = 0, x = a, y = 0, y = b, z = 0, z = c .
(a) Evaluate the integral
∫
E ⋅ dA over the surface of the oven.
surface
( dA points outwards; the six faces have dA = ± xˆ dydz, ± yˆ dzdx, and ± zˆ dxdy. )
(b) Evaluate the integral
∫
∇ ⋅ E dV over the interior volume of the oven.
volume
(c) How might the divergence theorem helped you in part (b)?
Problem 3 (10 points): The point is…
Anthrax, the Urkin’s pet cat, is resting comfortably at the origin after eating an electrically
charged canary. He exudes the (radial) electric field
E ( x, y , z ) =
x xˆ + y yˆ + z zˆ
(x
2
+ y2 + z2 )
3
2
where xˆ , yˆ , zˆ are unit vectors parallel to the x, y, and z axes. (Assume this expression is accurate
for all points outside the cat.)
Prove that the density of electric charge ρ ( x, y, z ) is zero for all points (x, y, z) external to the
cat.
Physics 212, Spring 2003. Unit 3
A microscopic version of Gauss’ law
3.11
44
Physics 212
Honors topics in electrodynamics
Unit 4
Why superposition works; Green’s
functions
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.1
45
Unit 4: Why superposition works; Green’s functions
Recap of last week
Introduction: superposition
One of the most important/useful/convenient/fundamental properties of the equations which
characterize electric and magnetic fields is that all the fields and potentials obey the
superposition principle. Imagine that we are able to solve for the electric field (call it E1 ) in a
universe which contains nothing but charge Q1 at position r1 and also that we can solve for E2 in
a (different) universe which contains nothing but charge Q2 at r2 . Because of superposition we
can be sure that the electric field in a universe which contains both charges will be the sum of the
fields we found doing our pair of lone-charge-universe field calculations:
Eboth charges present = E1 + E2 .
Superposition allows us to break a complicated problem up into a number of simple problems
which we can solve, then to sum the simple problems’ solutions to generate the solution to the
complicated problem.
In the two-charge example mentioned above Q1 and Q2 are sources of the electric field.
Naturally, they don’t contribute equally to the electric field everywhere: the closer an observer is
to one of the charges, the more important that charge’s contribution to the measured electric field
will be. If we placed unit charges at r1 and r2 we would need to weight their contributions to the
(
field at r by 1 4πε 0 r − ri
2
) . (For sources which are different from one Coulomb we will also
need to scale their contributions according to how much charge each particle carries.)
(
The unit-charge weighting function 1 4πε 0 r − ri
2
) is a Green’s function (named for the British
miller George Green (1793-1841), whose hobby was mathematics). A Green’s function is a
“helper” function which tells us how important each source’s contribution is when we superpose
the effects (fields) of all the individual sources.
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.2
46
Discussion: superposition and Poisson’s equation
Last week we cooked up a pair of powerful differential equations which can be used (at least in
principle) to generate the electrostatic potential (and from it the electric field) if we know the
charge distribution:
∇ 2V = −
ρ ( x, y , z )
ε0
(with solution V ( r ) =
∫
all space
1
ρ ( r ′)
4πε 0 r − r ′
d3r ′ )
and
E = −∇V .
The ∇2V expression is Poisson’s equation, named for the French mathematician Simeon Denis
Poisson (1781 - 1840). It is, in a sense, the master equation for the electrostatic potential: given
ρ, there will always exist a solution to Poisson’s equation which we can construct (perhaps
numerically, as opposed to analytically) by calculating an integral.
An important property of Poisson’s equation is that the solutions superpose. Imagine that we
have already calculated the potential V1 which corresponds to a charge distribution ρ1, and also
the potential V2 which corresponds to a different charge distribution ρ2. If we want to know the
potential Vsum which corresponds to a charge distribution ρsum = ρ1 + ρ2 all we have to do is add
the potentials we previously found: Vsum = V1 + V2. It’s not necessary to crank through an
integral over the combined charge distribution ρsum. From Physics 112 you already know about
the simplest version of this: a pair of charges q1 and q2 at positions r1 and r2 will yield a
potential
1  q1
q2 
+
V (r ) =

.
4πε 0  r − r1 r − r2 
It is obvious that two identical charges (q1 = q2) would not, in general, contribute equally to the
potential except at points which are equidistant from both charges. The weighting function
which determines the size of the ith charge’s contribution to the overall potential is the
(Coulomb Green’s) function 1 ( 4πε 0 r − ri ) .
Here’s what you’ll find/work on today:
• Proving that potentials which satisfy Poisson’s equation really do obey the principle of
superposition.
• Investigating superposition for some other systems.
Exercise 4.1: Proving that Coulomb potentials really do obey the principle of superposition
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.3
47
Poisson’s equation says that ∇ 2V = −
ρ ( x, y , z )
1 ρ ( r ′) 3
d r′ .
with solution V ( r ) = ∫
4πε 0 r − r ′
ε0
all space
Imagine that you’ve solved for the potential V1 corresponding to a particular charge distribution
ρ1, and also for the potential V2 which corresponds to a different charge distribution ρ2. Prove
ρ ( x, y , z )
that ∇ 2Vsum = − sum
where Vsum = V1 + V2 and ρsum = ρ1 + ρ2 .
ε0
discussion
Exercise 4.2: Another superposition proof
The net force on a forced, damped harmonic oscillator is given by ma = − kx − β v + f ( t ) . The
damping, described by the –βv term to the right of the “=” sign, is velocity-dependent and tends
to slow the oscillator down. The external driving force is f(t). We can rewrite the equation this
way:
d 2x
dx
+ kx = f ( t ) .
m 2 +β
dt
dt
Imagine that our oscillator is being driven by a particular driving force f1(t) and that we have
already solved for its position as a function of time x1(t). A different oscillator with identical
mass and damping is driven by a driving force f2(t), and we have already solved for this
oscillator’s position as a function of time x2(t). Prove that when one of the oscillators is
subjected to a force fsum(t) = f1(t)+ f2(t), it will move so that its position is xsum(t) = x1(t) + x2(t) .
discussion
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.4
48
Exercise 4.3: Yet again another proof
The “Karma” K ( x) of central Illinois is described by the differential equation
A
d 3 K ( x)
d 2 K ( x)
dK ( x)
+
B
+C
+ DK ( x) = U ( x) .
3
2
dx
dx
dx
The function U(x) (incorrectly described in the Bhagavad-Gita as “the source of all visible
kinetic energy in the Universe”) depends on the local weather. Imagine that you have already
solved for the Karma on rainy days (Krain(x) corresponding to Urain(x)) and on days with hail
(Khail(x) corresponding to Uhail(x)). During an especially unpleasant day in which it rains and
hails simultaneously, the central Illinois U(x) function is Urain+hail(x) = Urain(x) + Uhail(x).
Does the mid-state Karma function obey the superposition principle? (Is it true that
Krain+hail(x) = Krain(x) + Khail(x))? Prove your answer.
discussion
Exercise 4.4: Yet again another (and not, perhaps, the last) proof
The “Karma” K ( x) of Indiana behaves differently from that in Illinois, obeying the differential
equation
A
d 2 K ( x)
2
+ B ( K ( x) ) = U ( x) .
2
dx
As in Illinois, the function U(x) depends on the weather. Imagine that you have already solved
for the Indiana Karma on rainy days (Krain(x) corresponding to Urain(x)) and on days with hail
(Khail(x) corresponding to Uhail(x)). Once again, when it rains and hails simultaneously, the
function U(x) is given by Urain+hail(x) = Urain(x) + Uhail(x).
Does the Indiana Karma function superpose? (Is it true that Krain+hail(x) = Krain(x) + Khail(x))?
Prove your answer.
discussion
Exercise 4.5: Probably the last superposition proof
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.5
49
The Iowa Karma function K ( x) obeys the differential equation
AK ( x)
dK ( x)
+ BK ( x) = U ( x) .
dx
As before, U(x) depends on the weather. Imagine that you know Krain(x), corresponding to
Urain(x), and Khail(x), corresponding to Uhail(x). When it rains and hails simultaneously, the
function U(x) is again given by Urain+hail(x) = Urain(x) + Uhail(x).
Does the Iowa Karma function superpose? (Is it true that Krain+hail(x) = Krain(x) + Khail(x))? Prove
your answer.
discussion: linear vs. nonlinear differential equations
Linear equations will obey superposition, while nonlinear ones (often, usually, always??) will
not.
discussion: cheap audio equipment
“Total harmonic distortion:”
In
1
2
3
4
5 (kHz)
Out
1
2
3
4
5 (kHz)
Pure frequency in yields various frequencies out if amplifier is (slightly) nonlinear.
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.6
50
Summary and wrap-up discussion
•
•
The principle of superposition generally holds for linear differential equations.
The Green’s function for a linear system (which obeys superposition) is the “helper function”
which indicates how to sum the effects caused by various sources. For example, the Coulomb
potential’s Green’s function is G ( r , ri ) ≡ 1 ( 4πε 0 r − ri ) .
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.7
51
Reading and homework should be completed by 4pm Friday.
Required reading:
The Feynman Lectures on Physics (volume I), chapter 25, sections 1 and 2.
Read over (but don't worry about understanding) the Unit 5 material we'll tackle in class next
week.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): Anthrax’s personal MP3 player sounds funny
Anthrax, the Urkins’ unpleasant feline, enjoys listening to music on a defective MP3 player. The
relationship between the player’s input signal Ain ( t ) and its output signal Aout ( t ) is described
by the equation
Aout ( t ) = 10−3
dAin ( t )
.
dt
Notice that this equation is linear: superposition works for it.
(a) The closing note of Grunt Like a Chicken Thief is a sustained 500 Hz triangle wave whose
amplitude vs. time is shown in the following figure:
V
+0.5 V
t
-0.5 V
1 ms
Ain is a 500 Hz triangle wave
Draw an accurate plot of the defective MP3 player’s output signal, Aout.
(b) Any periodic function can be well-represented by a (possibly infinite) sum of sines and
cosines. In particular, the triangle-wave final wail of Grunt Like a Chicken Thief can be
represented this way:
4 k =∞ 1
Ain ( t ) = − 2 ∑ 2 cos ( k ⋅ π ×103 t ) .
π k =1,3,5,... k
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.8
52
An expression of this sort is called a Fourier series. Using this, along with the knowledge
gleaned from your answer to part (a), derive an expression for the Fourier series which
represents the following 500 Hz square wave:
+1 V
V
t
-1 V
1 ms
500 Hz square wave
.
Notice that superposition allows you to draw conclusions about the behavior of the input and
output signals based on an understanding of how the individual Fourier components (the sines
and cosines) behave when passed through the defective MP3 player.
Problem 2 (10 points): Starburn
In this problem you will find it convenient to use (struggle with??) a volume integral in spherical
coordinates (r,θ,φ) instead of the more familiar Cartesian (x,y,z) coordinates. Please find Brian
or me if you need help setting up the problem’s solution. Here’s how spherical coordinates work:
z
dV = dr × rdθ × r sinθ dφ
r
θ
dr
θ
φ
y
φ
dθ
dV
dφ
x
The lengths of the sides of an almost-rectangular volume element dV at (r,θ,φ) are dr, r dθ, and
r sinθ dφ. Its volume is therefore dV = dr × rdθ × r sinθ dφ = r2 sinθ dr dθ dφ.
A volume integral of a function F can be written in the two coordinate systems as
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.9
53
rf θ f φ f
 y f  x f
 

 

= ∫  ∫  ∫ F ( x, y, z ) dx  dy  dz or I spherical = ∫  ∫  ∫ F ( r ,θ ,φ ) r sin θ dφ  rdθ  dr .

 

zi 
ri 
 θi  φi
 yi  xi
zf
I cartesian
You will derive an argument against a model of our universe as infinite and static, in which new
stars appear periodically to replace old stars as they burn out.
It is thought that there are roughly 1020 stars inside the 15 billion lightyear diameter observable
universe. (This number could be wrong by a factor of 1000!) Since a lightyear is approximately
1016 meters, let’s assume that the average density of stars in our universe is 1 star per 1058 cubic
meters, and that we can treat a distant star as a point source of light. If all stars shine with the
same luminosity as our sun, each one is pouring 3.8×1026 watts of power into space. (Recall that
a watt is the same as a joule per second.) Naturally, the intensity of power per square meter a
distance r from a star is just the fraction of the area of a sphere of radius r subtended by one
square meter, namely 1 ( 4π r 2 ) .
(a) Imagine that starlight falls on a sphere of radius 1 π floating in space. (The cross-sectional
area of the sphere is a square meter.) Using the assumptions mentioned above, derive an
expression for the power, in watts, striking the sphere from all stars within a distance R from the
sphere. Since it will be necessary for you to sum the intensity from lots of stars, you will find it
convenient to recall how to do an integral in spherical coordinates using the volume element
dV = r 2 sin θ drdθ dφ with θ ∈ ( 0, π ) but φ ∈ ( 0, 2π ) . (Neglect light from the sun!)
(b) The intensity of sunlight reaching the ground at noon in Champaign on a clear spring day is
about 1000 watts per square meter. What value (if any) of R would make the starlight
illuminating the sphere in part (a) be just as bright as this? (Express you answer in lightyears,
please.)
(c) Using the same assumptions, estimate the intensity of starlight at the Earth's surface.
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.10
54
Problem 3 (10 points): New technology ice cream
Research scientists at Urkin’s D&HE contemplate upgrading their sauerkraut mill so that it can
be used to prepare gallons of ever-popular brussels sprout ice cream. The complication is that the
ice cream matrix needs to be agitated at frequency 4 Hz in order to set properly, while the
brussels sprouts need to detect vibrations at 3 Hz in order to develop the correct flavor. Can the
modified kraut mill handle the task? This is, the scientists, realize, a problem of superposition.
Will the simultaneous installation of new 4 Hz and 3 Hz drivers in the kraut mill manage to
deliver both frequencies to the ice cream under preparation?
Sophisticated computer simulations of the sauerkraut mill show that the mixing amplitude x ( t )
delivered to the ice cream in the mill’s crucible by a single driver supplying force f ( t ) obeys
one of the following equations:
(a)
(b)
(c)
d 4x
d 2x
+
B
+ C = f (t )
dt 4
dt 2
d 2x
dx
A 2 + Bx + Cx 2 = f ( t )
dt
dt
3
d x
d 2x
dx
A 3 + B 2 + C + Dx = f ( t ) .
dt
dt
dt
A
The coefficients A, B, C, and D are constant. The researchers realize that the question is this:
when two drivers exert a combined force f ( t ) = f 4 Hz ( t ) + f 3 Hz ( t ) will the response be the same
as the sum of the single-driver responses (so that x ( t ) = x4 Hz ( t ) + x3 Hz ( t ) )? If this is the case,
the sauerkraut mill upgrade will work nicely.
Which (if any) of the possible response functions (a), (b), and (c) will allow a successful
upgrade? Please prove, or otherwise justify, your answers.
Physics 212, Spring 2003. Unit 4
Why superposition works; Green’s functions
4.11
55
Physics 212
Honors topics in electrodynamics
Unit 5
Pinwheels (curl and the vector potential);
le souvenir du temps passé
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.1
56
Unit 5: Pinwheels (curl and the vector potential); le souvenir du temps passé
Recap of last week
Introduction: Does it spin?
We’ve been looking at some of the characteristics of static electric potentials and fields with the
help of a variety of tools such as Poisson’s equation. As you’ve seen in Physics 112, the
connection between the electric potential and electric field is closely tied to the idea that the
work done moving a charge from one place to another is independent of the path taken. If our
charges are not static (in particular, if they are accelerating) all bets are off.
We’re going to investigate another kind of differential operator today: the curl of a vector
function. Curl will prove useful when we discuss systems in which currents are flowing, and
when we investigate (derive/discover) the existence of electromagnetic radiation.
We will also look at a vector analog of the scalar potential. This is less mysterious than it
sounds: the scalar potential at r created by a point charge q at position r ′ moving with velocity
v is approximately
V (r ) ≈
1
q
4πε0 r − r ′
while the vector potential A is approximately
A( r ) ≈
1
q
µ0 q
v =
v
2
4πε0c r − r ′
4π r − r ′
since µ 0 =
1
.
ε 0c 2
.
(There are some subtleties concerning what we need to do to make these equations exactly true.)
Here’s what you’ll find/work on today:
• properties of the electric field that are impossible to find in fields generated by static charges
• curl and divergence of electric fields
• Stokes’ Theorem
• The vector potential
• Time retardation’s influence on the vector potential
Exercise 5.1: Pinwheel statics (or should we say dynamics?)
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.2
57
Imagine we make a tiny pinwheel with point charges q on each of its four blades as drawn in the
following figure. Electric field lines in the vicinity of the pinwheel, caused by (static) charges
placed elsewhere are shown in the diagram. The pinwheel’s center bearing is kept fixed. Let’s
see if it is possible for the electric field to cause the pinwheel to spin faster and faster.
q
y
∆y δ
=
2
2
q
q
q
∆x δ
=
2
2
z
pinwheel
center
at (x,y)
x
∆x
∆y
∆x
∆y
, y, 0) , ( x, y +
, 0) , ( x −
, y, 0) , ( x, y −
, 0) .
2
2
2
2
Recall that torque is r × F . Assuming that the pinwheel’s arms are of equal lengths (so that
∆x = ∆y = δ), write an expression for the torque on the pinwheel in terms of the x and y
components of the electric field. (Assume the z component of the electric field is zero.)
(a) The charges are at positions ( x +
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.3
58
(b) If you haven’t already done so, transform your answer to part (a) into an expression
involving partial derivatives of the fields by taking the limit that ∆x and ∆y go to zero.
discussion
You should end up with something like this: τ = q
δ 2  ∂E y
∂E x 

 zˆ .
∂y 
2  ∂x
y
If our pinwheel also had had a pair of arms parallel to the z axis, as drawn
here, we could have calculated torque components along all three axes to find
 ∂E y ∂Ex 
 ∂Ez ∂E y 
 ∂Ex ∂Ez
 zˆ + 
 xˆ + 
∂y 
∂z 
∂x
 ∂z
 ∂y
 ∂x
τ = 
≡
2
  δ
ˆ
y


  2
δ2
x
z
(∇ × E ) 2 .
∇ × E is called the “curl of E .” If the electric field has a nonzero curl, it’ll make the pinwheel
spin.
How might such electric fields look? Here are two examples.
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.4
59
Exercise 5.2: Can the electric field from a charge distribution do this?
If no charges are moving, we can use Poisson’s equation to calculate the potential, and then take
its gradient to solve for the electric field:
∇ 2V = −
ρ ( x, y , z )
1 ρ ( r ′) 3
d r ′ , and E = −∇V .
so that V ( r ) = ∫
4
r
r
πε
−
ε0
′
0
all space
( )
When the electric field is the gradient of a scalar field is it possible for ∇ × E = −∇ × ∇V to be
non-zero? (Prove your answer.)
Exercise 5.3: What’s the divergence of a curl?
Let’s say we find a different vector field (call it A ) whose curl is non-zero. Can the divergence
of the curl of A ( ∇i ∇ × A ) also be different from zero? (Prove your answer.)
(
)
discussion
Path independence, work done along a closed path, line integrals, and curl...
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.5
60
Exercise 5.4: Line integrals around microscopic loops; curl
Imagine that an electric field (with non-zero curl) does work moving a charged particle
counterclockwise around a tiny rectangular loop, as shown in the diagram. The midpoints of the
∆x
∆y
∆x
∆y
four sides are at positions ( x +
, y, 0) , ( x, y +
, 0) , ( x −
, y, 0) , and ( x, y −
, 0) ;
2
2
2
2
the work done by the electric field is ∫ qE idl .
loop
y
∆y
∆x
Rectangular loop is centered at (x,y)
z
x
(a) Calculate the work done by the electric field, and then re-express it in terms of ∆x, ∆y, and
various partial derivatives of the electric field. (You should assume that the electric field along
each side of the loop is well-represented by its value at the midpoint of the side, and write the
field explicitly in terms of its x and y components: E ( x, y, z ) = Ex ( x, y, z ) xˆ + E y ( x, y, z ) yˆ .)
(b) Re-express your answer in terms of ∇ × E if you haven’t already done so.
Discussion
dA vs. dA ; n̂ (normal to the loop via the right hand rule).
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.6
61
Exercise 5.5: Line integrals around lots of microscopic loops
Now imagine that we tile part of the x-y plane with several small loops as shown below. Prove
that the work done by the field when a charge moves around the outer perimeter of the tiled
region is equal to the sum of the work done moving a charge around each of the small loops:
(
∑ ∫
small loops
i th loop
)
qE idl =
∫
big loop
qE idl .
∆x
y
z
∆y
1
2
3
4
x
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.7
62
Discussion: Stokes’ theorem
From an earlier exercise, you should have concluded that
∫
=
E idl
tiny loop
( ∇ × E ) inˆ dx dy
=
( ∇ × E ) idA
where n̂ is a unit vector normal to the surface of the loop (its direction is given by the right hand
rule) and dA = n̂ dx dy . We can use this to rewrite
(
∑ ∫
small loops
i th loop
)
E idl =
∫
big loop
E idl
as
∑
small loops
( ∇ × E ) idA
i
⇒
∫
( ∇ × E ) idA
surface surrounded
by big loop
=
∫
E idl .
big loop
This is Stokes’ Theorem, named for the Irish mathematical physicist George Gabriel Stokes
(1819-1903); it is true for any vector field.
Physics 212, Spring 2003. Unit 5
Pinwheels (curl and the vector potential); le souvenir du temps passé
5.8
63
Discussion: the Helmholtz theorem
Suppose we are told what the divergence and curl are for some vector function F ( x, y, z ) , but
don’t know the explicit form of F itself:
∇ i F ( x , y , z ) = D ( x, y , z )
and
∇ × F ( x, y , z ) = C ( x, y , z ) .
(Naturally, ∇iC will be zero since the divergence of a curl is always zero.)
Is there a general technique we can use to determine F if we know C and D? (Being told that
∇i E = ρ ε 0 and ∇ × E = 0 for a static charge distribution is an example of this.)
According to something called the Helmholtz theorem, we can always write any well-behaved
vector function as the sum of a gradient and a curl as long as C and D go to zero sufficiently
rapidly at large distances. The exact form is
 1
F ( r ) = −∇ 

 4π
 1
D ( r′ ) 3 
′
+
∇
×
d
r


∫ r − r′ 

all space

 4π
C ( r′ ) 3 
∫ r − r′ d r′  .
all space

If you use this to calculate something, keep in mind that the partial derivatives in the gradient
and curl are taken with respect to x, y, z and not with respect to x′, y′, z ′ . As a result, terms
which depend only on the primed variables act like constants when you evaluate the partial
derivatives.
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Discussion: The vector potential
The scalar potential for a stationary point charge at position r ′ is V ( r ) =
1
q
.
4πε0 r − r ′
If we have a continuous distribution of static charge so that the amount of charge contained in a
small volume dV is ρ dV , we can calculate the potential at r by doing an integral ( d 3 r ′ is just a
volume element):
V (r ) =
1
∫
ρ ( r ′)
4πε 0 r − r ′
all space
d 3r ′ .
Are there ever times when we might want to superpose something that is a vector instead of a
scalar? Certainly! It is possible to determine the electric field by calculating the gradient of the
potential; since the potential obeys the superposition principle when we have several charges in
our universe, the electric field also obeys superposition.
We know that the electric field at r caused by a small amount of stationary charge ρ dV at r ′ is
1 ρ dV r − r ′
.
E (r ) =
4πε 0 r − r ′ 2 r − r ′
(The ( r − r ′ ) r − r ′ piece is just a unit vector pointing away from the charge.)
As a result, we can calculate the electric field (without referring explicitly to the potential) this
way:
E (r ) =
∫
all space
1
ρ ( r ′) r − r ′
4πε 0 r − r ′
2
r − r′
d 3r ′ .
( d 3 r ′ is just the volume element dV ′ .) An integral is really just a big sum. The presence of the
vector’s direction as part of the integrand may make the integral harder to evaluate, but that’s
just a practical matter-- it doesn’t make the integral’s meaning (more??) obscure.
Remember that we're only dealing with the case right now that the charge density is unchanging:
the number of Coulombs per cubic meter at a particular point never changes, no matter how long
we wait. We can have constant ρ when charges are moving: as long as the amount of charge
dq = ρdV flowing into a small volume element equals the amount flowing out, we'll never have a
change in the amount of charge contained by dV.
Here’s another quantity that has a direction associated with it. Let's say our charge distribution
is swirling around so that the charge dq = ρ ( x, y, z ) dV at the point (x,y,z) has velocity
v ( x, y, z ) . By the way, ρ ( r ) v ( r ) is called the current density J ( r ) : it corresponds to the
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number of amperes per second per square meter flowing through a surface at r which is
perpendicular to J ( r ) . I'm going to impose the constraint that J ( r ) never changes with time,
even though its value can differ from place to place.
For grins, let’s define the vector potential associated with this sort of current flow as follows:
A( r ) ≡
1
1 J ( r ′) 3
µ0 J ( r ′ ) 3
d
r
=
′
2
∫ 4πε 0 r − r ′
∫ 4π r − r ′ d r ′
c all space
all space
since µ0 =
1
.
ε 0c 2
Keep in mind that this equation is really shorthand for three distinct equations:
Ax ( r ) =
µ0 J x ( r′) 3
µ0 J y ( r′) 3
µ0 J z ( r′ ) 3
d r′, Ay ( r ) = ∫
d r ′, Az ( r ) = ∫
d r′ .
4π r − r ′
4π r − r ′
4π r − r ′
all space
all space
all space
∫
Of what use could this possibly be? Think about the scalar potential: the reason that it is useful
to work with the function
1 ρ ( r ′) 3
V (r ) = ∫
d r′
4πε 0 r − r ′
all space
is that it is the solution to Poisson’s equation
∇ 2V = −
ρ ( x, y , z )
.
ε0
Imagine that we encounter an equation which resembles Poisson’s equation, except that the
function of x,y,z on the right side is f ( x, y, z ) = ρ ( x, y, z ) vx ( x, y, z ) instead of ρ ( x, y, z ) , and
the constant on the right side is -µ0 instead of -1/ε0. Our equation would be
∇ 2 Ax = − µ0 f ( x, y, z ) = − µ0 ρ ( x, y, z ) vx ( x, y, z ) .
with solution
Ax ( r ) =
µ0 f ( r ′ ) 3
µ0 ρ ( r ′ ) v x ( r ′ ) 3
d r′ = ∫
d r′ .
4π r − r ′
4π
r − r′
all space
all space
∫
(This would correspond to a charge distribution ρ ( x, y, z ) swirling around, with the velocity of
the charge differing from place to place, so that the “local velocity” of charge at ( x, y, z ) is
vx ( x, y, z ) . Neither function is time dependent.)
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If we encounter similar equations involving the y and z velocity components, we could multiply
both sides of each equation by the appropriate unit vector, then add them:
∇ 2 Ax xˆ = − µ0 ρ ( x, y, z ) vx ( x, y, z ) xˆ 

∇ 2 Ay yˆ = − µ0 ρ ( x, y, z ) v y ( x, y, z ) yˆ 

∇ 2 Az zˆ = − µ0 ρ ( x, y, z ) vz ( x, y, z ) zˆ 
⇒ ∇ A = − µ ρ ( x, y , z ) v ( x, y , z ) .
2
0
Why might we ever bump into an equation of this sort? How could something as abstract as the
vector potential possibly matter? Who knows? (Who cares?) In any event, if for some reason
we do encounter this ∇2 A equation, we’ll already know how to solve it because of our
experience with Poisson’s equation:
µ0 ρ ( r ′ ) v ( r ′ ) 3
A( r ) = ∫
d r′ .
−
4
π
r
r
′
all space
(There actually is a good reason for messing with the vector potential: the magnetic field is
exactly the same as the curl of the vector potential: B = ∇ × A . We’ll deal with this later.)
We might want to discuss the vector potential for a point charge in motion, but this sort of
charge distribution does not satisfy the constraint that ρ ( x, y, z ) is constant. (As the point charge
moves, the charge density changes a great deal at each point the charge occupies during its
travels.) However, as long as the charge is moving slowly compared to the speed of light, it's not
a bad approximation to use
A( r ) =
µ0 ρ ( r ′ ) v ( r ′ ) 3
µ0 qδ ( r ′ − rq ( t ) ) v ( t ) 3
µ qv ( t )
d r′ ≈ ∫
d r′ = 0
r − r′
r − r′
4π
4π
4π r − rq ( t )
all space
all space
∫
where rq ( t ) and v ( t ) represent the charge's position and velocity.
Exercise 5.6: Sketching the vector potential
A slowly moving point charge is passing through the origin at time t = 0 with constant velocity
v = v0 zˆ . Ignoring any effects associated with the difference between t and t ′ mentioned earlier,
draw a reasonably good sketch of the vector potential associated with this charge’s behavior at
t = 0.
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Exercise 5.7: Sketching another vector potential
Two slowly moving point charges are passing through the points ( x, y, z ) = (0, −1,0) and
( x, y, z ) = (0, +1,0)
with constant velocities v = v0 zˆ . Ignoring any effects associated with the
difference between t and t ′ mentioned earlier, draw a reasonably good sketch of the vector
potential at t = 0.
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Exercise 5.8: Sketching yet again another vector potential
Two slowly moving point charges are passing through the points ( x, y, z ) = (0, −1,0) and
( x, y, z ) = (0, +1,0)
with constant velocities v = v0 zˆ and v = −v0 zˆ respectively. Ignoring any
effects associated with the difference between t and t ′ mentioned earlier, draw a reasonably
good sketch of the vector potential at t = 0.
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Discussion: retardation
The correct expressions for the scalar and vector potentials include the fact that an observer at a
position r sees the charge as it was at an earlier time: the information about the position of the
charge takes a finite amount of time to propagate from the charge to the observer. (We see the
sun as it was eight minutes ago; we see Alpha Centauri as it was 4.35 years ago.) If the charge
density and/or current density changes with time, the integrals used to calculate the potentials
will need to take this into account.
If the observer at position r (at observer-time t) sees charge at position r ′ , the time at which the
charge was really at that point was at a time earlier than t by ∆t = t − t ′ = r − r ′ c . We need to
modify our equations like this:
V (r ) =
∫
all space
A( r ) =
1
ρ ( r ′)
4πε 0 r − r ′
d 3r ′
⇒ V (r ,t) =
∫
all space
1
ρ ( r ′, t − r − r ′ c )
4πε 0
r − r′
d 3r ′
µ0 J ( r ′ ) 3
µ 0 J ( r ′, t − r − r ′ c ) 3
,
⇒
=
d
r
A
r
t
d r′
′
(
)
∫ 4π r − r ′
∫ 4π
r − r′
all space
all space
What happens to the potentials from a finite charge distribution with total charge Q when we
1
Q
include retardation? It's tempting to think that V and A will satisfy V ( r , t ) ≈
4πε 0 r − rQ ( t ′ )
and A ( r , t ) ≈
µ0 Qv ( t ′ )
if we're far from the charge. (The retarded time t ′ satisfies
4π r − rQ ( t ′ )
t ′ = t − r − rq ( t ′ ) c ). However, this is wrong!
The problem arises because Q ≠
∫
ρ ( r ′, t − r − r ′ c ) d 3r ′ . As we integrate over the volume
all space
containing the charge cloud, the integral passes through different regions at different times due
to retardation. (It will pick up contributions to V from distant regions further in the past than
from close regions.)
Here's a concrete example. Imagine that a small charge, shaped in its rest frame like a tiny cube
with side a, is moving towards an observer. The observer will evaluate the potential (scalar or
vector) caused by this charge as a sum of contributions from small slices in space as the cube
moves towards him, as shown in the figure.
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a 1 − v2 c2
1− v c
29
1 3 5 7 9
30
2 4 6 8
a
v
distant
observer
a 1 − v2 c2
cube is here when
information from
first slice reaches
observer
cube is here when
information from
last slice reaches
observer
The observer will take a "snapshot" of the amount of charge filling different regions of space and
will calculate the net potential as the sum of the contributions from all the small regions of space
in his snapshot. Because of retardation, the "local time" (the retarded time) in each region of
space in the snapshot will be different: the further away a region is, the earlier in time he'll be
seeing it. The slice of space labeled "1" in the observer's snapshot just happened to be holding
the back of the Lorentz-contracted cube. Since the cube is moving, the front of the cube just
happened to be occupying slice 30 at the correct time for it to appear in the snapshot: slice 30 is
seen at a less retarded time than slice 1 since it is closer to the observer. As far as the observer is
concerned, the potential must be the same as it would be with slices 1 through 30 all containing
electric charge! (If the charge were moving away from the observer, there would seem to be less
charge there.)
With some algebra, it's easy to show that the length of the region in space which seems to hold
charge is a factor of 1 (1 − v c ) longer than the side of the charge which is parallel to its line-ofsite to the observer. If the charge were not moving directly towards/away from the observer, the
correction would involve the component of velocity parallel to the observer's line-of-site:
1 (1 − v ⋅ εˆ c ) where εˆ is the unit vector from the object to the observer, as shown in the
following figure.
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71
line-of-sight
to observer
observer
εˆ
v
As a result, our expressions for the potentials associated with a point charge need to be written
V (r ,t) =
Q
1
×
4πε 0 r − rQ ( t ′ ) (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
A( r , t ) =
µ0 Qv ( t ′ )
1
×
.
4π r − rQ ( t ′ ) (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
and
1
These are know as the Liénard-Wiechert potentials. (A.Liénard, 1898; E.Wiechert, 1900.)
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Exercise 5.9: Retardation
Consider a point charge which oscillates along the z axis so that its position is
( x, y, z ) = (0,0, a sin(ω t )) . The amplitude and frequency of its motion are a=0.001 feet and
ω=2π×109 radians per second (so that the period of its motion is one nanosecond.)
Including effects associated with the finite speed of light (c ≈ 1 foot per nanosecond), sketch
how the vector potential looks along the line y = z = 0 within a few feet of the origin. (Note that
v ⋅ εˆ ≈ 0 so you can ignore the 1 (1 − v ⋅ εˆ c ) correction.)
Discussion
Summary and wrap-up discussion
•
•
•
•
•
•
properties of the electric field that are impossible to find in fields generated by static charges
curl and divergence
Stokes’ Theorem
The vector potential
Time retardation’s influence on the vector potential
Why did we do this, anyway?
Comments about homework assigned for next week
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Reading and homework should be completed by 4pm Friday.
Required reading:
Purcell, chapter 2, sections 13 - 16; chapter 6, section 3.
The Feynman Lectures on Physics (volume II), chapter 21, section 5.
Read over the remaining Unit 5 material we didn’t get to in class this week. Also read over (but
don't worry about understanding) the Unit 6 material we'll tackle next week.
Optional reading:
The Feynman Lectures on Physics (volume II), chapter 14.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): Anthrax’s litter box is exposed to an electric field
Mrs. Urkin creates an E field across the top Anthrax’s cat box in order to lay claim to the
invention of the field of electroscatology. (The top of the cat’s litter box is in the x-y plane at
z = 0 with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. ) She experiments with different field configurations as described
below.
(a) The first electric field configuration is E ( x, y, z ) = xyz ( xˆ + yˆ + zˆ ) at the top of the
aquarium. Is it possible for this particular electric field to be the gradient of a scalar potential
V ( x, y, z ) ? Be sure to prove, or otherwise justify your answer.
(b) The next electric field configuration is E ( x, y, z ) = yz xˆ + zx yˆ + xy zˆ at the top of the
aquarium. Can this electric field be the gradient of a scalar potential? Prove your answer.
(c) Another configuration is E ( x, y, z ) = x 2 xˆ - y 2 yˆ . Can this electric field be the gradient of a
scalar potential? Prove your answer.
(d) The final field configuration is E ( x, y, z ) = y 2 xˆ - x 2 yˆ . Can this electric field be the gradient
of a scalar potential? Prove your answer.
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Problem 2 (10 points): Being clever may help you avoid doing the line integrals
Mr. Urkin fabricates Christmas presents for his nephews from electrically charged marbles and
polyvinyl chloride sewer pipes as illustrated in the figure. One of the “toys” comprises a square
loop of (frictionless) sewer pipe in which a marble rolls. Another is a circular loop of pipe, also
frictionless, in which a second marble rolls. Each marble carries charge q.
1 meter
1 meter
plastic pipe
plastic pipe
y
y
x
1 meter
charged marble
x
charged marble
(a) Placed flat on the floor during tests with its center at ( x, y ) = ( 0, 0 ) and its sides parallel to
the x and y axes, the square loop is exposed to an electric field E ( x, y, z ) = x yˆ . (The floor is
in the plane z = 0.) Assuming the marble does not lose any energy as it rounds the corners,
calculate the change in its kinetic energy after the field drives it once around the loop.
(b) The round loop is tested in a similar fashion: it is placed in the same electric field, with its
center at the origin. How much kinetic energy does the marble gain traveling once around the
loop?
Problem 3 (10 points): Vector potential sketch
Read through the in-class material about retardation, digesting it as best as you can (see me with
questions) and then do Unit 5’s in-class exercise 5.9.
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Physics 212
Honors topics in electrodynamics
Unit 6
The origin of the magnetic field as a
consequence of special relativity
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.1
77
Unit 6: The origin of the magnetic field as a consequence of special
relativity
Recap of last week
Introduction: magnetic fields are just electric fields seen from a different perspective.
Magnetic fields and forces exist because of Coulomb’s law and special relativity. Though
observers in one frame may determine that a current-carrying wire is electrically neutral,
observers in another frame might conclude that the wire does carry a net electric charge. We’ll
derive this today, for the particular case of motion parallel (or antiparallel) to the current.
You’ll slog through much less vector calculus today than last week!
Here’s what you’ll find/work on today:
• analyzing the charge density in a wire loop which suddenly finds itself carrying a current,
according to observers in two different frames of reference
• describing how electric fields from a static charge transform when changing to a frame in
which the charge is in motion
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Exercise 6.1: Whap! They’re moving!
A rectangular loop is made from a thin filament of superconducting wire as drawn below.
Initially no current flows in the wire; there are as many (positive) metal ions in the wire’s crystal
lattice as there are conduction electrons. The spacing between adjacent electrons is a while the
spacing between positive metal ions is b. (Naturally, since the wire carries no net charge, a = b.)
before...
...after
synchronized
clocks
wire loop
u
electron flow
u
metal ions
electrons
b
a
When all (synchronized) clocks in the lab frame read zero, all the electrons are suddenly
accelerated from rest to a speed u so that the flow of electrons in the wire loop is in a counterclockwise sense.
(a) According to observers in the lab frame (the rest frame of the still-stationary metal ions),
what is the spacing between adjacent (but now-moving) electrons?
Discussion
...but what about Lorentz contraction?
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(b) An observer in a different frame of reference sees the wire loop glide past with speed v,
moving to the left as shown in the figure. When the observer’s clock reads zero, clocks close to
the right-most edge of the (moving) wire loop are also seen to read zero.
v
Sketch the readings on the other (moving) clocks’ faces at the instant the right-side clocks read
zero.
discussion
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(c) Keep in mind that an electron is accelerated when the lab-frame clock next to it reads zero.
From the perspective of the observer who sees the wire loop moving to the left, the densities of
electrons and metal ions are identical before any of the electrons begin circulating in the wire
loop.
Describe the relative densities of electrons and metal ions in each of the four sides of the loop
after all the electrons have been set in motion. It may help to think in terms of which electrons
begin moving when.
(d) What would happen to a test charge, stationary in the frame of the observer who sees the
wire loop moving to the left, if the test charge were placed close to the bottom side of the loop?
How would this charge behave when seen from the rest frame of the wire loop?
discussion
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The origin of the magnetic field as a consequence of special relativity
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Exercise 6.2: Quantitatively, now...
Let’s consider the bottom leg of the wire loop now, after the electrons have started moving.
You should have concluded that the electron spacing, viewed from the lab frame (the rest frame
of the wire), does not change after the electrons are accelerated. (How could it possibly change?
In this frame all electrons started moving simultaneously, so there was never a time interval
during which one electron was moving but its neighbor was not.)
a
metal ions
u
a
electrons
(a) Calculate (in terms of a and u) the spacing between electrons in the electrons’ rest frame.
(Even though u c for realistic currents you’ll need to consider the effects of special
relativity.)
n ( n − 1) 2
n
ε +
ε + …
1!
2!
(This works even when n isn’t an integer, and is useful for approximations.) Some examples for
−1 2
ε 1 : 1 + ε ≈ 1 + ε 2 ; 1 (1 + ε ) ≈ (1 − ε ) ; 1 1+ ε = (1 + ε ) ≈ 1 − ε 2 . In these exercises you’ll
(b) Recall that the binomial expansion for (1 + ε ) is (1 + ε ) = 1 +
n
n
only need to calculate things to second order accuracy in any of the velocities present in your
equations (terms proportional to v 3 , u 3 , uv 2 ... can be ignored).
Use the binomial expansion to calculate a version of the relativistic velocity addition formula
which is accurate to second order in the velocities.
Physics 212, Spring 2003. Unit 6
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(c) In another frame of reference the wire is seen to move to the left at speed v c . (By this I
mean that the metal lattice is moving with speed v). How fast are the electrons moving in this
frame? (Please use the second-order expression you just derived.)
(d) Calculate the spacing between metal ions in the frame in which the wire is moving. Do the
same for the spacing between electrons in this frame.
(e) Use the binomial expansion to generate an approximate expression for the spacing between
adjacent metal ions in the frame with the wire moving. Please include all terms up to second
order in the ratios of small velocities (u and v).
Do the same thing for the electrons.
(f) Calculate the apparent charge density in the wire in terms of the fundamental charge (e), a, u,
v, and c. Verify that your expression yields zero if v = 0 or u = 0.
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Discussion
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Exercise 6.3: Real numbers
Let’s calculate what happens in a real wire.
Imagine that we have a 1000 ampere current flowing in a heavy copper cable with crosssectional area 1 cm2. The density of copper is about 9 grams per cm3 while the mass of a single
copper atom is approximately 10-22 grams. As a result, every cubic centimeter of copper cable
holds 9 × 1022 copper atoms. Each copper atom gives up two conduction electrons so that there
are 1.8 × 1023 mobile electrons per cm3 of cable. The charge on each electron is 1.6 × 10-19
Coulombs.
(a) How fast are the electrons moving, on average?
(b) How large is the electric field generated by the (charged!!) wire at a distance of 10
centimeters according to a test charge which observes the wire moving to the left with speed
100 m/s? (Use Gauss’ law!)
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(c) If the test charge is an electron (with mass 9.1 × 10-31 kg), how large is the acceleration it
experiences due to the passage of the current-carrying wire?
(d) In the wire loop’s rest frame the charge moves with initial velocity 100 m/s parallel to the
wire. Its acceleration is nearly the same in this frame. What is the radius of curvature of its path
at the instant that it is moving parallel to the wire?
Discussion
Exercise 6.4: How the fields transform (let’s call this one “optional” in case we don’t have
enough time)
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Relativistic effects will also deflect a moving test charge which heads directly towards the
current-carrying wire.
Let’s investigate this now, starting with the fact that Gauss’ law is always true, even for charges
in motion. The flux of electric field through a closed surface is always
∫ E idA = Qenclosed ε0 , even when the enclosed charge is in motion. (An experimental proof
surface
of this: the electric field far from a hydrogen atom is nearly zero, even though its proton is nearly
at rest, while its electron is moving with v ∼ c 137 .)
We’ll need to determine how the electric field at a point in space transforms as we shift frames
of reference. Let’s say that observers in two frames of reference, O (the "unprimed" frame) and
O', measure the electric field at the origins of their respective coordinate systems just as the
origins coincide.
Frame O' (according to observers in O) is moving with velocity v along the x axis of O as
shown in the figure. In the figure, the field measurements have already been performed.
O
O'
y
y'
v
x
z
z'
x'
electric field
lines
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(a) The field lines have been created by a pair of capacitor plates which are stationary in O as
shown in the next figure. The surface charge densities are ±σ .
O
O'
y
−σ
y'
v
x
z
x'
capacitor plates
z'
+σ
What are the surface charge densities σ ′ measured by observers in O'?
(b) What is E ′ E ?
Discussion
Fields which are perpendicular to the direction of motion... (Note: in O there is no magnetic
field.)
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.12
88
(c) In a different universe, field lines have been created by a pair of
capacitor plates which are stationary in O as shown, but whose
faces are perpendicular to the direction of motion of O'. As before,
the surface charge densities are ±σ .
capacitor
plates
O
What are the surface charge densities σ ′ measured by observers in
O'?
O'
y
y'
+σ
x
z
−σ
v
x'
z'
(d) What is E ′ E ?
Discussion
Fields which are parallel to the direction of motion... (Note: in O there is no magnetic field.)
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.13
89
Discussion
You should have concluded that E ′ = E and E⊥′ =
(
)
1 − v 2 c 2 E⊥ = γ E⊥ where E and E⊥
represent electric field components parallel to and perpendicular to the relative velocities of the
two frames. An important point: this is only true if there is no magnetic field in the frame in
which the electric charge (the surface charge on the capacitors) is at rest.
Naturally, this applies to electric fields created by any charge distribution, not just the sheets of
charge on capacitor plates. In particular, the electric field around a moving point charge becomes
stronger in the plane perpendicular to its direction of travel. If we use the density of field lines to
represent the strength of the electric field, we’ll find our field lines seem to bunch up like this:
v
charge, at rest in O
charge, as seen from O'.
To a test charge moving towards the current-carrying wire, the ions’ and electrons’ velocities
will appear as follows (in the rest frame of the test charge):
metal ions
electrons
test charge
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.14
90
(e) On the following diagram (which shows one pair of conduction electrons and metal ions,
placed symmetrically about the test charge), sketch the field lines at the test charge associated
with the two electrons. (In the test charge’s rest frame, the wire is moving towards the test
charge.)
test charge
Discussion
Direction of the force, and comparison with qv × B ...
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.15
91
Summary and wrap-up discussion
•
Magnetic force is just a consequence of relativity: observers in different frames do not agree
about charge densities!
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.16
92
Reading and homework should be completed by 4pm Friday.
Required reading:
Purcell, chapter 5, sections 4 - 6; chapter 5, section 9.
Read over (but don't worry about understanding) the Unit 7 material we'll tackle next week.
Optional reading:
The Feynman Lectures on Physics (volume II), chapter 26.
The Feynman Lectures on Physics (volume II), chapter 42.
(Some of these problems require you to know how to multiply matrices, and to remember that
the transpose of a matrix is what you get when you swap its rows and columns. Please see me
for a quick lesson if you've never learned about this.)
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1: More on Lorentz transformations
Imagine observers in the frame O see another frame O' moving in the positive x direction with
velocity v so that the origins of the two coordinate systems overlap at t = t' = 0. We can use the
Lorentz transformations to relate the space-time coordinates of an event in one frame with those
in another frame. Defining β ≡ v c and γ ≡ 1 1 − β 2 , the Lorentz transformations become
x′ = γ ( x − β ct ) , y′ = y, z ′ = z, t ′ = γ ( t − β x c )
x = γ ( x′ + β ct ′ ) , y = y′, z = z′, t = γ ( t ′ + β x′ c ) .
Keep in mind that v is the velocity of the primed frame's coordinate axes as seen from the
unprimed frame.
We can write the Lorentz transformations in matrix form this way:
 ct ′  γ −γβ
 x′   −γβ γ
 =
 y′   0
0
  
0
 z′   0
0
0
1
0
0   ct 
0   x 
0  y 
 
1  z 
and
 ct   γ +γβ
 x   +γβ γ
 =
 y  0
0
  
0
z  0
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
0
0
1
0
0   ct ′
0   x′ 
.
0  y′ 
 
1  z′ 
6.17
93
More compactly, defining
ct ′
 ct 
 γ −γβ
 x′ 
x
 −γβ γ




x′ ≡
, x≡
, and Λ ≡ 
 y′ 
 y
 0
0
 
 

0
 z′ 
z
 0
0
0
1
0
0
0 
0

1
allows us to write the Lorentz transformations as
x′ = Λ x ,
x = Λ −1
1
0
−1
−1
x′ where ΛΛ = Λ Λ = I ≡ 
0

0
0
1
0
0
0
0
1
0
0
0 
.
0

1
Any quantity which transforms this way under changes of reference frame is called a 4-vector.
Another four vector can be built from a particle's energy E and momentum p :
E c
 p 
p≡ x .
 py 


 pz 
since it transforms like this: p′ = Λ p . Naturally, this is just shorthand for the four equations
p′x = γ ( px − β E c ) , p′y = p y , p′z = pz , E ′ c = γ ( E c − β px ) .
 γ −γβ
 −γβ γ
(a) For Λ defined as 
 0
0

0
 0
0
0
1
0
0
 γ +γβ
 +γβ γ
0 
, prove that its inverse Λ −1 is 
 0
0
0


1
0
 0
0
0
1
0
0
0 
.
0

1
(b) A proton at rest has energy E proton = m p c 2 ≈ 938 MeV . (Its momentum, not surprisingly, is
zero.) The protons which circulate inside the Fermilab Tevatron have energies close to 1,000
GeV. (1 GeV = 1,000 MeV.) What value of γ does a Tevatron proton have?
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.18
94
(c) Somehow a Tevatron proton captures an electron, becoming a fast-moving hydrogen atom
without changing its speed. An electron at rest has Eelectron = me c 2 ≈ 0.511 MeV . What is the
electron's energy in the rest frame of Fermilab's sedentary buffalo herd?
Problem 2: Field transformation laws
As you've seen, the electric and magnetic fields are intimately connected by relativity. We can
construct the "electromagnetic field strength tensor" out of the fields' components like this:
 0 − Ex c − E y c − Ez c 
E c
0
By 
− Bz
x

.
F≡
 E y c Bz
0
− Bx 


0 
Bx
 Ez c − By
What makes this a "second rank tensor" (a 4-vector is a "first rank tensor") is that the electric
and magnetic fields transform when one changes frames this way:
 0
− E x′ c − E y ′ c − E z ′ c 


E ′ c
′
′
− Bz
0
By 
x

 = ΛF Λ T = ΛF Λ
′
F =
E ′ c B ′
− Bx′ 
0
z
 y

 E ′ c −B ′
′
Bx
0 
 z
y
If an observer in one frame of reference measures E and B at the point (x,y,z) at time t, that
observer can predict what an observer in a different frame would measure for E ′ and B′ should
their field-measuring devices glide past each other just as the measurements are being
performed.
(a) The electric and magnetic fields of a 1 Coulomb charge Q are measured by a pair of gizmos
as shown in the figure. From the perspective of observers in frame O, the charge is at rest at the
origin and one of the field-measuring devices is also at rest, with position (x,y,z) = (0,1,0).
Observers in O see the second field-measuring gizmo moving at high speed, with velocity
v = β cxˆ .
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.19
95
E
B
v=0
E
B
v=βc
1 meter
y
Q
x
z
At time t = 0 this gizmo passed very close to the stationary device at (x,y,z) = (0,1,0). The
readings on the dials of the moving device were photographed by observers in O as it was
illuminated by the lights from the stationary gizmo's panel lights.
•
Calculate the components of the electric and magnetic field vectors measured by the
stationary device at the time the photograph was taken.
•
Calculate the components of the electric and magnetic field vectors measured by the moving
device.
•
According to observers in the rest frame of the moving device, how far away is the charge
Q?
(b) The same gizmos are now placed on the x axis with the stationary device at position
(x,y,z) = (1,0,0) and the moving device traveling along the x axis with velocity v = β cxˆ . As the
moving device brushes past the stationary device, photographs are taken.
•
Calculate the components of the electric and magnetic field vectors measured by the
stationary device at the time the photograph was taken.
•
Calculate the components of the electric and magnetic field vectors measured by the moving
device.
•
According to observers in the rest frame of the moving device, how far away is the charge
Q?
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.20
96
Problem 3: And now for something completely different
The idealized “operational amplifier” shown in the figure has the following properties:
VV+
•
•
•
Vout
+
Vout = G(V+ - V-)
No current enters its inputs, regardless of how large the input voltages V+ and V- become.
The output voltage is always exactly Vout = G(V+ - V-) where G is the (constant) gain of the
amplifier.
The gain G is “nearly” infinite.
The inputs to which V+ and V- are connected are called the amplifier’s “non-inverting” and
“inverting” inputs, respectively. Note that voltages can be bipolar: both positive and negative
values are fine.
(a) In the following circuit, what is Vout as a function of Vin?
Vin
Vout
+
(b) What is Vout as a function of Vin for this circuit? Keep in mind that no current flows into the
amplifier’s input terminals and that Ohm’s law requires the voltage drop across a resistor to be
equal to the product of its resistance and the current flowing through it. .
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.21
97
R2 = 20kΩ
-
Vin
Vout = ?
R1 = 10kΩ
+
Physics 212, Spring 2003. Unit 6
The origin of the magnetic field as a consequence of special relativity
6.22
98
Physics 212
Honors topics in electrodynamics
Unit 7
Operational amplifier circuits
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.1
99
Unit 7: Operational amplifier circuits
Recap of last week
Introduction: opamps and other circuit elements
We’re going to build toys today. It’s a lot more fun to build electronic circuits with active
components (amplifiers and the like) than it is to be stuck in the passive-only world of electrical
circuits built only with resistors and capacitors. Our principal building block will be the
operational amplifier (opamp), a differential amplifier whose “open loop” gain is so high, and
input impedance so large, that it is always used with feedback to control its behavior.
Let’s discuss opamps using a particular example, the venerable LF411. This opamp is built on a
single sliver of silicon, and comprises twenty-two transistors, one diode, one capacitor, and
eleven resistors. You can find a description of it at the Texas Instruments web site
http://www.ti.com/: run a search on “LF411” to find the data sheets.
One of the many ways the chip is packaged is in an 8-pin dual-inline package (DIP):
IN- and IN+ are the “inverting” and “non-inverting” inputs to the opamp. OUT is, of course, the
opamp’s output pin. VCC+ and VCC- are the power supply pins, held somewhere between ±3.5
and ±18 volts. The pins BAL1 and BAL2 may be used to fine-tune the amplifier’s output when
the inputs IN+ and IN- are held at exactly the same voltage. We will leave them disconnected.
We’ll use a symbols like the following to represent the opamp, omitting the power and balance
pins.
or
+
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.2
100
(It’s OK to draw them with the non-inverting input at the top instead of the bottom.)
The output voltage is Vout = G (V+ − V− ) , where G is the opamp’s “open loop” gain. According to
Texas Instruments, the typical open loop gain for an LF411 is 200,000. (However, the output
cannot go higher or lower than the power supply voltages!) The LF411’s input impedance is
typically 1012Ω: increasing the input voltage by a volt will only change the current flowing into
the input pin by a picoamp. The device can drive output currents as large as 20 milliamps, and
output voltages within a few volts of VCC+ and VCC-, so it has enough “oomph” to drive a pair of
headphones without additional amplification. (Headphones have input impedances of several
dozen ohms and can handle input voltages of a couple of volts before melting.)
We NEVER use an opamp without feedback between its output and its inputs: its gain, “input
offset voltage,” and other parameters are not particularly stable against changes in power supply
voltage, temperature, and so forth. The following circuit is a BAD BAD THING!
~
Vin
-
Bad circuit!!!
+
Vout = -GVin
Here’s how to think about an opamp:
1. An opamp’s gain is (almost) infinite;
2. An opamp’s input impedance is (almost) infinite.
This sounds crazy, but there are three very useful rules of thumb which follow from this:
1. In a properly designed circuit (which won’t “pin” its output), the opamp does whatever is
necessary to make its inputs (almost) exactly equal;
2. (almost) Absolutely no current flows into the opamp’s input pins. (almost) As much current as
needed flows from (or to) the output pin to maintain its output voltage.
3. There must always be a DC path between the opamp output and (at least) one input, even
if it’s made with a huge resistor that does very little in setting the circuit’s behavior.
Here’s what you’ll find/work on today:
• Analysis (and construction!) of a few simple opamp circuits
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.3
101
Exercise 7.1: Opamp voltage follower
You analyzed this circuit as a homework problem. Be sure that you understand how the circuit
works. (Do you see why Vout as a function of Vin behaves the way it does?)
Vout = ?
Vin
+
Exercise 7.2: Mystery amplifier #1
This was another homework problem, and is the circuit that we have setup for you on the
breadboard at your lab bench. Make sure you are clear about how the circuit works.
R2 = 20kΩ
Vin
Vout = ?
R1 = 10kΩ
+
Discussion
What happens to Vout if we load the above circuit’s output, as shown?
Vin
+
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
Vout
load resistance
7.4
102
Hands-on time:
We’ve already assembled the simple “inverting amplifier” circuit in the previous exercise for
you, provided power to the opamp chip, attached a signal generator to its input, and set up the
oscilloscope to trigger sensibly. Note that we’ve put some capacitors on the board between
ground and the power supply voltages to smooth out noise and glitches on the power “rails.” The
capacitors are probably the kind that care about the polarity they see at their input leads: one
particular lead must always be more positive than the other. (Reversing the polarity can cause the
cap to explode, or catch fire, in a most dramatic fashion.)
Investigate the following with your circuit:
1.
2.
3.
4.
How close is the gain to what you’d expect?
How close to being equal are the voltages on the two input pins?
Change one of the resistors. Does the gain change in the way you expect?
Change (increase) the frequency of the input signal. At what frequency do you begin to
notice the amplifier’s gain dropping?
5. Change the input signal frequency again, keeping an eye on the phase difference between the
input and output signals. When does the phase shift begin to change noticeably?
Exercise 7.3: Mystery amplifier #2
What does the following (idealized) circuit do? (By this I mean what is Vout in terms of Vin?)
C = .01µF
Vin
Vout = ?
R = 100kΩ
+
Note that this circuit violates our third rule of thumb and is NOT A GOOD CIRCUIT!
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.5
103
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.6
104
Discussion and hands-on time:
An important design rule: due to imprecision in manufacturing, the point at which an opamp
produces zero volts at its output will be when its inputs are held at slightly unequal voltages.
This input offset voltage can be as large as several millivolts. Generally, it is necessary to have a
DC feedback path from the opamp’s output to an input. As a result, a better version of the
integrator circuit shown above would be the following:
22MegΩ
.01µF
Vin
Vout
100kΩ
+
The 22Meg resistor is small compared to the LF411’s input impedance, but large compared to
the 100k input resistor. The circuit’s behavior as an integrator will change to that of an ordinary
inverting amplifier for times which are not short compared to RC = 22Meg × .01µF = 0.22
seconds.
Build this circuit and see how it responds to sinusoidal, square, and triangle wave inputs. Vary
the signal generator’s frequency and observe how the circuit’s response to a sinusoidal input
changes (the circuit’s gain, in particular).
Exercise 7.4: Mystery amplifier #3
What does this circuit do?
R = 100kΩ
Vin
Vout = ?
C = .01µF
+
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.7
105
Discussion
The above circuit is vulnerable to high frequency noise (some of which is generated by the
amplifier itself) since its gain increases with frequency. If you need to use one of these, a better
version includes another (smaller) resistor and capacitor, as in the following circuit.
50pF
1kΩ
Vin
-
100kΩ
Vout
.01µF
+
Build one of these and see if it behaves as you’d expect.
Exercise 7.5: Mystery amplifier #4
A diode’s current-vs.-voltage behavior is I (V ) ≈ I S e 40V . The parameter IS (the “reverse saturation
current) depends on details of the diode’s construction and changes rapidly with temperature.
Here are a couple of voltage vs. current curves for a 1N914 silicon junction diode. Since the
current axis scale is logarithmic, the V vs. I curves are straight lines.
Let’s assume that IS is 10-15 amperes for the diode in the following circuit. As drawn, the diode
likes to pass current from left to right.
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.8
106
In the following circuit what is Vout as a function of Vin? (Assume that Vin ≥ 0.)
current flows in this direction
diode
-
Vin
Vout = ?
R = 1kΩ
+
Discussion
Since the diode will only pass current in one direction, a better scheme might be to use a pair of
diodes in parallel, with the second diode passing current from right to left.
Build the circuit with the pair of diodes as described above, and see how it behaves in response
to triangle waves.
Exercise 7.6: Do-it-yourself summing amplifier
How would you design an opamp circuit which sums two voltages so that Vout = -(Vin1 + Vin2)?
(Hint: one version of the circuit requires three resistors and one opamp.)
Discussion
Build an example of your circuit and make it work. (Figure out a clever way to have your
function generator provide a second input to your circuit.)
Summary and wrap-up discussion
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.9
107
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.10
108
Reading and homework should be completed by 4pm Friday.
Required reading:
Read over (but don't worry about understanding) the Unit 8 material we'll tackle next week.
Optional reading:
Horowitz & Hill, chapter 4 through section 4.06. (The book is on reserve in the Physics
Department library.)
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): Diode circuit
IS is 10-15 amperes for the diode in the following circuit. (As drawn, the diode likes to pass
current from left to right.) What is Vout as a function of Vin? (Assume that Vin ≥ 0. There is a
description of the current-voltage behavior of a diode in the in-class materials.)
R = 1kΩ
diode
Vin
Vout = ?
+
Problem 2 (10 points): Multiplying voltages
You now know about opamp circuits which incorporate diodes and resistors. (You will need the
results of the previous problem in order to work this problem.) Keeping in mind that
V1 × V2 = e
ln (V1 ×V2 )
= eln V1 + ln V2 ,
design a circuit whose output is
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.11
109
Vout = Vin1 × Vin2
where Vin1 > 0 and Vin2 > 0 . You should assume that the (ideal) diode behaves as described
previously, with IS = 10-15 amperes. (Your circuit will probably require several opamps.)
Problem 3 (10 points): Mrs. Urkin commissions a prototype for something more terrible
Anticipating a possible upturn in the market for horrible instruments of torture and punishment,
scientists in the UD&HE Research Division are charged with the responsibility of developing a
current-source poultry incinerator based on a high-voltage, high-power operational amplifier.
The potentiometer (a variable resistor) at the opamp's input controls the amount of current
passing through the dressed bird in a species-independent fashion. (For a given potentiometer
setting the same current will flow through a chicken as a turkey.)
1000 V
-
pin 1
pin 2
+
pin 3
potentiometer:
Rpin 1 to pin 3 fixed at 1000 Ω
Rpin 2 to pin 3 varies from 0 to 1000 Ω
Rpin 1 to pin 2 = Rpin 1 to pin 3 - Rpin 2 to pin 3
10 Ω
Determine the current I which flows through the bird as a function of the resistance Rpin 2 to pin 3
dialed into the potentiometer. (The utility of this circuit is that the current which flows through
the "load" [the bird] is independent of the properties of the load.)
Physics 212, Spring 2003. Unit 7
Operational amplifier circuits
7.12
110
Physics 212
Honors topics in electrodynamics
Unit 8
Electronic models of mechanical
systems; analog computers 1
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.1
111
Units 8 & 9: Electronic models of mechanical systems; analog computers
Recap of last week
Introduction: using opamp circuits to represent (differential) equations
Last week you analyzed (and built) a number of circuits which can be used to add voltages, take
logarithms, multiply voltages by constants, and so forth. I have installed a table of the circuits
(some will probably be new to you) on the last page of the unit 8 material.
As you may already know, it is possible to build circuits whose behavior mimics that of nonelectronic systems. I think it’s easiest to show this with an example.
Imagine we apply a driving force F(t) to a mass immersed in a viscous liquid so that the drag
force it experiences is proportional to velocity. We’ll have Fnet = F ( t ) + Fdrag or
ma = F ( t ) − β v (β is a positive constant, and the drag force opposes the velocity) so that
m dv dt = F ( t ) − β v . Rewrite this slightly so that the highest derivative is by itself on the left:
dv 1
β
= F (t ) − v .
dt m
m
A circuit which “solves” dv dt = F ( t ) m − β v m can be arranged like this:
1.
2.
3.
4.
5.
6.
7.
assume we have a voltage which correctly represents dv dt coming in from the left.
integrate it to produce a voltage which represents -v.
multiply the voltage corresponding to –v by the constant β/m.
receive (from a function generator) a voltage which corresponds to F(t)/m.
add the voltages to produce a voltage representing –(F(t)/m - vβ/m).
invert –(F(t)/m - vβ/m) to produce (F(t)/m - vβ/m).
bring this signal out from the right, inject it into the input of step 2’s integrator.
For example, this circuit produces an integral of its input signal:
22 MegΩ
Vin
R
C
+
Vout
Vout = −
1
Vin dt ( R
RC ∫
22 MegΩ)
.
Other functions are shown in the table at the end of this unit.
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.2
112
Here’s what we can do when β/m < 1:
22 MegΩ
R
C
-
R1
-v
+
+
R2
Integrate with RC=1
-vβ/m
F(t)/m
-vβ/m
-
Scale with
R2
( R1 + R2 ) = β
m
R
R
-F(t)/m + vβ/m
-
R
+
R
R
-
F(t)/m - vβ/m (= dv/dt)
dv/dt
+
Nillog
Invert
Add
We could have combined steps 2 and 3 ( reducing our component count) by picking the
integrator’s RC value to be m/β instead of 1:
22 MegΩ
dv/dt
R
C
-
-vβ/m
Integrate with RC=m/β
-vβ/m
R
R
F(t)/m
-
-F(t)/m + vβ/m
+
R
Nillog
Add
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
R
R
-
F(t)/m - vβ/m (= dv/dt)
+
+
Invert
8.3
113
If we are interested in seeing a representation of the object’s position we can add an extra
integrator to calculate x = ∫ v dt :
22 MegΩ
C
-
R
-vβ/m
R
+
C
-
x
+
Integrate with RC=m/β
Integrate with RC=β/m
-vβ/m
R
R
F(t)/m
-F(t)/m + vβ/m
-
R
+
R
R
-
F(t)/m - vβ/m (= dv/dt)
dv/dt
22 MegΩ
+
Nillog
Invert
Add
By watching the circuit’s voltages on an oscilloscope we’ll see a good representation of the
behavior of the mechanical system’s position and velocity.
To make scope work easier, let’s use one millisecond as the unit of time. This is easy to do:
instead of building the circuit so that RC = 1, or β/m, or m/β, we’ll use RC = 10-3, or β/m × 10-3,
or m/β × 10-3 instead. (For example, use R = 100 kΩ and C =.01 µF to obtain RC = 10-3.)
In addition, let’s imagine we’ve chosen a system where β/m ≈ 0.2. Defining f ( t ) ≡ F ( t ) m ,
our equation becomes
dv
= f ( t ) − 0.2v
or
dt
d 2x
dx
= f ( t ) − 0.2
2
dt
dt
if you prefer. The circuit becomes
F(t)/m – 0.2v (= dv/dt)
dv/dt
22 MegΩ
500kΩ
.01µF
100kΩ
-0.2v
+
100kΩ
Integrate
-
-F(t)/m + 0.2v
+
Add and flip sign
F(t)/m
100kΩ
Nillog
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
100kΩ
100kΩ
+
Invert
.
8.4
114
Exercise 8.1: Analytic solution
Imagine that you drive your mass-in-viscous fluid with a time-dependent driving force as shown
in the following figure.
f(t)
f0
t
The equation of motion for the velocity of the mass is dv dt = −0.2v when the force is zero and
dv dt = f 0 − 0.2v when the force is f0.
(a) Let’s call the time at which the force goes to zero t = 0. If the velocity at the instant the
force drops from f0 to zero is v0, prove that v(t ) = v0e −0.2t is a solution to the equation of motion
during the time when the force remains zero.
(b) Now let’s call the time at which the force changes from zero t = 0. If the velocity at the
instant the force increases from zero to f0 is v1, prove that v(t ) = ( v1 − 5 f 0 ) e −0.2t + 5 f 0 is a
solution to the eqution of motion as long as the force remains f0.
Exercise 8.2: Build it!
Build the three-opamp-circuit drawn above. Try driving it with 1 volt square waves of various
frequencies. How does your result for the voltage corresponding to -vβ/m correspond to the exact
solution from the prevuous exercise?
Discussion
You’ve just built an analog computer! Why might one want to do (to have done) that?
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.5
115
Optional material: Modeling a damped, driven simple harmonic oscillator
Introduction
A damped, driven harmonic oscillator experiences a net force with contributions from a spring
( Fspring = −kx ), a velocity-dependent drag ( Fdrag = − β v ), and a driving force (Fdrive). We can
write its equation of motion as Fnet = Fspring + Fdrag + Fdrive . After a small amount of manipulation,
the equation becomes
d 2 x (t )
Fdrive ( t )
β dx ( t )
k
=
−
−
+
.
x
t
(
)
dt 2
m dt
m
m
You may remember that the “natural frequency” of this oscillator is ω = k m radians/sec. Its
natural frequency in Hz (cycles/sec) is ω ( 2π ) .
For homework you designed a circuit which models this oscillator using the values k/m = 1 and
β/m = 0.2. Defining f ( t ) ≡ F ( t ) m , your oscillator’s equation becomes
d 2 x (t )
dx ( t )
=
−
0.2
− x (t ) + f (t ) .
dt 2
dt
Build it!
To make scope work easy, let’s use milliseconds as the unit of time so that your integrators will
have RC values in the vicinity of 10-3 instead of 1. (R = 100 kΩ and C =.01 µF gives RC = 10-3.)
The oscillator’s natural frequency is 1 ( 2π ) cycles per millisecond, or about 159 Hz.
Build your circuit, and try driving it with 100 millivolt sine waves of various frequencies.
Describe how the amplitudes of the voltages which correspond to x(t) and v(t) behave as you
change the frequency of the driving “force.” Also take note of the relative phases of x(t) and f(t).
Configure your oscilloscope to display an x-y plot of the voltages representing x(t) and v(t).
What do you see?
Discussion
Summary and wrap-up discussion
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.6
116
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.7
117
Opamp circuit function blocks
add voltages and flip sign (R1=R2=R3=R4)
R1
V1
-
R2
R1
V1
Vout
+
R2
-
V2
R4
 ( R + R2 ) R4 
 R2 
Vout =  1
 V2 −   V1
 R1 
 ( R3 + R4 ) R1 
R
R
= −  3 V1 + 3 V2
R2
 R1

Vout


Multiply by a positive constant < 1
Vin
-
R1
multiply by a positive constant > 1
R1
Vout
+
R2
Vout = +
R2
Vin
R1 + R2
Vin
R2
-
Vin
R1 + R2
Vin
R1
exponentiation
R2
diodes
-
Vin
Vout
+
Vin
Vout = −
differentiate
− sign (Vin )
40
C
Vin
R
-
Vout
⋅ ln Vin − ln ( I s R ) 
Vout = − RC
•
•
diodes
+
1 kΩ
Vout
1
Vin dt ( R 22 MegΩ)
RC ∫
take logarithm
Vin
Vout ≈
+
Vout
+
22 MegΩ
C
-
R
-
Vout ≈ − sign (Vin ) ⋅ I s R ⋅ e 40Vin
R
Vout = − 2 Vin
R1
integrate
R
Vout
+
Vout = +
Multiply by a negative constant
R1
Vout
+
R3
...
V2
subtract two voltages (R1=R2=R3=R4)
R3
( I s ∼ 10−15 )
dVin
(R
dt
50 pF
+
R
Vout
1 kΩ, C
50 pF)
Useful facts
RC has units of time.
If you’d like to have a millisecond act
as your unit quantity of time (this is
convenient for oscilloscope work),
chose RC to be 10-3. For example,
R = 100 kΩ and C =.01 µF should
work fine
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.8
118
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.9
119
Reading and homework should be completed by 4pm Friday.
Required reading:
Read over (but don't worry about understanding) the Unit 9 material we'll tackle next week.
Optional reading:
Horowitz & Hill, chapter 4 through section 4.06. (The book is on reserve in the Physics
Department library.)
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): No Venusians were found
Research electricians at Urkin’s have shown the following circuit to be unable to detect signs of
extraterrestrial life, though it has proved useful for other purposes. Calculate VA and VB in terms
of Vin.
Vin
VA
+
R
R
VB
+
Problem 2 (20 points): Doing A/B electronically
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.10
120
Design a circuit whose output is Vout = Vin1 / Vin2 where Vin1 > 0 and Vin2 > 0 . You should assume
that (ideal) diodes behave as described previously, with IS = 10-15 amperes. (Your circuit will
probably require several opamps.)
Physics 212, Spring 2003. Unit 8
Electronic models of mechanical systems; analog computers 1
8.11
121
Physics 212
Honors topics in electrodynamics
Unit 9
Electronic models of mechanical
systems; analog computers 2
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 9
Electronic models of mechanical systems; analog computers 2
9.1
122
Unit 9 is now included with the unit 8 writeup which covers both weeks.
Physics 212, Spring 2003. Unit 9
Electronic models of mechanical systems; analog computers 2
9.2
123
Reading and homework should be completed by 4pm Friday.
Required reading:
Read over (but don't worry about understanding) the Unit 10 material we'll tackle next week.
Optional reading:
Horowitz & Hill, chapter 4 through section 4.06. (The book is on reserve in the Physics
Department library.)
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (20 points): We are so clever, and so frugal
Design an opamp circuit which “solves” the equation
d 2 x (t )
dx ( t )
= − 0.2
− x (t ) + f (t )
2
dt
dt
using only four opamps. (f(t) [or -f(t) if it should prove convenient] is provided by a function
generator.) Be sure to include explanatory comments on your circuit diagram so we can tell
what you had in mind.
Problem 2 (10 points): Yep, it subtracts them!
Prove that the following circuit
V1
V2
R1
R3
R2
-
Vout
+
R4
produces an output voltage
 ( R + R2 ) R4 
 R2 
Vout =  1
 V2 −   V1 .
 R1 
 ( R3 + R4 ) R1 
Physics 212, Spring 2003. Unit 9
Electronic models of mechanical systems; analog computers 2
9.3
124
Physics 212
Honors topics in electrodynamics
Unit 10
Magnetic fields and the vector potential
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.1
125
Unit 10: Magnetic fields and the vector potential
Recap of last week
Introduction: A and B
Today we’ll investigate the connection between the vector potential and the magnetic field. It
should be clear that there’s some sort of link between the two: if no currents are flowing, both
the vector potential and the magnetic field are zero. In addition, both grow linearly with the
magnitude of the current. (In addition, recall the problem set for unit 6 in which you used the
Lorentz transformations to calculate the electric and magnetic fields of a moving charge by
transforming the electric field of a stationary charge.)
Since the vector potential is defined in terms of an integral over the current density J , there’s
also some sort of direct connection between the B and J . You’ll discover that it is the curl of
the magnetic field which is proportional to the current density.
Here’s what you’ll do today:
• Calculate the curl of the vector potential associated with a current-carrying wire
• Compare ∇ × A with B
• Calculate ∇ × A for a point charge moving with constant velocity
• Discover that ∇ × B = µ0 J
• Derive the Biot-Savart law
• Derive Ampère’s law
Discussion: A definition of the magnetic field, and a recap of your previous results
From Physics 112 you know that the (Lorentz) force on a moving charge is F = qE + qv × B .
Let’s use this as the definition of the magnetic field: it is what is responsible for the velocitydependent portion of the force acting on a charged particle.
You can immediately see that the observers in different frames will disagree about the values of
E and B since all the force in the rest frame of a particle must come from E . Both E and B
contribute to the force in a frame in which the particle is moving.
A few units back you worked up a calculation of the apparent charge density of a currentcarrying wire as seen by a moving test charge. Your calculation concerned a test charge moving
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.2
126
to the right with speed v parallel to a wire in which electrons flow to the right (as seen from the
wire’s rest frame) with speed u .
Your calculations should have revealed that the apparent net charge density on the wire
(according to the test charge) was λ = ( quv ) ( ac 2 ) C/m. q is the charge on a metal ion:
+1.6×10-19 C.
Further calculation would/should have shown that the force on the test charge Q was
F =λ
Q
quv Q
.
=
ac 2 2π rε 0
2π rε 0
The magnitude of the current flowing in the wire is the amount of charge per second that flows
past one point in the wire. Since the spacing between electrons (in the wire’s rest frame) is a
while the average electron velocity is u, there will be u/a electrons per second flowing past one
position in the wire. As a result, the magnitude of the current is I = qu a . We can use this to
rewrite the force as
F = Qv
1 1 I
≡ Qv × B .
ε 0c 2 2π r
From our definition of the magnetic field, we conclude that its magnitude a distance r from a
straight current-carrying wire must be
1 1 I
µ I
B =
= 0
2
ε 0c 2π r 2π r
since µ 0 = 1 ( ε 0c 2 ) . This agrees with the Physics 112 result. (The direction also agrees if you
work it all out correctly.)
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.3
127
Exercise 10.1: Line current vector potential
A few weeks ago we found that the vector potential for a non-relativistic charge is
µ
qi vi
µ
q
.
A( r ) ≈ 0
v . Use superposition for several charges: A ( r ) ≈ 0 ∑
4π r − r ′
4π all charges i r − ri
For a distribution of charge ρ ( r ) swirling around so that the distribution’s velocity v ( r )
depends on position, we can superpose the contributions from cells of volume dV containing
charge dq = ρdV this way:
A( r ) ≈
µ0 ρ ( r ′) v ( r ′) 3
µ0 J ( r ′) 3
d
r
d r ′ (d 3 r ′ is the same as dV ) .
=
′
∫
∫
4π
r − r′
4π r − r ′
all space
all space
The current density J is the number of Coulombs per second per square meter which flow
through a surface perpendicular to J . (The above equation is exact only if J does not change
with time.)
If the current flow is confined to a wire, it is convenient to “do the integral” along the two
coordinates perpendicular to the direction along the wire. For example, if we have a current I
flowing in a thin wire along the z axis…
z
y
x
the vector potential created by the current is A ( r ) =
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
z =+∞
µ0 I zˆ
dz ′ .
4π r − r ′
z =−∞
∫
10.4
128
(a) In Cartesian coordinates, the vector potential created by this current is
A ( x, y , z ) =
µ0
4π
+∞
∫
−∞
I dz ′
x 2 + y 2 + ( z − z ′)
2
.
Note that the integral diverges: for large z′ it behaves like ∫ dz′ z′ = ln ( z′ ) .
We encounter the same difficulty when we calculate the electrostatic potential from a line
charge:
+∞
1 ρ ( r ′) 3
1
λ
V ( x, y , z ) = ∫
d r′ =
dz ′ .
∫
2
2
2
r
r
4
4
−
πε
πε
′
0
0
all space
−∞
x + y + ( z − z ′)
One way around this would be to calculate the electric field instead (which behaves like 1/r),
then to do the integral to work with the potential difference between two points which are a finite
distance from the line charge.
Since we’re interested in showing that the curl of the vector potential is the same as the magnetic
field, we can circumvent the difficulty here by taking the curl of A , and moving the curl inside
the integral. This will yield a convergent integral:

µ
∇ × A ( x, y , z ) = ∇ ×  0
 4π
+∞
∫
−∞
Idz ′
x 2 + y 2 + ( z − z ′)
2

µI
= 0
 4π

1
∫ ∇ ×  2 2
2
−∞
 x + y + ( z − z ′)
+∞

zˆ  dz ′ .

Note that the partial derivatives of the curl act only on the unprimed variables x, y, z and not on
the primed variables x′, y′, z ′ .

1
Calculate ∇ × 
2
2
2
 x + y + ( z − z ′ )

ẑ  .

Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.5
129
(b) The vector potential is symmetric under rotations about the z axis, and is independent of the
value of z. As a result, we can work with points on the x axis (so that y = z = 0) without loss of
generality in order to simplify things. (We could do the calculation in cylindrical coordinates
instead.)
µI
Calculate ∇ × A ( x, y, z ) = 0
4π
(You may find that
∫
(x
dz′
2
+ z′

1
∫ ∇ ×  2 2
2
−∞
 x + y + ( z − z ′)
+∞
)
2 32
=
z′
x
2
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
x 2 + z′2

zˆ  dz ′ for points in the x axis.

is useful.)
10.6
130
(c) Compare your answer in part (b) with the magnetic field found using the relativity-based
calculation described earlier. (The results should agree!)
Discussion: is the magnetic field always the curl of the vector potential?
It looks like the vector potential as we’ve defined it is a good candidate for what we want! (In a
more advanced course you’ll probably encounter a proof that the definition we’re using for the
vector potential is correct. I’ll just assert that it’s true and forget about the formalism.)
Can we always write the magnetic field as the curl of the vector potential?
There seem to be no isolated sources or sinks of magnetic field lines anywhere in the universe.
(These would be the magnetic analogs of positive and negative electric charge.) Think of the
field lines created by a solenoid: they don’t begin (or end) anywhere. Consequently, the
magnetic version of Poisson’s equation, ∇i B = ρmagnetic ε 0 , becomes ∇i B = 0 .
Remember Helmholtz’s theorem? We didn’t get to it during unit 5, but you may have read about
it if you looked through the notes. Here’s how we can use it:
For any vector field B which dies away very far from the origin (not just the magnetic field), we
begin by defining the scalar and vector fields D and C so that D ( x, y, z ) ≡ ∇i B ( x, y, z ) and
C ( x, y, z ) ≡ ∇ × B ( x, y, z ) . According to Helmholtz’s theorem we can then reconstruct B from
D and C this way:
 1
B ( r ) = −∇ 

 4π
 1
D ( r ′) 3 
′
d
r
+
∇
×


∫ r − r′ 

all space

 4π
C ( r′) 3 
∫ r − r ′ d r ′  .
all space

If ∇i B ( x, y, z ) = 0 (so that D = 0), this reduces to
 1
 1
C ( r ′) 3 
∇′ × B ( r ′ ) 3 
′
B (r ) = ∇ ×
d
r
=
∇
×


∫ r − r′ 
∫ r − r ′ d r ′  ≡ ∇ × A .
 4π all space
 4π all space




The prime on the curl inside the integral indicates that its derivatives are with respect to primed,
not unprimed, coordinates.
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.7
131
We can always write a divergenceless function as the curl of some vector field. As a result, we
can always find a vector potential A which satisfies the equation B = ∇ × A :
A(r ) =
1 ∇′ × B ( r ′ ) 3 .
d r′
4π r − r ′
all space
∫
( )
One last point: since the curl of a gradient is always zero (you showed that ∇ × ∇V = 0 and
(
)
∇i ∇ × A = 0 a few weeks ago), we are always free to add the gradient of any old scalar
function to the vector potential without changing the magnetic field which results since
∇ × A + ∇V = ∇ × A + ∇ × ∇ V = ∇ × A = B .
(
)
( )
It turns out that this allows us to impose the condition on the vector potential that it has zero
divergence: ∇i A = 0 . (This isn’t a requirement for A , but it is sometimes convenient to
calculate a vector potential which satisfies this condition to simplify other calculations.)
Exercise 10.2: Magnetic field caused by a point charge moving with constant velocity
A point charge moves with constant velocity along the z axis so that its position is
( x′, y′, z′) = ( 0,0, vt ) . An observer positioned at ( x, y, z ) = ( x,0,0 ) measures the magnetic field
associated with the particle. (Assume that the particle is moving slowly so that retardation
effects can be ignored: during the time it would take light to travel from the charge to the
observer the charge’s position doesn’t change significantly.)
Make use of the vector potential to calculate the magnetic field as a function of time at the
observer’s position.
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.8
132
discussion
Note the 1/r2 dependence to the field as the charge passes through the origin.
discussion: cylindrical coordinates
We could have calculated B = ∇ × A using a non-Cartesian coordinate system. In fact, since the
problems we’ve done so far have all been symmetric under rotations about the z axis, it would
have made sense to use cylindrical coordinates, illustrated in the figure. Naturally, we’d find the
same values for B , but they would be parameterized in terms of ( r ,θ , z ) instead of ( x, y, z ) .
The unit vectors rˆ, θˆ, zˆ point in the direction of increasing r, θ, and z.
z
x
r
θ
y
In cylindrical coordinates the curl of the vector potential is
 1 ∂Az ∂Aθ
∇× A = 
−
∂z
 r ∂θ
1  ∂ ( rAθ ) ∂Ar

 ∂Ar ∂Az  ˆ
−
−
θ + 
 rˆ + 
∂r 
∂θ
r  ∂r

 ∂z
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential

 zˆ .

10.9
133
discussion: retarded time
We’ve concocted a couple of ways of calculating the magnetic field. The first uses relativity to
describe how the effects of Lorentz contraction and non-simultaneity make observers in different
frames disagree about charge density. The second involves a global calculation if we know the
“current density” ρ ( r ) v ( r ) everywhere. It is common to use J ( r , t ) ≡ ρ ( r , t ) v ( r , t ) to
represent the current density so that
 J ( r ′, t ′ )  3
µ
B ( r ,t) = ∇ × A( r ,t) = 0 ∫ ∇ × 
d r ′ .
4π all space
 r − r′ 
Note that the curl only acts on unprimed coordinates.
One subtle point is that the time t ′ inside the integral is the retarded time: at each value of r ′
that the integral touches, t ′ represents the present time t set back by the amount of time a light
beam would have required to arrive at the observer’s position ( r , no prime). The connection
between the two times is t ′ = t − r − r ′ c .
Sometimes the need to use the retarded time inside the integral can make it very difficult to
evaluate! Since the curl acts on unprimed coordinates, any time dependence in the current
density will cause it to “react” to the curl’s partial derivatives due to the presence of unprimed
spatial coordinates in the retardation term.
If the current density J depends on position but not on time (the currents never change), things
simplify somewhat.
Exercise 10.3: a microscopic version of some of this
Starting with the global version of Gauss’ law
∫
E idA =
surface
Qenclosed
ε0
=
1
∫
ε 0 volume
ρ ( r ) dV ,
we were able to construct a microscopic (local) version
∇i E =
ρ
ε0
several weeks ago.
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.10
134
Let’s do the same thing for the magnetic field now: we’ll establish a connection between its curl
and the current density, assuming the current density is constant in time. There’s a hard way to
do this, which proceeds as follows:
•
(
)
(
∇ × B ( r ) = ∇ × ∇ × A ( r ) . Show that, for example, the x component of ∇ × ∇ × A
(
)
( )
)
satisfies  ∇ × ∇ × A  = ∂ ∇i A ∂x − ∇ 2 Ax .

x
•
Since we are able to work with vector potentials for which ∇i A = 0 , ∇ × B ( r ) x = −∇ 2 Ax .
There are similar equations for the y and z components which can be written this way:
( ∇ × B)
( ∇ × B)
( ∇ × B)
xˆ = −∇ 2 Ax xˆ 


yˆ = −∇ 2 Ay yˆ 
y

2
zˆ = −∇ Az zˆ 
z

x
⇒ ∇ × B = −∇ A .
2
Rewrite ∇ × B = −∇ 2 A so that the right side if the equation is expressed in terms of the
current density J ( r ) . (You could find what you need in a previous unit.)
(a) You should demonstrate this the easy way instead: compare the expression for the vector
potential in the earlier discussion (which refers to the Helmholtz formula) with our original
definition of the vector potential.
From Helmholtz:
A(r ) =
1 ∇′ × B ( r ′ ) 3 .
d r′
4π r − r ′
all space
∫
From our original definition of the vector potential:
A( r ) =
µ0 ρ ( r ′) v ( r ′) 3
1 µ0 J ( r ′) 3
d
r
=
′
∫ 4π r − r ′
∫ 4π r − r ′ d r ′ .
all space
all space
By inspection, what’s ∇ × B in terms of the current density J ?
Exercise 10.4: Biot-Savart law
Let’s consider a current density which is confined to a thin wire. Earlier, when you calculated
the vector potential for a wire running along the z axis, you found that
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.11
135
A ( x, y , z ) =
µ0 J ( r ′ ) 3
µ
d r′ = 0
∫
4π r − r ′
4π
all space
+∞
∫
−∞
I dz ′
.
r − r′
Define dl to be a short distance along the wire in the direction that current is flowing so we can
rewrite the integral more generally, for arbitrary wires:
A ( x, y , z ) =
∫
along
wire
µ0 Idl
.
4π r − r ′
This suggests that we write
A( r ) =
∫
dA =
so that
dA
contributions
from entire
wire
µ0 Idl
4π r − r ′
and
B = ∇× A =
∫
∇ × dA
so that
dB =
contributions
from entire
wire
(a) Evaluate dB =
 Idl 
µ0
∇×
.
4π
 r − r′ 
 Idl 
µ0
∇×
 for a small segment of current at the origin flowing in the z
4π
 r − r′ 
direction: Idl = I zˆ dz . Assume that the observer is located at r as shown in the diagram (in the
y-z plane), and that r ′ = 0 . (Keep in mind that the curl acts only on unprimed coordinates.)
After grinding through the differentiation, reexpress your answer in terms of θ and r .
z
observer
θ
r
I dl
y
x
(b) Rewrite your expression for dB in terms of dl × rˆ , where r̂ is a unit vector which points
from the small current element Idl (at the origin here) towards the observer’s location.
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.12
136
discussion
•
Written this way (using a cross product), your expression is independent of the choice of
coordinate system: very convenient!
•
You should have obtained dB =
µ0 I dl × rˆ
. This is the Biot-Savart law, named for Jean4π r 2
Baptiste Biot (1774-1862) and Felix Savart (1791-1841), two French mathematicians. This
is the magnetic equivalent of Coulomb’s law: the contribution to B from a small amount of
current somewhere follows an inverse-square law.
•
If we have to contend with current densities instead of currents in thin wires, we’ll need to
replace I dl with JdV in our expression for dB .
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
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137
Exercise 10.5: Ampère’s law
A short time ago you found that ∇ × B = µ 0 J . Using Stokes’ theorem, you’ll now derive the
macroscopic version of this, known as Ampère’s law (named for the French mathematician and
physicist André Ampère [1775-1836]). Imagine we have a current density J flowing through a
loop, as shown in the diagram. We can integrate both sides of the equation ∇ × B = µ 0 J over the
surface enclosed by the loop:
z
∫
∇ × BidA
surface
enclosed
=
∫
µ0 J idA .
J
surface
enclosed
y
x
Use Stokes’ theorem to rewrite the left side as a line integral of the magnetic field around the
loop. Note that the details of how the surface extends across the loop do not matter: it can be as
lumpy as we want without spoiling the connection between the line integral and the surface
integral guaranteed by Stokes’ theorem.
Now rewrite the right side as an expression involving the total current I which flows through the
loop. Ampère’s law!
discussion
Use of Ampere’s law with cylindrical symmetry
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.14
138
Summary and wrap-up discussion
•
•
•
•
•
Calculating the curl of the vector potential for a current-carrying wire
∇ × A is the same as B
∇ × A for a point charge moving with constant velocity
µ I dl × rˆ
The Biot-Savart law: dB = 0
4π r 2
Ampère’s law:
∫ B ⋅ dl = µ0 Ithrough
perimeter
surface
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.15
139
Reading and homework should be completed by 4pm Friday.
Required reading:
Purcell, chapter 2, sections 13 - 16; chapter 6, section 3.
Optional reading:
The Feynman Lectures on Physics (volume II), chapter 14.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (10 points): Les belles choses qu’on peut dessiner
You sketched a vector potential, including retardation effects, as the last problem of unit 5’s
homework assignment. Reproduce your sketch, and then draw on the sketch the vectors
corresponding to the magnetic field associated with (the curl of ) this vector potential. (As in the
unit 5 problem, only indicate the field on the x axis.)
Problem 2: Mrs. Urkin’s industrial jell-o dispenser
To speed its distribution of electrically charged jell-o, Urkin’s D&HE builds a pipeline from the
Emporium to the central New Jersey rail head at which tank cars are to be filled with the
product. You may approximate the pipe as a 2 kilometer long straight cylinder of radius 0.1
meter. The jell-o flows due north through the pipe at 10 m/s and carries uniform charge 1 µC per
cubic meter.
(a) Calculate the current density J inside and outside the pipe.
(b) Calculate the curl of the magnetic field ∇ × B inside and outside the pipe. (Note that you do
not need to know B to calculate ∇ × B here! See exercise 9.3 in the class materials if you’re
stuck.)
(c) Calculate the value of the vector potential A along the centerline (r = 0) of the pipe midway
between its ends. (It’s easiest to do this problem in cylindrical coordinates. Keep in mind that
the volume element in your integral will be dV = rdrdθdz, and not drdθdz.)
(continued next page...)
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
10.16
140
(d) Recall Stokes’ theorem, true for any vector field B (not just a magnetic field):
∫ ( ∇ × B ) ⋅ dA = ∫
surface
B ⋅ dl .
perimeter
of surface
Use this to determine the magnetic field B as a function of distance from the center of the pipe.
Physics 212, Spring 2003. Unit 10
Magnetic fields and the vector potential
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141
Physics 212
Honors topics in electrodynamics
Unit 11
Maxwell’s equations
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.1
142
Unit 11: Maxwell’s equations
Recap of last week
Introduction: Establishing a connection between E and B
We’ve worked up a number of connections between electric fields and charge distributions,
magnetic fields and current distributions, potentials and fields, and so forth. In many ways our
equations are still incomplete, however: some of them are only true for static charge
distributions, or for currents which never change. In addition, we don’t have anything which
provides a direct link between the electric and magnetic fields, even though we know that special
relativity requires a connection between them. Here’s a table which summarizes what we know
so far.
Scalar and vector potentials in terms of charge and current distributions
static charges, currents
1 ρ ( r ′) 3
V (r ) = ∫
d r′
4πε 0 r − r ′
all space
A( r ) =
µ0 J ( r ′ ) 3
d r′
∫
4π r − r ′
all space
static charges, currents
Electric and magnetic fields in terms of charge and current distributions
1 ρ ( r ′ ) r − r ′ 3 (Coulomb’s law)
static charges, currents
E (r ) =
d r′
2
∫
all space
B ( r1 ) =
4πε 0 r − r ′ r − r ′
µ0 I J × rˆ2→1 3
d r2 (Biot-Savart law)
4π r22→1
space
∫
static charges, currents
Electric and magnetic fields in terms of potentials
static charges, currents
E = −∇V
always true
B = ∇× A
Divergences (flux integrals) of electric and magnetic fields
Q
ρ
always true
∇i E =
or ∫ E idA = enclosed (Gauss’ law)
ε0
ε0
surface
always true
∇i B = 0 (No magnetic charge)
Curls (line integrals) of electric and magnetic fields
static charges, currents
∇× E = 0
static charges, currents
∇ × B = µ0 J or ∫ ∇ × BidA = µ0 I through
surface
surface
Today we’ll work on “correcting” the curl equations to make them accurate even when charges
move and currents are time-dependent.
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.2
143
Here’s what you’ll be working on:
• using Stokes’ theorem and Ampère’s law to investigate the connection between changing
electric fields and the existence of non-zero magnetic fields
• developing the connection between changing magnetic fields and the existence of non-zero
electric fields.
• Maxwell’s equations!
Exercise 11.1: Stokes’ theorem, applied to two different surfaces
A current I flows through a wire, as shown in the following figure. The wire passes through two
different surfaces: the first is a flat circle, while the second resembles an already-opened can of
tuna.
first surface
“tuna can” is
open here
I
this loop encloses
the (entire) surface
surface B
this loop encloses
the (entire) surface
surface A
The two loops which bound the surfaces are identical in size and shape. The part of the second
surface labeled “surface A” is a cylinder, while “surface B” is a flat circle.
(a) Evaluate
∫ ∇ × BidA
for surface A only in terms of I. (Divide it into two pieces to obtain a
surface
A
pair of surfaces with simply-connected boundaries if you like. Stokes’ theorem will be helpful to
you here.)
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.3
144
(b) Prove
∫ ∇ × BidA
first
surface
(c) Evaluate
∫
=
∫ ∇ × BidA . (Making use of Stokes’ theorem should help.)
second
surface
Bidl in terms of I. Do the same for
first
loop
∫
Bidl .
second
loop
discussion
The integrals are independent of details of shapes of surfaces and bounding curves.
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.4
145
Exercise 11.2: Another version of the last two surfaces
A current I flows through a wire, charging a parallel-plate capacitor, as shown in the following
figure. The surfaces shown are similar to those in the previous exercise: the first surface is a
circular disk, perpendicular to the wire, seen edge-on. The second looks like a soup can with the
top removed: the open end is on the left. The closed end of the soup can passes between the
capacitor plates. As before, line integrals around the boundaries of the two surfaces involve
traveling along circles of identical radii, indicated in the figure. You should assume that the
capacitor plates are very close together so that the magnetic fields on the circles labeled
“boundaries of surfaces” are identical.
boundaries of surfaces
parallel-plate
capacitor
I
I
first surface
(a) Evaluate
∫
second surface
Bidl in terms of I. (Since the magnetic field is parallel to the boundary
boundary
of first
surface
used in your integral, you could have used
∫
Bidl = 2π r B if you had known the value of
boundary
of first
surface
the magnetic field.)
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.5
146
(b) It is certainly true that
∫
Bidl =
boundary
of first
surface
∫
Bidl since the magnetic fields are the same along
boundary
of second
surface
the two boundaries. Must it still be true that
∫ ∇ × BidA
first
surface
=
∫ ∇ × BidA ?
(Remember Stokes’
second
surface
theorem!)
discussion
But I through the surface is zero! What’s going on?
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.6
147
Exercise 11.3: The electric field gets into the act
There must be a magnetic field between the capacitor plates while it is charging up. There’s no
way to have a non-zero line integral along the boundary of that soup can without having a nonzero curl to the magnetic field: that would violate Stokes’ theorem, which is true for any vector
field. If the magnetic field were zero over a finite (non-pointlike) volume of space, its curl would
also be zero. We are forced to conclude that something going on between the capacitor plates is
creating a magnetic field so that Stokes’ theorem is satisfied.
This is an especially nice example of how mathematics can lead us to a better understanding of
physical reality: it is impossible for Stokes’ theorem to be wrong. As a result, it is impossible for
the magnetic field between the capacitor plates to be zero, even though there is no current
flowing across the gap between the plates.
We know there’s no current between the plates of an already-charged capacitor: with no charges
in motion there can be no relativistic effects giving rise to a magnetic field due to different
Lorentz contraction effects on (stationary) positive and (stationary) negative charges. It must be
something associated with the change in the electric field between the plates that is giving rise to
a magnetic field.
(a) Let’s assume that our capacitor plates each have area A and are uncharged at time t = 0.
Assume the current flowing onto the left-hand plate is I, and is constant. Calculate the electric
field between the plates as a function of time, E(t). (If you don’t remember how to do this from
Physics 112, use Gauss’ law, and keep in mind that the electric field is zero outside the plates.)
(b) Now calculate
∂E
.
∂t
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.7
148
(c) We know that:
∫
∫
Bidl =
boundary
of second
surface
∇ × BidA = 2π r B = µ 0 I
second
surface
and also that
∂E
I
=
∂t Aε 0
so that
∫
second
surface
∂E
I
idA =
ε0
∂t
and also that
ε 0 = 8.854 × 10−12 ,
µ 0 = 4π × 10−7 ≈ 1.257 ×10−7 , ε 0 µ 0 = 1 c 2 .
Fill in the missing terms in curly brackets:
∫
∇ × BidA =
second
surface

∂E 
expression
in
terms
of

idA
∫
∂t 
second 
surface
so that

∂E 
∇ × B = expression in terms of
.
∂t 

Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.8
149
Discussion
In the absence of currents, you’ve just proposed a connection between the curl of the magnetic
field and the time derivative of the electric field:
∇ × B = µ 0ε 0
∂E 1 ∂E
.
=
∂t c 2 ∂t
In integral form this can be written
∫
∇ × BidA =
surface
∫
Bidl =
boundary
of surface

1 ∂E
1 ∂ 
idA = 2  ∫ E i dA
2
c ∂t
c ∂t  surface

surface
∫
so that
∫
boundary
of surface
Bidl =
1 ∂Φ E
c 2 ∂t
where ΦE is the flux of electric field through the surface.
Imagine that we had the following situation, in which a current passes through part of our
surface, while a changing electric field crosses another part of the surface, as shown in the
figure.
I1
I2
surface
parallel-plate
capacitor
In this case we’ll expect the curl to contain contributions from both the current and the changing
electric field:
1 ∂E
.
∇ × B = µ0 J + 2
c ∂t
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.9
150
Exercise 11.4: Now to investigate the curl of E ...
This one’s easier to get. We’ll start with a Physics 112 approach, then turn the integral
expression into a differential one.
(a) A rectangular conducting loop slowly glides with speed v into a region in which there is
magnetic field B , pointing in the positive z direction as shown in the figure. The mobile charge
carriers inside the conductor each have positive charge q.
magnetic field
z
B
a
y
b
x
v
loop
no magnetic field here:
B=0
Calculate the force (magnitude and direction) on a charge q (due to its motion through the
magnetic field) in the side of the loop which has just entered the magnetic field..
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.10
151
(b) Somewhat later, half the moving rectangular loop has entered the magnetic field as shown.
z
y
v
x
Calculate the direction of the magnetic force on charges q in the sides of the loop which are
parallel to the loop’s direction of motion. (Assume that the charges are almost at rest with
respect to the wire of the loop.)
(c) In the rest frame of the loop, each mobile charge will ascribe the force it experiences to the
presence of an electric field. Calculate ∫ E idl in the loop’s rest frame when it has
conducting
loop
partially entered the magnetic field. Note that the line integral involves traveling around the
loop in the positive sense, as given by the right hand rule. (The path around the loop is
counterclockwise, when seen from above.)
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.11
152
(d) Notice that the effects of the induced electric field disappear as soon as the entire loop is
inside the field: when d Φ B dt = 0 , ∫ E ⋅ dl = 0 . (The magnetic flux is Φ B = ∫ B ⋅ dA : positive
loop
flux is in the positive z direction.)
Prove that
∂Φ B
= − ∫ E idl for this loop. (This is Faraday’s law, named for English chemist
∂t
loop
Michael Faraday [1791 – 1867]. It is true in general, not just for our rectangular loop moving
with constant velocity.)

∂Φ B ∂ 
=  ∫ B ⋅ dA = − ∫ E idl . This also works in the rest frame of the
∂t
∂t  loop

loop
loop, where the magnetic field sweeps across the (stationary) loop. In fact, it also holds if the
magnetic field is spatially constant across the loop, but varies with time. As a result, we can pull

∂Φ B ∂ 
∂B
=  ∫ B ⋅ dA = ∫
⋅ dA = − ∫ E idl .
the time derivative inside the integral to write
∂t
∂t  loop
loop
 loop ∂t
(e) You know that
Use Stokes’ theorem to rewrite the rightmost integral as a surface integral of the curl of the
electric field:
−
∫
{
E idl = something involving ∇ × E
}
loop
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.12
153
(f) What should go between the curly brackets?
∂B
= something involving ∇ × E
∂t
{
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
}
11.13
154
discussion
We can correct the last several rows in the table at the start of this unit. The following two tables
hold equivalent statements: the first are differential, the second integral equations which describe
exactly the same physics. These are known as Maxwell’s equations, named for James Clerk
Maxwell (1831 – 1879), Scottish physicist and mathematician. They appeared in his 1873 work
Treatise on Electricity and Magnetism. They represent one of the most profound intellectual
achievements that the human species has yet produced. Fully relativistic, these equations (and
extensions of them) have served as the basis for much of modern physics.
Divergences of electric and magnetic fields
ρ
always true (Gauss’ law)
∇i E =
ε0
always true (No magnetic
charge)
Curls of electric and magnetic fields
always true (Faraday’s law)
∂B
∇× E = −
∂t
always true (Ampère’s law)
1 ∂E
∇ × B = µ0 J + 2
c ∂t
∇i B = 0
∫
Flux integrals of electric and magnetic fields
Q
always true (Gauss’ law)
E idA = enclosed
ε0
surface
∫
always true (No magnetic
charge)
B ⋅ dA = 0
surface
Line integrals of electric and magnetic fields
∂Φ B
always true (Faraday’s law)
E ⋅ dl = −
∫
∂t
perimeter
∫
perimeter
B ⋅ dl = µ0 I through +
surface
1 ∂Φ E
c 2 ∂t
always true (Ampère’s law)
The differential versions of Maxwell’s equations relate the behavior of the electric and magnetic
fields (and their derivatives) at all points in space. It does not matter what is done to produce the
fields: once they are known to exist they MUST obey the Maxwell equations.
Often one works with electric and magnetic fields external to the region where there are currents
or free charges. (This way ρ = 0 and J = 0 .) When this is the case, Maxwell’s equations
simplify to
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.14
155
Maxwell’s equations in the absence of charges and
currents
∇i E = 0
∇i B = 0
∇× E = −
∂B
∂t
∇× B =
1 ∂E
c 2 ∂t
Note the symmetry between the electric and magnetic fields in Maxwell’s equations. The
presence of 1/c2 in only one of the curl equations is a consequence of our ill-advised choice of
units for the magnetic field. If we had used, say, the “Nillog” (where 1 Nillog = c × 1 Tesla), the
revised curl equations would each have a 1/c coefficient in front of the time derivative.
The negative sign in the ∇ × E equation is important: it’s why Lenz’s law works. Without it, a
change in magnetic flux through a wire loop would induce a current in the wire which would
increase the change in magnetic flux, inducing an even stronger current. The current would rise
exponentially, vaporizing the wire.
The positive sign in the ∇ × B equation is needed so that the changing electric field between
capacitor plates will create a magnetic field whose curl matches that of the field made just
upstream/downstream of the capacitor (where there is a current flowing in the circuit).
As long as we include time retardation, we can still calculate the vector potential and use it to
extract the magnetic field:

r − r′ 
J  r ′, t −
c  3
µ0 
A( r , t) = ∫
d r ′ and B = ∇ × A .
4π
r − r′
all space
When there are free charges in motion we can calculate a scalar potential in a similar fashion:
V (r, t) =
∫
all space
1
4πε 0


r − r′ 
c  3
d r′ .
r − r′
ρ  r ′, t −
Note that the vector potential also lets us calculate the electric field when ρ = 0 (so that
∇ ⋅ E = 0 ) in our part of the universe:
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.15
156
∇× E = −
∂∇ × A
∂A
∂B
= −∇ ×
, yielding
so ∇ × E = −
∂t
∂t
∂t
E=−
∂A
.
∂t
The more general case (ρ ≠ 0) requires the inclusion of the −∇V term.
To summarize:
Scalar and vector potentials in terms of charge and current distributions


r − r′ 
r − r′ 
ρ  r ′, t −
J  r ′, t −

c  3
c  3

µ0 
1
V (r, t) = ∫
d r′
A( r , t) = ∫
d r′
4
πε
−
−
π
r
r
4
r
r
′
′
0
all space
all space
Electric and magnetic fields in terms of potentials
∂A
B = ∇× A
E = −∇V −
∂t
Divergences of electric and magnetic fields
ρ
∇i B = 0
∇i E =
ε0
Curls of electric and magnetic fields
1 ∂E
∂B
∇ × B = µ0 J + 2
∇× E = −
c ∂t
∂t
Summary and wrap-up discussion
Here’s what you were working on:
• using Stokes’ theorem and Ampère’s law to investigate the connection between changing
electric fields and the existence of non-zero magnetic fields
• developing the connection between changing magnetic fields and the existence of non-zero
electric fields.
• Maxwell’s equations!
Comments about homework assigned for next week
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.16
157
Reading and homework should be completed by 4pm Friday.
Required reading:
nothing this week.
Optional reading:
Purcell, chapter 9 (you’ll have to borrow it from the library).
The Feynman Lectures on Physics (volume II), chapter 4, section 1.
The Feynman Lectures on Physics (volume II), chapter 18.
Recall that the Maxwell equations are
Sources (charges and currents)
are present
ρ
ε0
∂B
∂t
1 ∂E
∇i B = 0
∇ × B = µ0 J + 2
c ∂t
Sources are absent (ρ = 0, J = 0)
∂B
∇i E = 0
∇× E = −
∂t
1 ∂E
∇i B = 0
∇× B = 2
c ∂t
∇i E =
∇× E = −
They are always true for electric and magnetic fields. It is impossible to find a time-varying
electric field which is not accompanied by a magnetic field with non-zero curl.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as
last week’s problem set and write a (true!) statement on your problem set asserting that you have
successfully printed your personal copies.
Problem 1 (15 points): How can we make these fields satisfy Maxwell’s equations?
The electric and magnetic fields that exist in some universe which is empty of charges and
currents are known to be given by
E ( x, y, z , t ) = E0 f ( kz − ω t ) xˆ and B ( x, y, z, t ) = B0 f ( kz − ω t ) yˆ
where E0, B0, k, and ω are constants, and f is a function. Since these fields do exist, they must
satisfy all four of the Maxwell equations.
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.17
158
(a) Prove that these fields satisfy the divergence equations ∇i E = 0 and ∇i B = 0 .
(b) Show that it is also possible for the fields to satisfy the curl equations ∇ × E = −
∂B
and
∂t
1 ∂E
if the ratios E0 B0 and ω k are chosen properly, and determine the required
c 2 ∂t
values of E0 B0 and ω k .
∇× B =
(c) Imagine that the function f has a peak when its argument is zero (so that the peak is at z = 0
at time t = 0, but is located elsewhere when t ≠ 0 ). Calculate the speed with which this peak
travels, making use of your answers to part b.
In an entirely different universe, Mrs. Urkin realizes that she had only partially folded the egg
whites into the chocolate mousse that she has just placed before her dinner guests. Fortunately,
the circular traces of egg in the dark chocolate background remind everyone of a simple electric
field with non-zero curl, namely
E ( x, y, z ) = E0 [ yxˆ − xyˆ ] .
Her guests notice that the local magnetic field is zero at the instant that dessert is placed on the
table and that, to their amazement, they are actually immersed in the time-independent electric
field E ( x, y, z ) = E0 [ yxˆ − xyˆ ] . (There are no charges or currents in the vicinity of the dining room
table.)
(d) Sketch the electric field and calculate its curl ∇ × E .
(e) Solve for the magnetic field B ( x, y, z , t ) in Mrs. Urkin’s dining room. Keep in mind that the
Maxwell equations are always satisfied by real electric and magnetic fields.
Problem 2 (15 points): Deriving the wave equation
The Maxwell equations couple the electric and magnetic fields: a changing electric field creates
a magnetic field with non-zero curl; a changing magnetic field creates an electric field with nonzero curl. In this problem you’ll manipulate the Maxwell equations to produce a pair of
decoupled wave equations for the electric and magnetic fields.
Imagine that an electric field E ( x, y, z , t ) = Ex ( x, y, z , t ) xˆ + E y ( x, y, z , t ) yˆ + Ez ( x, y, z, t ) zˆ is known
to exist in space.
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.18
159
(
)
(a) With the help of the Maxwell equations, prove that ∇ × ∇ × E = −
(
)
(
∂ ∇× B
∂t
) =− 1 ∂ E.
2
c 2 ∂t 2
(b) Show that the z component of ∇ × ∇ × E satisfies the following:
(
)
 ∇ × ∇ × E  = −∇ 2 ( Ez ) +

z
(
)
(
∂ ∇⋅ E
∂z
).
(The equations for the x and y components of ∇ × ∇ × E are similar to this, allowing us to
(
)
( )
(
)
combine the three equations to write ∇ × ∇ × E = −∇ 2 E + ∇ ∇⋅ E .)
(c) If the charge density ρ is zero we’ll have ∇ ⋅ E = ρ ε 0 = 0 so that
−
(
)
2
1 ∂ 2 Ez 
 = −∇ 2 ( Ez ) , yielding the wave equation ∇ 2 ( Ez ) − 1 ∂ Ez = 0
=
∇
×
∇
×
E

z
c 2 ∂t 2
c 2 ∂t 2
for Ez. (If you like, you can combine this with the wave equations for the x and y components
1 ∂2 E
2
into the single equation ∇ E − 2 2 = 0 .)
c ∂t
( )
Derive the relationship between k and ω necessary to make E ( x, y, z , t ) = E0 cos ( kz − ω t ) xˆ a
solution to the wave equations. (It should look familiar to you if you’ve already done the first
problem!)
Note that you can derive the wave equation for the magnetic field in a similar fashion:
•
•
•
( )
⇒
( )
∇ × ( ∇ × B ) = −∇ ( B ) + ∇ ( ∇ ⋅ B ) = −∇ ( B ) ⇒
∇× ∇× B = +
1 ∂ ∇× E
1 ∂2 B
=
−
c2
∂t
c 2 ∂t 2
2
2
1 ∂2 B
∇ B − 2 2 = 0.
c ∂t
2
( )
(d) Recall where that 1/c2 factor in front of the time derivative came from: 1/c2 is really just the
product of µ0 and ε0. The constant ε0 comes from Coulomb’s law: it sets the scale for the
strength of the Coulomb force between a pair charges. The constant µ0 comes from the
connection between the amount of current flowing in a wire and the magnitude of the resulting
magnetic field a fixed distance from the wire.
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
11.19
160
Imagine a different universe in which the speed of light is ten times greater, but in which the
Coulomb force is the same as it is for us. How would magnetic fields created by currentcarrying wires compare in the two universes?
Physics 212, Spring 2003. Unit 11
Maxwell’s equations
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Physics 212
Honors topics in electrodynamics
Unit 12
Visualizing fields and potentials with
Mathematica
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 12
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Unit 12: Visualizing fields and potentials with Mathematica
Recap of last week
Introduction: Visualization and Mathematica
We’ve been working with fairly abstract quantities—electric and magnetic fields and their
derivatives, as well as scalar and vector potentials (and their derivatives).
Let’s look into using the software package Mathematica to generate plots of potentials and fields
in order to better visualize them. This week will serve as an introduction to Mathematica. Next
week we’ll use the program as a tool to investigate the nature of the fields associated with a
moving charge.
We’ll be working in Loomis 257 where a large number of personal computers are available to
run Mathematica. It’s likely that the PC’s are running Windows NT, and are loaded with
Mathematica version 4.
If you’ve never used Mathematica, you’re likely to find that its interface is fairly intuitive, and
that the software behaves quite sensibly once you’re used to it. It’s a good tool, and you may
find it useful in a number of other courses. It is able to do symbolic manipulation of equations
and all sorts of other good stuff.
Most of your interactions with Mathematica are through a window displaying a notebook, which
will contain your instructions to Mathematica and the results of its efforts.
I’ve been noticing a certain tendency for Mathematica to spit out a dozen error message when it
corrupts some of its internal settings files. According to Wolfram Research, sometimes this can
be fixed by restarting the program while holding down the ctrl and shift keys simultaneously. (I
find that this doesn’t work.) In addition, the program has a tendancy to insert non-displaying
formatting characters into notebook files in a manner which corrupts the files occasionally. Try
this: copy the suspect text from your notebook into a text-only editor (MicroSoft Notepad, for
example), remove things which don’t belong, then paste it back into your notebook.
Alternatively, retype the misbehaving lines, then delete the original text. This is an annoying
problem, and should not be present in software judged ready for commercial release. Sometimes
stopping the kernel (do this from the kernel menu), then restarting it will fix problems.
Screen shots are taken from my PC’s, so what you see will be slightly different.
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Discussion: Introduction to Mathematica
Fire up Mathematica on your PC. The path probably is something similar to the following:
Start menu → Programs → Mathematica → Mathematica 4
You’ll see something like this:
The large window labeled “Untitled-1” is a Mathematica notebook in which you can enter
calculations and instructions to the Mathematica program. The palette to the right is useful for
entering various quantities (Greek letters, integrals, and so forth) into your notebook.
Mathematica’s online help facility is pretty good—you can get to it using the “Help browser...”
item in the “Help” menu, as shown below.
The Mathematica program is split into two parts: a “front end,” which manages the user
interface, and a “kernel,” which actually performs various computations for you. You do not
interact directly with the kernel.
Physics 212, Spring 2003. Unit 12
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Many of your interactions with Mathematica require you to type instructions into the notebook,
then request Mathematica to evaluate them. To cause the program to evaluate your inputs, often
you’ll type keypad-enter or shift-enter.
Have Mathematica calculate 2 + 2 for you...
...then hit keypad-enter (or shift-enter) to tell it to evaluate what it sees.
Notice that Mathematica has added the In[x]: and Out[x]: labels in the example below. In
addition, the strange brackets at the right mark “cells,” subunits in the Mathematica notebook.
Physics 212, Spring 2003. Unit 12
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You can define constants like this:
They can then be used in other expressions:
The palette is useful for building a variety of functions. For example, the tile labeled
“π” represents 3.141592653...
Mathematica contains loads of functions, all of whose arguments must be placed between square
brackets ( [ and ] ). Parentheses will not work!
For example (the “π” came from clicking the appropriate tile in the palette):
Physics 212, Spring 2003. Unit 12
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You can define your own functions this way:
Note the use of the underscore at the end of the argument name ( x_ ), the use of the palette to
enter x3 (use the tab key to move from one input box to the next), and the := assignment
symbol in the definition.
Here are some other useful Mathematica symbols. Since Mathematica also does symbolic
manipulation of equations, it sometimes does surprising things: take note of the difference
between 8/5 and 8./5. for example.
A few obscure things:
• % means “the result of the previous calculation”
• == means “logical test”
• >= prints as ≥
I always seem to get burned by this: ALL (nearly all??) Mathematica built-in functions begin
with an uppercase letter: Sin[π] works, but sin[π] doesn’t. The program is case-sensitive.
It is easy to plot stuff in Mathematica but keep in mind that sometimes you’ll need to tell it to
have the kernel load some “add-on” definitions.
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No add-on stuff was needed here. Pay attention to the format: x2 Exp[-x] is the function to
be plotted, while {x, 0, 10} is an argument list (lists are always enclosed in {} curly
brackets). The arguments, as you can surely tell, inform Mathematica of the variable to plotted
along the abscissa, as well as the minimum and maximum values to be plotted.
Mathematica is good at symbolic manipulations. For example, the palette symbol
represents a partial derivative. The two boxes are for you to fill in: the hollow one is requested
first, the solid one is requested second. Use the tab key to move from one box to the next.
( ∂ x f ( x ) means ∂f ( x ) ∂x ; it’s a common, and convenient, notation used frequently in physics.)
Here’s how this looks in your notebook (but note one way in which you must actually make use
of a function defined this way):
This is very convenient! (Note the upper case first letters in Tan[x] and Sec[x].) Because a
reference to f[x] asks Mathematica to take a derivative of Sin[x], it is necessary to let the
program do the differentiation before replacing x with a numerical value: asking it to evaluate
d sin(π )
d sin( x)
f[Pi] is the same as asking it to calculate
instead of
. (Thus the
dπ
dx x =π
f[Pi]/.x->Pi format. There are other ways to do this also...)
Physics 212, Spring 2003. Unit 12
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Mathematica also does integrals, using the indefinite integral palette tile
. Note the two
ways I’ve used to enter x2: the first is with a “^” to indicate an exponent, while the second is
done using the palette.
Not surprisingly, Mathematica notebooks can be stored so that you can resume/correct/extend
your work in a convenient fashion.
To find more information about Mathematica functions that you might find useful, use the Help
menus:
There are lots of cool things available. A far-from-complete list:
Plot[]
Plot3D[]
PlotVectorField[]
PlotVectorField3D[]
ContourPlot[]
ContourPlot3D[]
PlotGradientField[]
PlotGradientField3D[]
SphericalPlot3D[]
Here are some common mistakes/problems:
•
•
•
Leaving out the underscore at the end of an argument in a function definition:
Wombat[x]:=x^1.5; instead of Wombat[x_]:=x^1.5;
Forgetting that built-in functions begin with upper case letters: sin[x] instead of
Sin[x]
Forgetting that arguments are surrounded by square brackets instead of parentheses:
Sin(x) instead of Sin[x]
Physics 212, Spring 2003. Unit 12
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•
•
Forgetting to load a required software package, such as Graphics`PlotField` or
leaving off the back-apostrophe(s) in the package name.
Having Mathematica insert non-printing characters in places where it shouldn't. Retype, then
delete the offendin text.
We’ll work through some of the notebooks I’ve already set up so you can see some examples of
Mathematica at work.
Physics 212, Spring 2003. Unit 12
Visualizing fields and potentials with Mathematica
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Exercise 12.1: Plotting potentials with Mathematica
Let’s start with potentials for linear charges which are oriented parallel to the z axis. (This way
we’ll be dealing with a system which can be drawn nicely in two dimensions since everything
we’ll calculate is independent of z.) To avoid problems with singularities, we’ll make each of our
charges a hollow cylinder. Recall that the potential varies logarithmically with distance from an
infinite line charge or cylinder.
Make a local copy of the file p212_unit12_mathematica_1.nb for yourself, then open it in
Mathematica.
Have Mathematica evaluate the entire notebook, then read through it to make sense of what’s
there. If there are headphones available, plug them into the PC to hear the sounds generated by
the cells at the end of the notebook.
When you think you understand what’s going on, change the charge and/or position of one of the
cylinders and reevaluate the cell which generates a contour plot.
Here is a partial list of what’s in (an earlier version??) of the notebook.
(* Comments are placed between (* and *). They can span multiple lines
and be nested. Notice the long brace near the right edge of this
window. A brace indicates the extent of a "cell," a subunit of a
notebook. *)
(*Generally good programming practice is to use lots of comments
and to define functions only once, then refer to them in your work *)
(* To cause part, or all, of the notebook to be evaluated go to
the Kernel -> Evaluation -> Evaluate Cells (or Evaluate notebook)
menu path. Another way is to select a cell (or cells), then hit
keypad - enter. *)
(*
RealTime3D` will let us use the mouse to rotate some of our
plots. An important thing to note is the back - apostrophe
as the last character in"RealTime3D`" I'll also load other
packages here. *)
Needs["RealTime3D`"]
Needs["Graphics`PlotField`"]
Needs["Calculus`VectorAnalysis`"]
Off[General::spell1]
(* Define some constants here. The semicolons suppress printing of
the values. Note that quantities separated by blank spaces are
multiplied together : 4 and Pi for example (see below). *)
Physics 212, Spring 2003. Unit 12
Visualizing fields and potentials with Mathematica
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171
EpsilonNought = 8.854 10^-12;
OneOver4PiEpsilonNought = 1. / (4 Pi EpsilonNought);
(* Define the square of distance between {x1, y1, z1} and {x2, y2, z2}.
Note the use of := and also the syntax concerning underscores at
the ends of argument names. *)
DistanceSquared[x1_, y1_, z1_, x2_, y2_,
z2_] := (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2;
(* Now define the distance between {x1, y1, z1} and {x2, y2, z2} *)
Distance[x1_, y1_, z1_, x2_, y2_, z2_] :=
Sqrt[DistanceSquared[x1, y1, z1, x2, y2, z2]];
(* Now define the Green's function for the potential at (x, y, z)
associated with a hollow line charge of radius rmin at
(xprime, yprime, zprime). Recall that the Green's function is the
same as the potential caused at position r by a unit
whatever - it - is placed at rprime. *)
GreensFunctionLinePotential[x_, y_, z_, xprime_, yprime_, zprime_, rmin_] :=
2 * OneOver4PiEpsilonNought *
Log[Max[rmin, Distance[x, y, z, xprime, yprime, zprime]]];
(* specify the coordinates, radius, and charge per meter for
the first cylinder *)
x1 = -1.;
y1 = 0.;
rmin1 = 0.05;
lambda1 = -1./ OneOver4PiEpsilonNought;
(* Now do the same for the second cylinder *)
x2 = 1.;
y2 = 0.;
rmin2 = 0.05;
lambda2 = +1./ OneOver4PiEpsilonNought;
(* Here's the contribution to the potential at {x,
y} due to the first cylinder ... *)
v1[x_, y_] :=
lambda1 * GreensFunctionLinePotential[x, y, 0., x1, y1, 0., rmin1] ;
(* ... and due to the second cylinder *)
v2[x_, y_] :=
lambda2 * GreensFunctionLinePotential[x, y, 0., x2, y2, 0., rmin2] ;
(* The potential due to both of them is just the sum (superpositon) of the
individual contributions *)
v[x_, y_] := v1[x, y] + v2[x, y];
Physics 212, Spring 2003. Unit 12
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(* Let's put the requests for plots into separate cells. Since you
can ask the kernel to reevaluate a single cell instead of the entire
notebook this'll save some execution time when we want to redo
only one of the plots. *)
(* Make a contour plot. None of the options (PlotPoints, and so on)
are required. *)
ContourPlot[v[x, y], {x, -5, 5}, {y, -5, 5}, PlotPoints -> 50,
ColorFunction -> Hue, Contours -> 20, PlotLabel -> "Contour Plot"];
Contour
Plot
4
2
0
-2
-4
-4
-2
0
2
4
(* Now make a 3 - d plot. You can click on the plot, then resize it,
and rotate it with the mouse to inspect it from different angles.
Note that loading the RealTime3D package will suppress the axis labels. *)
Plot3D[v[x, y], {x, -2, 2}, {y, -2, 2}, ColorFunction -> Hue,
PlotPoints -> 50, PlotLabel -> "3D plot",
AxesLabel -> {"x", "y", "potential"}];
Physics 212, Spring 2003. Unit 12
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(* Now make a field vectors plot. Note that we need to have
already loaded the package Graphics`PlotField`. Also, notice
that we're not getting field lines--only a bunch of vectors
whose lengths and directions tell us about the electric field
strength and direction. They are placed on a grid rather than
being drawn so that the density of them would indicate the
field strength. *)
PlotGradientField[v[x, y], {x, -2., 2.}, {y, -2., 2.}, Frame -> True,
PlotLabel -> "Electric field vectors", AxesLabel -> {"x", "y"}];
Electric
field vectors
2
1
0
-1
-2
-2
-1
0
1
2
(* Now redraw that field vectors plot, but also draw in the
positions of the charged cylinders. The use of % indicates
"the previous thing." *)
Show[{%, Graphics[Disk[{x1, y1}, rmin1]], Graphics[Disk[{x2, y2}, rmin2]]}];
Electric
field vectors
2
1
0
-1
-2
-2
-1
0
1
2
Physics 212, Spring 2003. Unit 12
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174
(*** * Another way to do this would have been as follows-MyPlot =
PlotGradientField[v[x, y], {x, -2., 2.}, {y, -2., 2.}, Frame -> True];
Show[{MyPlot, Graphics[Disk[{x1, y1}, rmin1]],
Graphics[Disk[{x2, y2}, rmin2]]}];
so that we wouldn't need to put this request to redisplay the plot
immediately after the previous plot was generated. ****)
(* Let's read a sound from a file. *)
Needs["Miscellaneous`Audio`"]
(* Construct the full file pathname for where the
current notebook file lives. There has to be
an easier way to do this but I haven't yet
discovered it! *)
thingone
thingone
thingone
thingone
thingone
=
=
=
=
=
Options[$FrontEnd, NotebookDirectory];
ToString[thingone[[1]][[2]][[1]]];
StringDrop[thingone, 3];
StringDrop[thingone, -1];
thingone <> ", ";
NotebookDirectoryPath = StringReplace[thingone, {", " -> "\\"}];
(* Now concatenate the directory path onto the sound file name. *)
SoundFile = NotebookDirectoryPath <> "sound1.wav"
noise = ReadSoundFile[ SoundFile, PrintHeader -> True];
ListPlay[noise, PlayRange -> {-2^15, 2^15}, SampleRate -> 11025];
(* Now unload the audio package since it may conflict with the
music package if that is requested later. *)
Remove[Audio]
Out[37]=
"H:\\Physics_education\\PHYCS_212\\p212_unit12\\sound1.wav"
From In[37]:=
"Format: "\[InvisibleSpace]"Microsoft PCM WAVE RIFF"
From In[37]:=
"Duration: "\[InvisibleSpace]2.61279\[InvisibleSpace]" seconds"
From In[37]:=
"Channels: "\[InvisibleSpace]1
From In[37]:=
"Sampling rate: "\[InvisibleSpace]11025
From In[37]:=
"Bits per sample: "\[InvisibleSpace]8
From In[37]:=
"Data size: "\[InvisibleSpace]28806\[InvisibleSpace]" bytes"
From In[37]:=
"Number of samples: "\[InvisibleSpace]28806
Physics 212, Spring 2003. Unit 12
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Needs["Miscellaneous`Music`"]
(* Names and frequencies (in Hz) for notes are available : for example,
A4 is 440, C4 is 523.251, Bflat4 is 466.164.
Lowest note is A0 (27.5 Hz), highest is Gsharp7 (6644.88 Hz). *)
(* Define a wacko time - varying frequency (frequency in radians/sec) *)
WeirdFrequency1[t_] := 2 Pi A3 Sin[5. t];
WeirdFrequency2[t_] := 2 Pi Aflat3 Sin[5. t];
(* Define left and right channel signal strength vs time *)
LeftChannel[t_] := Sin[2 t]*Sin[WeirdFrequency1[t]];
RightChannel[t_] := Cos[2 t]*Sin[WeirdFrequency2[t]];
(* Now play it! Also, assign it to a sound object so we can replay
it quickly.*)
WeirdSound = Play[{LeftChannel[t], RightChannel[t]},
{t, 0, 5}, PlayRange -> {0, 4.}];
(* Now let's unload the package for music routines *)
Remove[Music]
discussion: grids of field vectors vs. display of field lines
Physics 212, Spring 2003. Unit 12
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176
Local vs. global concerns in graphical representations...
If it’s not 1/r2, field lines don’t make sense (they fade out, disappear).
Physics 212, Spring 2003. Unit 12
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Exercise 12.2: Potentials in 3D
It’s harder to represent fields and potentials in three dimensions: there are problems with
surfaces obscuring other surfaces, making it hard for you to see what you’re working with, and
calculations running slowly.
Make a local copy of the file p212_unit12_mathematica_2.nb for yourself, then open it in
Mathematica. Take a look at what it does: there are a few 3D plot-generating cells, as well as a
couple of animations.
Play with code in the notebook to familiarize yourself with how it works. Here’s a listing of (a
preliminary version of) the notebook:
(* Load required packages here. *)
Needs["RealTime3D`"]
Needs["Graphics`ContourPlot3D`"]
Needs["Graphics`PlotField3D`"]
Needs["Calculus`VectorAnalysis`"]
(* Define some constants. *)
EpsilonNought = 8.854 10^-12;
OneOver4PiEpsilonNought = 1. / (4 Pi EpsilonNought);
(* Define the square of distance between {x1, y1, z1} and {x2, y2, z2}.
Note the use of :=
and also the syntax concerning underscores at the ends of argument names *)
DistanceSquared[x1_, y1_, z1_, x2_, y2_,
z2_] := (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2;
(* Now define the distance between {x1, y1, z1} and {x2, y2, z2} *)
Distance[x1_, y1_, z1_, x2_, y2_, z2_] :=
Sqrt[DistanceSquared[x1, y1, z1, x2, y2, z2]];
(* Now define the Green's function for the potential at (x, y, z) associated
with a hollow spherical charge of radius rmin at (xprime, yprime, zprime).
Recall that the Green's function is the same as the potential caused
at position r by a unit whatever - it - is placed at rprime. *)
GreensFunctionPointPotential[x_, y_, z_, xprime_, yprime_, zprime_, rmin_] :=
OneOver4PiEpsilonNought
/ Max[rmin, Distance[x, y, z, xprime, yprime, zprime]];
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(* specify the coordinates, radius, and charge for the first charge *)
x1 = -2.;
y1 = 0.;
z1 = 0.;
rmin1 = 0.05;
charge1 = -1. / OneOver4PiEpsilonNought;
(* Now do the same for the second charge *)
x2 = 2.;
y2 = 0.;
z2 = 0.;
rmin2 = 0.05;
charge2 = +1. / OneOver4PiEpsilonNought;
(* Here's the contribution to the potential at {x, y} due to the first
charge ... *)
v1[x_, y_, z_] :=
charge1 * GreensFunctionPointPotential[x, y, z, x1, y1, z1, rmin1] ;
(* ... and due to the second charge *)
v2[x_, y_, z_] :=
charge2 * GreensFunctionPointPotential[x, y, z, x2, y2, z2, rmin2] ;
(* The potential due to both of them is just the sum (superpositon) of the
individual contributions *)
v[x_, y_, z_] := v1[x, y, z] + v2[x, y, z];
(* Let's put the requests for plots into separate cells.
Since you can ask the kernel to reevaluate a single cell instead of the
entire notebook this'll save some execution time when we want to redo only
one of the plots. *)
(* Make a contour plot. None of the options (PlotPoints, and so on) are
required. After the plot is finished, click on it, resize it, and rotate
it around using the mouse. The list of values for the "Contours" option
says for which values of the potential we'll draw our contours. *)
ContourPlot3D[v[x, y, z], {x, -5, 5}, {y, -5, 5}, {z, -5, 5},
PlotPoints -> {5, 5}, Contours -> {-0.1, -0.5, 0., 0.5, 0.1},
Axes -> True ];
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(* Now make a field vectors plot. Note the z range I have chosen.
You can resize the plot, also rotate it around.*)
PlotGradientField3D[v[x, y, z], {x, -5., 5.}, {y, -5., 5.}, {z, 4., 10.},
VectorHeads -> True];
(* Now do an animation. *)
Needs["Graphics`Animation`"];
Plot3D[Sin[x*y], {x, 0, 3}, {y, 0, 3}, Axes -> None,
Mesh -> False, ColorFunction -> GrayLevel];
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SpinShow[%, RotateLights -> True];
"Double-click on a plot to see something spin!"
(Lots of these get drawn)
(* Now do an animation. *)
dn = 0.1;
nstart = 1;
nstop = 3;
Do[
Plot3D[Sin[n*x]*Sin[n*y], {x, 0, 2 Pi}, {y, 0, 2 Pi}],
{n, nstart, nstop, dn}];
Do[
Plot3D[Sin[n*x]*Sin[n*y], {x, 0, 2 Pi}, {y, 0, 2 Pi}],
{n, nstop - dn, nstart + dn, -dn}];
"Double-click on a plot to see something move!"
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Physics 212, Spring 2003. Unit 12
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Exercise 12.3: Going solo with Mathematica
Go play! I suggest you try doing something of this sort:
•
•
animate a scalar potential contour plot, also a field vectors
plot for a pair of charges whose separation oscillates
cook up the strangest sound you possibly can using
Mathematica’s sound-generating tools
Summary and wrap-up discussion
Next week we’ll work on using Mathematica to plot vector potentials, and to investigate the
associated magnetic field.
Comments about homework assigned for next week
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Reading and homework should be completed by 4pm Friday.
Optional reading:
The Mathematica Book, S. Wolfram, or any other Mathematica reference you might happen to
find useful.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to last week’s problem set and write a
(true!) statement on your problem set asserting that you have successfully printed your personal
copies.
Problem 1 (5 points): Some symbolic manipulations
Have Mathematica evaluate the following integrals and derivatives for you. Please submit a
hardcopy of an evaluated Mathematica notebook to show us your results for this problem. (I can
give you a hand if you have problems generating the hardcopy or moving your files off the PC in
Loomis 257.) To display useful palettes, follow this sort of menu path: File → Palettes →
Complete Characters → Letter-like Forms → Technical Symbols (you’ll find ∞ there)
(a)
∫ [ tan( x)]
2
dx (remember that Tan[x]and not tan(x) is what Mathematica uses to
represent a tangent.)
+∞
(b)
−x 2
∫ e dx
2
−∞
(c)
(d)
d ( ln x )
dx
4
∂ 2 ( x3 y 4 z 2 )
∂x∂y
Problem 2 (10 points): An animation
For the x range 0 ≤ x ≤ 2π please use the Plot[…] function inside a Do loop to plot sin(kx)
using k as the animation variable, with k stepping over values from 0 to 10 in increments of 0.1
unit. When you’ve done it correctly, Mathematica will generate a large number of plots for you;
double-clicking on one of them will cause the program to play all the plots as an animation. In
order to keep Mathematica from adjusting the minimum, maximum values of the y axis you can
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include the expression PlotRange->{{0, 2*Pi}, {-1, 1}} as a last argument to the
Plot[…] function.
Problem 3 (5 points): Plotting an implicit function
One can define a function (or relation) implicitly by specifying a rule that is satisfied by the all
points (x, f(x)). For example, the rule x2 + y2 = 4 is an implicit definition of a circle of radius 2
centered at the origin. Mathematica is able to plot functions defined this way, although you’ll
need to load one of its “packages” using the statement
Needs["Graphics`ImplicitPlot`"]
at the start of your notebook. (Be careful about the exact form for this: the single quotes are the
back-apostrophes usually found near the upper-left corner of a standard keyboard.)
Look up the ImplicitPlot[…] function in Mathematica’s online help facility and use it to
generate plots of:
(a) the Folium of Descartes, defined implicitly through the equation x3 + y 3 = 3xy
2
(a) a cardioid, defined implicitly through the equation  x 2 + y 2 − x  = x 2 + y 2
Your plots should look something like this:
Folium
of Descartes
Cardiod
2
1
1
0.5
-3
-2
-1
1
2
3
-1
-2
0.5
1
1.5
2
-0.5
-3
-1
-4
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Problem 4 (10 points): Solving an algebraic equation
Mathematica can solve a variety of algebraic equations, both through symbolic manipulation,
and through numerical means. See the program’s online help documentation concerning the
Solve[…] function. Here are a couple of examples:
Note the required pair of equal signs “==” used to define the equation.
(a) Use the Solve[…] function to generate the roots of the most general form of a cubic
equation in one variable, ax 3 + bx 2 + cx + d = 0 .
(b) Some equations do not lend themselves to solution through symbolic techniques built into
Mathematica. If you encounter one of these, try using the FindRoot[…] function instead to
search for roots to the equation. (See the online help information for the format of
FindRoot’s arguments.) Use FindRoot to determine numerical values of the solutions to
sin θ = θ 2 .
plots of sin Htheta L and theta ê2
2
1
-4
-2
2
4
-1
-2
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Physics 212
Honors topics in electrodynamics
Unit 13
Studying electromagnetic radiation with
Mathematica
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
Physics 212, Spring 2003. Unit 13
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Unit 13: Studying electromagnetic radiation with Mathematica
Recap of last week
Introduction: Those Liénard-Wiechert potentials for a point charge
When a point charge is in motion, the scalar potential it creates is modified in two fundamental
ways:
V (r ) =
1
Q
4πε 0 r − rQ
⇒ V (r,t) =


Q
1


4πε 0 r − rQ ( t ′ )  (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) 
1
1. The potential depends, in part, on the charge’s retarded position-- its location at the retarded
time t ′ , defined through the relation t ′ = t − r − rQ ( t ′ ) c .
2. The 1/r form of the potential is modified by a velocity-dependent factor 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
where εˆ ( t ′ ) is a unit vector pointing from the charge’s retarded position to the observer’s
position as shown in the following figure.
line-of-sight
to observer
observer
εˆ
observer sees Q here
(at its retarded position)
Q
v
Q is really here now
The charge’s vector potential is modified in similar fashion so that
A( r , t ) =
(
v ( t ′) 1
v ( t ′)
1
Q
×
= 2 V ( r ,t) .
2
c 4πε 0 r − rQ ( t ′ ) (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
c
)
(Recall that µ0 = 1 ε 0c 2 .)
This complicates things somewhat—the electric field is no longer guaranteed to be conservative,
so it is no longer the case that we can always define it as the gradient of a potential. The scalar
and vector potentials are still useful, however, since E = −∇V − ∂A ∂t and B = ∇ × A , so we can
use them to calculate the fields if we’re willing to do all the differentiation.
Here’s what you’ll work on, with the help of Mathematica:
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•
•
generating contours of constant retarded time for observers watching a charged particle
moving with constant velocity, and a separate particle which stops suddenly at the origin;
graphing the behavior of the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) factor as a function of observer position for
•
the constant-velocity and stop-at-the-origin cases just mentioned
studying the scalar potential including retardation, but neglecting the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
•
•
correction factor (there’ll be some surprises here)
graphing the vector potential
plotting the relative contributions to E of −∇V and − ∂A ∂t (more surprises!)
Exercise 13.1: Retarded time for a moving charge
Since Alpha Centauri is 4.3 light years away, something we see happening there at the present
time actually took place several years ago. The retarded time we would use in calculating the
potentials at Earth induced by electric charges on Alpha Centauri is set back from the present
time by 4.3 years.
It works the same way for something in motion: imagine that a moving object emitted a flash of
light nine nanoseconds ago. Since c ≈ 0.3 m/nsec, observers who are three meters from the
charge’s position at the time it emitted the flash will see the flash now. (Since emitting the flash
the charge may have moved elsewhere.)
Here’s a graphical example, involving an observer watching a charge moving at half the speed of
light up the y axis. The true position of the charge is 0.5ct yˆ (it passes through the origin at
t = 0) while the observer is stationary at x = 1.5 meters. At t = 5 nsec in the observer’s frame,
the charge is really at y = 0.75 meters but the observer is seeing light that left it earlier, when it
was at y = 0.
y
charge’s true position: x=0.75 m when t=5 nsec
charge’s retarded position: x=0
x
present time is t=5 nsec
observer at x=1.5 m
In this particular case, the retarded time corresponding to this observer’s t = 5 nsec is
t ′ = t − robs − rQ ( t ′ ) c = 0 . It is a simple algebra problem to solve for the retarded time as a
function of the observer’s time when the charge moves with constant velocity: to find the form
of the function t ′ ( t ) . When the charged particle’s velocity isn’t constant, though, it can be
nearly impossible to solve for t ′ ( t ) .
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Consider one particle moving with constant velocity 0.5ct yˆ , and another moving with velocity
0.5ct yˆ for t < 0, which stops at the origin at t = 0. Graphs of the particles’ y positions as
functions of time are shown below. (Remember that c ≈ 0.3 m/nsec.)
y (0.3 m per division)
y
rvst1y
rvst2y
t (nsec)
t
charge is moving
with v = 0.5c
charge is moving
with v = 0.5c
until t = 0
(a) Imagine that the present time is now t = 10 nsec and that observers are watching the charge
which moves with constant velocity. Sketch possible locations of observers who are seeing (at
t = 10 nsec) light which had illuminated the moving particle at the earlier times t = 0 nsec and
t = 5 nsec. (These observers would use retarded times t ′ = 0 nsec and t ′ = 5 nsec in their
calculations of the potentials associated with the charge.)
y
charge here now (t = 10 nsec)
charge was here at t = 0 nsec
x
charge was here at t = -10 nsec
charge is moving with v = 0.5c
grid cell (∆x, ∆y) is (0.3m, 0.3m)
present time is
t = 10 nsec
(b) Imagine that the present time is now t = 5 nsec (not 10 nsec). Observers are watching the
charge which moved with constant velocity for t < 0 nsec, but then stopped at the origin. Sketch
possible locations of observers who are seeing (at t = 5 nsec) light which had illuminated the
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moving particle at the earlier times t = 4, 2, 0, -5, -10 nsec. (These observers would use retarded
times t ′ = 4, 2, 0, − 5, − 10 nsec in their calculations of the potentials associated with the charge.)
y
charge here since t = 0 nsec
x
charge was here at t = -10 nsec
charge was moving with v = 0.5c
but stopped at (0,0) 5 nsec ago
present time is
t = 5 nsec
(c) Let’s do this with Mathematica now. I’ve set up a file of useful functions for you in a utility
file p212_mathematica.m in the Physics 212 area. (They are automatically loaded when you
evaluate the first cell in the notebook file p212_unit13_mathematica_1.nb, which you should use
as a template for your work.)
Here is a list of the functions defined for you:
Distance[{x1,y1}, {x2,y2}]
distance between points 1 and 2
Norm[{x,y}]
length of the vector extending from {0,0} to {x,y}
DotProd[{x1,y1}, {x2,y2}]
dot product between two (2D) vectors
CrossProd[{x1,y1}, {x2,y2}]
cross product between two (2D) vectors
UnitVector[{x,y}]
unit vector parallel to the line from {0,0} to {x,y}
rvst1[t]
{x,y} at time t for a particle moving with constant velocity
up the y axis so that x=0, y=0.5ct
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rvst2[t]
like rvst1[t], but the particle stops in 0.1 nsec at the origin at t=0.
rvst3[t]
like rvst2[t], but the particle takes 10 nsec to stop at the origin.
NOTE: you must select one of these, or define your own function so that rvst[t]
is defined. For example, to select rvst1 do this:
rvst[time_] := rvst1[time]
being careful to include the underscore at the end of time_ to the left of
the “:=” symbol, but NOT to the right of the “:=”
vvst[t]
time derivative of rvst[t]. (assumes user has defined rvst)
avst[t]
time derivative of vvst[t]. (assumes user has defined rvst)
TRetard[{x,y}, t]
returns retarded time at charge's position for observer
at {x,y}, t (uses rvst function to calculate this)
VDotETerm[{x,y}, t]
value of 1/(1 - v dot E) evaluated using the
retarded velocity and line-of-sight vector e (not
the electric field!)
ScalarPotential[{x,y}, t] scalar potential when charge is at rvst[t] and
observer is at {x,y} at time t. Calculation includes
retardation effects.
ScalarPotentialNoVDotE[{x,y}, t]
GradV[{x,y}, t]
like ScalarPotential, but omitting the
velocity-dependent correction factor.
gradient (in 2 dimensions) of scalar potential when
charge is at rvst[t] and observer is at {x,y} at time t.
Includes retardation effects.
VectorPotential[{x,y}, t] vector potential when charge is at rvst[t] and observer
is at {x,y} at time t. Calculation includes retardation
effects.
dAdt[{x,y}, t]
ECoulomb[{x,y}, t]
time derivative of vector potential when charge is at
rvst[t] etc.
Coulomb (1/r^2) portion of the electric field
ERadiation[{x,y}, t]
Radiation (1/r) portion of the electric field
ETotal{x,y}, t]
Total electric field (Coulomb + radiation)
NonRadialAngle[{xobs,yobs}, {fx, fy}, t]
returns the angle (in degrees) between
the vector {fx, fy} and the line from the
instantaneous (not the retarded) position
of the charge to the observer’s location.
Here's an example of how to use the predefined functions:
(* As an example, assign rvst to be the same as rvst2, then
plot its y component. *)
(* define the (new) function rvst here *)
rvst[timearg_] := N[rvst2[timearg]]
(* print one value of it to remind everyone that rvst returns a
vector, not a number *)
Print["rvst at -20 nsec is ", rvst[-20*10^-9]];
(* Now plot its y component (thus the [[2]] index) from - 20 nsec to +
20 nsec *)
Plot[rvst[t][[2]], {t, -20*10^-9, 20*10^-9}];
Physics 212, Spring 2003. Unit 13
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rvst at -20 nsec is {0., -2.99792}
-2 ×10 - 8
-1 × 10 - 8
1×10 - 8
2× 10 - 8
-0.5
-1
-1.5
-2
-2.5
-3
Graphs of the y components of rvst1 and rvst2 appeared in the writeup a few pages ago.
Do the following:
1. Assign the constant velocity function rvst1 to rvst.
2. Assume the present time is 30 nsec after the charge has reached (or passed through) the
origin in our frame of reference.
3. Make a contour plot of the retarded time in the region −20 ≤ x ≤ +20, − 20 ≤ y ≤ +20 . (The
contour closest to the charge’s t = 30 nsec position corresponds to observers who see the
charge as it is at t = 30 nsec. Contours further away correspond to observers who see the
charge as it was at progressively earlier times.)
4. Assign the stop-at-the-origin velocity function rvst2 to rvst.
5. Make a contour plot with the same x,y limits.
6. Comment sensibly on what you see, in particular about the differences in the two plots.
Your plots should look something like these:
Physics 212, Spring 2003. Unit 13
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d
time vs. position , constant
velocity
0.5 c, time now 3
20
10
0
10
20
-20
Retarded
-10
0
10
20
time vs. position , stop -at -origin , time now 30 nsec
20
10
0
-10
-20
-20
-10
0
10
20
The concentric region in the second plot extending 9 meters ( = c × 30 nsec ) out from the origin
expands as time passes: observers at greater distance become aware at later times that the charge
is stopped.
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Exercise 13.2: That v ( t ′ ) ⋅ εˆ ( t ′ ) correction factor
Let’s investigate the effects of retardation and that 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) factor on the scalar
potential now. We’ll work with two different cases, both involving a point charge. The first
charge is moving up the y axis at half the speed of light so that its position is rQ ( t ) = 0.5ct yˆ .
The second is also moving up the y axis, but decelerates rapidly with constant acceleration after
it passes through the origin, coming to rest in 0.1 nanosecond at rQ = 0.0075 ŷ . (These are the
two cases rvst1,
rvst2
used in the previous exercise.)
While moving, our charge has v = 0.5c so the correction factor is 2 in forward direction, 0.5 in
the backwards direction, and 1 at 90 degrees.
Use Plot3D to make graphs of the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) factor using the same time (30 nsec) and
x,y ranges for the plots. (The function is defined in the utility file as VDotETerm[{x,y}, t].) Do it
for both velocity cases we’ve been using. Your plots should look something like the two which
follow.
Bear in mind that the boundary of the flat region within 9 meters ( = c × 30 nsec ) of the origin in
the second plot expands outwards at c as more distant observer become aware that the charge has
stopped. This means that the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) correction factor included in the scalar
potential will change suddenly for observers as they realize that the charge is no longer moving.
V dot E correction
term vs. position , uniform
velocity , t=30 nsec
2
20
1.5
10
1
-20
0
-10
-10
0
10
20
-20
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V dot E correction
term vs. position , stop -at -origin , t=30 nsec
2
20
1.5
10
1
-20
0
-10
-10
0
10
20
-20
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Exercise 13.3: Scalar potential with retarded time, but no v dot e correction
When retardation is included, but the velocity-dependent correction factor is omitted, the scalar
caused by a moving charge at an observer's position is the same as that caused by a stationary
charge at the moving charge's retarded position. The equipotential contours for a charge moving
with constant velocity are circles centered on the retarded positions of the charge.
Use ContourPlot to make graphs of the scalar potential without the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) factor
using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in
the utility file as ScalarPotentialNoVDotE[{x,y}, t].) Do it for both velocity cases we’ve been
using. Your plots should look something like the two which follow.
Note the bunching of field lines towards the tops of the plots: this means the potential would
change more rapidly in front of the moving charges if this were the correct expression for the
scalar potential. (It's not, since we've left off the necessary 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) factor.)
As before, take note in the second plot of the change that occurs at the boundary 9 meters from
the origin.
calar
potential
NOT including
V dot E correction
term , constant
v, t=30 nsec
20
10
0
-10
-20
-20
-10
0
10
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13.11
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calar
potential
NOT including
V dot E correction
term , stop − at −0, t= 30 nsec
20
10
0
-10
-20
-20
-10
0
10
20
Exercise 13.4: Scalar (and vector) potential with retarded time and v dot e correction
Now use ContourPlot to make graphs of the scalar potential including the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
factor using the same time (30 nsec) and x,y ranges for the plots as before. (The function is
defined in the utility file as ScalarPotential[{x,y}, t].) Do it for both velocity cases we’ve
been using. Your plots should look something like the two which follow.
Note the surprising fact that the equipotential contours in the first plot are concentric, and are
centered on the instantaneous position of the charge! Distant observers cannot know that the
charge is really 4.5 meters up the y axis, since it might have stopped recently. This should be
clear from the second plot-- observers far from the (now stopped) charge still see the same
equipotentials as are seen by distant observers in the constant velocity plot shown first.
The effects at the 9 meter boundary are especially dramatic in the second plot in the forward
direction where the velocity correction factor changes from 2 to 1 suddenly. The rapid change in
the potential at this point causes an impressive bunching of the equipotential contours.
We haven't done anything with the vector potential yet. The constant velocity case is easy, since
v ( t ′)
A ( r , t ) = 2 V ( r , t ) . Except for the direction (along y) associated with the vector potential, it
c
behaves just like the scalar potential. The second case is also a snap-- the vector potential is zero
close to the (stopped) charge where all observers realize that it is stationary.
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potential
including
V dot E correction
term , uniform
velocity , t= 30
20
10
0
10
20
-20
Scalar
potential
-10
including
0
V dot E correction
10
20
term , stop − at −0, t= 30 nsec
20
10
0
-10
-20
-20
-10
0
10
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Exercise 13.5: Vector potential for the stop-at-0 charge
Use PlotVectorField to make a graph of the vector potential using the same time (30 nsec) and
x,y ranges for the plots as before. (The function is defined in the utility file as
VectorPotential[{x,y}, t].) Do it only for the stop-at-zero velocity case.
Your plot should look something like the one which follows.
Note that vector potential is always along y, and that the forward-backward symmetry has been
broken by the velocity-denominator term. The boundary at 9 meters moves outwards at the
speed of flight.
Vector
potential , stop − at −0, t = 30 ns
20
10
0
-10
-20
-20
-10
0
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Studying electromagnetic radiation with Mathematica
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20
13.15
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Discussion: Liénard-Wiechert fields
It’s always true that E = −∇V − ∂A ∂t and that B = ∇ × A . Calculating E and B is “just” a
matter of taking some derivatives!
After a fair amount of messy calculation it can be shown that:
(
)
 εˆ ′ 1 − v ′ 2 c 2 − v ′ c (1 − εˆ ′ ⋅ v ′ c ) 


−∇V ( r , t ) =
2
3
4πε 0
ˆ


1
ε
v
c
−
⋅
′
′
(
)
′
r − rQ 
+
1
Q
1
Q
1
4πε 0 r − r ′ c 2
Q
 εˆ ′ ( a ′ ⋅ εˆ ′ ) 


3
 (1 − εˆ ′ ⋅ v ′ c ) 
Primed quantities are evaluated at the appropriate retarded time; a′ is the (retarded) acceleration.
(The acceleration enters because of the subtleties involving functions of t ′ being differentiated
with respect to t.)
Note that the first term in the gradient is independent of acceleration, and is proportional to 1/r2.
(This is the “Coulomb term.”) However, the second term is proportional to acceleration, and
falls off as 1/r.
It is possible to pound out the derivatives to calculate − ∂A ∂t too: the result is
−
∂A ( r , t ) − v ′ 1
=
c 4πε 0
∂t


2
r − rQ ′ 
Q
( v ′ c ) ⋅ ( εˆ ′ − v ′ c) 
(1 − εˆ ′ ⋅ v ′ c ) 3

+
1
Q
1
4πε 0 r − r ′ c 2
Q
 − a ′ + εˆ ′ × ( a ′ × v ′ c ) 

.

(1 − εˆ ′ ⋅ v ′ c ) 3 
As with the −∇V equation, the first (Coulomb) term in − ∂A ∂t is independent of acceleration,
and is proportional to 1/r2. The second term, proportional to acceleration, falls as 1/r.
Let’s study the “Coulomb terms” of −∇V and − ∂A ∂t first by looking at a charge moving with
constant velocity.
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Exercise 13.6: -Grad V: constant velocity
(a) Use PlotVectorField to make a graph of the negative of the gradient of the scalar potential
using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in
the utility file as GradV[{x,y}, t]; do it only for the constant velocity case.) To make it easier to
see the directions of the vectors, also use the utility function UnitVector to replot −∇V .
Your plots should look something like the first two which follow.
It looks like the vectors point radially away from the instantaneous position of the charge,
without retardation. How close is this to the truth?
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Vectors...
− grad V, constant
v, t = 30 nsec
20
10
0
-10
-20
-20
-10
Unit vectors...
Unit vetors
0
of − grad V, constant
10
20
v, t = 30 nsec
20
10
0
-10
-20
-20
-10
0
10
20
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(b) Is −∇V really radial? (Does it really point directly away from the instantaneous position of
the charge?) Use Plot3D and the utility function NonRadialAngle to investigate, using the same
observer time and plot boundaries as before.
Your plot (which may take two or three minutes to appear) will look something like this:
non − radialness
angle
H degrees L of − grad V, uniform
velocity , t = 30 nsec
5
20
0
10
-5
-20
0
-10
-10
0
10
20
-20
What do you conclude?
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Exercise 13.7: -dA/dt for constant velocity
Use PlotVectorField to make a graph of the negative of the time derivative of the vector
potential using the same time (30 nsec) and x,y ranges for the plots as before. (The function is
defined in the utility file as dAdt[{x,y}, t]; do it only for the constant velocity case.) To make
it easier to see the directions of the vectors, also use the utility function UnitVector to replot
−∂A ∂t .
Your plots should look something like the first two which follow.
Not surprisingly, −∂A ∂t can only point along y.
Vectors...
− dA ê dt , constant
Unit vectors...
v, t= 0
unit vactors
20
20
10
10
0
0
-10
-10
-20
along
− dA êdt, constant
v, t= 0
-20
-20
-10
0
10
20
-20
-10
0
10
20
Note the sign flip at instantaneous charge position
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Exercise 13.8: E = -Grad V - dA/dt for constant velocity
Since E = −∇V − ∂A ∂t , we can investigate the electric field by combining the last few functions
we've been using. Do it-- generate vector, and unit vector plots of E = −∇V − ∂A ∂t , then plot
the degree to which the electric field points radially away form the instantaneous position of the
charge.
Here are my versions of the plots. Don't be mislead by the round-off errors which fool
Mathematica's arrow-drawing routine when the fields are too weak.
Vectors...
− grad V − dA ê dt, constant
v, t = 30 nsec
20
10
0
-10
-20
-20
-10
0
10
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Unit vectors...
unit vector
along
− grad V − dA ê dt , constant
v, t = 30 nsec
20
10
0
-10
-20
-20
-10
"Radialness..."
non − radialness
angle
0
10
20
Hdegrees L of − grad V − dA ê dt , constant
velocity , t = 30
0.0002
0.0001
20
0
-0.0001
10
-0.0002
-20
0
-10
-10
0
10
20
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Well, look at that! What can we conclude?
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Discussion: Electromagnetic radiation!
When do the terms in E = −∇V − ∂A ∂t proportional to acceleration dominate E ? Since the
Coulomb terms are proportional to 1/r2, while the acceleration terms are proportional to 1/r, it's
clear that at large distances the acceleration terms must dominate. But what constitutes a "large"
distance? Very roughly, the ratio of the "radiation" (1/r) and Coulomb terms for E is
ra c 2 = r c 2 a , where r represents the distance to the observer and a is the charge's
(
)
acceleration. As a result, when the observer's distance is large compared to c 2 a the accelerationbased term will dominate.
Exercise 13.9: Plotting the radiation terms
Including only the terms proportional to 1/r gives
E
1 r only
=−
1
Q
r − rQ ′ 4πε 0 c (1 − εˆ ′ ⋅ v ′ c )
2
3
{ a ′ − εˆ ′ ( a ′ ⋅ εˆ ′) − εˆ ′ × ( a ′ × v ′ c)} .
To simplify matters, let’s only consider acceleration a (anti)parallel to v . (Clearly this will not
work for circular motion!). In this case, ( a ′ × v ′ c ) = 0 so we can ignore the term with the cross
products.
{
}
Let’s look at the remaining terms in the curly brackets: a ′ − εˆ ′ ( a ′ ⋅ εˆ ′ ) . Here’s a diagram.
observer
a′ − ( a′ ⋅ εˆ′ ) εˆ′
a′
Q
εˆ′
( a′ ⋅ εˆ′) εˆ′
{
}
With some thought you should be able to see that a ′ − εˆ ′ ( a ′ ⋅ εˆ ′ ) is just the component of the
(retarded) acceleration which is perpendicular to the observer’s line of site towards the (retarded)
position of the charge. Let’s give this component the name a⊥′ (“a prime perp”). (Even if we
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hadn’t asked that the acceleration be along the direction of the velocity we’d still have the field
perpendicular to εˆ ′ since the ignored term εˆ ′ × ( a ′ × v ′ c ) is perpendicular to εˆ ′ due to the
cross products.)
With these simplifications, we can write E
1 r only
=−
Qa⊥′
.
3
r − rQ ′ 4πε 0 c (1 − εˆ ′ ⋅ v ′ c )
1
2
Keep in mind that most everything in this expression is a retarded variable, and that the velocitycorrection denominator is cubed.
Things to notice:
1. The electric field is perpendicular to the line-of-sight vector εˆ′ .
2. The electric field is proportional to the charge’s (retarded) acceleration.
3. The field falls off like 1/r.
4. There’s a negative sign in front: the radiation electric field of a positive charge with upwards
acceleration points down.
Use ContourPlot to make graphs of the magnitide of the radiation electric field using the same
time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as
ERadiation[{x,y}, t].) Do it for the as-yet unused position-vs.-time function rvst3 which
describes a particle which begins decelerating from v = 0.5c yˆ at t = -10 nsec, coming to a
complete stop at t = 0.
Your plot should look something like the sample plot which follows. The spikes are an artifact of
the granularity of the plot I’ve chosen.
Notice the following:
• Since the radiation fields are blowing past an observer only while that observer sees the
particle’s (retarded) acceleration as different from zero, the region in space containing a nonzero radiation field is only three meters thick.
• The fields would be strongest in directions perpendicular to the particle’s acceleration if the
3
velocity-correction factor 1 (1 − εˆ ′ ⋅ v ′ c ) weren’t present. Because of it, the fields are
somewhat stronger in the forward direction. (“Forward” means parallel to the particle’s
velocity during acceleration.)
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Radiation
E field , stop -at -zero , 10 ns deceleration
, t=30 ns
6×10 8
20
4×10 8
2×10 8
10
0
-20
0
-10
-10
0
10
20
-20
Now do the same thing, but using the position-vs.-time function rvst4 which describes a particle
which begins decelerating from v = 0.01c yˆ at t = -10 nsec, coming to a complete stop at t = 0.
Your plot should look something like the sample plot which follows. Because the particle is
nonrelativistic, the radiation electric fields are nearly forward-backward symmetric, and are
strongest perpendicular to the particle’s acceleration. In addition, the fields are considerably
weaker due to the reduced acceleration.
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Radiation
E field , non -relativistic , 10 ns deceleration
, t=30 ns
6×10 6
4×10
20
6
2×10 6
10
0
-20
0
-10
-10
0
10
20
-20
Discussion: What about the magnetic field?
We haven’t said anything about the magnetic field! It’s just a matter of calculation to take the
curl of the vector potential; what one finds is that the magnetic and electric fields obey
B = E c for electromagnetic radiation. The magnetic field is perpendicular to the electric field,
and also perpendicular to the line-of-sight direction.
Summary and wrap-up discussion
We’ve only made plots for particles which move with constant velocity, or stop rapidly.
However, everything we’ve done would work for a particle moving with arbitrary acceleration.
(Mathematica might gag, but that doesn’t mean that there’s anything wrong with our equations.)
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Here are the main points.
•
When a point charge is in motion, the scalar and vector potentials it creates are modified in
two fundamental ways:
1.
The potential depends, in part, on the charge’s retarded
position-- its location at the retarded time t ′ , defined through the relation
t ′ = t − r − rQ ( t ′ ) c .
2.
The 1/r form of the potential is modified by a velocitydependent factor 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) where εˆ ( t ′ ) is a unit vector pointing from the
charge’s retarded position to the observer’s position.
•
The electric field depends on both the gradient of the scalar potential and the time derivative
of the vector potential. These two contributions conspire in a remarkable fashion so that the
electric field of a charge moving with constant velocity is radial, always pointing exxactly
away from the instantaneous position of the charge.
•
Observers far from a charge which has stopped cannot know instantaneously that it has
stopped moving. As a result, they will continue to see the fields associated with a charge in
motion until enough time has passed so that they become aware the charge has stopped.
•
The sudden change in the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) term in the potentials (which corresponds to
the 1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) term in the fields) induces dramatic changes in the derivatives of
3
the potentials, giving rise to electromagnetic radiation: electric fields which are proportional
to (retarded) acceleration, and propagate away from the charge, falling off like 1/r instead of
1/r2.
Comments about homework assigned for next week
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Reading and homework should be completed by 4pm Friday.
Required reading:
Read all of the in-class exercise material for this unit.
Purcell, chapter 5, section 7.
Optional reading:
Purcell, chapter 9 and appendix B.
The Feynman Lectures on Physics (volume II), chapters 20 and 21.
Radiation review/reminder:
When a point charge is in motion, the scalar and vector potentials it creates are modified in two
fundamental ways:
3. The potentials depend, in part, on the charge’s retarded position-- its location at the retarded
time t ′ , defined through the relation t ′ = t − r − rQ ( t ′ ) c .
4. The 1/r forms of the potentials are modified by a velocity-dependent factor
1 (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) where εˆ ( t ′ ) is a unit vector pointing from the charge’s retarded
position to the observer’s position as shown in the following figure. (Sometimes I’ll use εˆ′
as an abbreviation for εˆ ( t ′ ) .)
line-of-sight
to observer
observer
εˆ
observer sees Q here
(at its retarded position)
Q
v
Q is really here now
The changes to the potentials work like this:


Q
1


4πε 0 r − rQ ( t ′ )  (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c ) 
v ( t ′) 1
v ( t ′)
1
Q
×
= 2 V ( r , t ) . (Recall that µ0 = 1 ε 0c 2 .)
A( r , t ) = 2
c 4πε 0 r − rQ ( t ′ ) (1 − v ( t ′ ) ⋅ εˆ ( t ′ ) c )
c
V (r ) =
1
Q
4πε 0 r − rQ
⇒ V (r,t) =
1
(
)
It is still true that E = −∇V − ∂A ∂t and B = ∇ × A so we can use the potentials to calculate the
fields. The radiation terms in E and B will be proportional to 1/r. If you crank out the
derivatives you’ll find that the radiation term in the expression for the electric field is this:
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E
1 r only
=−
1
Q
2
r − rQ ′ 4πε 0 c (1 − εˆ ′ ⋅ v ′ c )
3
{ a ′ − εˆ ′ ( a ′ ⋅ εˆ ′) − εˆ ′ × ( a ′ × v ′ c)} .
The particle’s (retarded) acceleration is a′ . Don’t be intimidated by all the cross products! As
described in the in-class exercise material, if we only consider acceleration a (anti)parallel to v
we can write
Qa⊥′
1
E
=−
3
2
1 r only
r − r ′ 4πε 0 c (1 − εˆ ′ ⋅ v ′ c )
Q
where a⊥′ (“a prime perp”) is the component of (retarded) acceleration perpendicular to the lineof-sight to the observer. Keep in mind that most everything in this expression is a retarded
variable, and that the velocity-correction denominator is cubed.
Calculating the magnetic field is “just” a matter of taking the curl of the vector potential; what
one finds is that the radiation (1/r) magnetic and electric fields obey B = E c and that the
magnetic field is perpendicular to the electric field, and also perpendicular to the line-of-sight
direction.
Things to keep in mind for the radiation fields:
5. The electric and magnetic fields are perpendicular to the line-of-sight vector εˆ′ .
6. The electric and magnetic fields are proportional to the charge’s (retarded) acceleration.
7. The fields fall off like 1/r.
8. B = E c .
9. There’s a negative sign in front of the expression for E : the radiation electric field of a
positive charge with upwards (retarded) acceleration points down.
Problem 0 (5 points): Make your own hardcopies…
Print (from the web) your own copy of the solutions to last week’s problem set and write a
(true!) statement on your problem set asserting that you have successfully printed your personal
copies.
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Problem 1 (15 points): Non-relativistic dipole radiation
A particle with charge q oscillates about the origin with position rq = 10−3 sin(107 t ) zˆ . Since the
particle is always moving non-relativistically, the electric field it generates at r , the location of a
1 qa⊥′
distant observer, is well-approximated by E ( r , t ) ≈ −
. Note that a⊥′ = a ′ sin θ as
r 4πε 0c 2
shown in the following figure, and that the retarded time is t ′ ≈ t − r c .
z
observer
at r
a′⊥
a′
εˆ′
θ
q
a′
You learned in Physics 112 that the power flux (energy per second per square meter) transported
by an electromagnetic wave has magnitude S = E × B µ0 and that µ0ε 0 = 1 c 2 .
(a) Calculate the RMS power flux (in watts per square meter) arriving at an observer
located at the point (r, θ), as shown in the figure.
(b) What is the total RMS power radiated by the oscillating charge? (You’ll need to
calculate ∫ S ⋅ dA to answer this question. The integral is easiest to do in spherical
all
directions
coordinates (r, θ, φ) where S = rˆ S , dA = rˆ r 2 sin θ dθ dφ , and r̂ is a unit vector pointing
away from the origin.)
(c) Assuming q = 1 Coulomb, what is the numerical value (in watts) for your answer to
part (b)?
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Problem 2 (15 points): Mrs. Urkin builds her own x-ray machine
After seeing Marathon Man for the 97th time, Mrs. Urkin decides to teach herself dentistry.
Knowing that x-ray machines employ bremsstrahlung (German for “braking radiation”) to
produce short-wavelength electromagnetic radiation, she borrows the UD&HE electron linear
accelerator to use as the electron gun for her device. When a fast-moving electron slams into a
dense target its rapid deceleration will cause it to radiate x-rays.
Assume that an electron from the “linac” initially travels upwards along the z axis (so that its
initial velocity is v = v0 zˆ ), then decelerates with constant acceleration a = −azˆ to stop near the
origin at t = 0. The teeth of an unfortunate patient are located at (r, θ ), as shown. Please assume
that r is very large compared to the distance through which the electron moves while coming to
rest so that r − re− ′ ≈ r always.
z
θ
ex
patient
at r
εˆ′
y
a′
(a) Prove that the magnitude of the radiation electric field (for fixed r ) is greatest at θ = 90°
when v
c.
(b) For constant acceleration as the electron slows to rest from a relativistic initial velocity prove
that the intensity of x-rays at θ = 90° remains constant during the deceleration process, even
though v ∼ c in the early stages of deceleration.
(c) An observer whose teeth are at (r, θ ) sees the decelerating electron’s velocity as 0.5c at one
particular instant in time. What value of θ, expressed in degrees, will yield the maximum electric
field strength for this (retarded) velocity? (You will have to slog through a differentiation and a
small amount of algebra to derive a useful quadratic equation, one of whose roots is the cosine of
the desired angle. The fact that sin 2 θ = 1 − cos 2 θ might be helpful.)
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Physics 212
Honors topics in electrodynamics
Unit 14
Oops! It doesn’t really work like that!
(Quantum mechanics enters the picture)
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
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Unit 14: Oops! It doesn’t really work like that! (Quantum mechanics enters
the picture)
Recap of last week
Introduction: The fields aren’t really fields, and you cannot know as much as you’d like
You’ve probably heard of the Heisenberg Uncertainty Principle. Today we’ll investigate a
plausibility argument for why it is true, as well as some of its consequences.
We’re forced to head down this path because one of the reasonable assumptions we’ve made
about electrodynamics is completely, totally, utterly wrong: the electromagnetic radiation
streaming from an accelerating charge is not composed of fields which vary smoothly in space
and time. Einstein’s Nobel Prize was not awarded for his work on relativity (which certainly did
deserve one), but rather for his explanation of the photoelectric effect, one of the consequences
of the “granularity” of electromagnetic radiation.
Here’s what you’ll be working on:
• a classical analysis of the rate at which electrons “should be” ejected from a conductor which
is exposed to an intense light source.
• a classical analysis of the relationship between the maximum kinetic energy of electrons
ejected from a conductor and the intensity of the light source illuminating the conductor.
• photons! (photon energy and momentum)
• modeling the resolution of a simple magnifier
• analyzing the “Heisenberg’s Microscope” thought experiment
• estimating atomic sizes using the Uncertainty Principle
We’ll also discuss the master equation of quantum mechanics, known as Schrödinger’s equation.
Exercise 14.1: An estimate of the rate of photoelectron ejection from a metallic surface
A charge above a classical conducting surface is attracted back towards the surface due to the
fields created by the surface charge it attracts beneath it, as shown in the figure. The fields above
the conductor are exactly the same as they would be if an image charge of equal magnitude, but
opposite sign, were placed as shown, without the conductor being present.
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electron
+
+ + + ++ + + +
+
these field lines
don’t really exist
real surface charge
conductor
fictitious positive image charge
(a) Since atomic scales are typically a tenth of a nanometer, it shouldn’t be too crazy to model an
electron in the process of escaping from the conductor as sitting ~10-10 meters above the surface,
fighting the attractive force of its image charge as it tries to wander off. Estimate the energy
needed to fee an electron from its image charge, using this model. (Recall the expression for
electrostatic potential energy you learned in physics 112.) Please express you answer in
electron-volts (1 eV = 1.6×10-19 Joules).
(b) A bright light illuminates the metal surface with an intensity of 1 watt per square centimeter.
(This corresponds to approximately 6×1018 eV per second). Let’s make the entirely
unreasonable assumption that electrons at the surface absorb all the incident light until they have
accumulated sufficient energy to escape from the metal, and that there’s only one electron per
atom that is capable of absorbing energy. How long after the light is switched on will electrons
begin to escape the metal?
Discussion
Our ridiculous model has the virtue that it underestimates by many orders of magnitude how
long it should actually take for electrons to begin escaping from the metal. The reality is quite
different! It was known in 1928 that it took less than 3 nanoseconds to begin observing electrons
after illumination; in 1955 it was known to take less than 10-10 seconds. I’m sure the present
upper limit is substantially smaller.
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Exercise 14.2: An estimate of the dependence of maximum photoelectron kinetic energy on
illumination intensity
Let’s continue to work with our flawed classical model. We illuminate our metallic surface with
monochromatic light and crank up the intensity, measuring the kinetic energy of the ejected
photoelectrons. Imagine that we find something like the following distribution for the kinetic
energy spectrum of detected electrons:
light source
ejected electron
number detected
electron detector/spectrometer
maximum energy
conductor
kinetic energy (eV)
.
maximum kinetic energy (eV)
Propose a sensible-sounding (classical) model that relates the maximum kinetic energy of
detected photoelectrons to the intensity of the light source, then sketch what you expect to find
on the graph below.
intensity of light source
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Discussion: photons!
Our ridiculous model gets this wrong too: the maximum kinetic energy is independent of the
intensity of illumination. The number of ejected electrons does increase with intensity, but the
photoelectron kinetic energy spectrum does not change with intensity!
Einstein’s idea was that the energy of the electromagnetic field comes in discrete quanta, called
photons. The energy carried by a single photon depends only on the classical wavelength
associated with the light shining on our metal surface: the shorter the wavelength, the higher the
photon’s energy. The exact expression is
hc
Eγ =
λ
red
orange
yellow
green
blue
where h = 6.626 × 10 Joule-second so that hc = 1.986 × 10-25 Joule-meter. It is more useful to
express this in terms of electron volt-nanometers: hc = 1,240 eV-nm. For example, the energy
carried by a photon of green light (~600 nm wavelength) is about 2 eV. If you’re viewing this in
color (on the web, for example), the following spectrum will let you see the correspondence
between color and wavelength.
violet
-34
The photoelectric effect involves a single photon being absorbed by an electron, which uses the
acquired energy to escape from the surface of the metal. A brighter light source sends more
photons per second towards the surface, but each photon still carries the same amount of energy.
Photons carry momentum too. As is the case with classical electromagnetic waves, a photon with
energy E carries momentum p with p = E c = h λ . (See the brief discussion of relativistic
kinematics in the second week’s materials: since E 2 = p 2 c 2 + m 2 c 4 , anything massless must
satisfy the energy-momentum relation E = pc: if it carries energy, it also carries momentum.)
Exercise 14.3: Resolution of a simple magnifier
Imagine we build a simple optical system which images light from a small object onto a sensor.
How well can it determine the position of a small object? Let’s begin with a very simple
optical system, a pinhole camera. Note that determining the resolution isn’t a matter of
“geometrical optics” since we’ll have to think about diffraction effects associated with the finite
width of the pinhole. To further simplify the analysis, let’s use a horizontal slit instead of a
pinhole so that the camera can only image light along the vertical direction. Here’s a diagram:
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detector/sensor
Do
Dd/s
slit, width A
object emits light
of wavelength λ
image formed here
opaque screen
.
If the object is a point source, how well localized is its image on the detector? Can we tell to
arbitrary precision where the object was? Another way of asking this: is it possible to produce a
single Euclidian point (or in this case Euclidian line) image at our detector/sensor? If we can’t
(because of diffraction effects, for example), then the theoretical resolution of our optical system
is limited by physical principles.
It is tempting to think that the narrower the slit the better able we’ll be to localize the image of
our point source on the detector surface. This is a geometrical-optics way of thinking! The
problem is that we’ve neglected the role played by the opaque sheet in the propagation of the
electromagnetic waves from the point source to the detector.
The idea is simple: unless the opaque sheet does something to cancel the incident
electromagnetic waves, they’ll go sailing straight through without being affected. The electric
field in light waves from the point source accelerates electrons in the opaque screen, causing
them to radiate their own fields which cancel the incident fields. It’s not hard to see how this
works! The key is in the presence of the negative sign in front of the expression for the radiation
electric field:
Qa⊥′
1
E
=−
3
2
1 r only
r − r ′ 4πε 0 c (1 − εˆ ′ ⋅ v ′ c )
Q
An electron in the opaque screen is accelerated by the incident electric field so that it
experiences an acceleration a ∼ QEincident . Neglecting fine points such as the distinction between
a and a⊥ , the radiation field is produced along the direction −Qa ∼ −QEincident which is opposite
to the incident field's direction.
Think of the opaque screen with the slit as a solid plane of antennae which radiate
electromagnetic waves (perfectly canceling the incident electromagnetic wave downstream of
the screen) combined with a narrow strip of identical antennae which generate electromagnetic
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waves of the opposite polarity to cancel the perfect cancellation in the region where the actual
screen has a slit. Sounds confusing, yes? Here’s a diagram. To make it easier to draw, I’ll only
draw the polarization component of incident radiation parallel to the slit.
=
+
What this all comes down to is that we’ll do fine if think of the electromagnetic radiation passing
through the slit as if it had all been generated by a strip of antennae as wide as the slit which is
cut into the opaque screen, all radiating coherently (in phase). No other sources of light need
contribute to the signal observed by our detector. Another diagram:
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Dd/s
enlarged view of slit, width A
θ
A/2
extra distance for lower ray
opaque screen
detector/sensor
Since electromagnetic waves leaving all antennae are in phase, they can interfere constructively
or destructively at the detector/sensor if the path lengths are different from antenna to antenna to
a spot on the sensor.
(a) Assuming the sensor is very far from the opaque screen (compared to the width of the slit)
and that the wavelength of light is λ, at what angle θ will the two rays drawn in the diagram
arrive at the sensor out of phase? (This angle corresponds to a half-wavelength difference in path
A,
length for the two rays, as indicated in the diagram.) You may assume that Dd / s
Dd / s
λ , and that θ is small enough so that sin θ ≈ θ .
(b) In terms of λ, A, and Dd/s what is the half-width of the bright spot at the detector?
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Discussion
Notice that all pairs of antennae interfere destructively at the angle you calculated.
Discussion
Resolution of our optical system: it would be difficult to tell two point sources apart if their
bright spots overlapped at the detector. It's reasonable to identify the half-width of the bright spot
from a point source as the system's resolution.
This holds true even when we replace the slit with a lens of aperture A. Except for fine points
concerning difference in how the interference zeroes behave for circular and slit apertures, it
works the same way. Small wavelengths and large apertures give better position resolution.
Exercise 14.4: Heisenberg’s microscope gedanken (thought) experiment
Imagine we build a simple magnifier to locate a small object illuminated by a light source of
wavelength λ. Our sensor is so quiet and so efficient that we can determine the position of the
object when a single photon scatters from it, passes through the lens, and arrives at the sensor.
Naturally, we are unable to tell the path the photon took as it traveled through the lens to the
detector; all we know is where the photon ultimately arrived at the face of the detector.
photon’s path
light source
x
θ
z
A/2
D
Dd/s
detector/sensor
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(a) A photon, initially traveling along the z axis of our optical system, bounces off the object at
an angle θ as shown. Calculate the x component of momentum transferred to the object when
the photon scatters from it. (You may assume that the photon doesn’t lose any energy in the
collision.)
(b) Let's define the scale of momentum transfer to our object to be ∆px. The full range of possible
x momentum transfers to the object, for all photon paths which pass through the lens, is ±∆px .
Calculate ∆px assuming that D
A and that θ is small enough so that sin θ ≈ θ .
(c) Assuming that D = Dd/s for simplicity, calculate the position resolution ∆x for this optical
system.
(d) Calculate the product ∆px∆x for our locate-the-object optical system.
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Discussion
The Heisenberg Uncertainty Principle! It’s true in general, not just for this particular (optical)
system.
The exact expression, which would require us to define more precisely what we mean by a
position and momentum uncertainty is ∆px ∆x ≥ h 4π . Since h 2π appears so frequently in the
equations of quantum mechanics, the convention is to use a new symbol (pronounced "h-bar"):
≡ h 2π . = 1.0546 × 10−34 Joule ⋅ seconds = 6.5822 ×10−16 eV ⋅ seconds . Keep in mind that
1 Joule ⋅ second = 1 ( kg ⋅ m/sec ) ⋅ m . The smaller the value of the smaller the domain in which
quantum mechanics will differ noticeably from the (fundamentally incorrect) classical
description.
In three-dimensional systems we have three separate expressions: ∆px ∆x ≥
∆p z ∆ z ≥
2 , ∆p y ∆y ≥
2,
2.
The uncertainty principle is built into the structure of our quantum mechanical universe. It holds
true for ALL systems, not just those which employ photons in measurements.
Exercise 14.5: Atomic stability, thanks to the uncertainty principle.
An electron is known to be confined inside a hollow cubic of side length 2a so that a reasonable
measure of its position uncertainty is ∆x = ∆y = ∆z = a .
(a) In terms of a, the electron mass me, and , calculate the typical kinetic energy this electron
will have. (You now know how to calculate typical magnitudes for the electron's momentum
components. Use these to calculate the electron's kinetic energy.) You should find that
K ∼ 3 2 8me a 2 ≈ (.029 eV ⋅ nm ) a 2 when a is expressed in nanometers.
(
)
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(b) Here's a not-nearly-as-stupid-as-it-seems model for a hydrogen atom. Assume the electron in
a hydrogen atom is confined by the Coulomb potential to rattle around inside a cube of side
length 2a so that the electron's typical potential energy is − q 2 ( 4πε 0 a ) . (q is the fundamental
charge; − q 2 ( 4πε 0 a ) ≈ − (1.44 eV ⋅ nm ) a when a is expressed in nanometers.) As a result, our
simple model predicts that the electron's total energy should be
.029 eV ⋅ nm 1.44 eV ⋅ nm
.
a2
a
The size of the volume in which the electron resides will be that which minimizes its total
energy.
E = K +U ≈
Sketch a graph of the electron's total energy as a function of a.
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(c) Calculate the value of a which minimizes the electron's energy.
(d) Calculate the electron's total energy at this preferred value for a.
Discussion
Atoms don’t collapse: lower (more negative) potential energy is more than offset by the rising
kinetic energy associated with better localizing the electrons!
The full-blown quantum mechanical treatment of the hydrogen atom reveals that the probability
to find the electron near the point ( r ,θ , φ ) is proportional to e − r .053 nm . (The value of .053 nm is
called the Bohr radius.) The energy needed to remove a bound electron from a hydrogen atom is
13.6 eV.
Your results for the atom's size and the energy of the electron should have been surprisingly
close to these values!
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Discussion: what’s really going on
It is not possible to describe the world correctly if we insist on using a theory which describes a
system in terms of the positions and momenta of its constituent particles. The predictions of any
such theory are guaranteed to be wrong whenever the theory crashes into the limits on precision
set by the uncertainty principle. It’s a deeper problem than not just being able to measure the
position of a particle without disturbing its momentum: a variety of effects which are absolutely
impossible in a classical theory become allowed in quantum mechanics.
We are forced to abandon a theory which makes statements like these:
dp
);
dt
•
F = ma (which can be written −∇V =
•
p2
+ U = E ).
K + U = E (which can be written
2m
In Newtonian mechanics we have the habit of solving directly for functions which tell us the
values of the particle’s kinematic parameters. For example, in a uniform gravitational field,
1
x ( t ) = x0 + v0t + at 2 .
2
This doesn’t work for quantum systems! Our classical description assumes that we are able to
say exactly where the particle is at a particular time. No good!
We need new ways of thinking, described by a new equation.
The idea is this: we need to give up an approach in which we talk directly about the values
assumed by kinematic parameters as functions of position and time. Instead we need to work
with the probabilities that we might find particular values for the various parameters if we
happen to make a measurement. More exactly, we need to work with the probability amplitudes
associated with quantities like position, momentum, and energy.
We’ll use our new equation to calculate the form of a particles wave function: a function which
contains information about the particle. (It’s like an owner’s manual for a stereo: the manual
contains information about the stereo. The owner’s manual is NOT the stereo, just an
information source about the stereo.) The way we extract information about the particle from its
wavefunction ψ ( x, y, z , t ) is this:
•
ψ ( x, y, z, t ) dV is the probability to find the particle in a small volume element dV centered
2
at position ( x, y, z ) at time t.
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•

∫ (ψ )  i
∗
all
space
dψ 
 dV is the average energy we will obtain if we measure the energy of many
dt 
identical systems
•

∫ (ψ )  −i
∗
all
space
∂ψ 
 dV is the x component of the average momentum we will obtain if we
∂x 
measure the energy of many identical systems
In general, ψ ( x, y, z , t ) can be complex: ψ ∗ is the complex conjugate of ψ .
∂
∂
as the “energy operator” and −i
as the “x-momentum operator.”
∂x
∂t
 ∂ ∂ ∂
For three-dimensional systems, the momentum operator is −i  + +  = −i ∇ .
 ∂x ∂y ∂z 
We refer to i
We can replace momentum and energy in our K + U = E equation:
2
2
d
p2
∂2
d
−
∇2 + U = i
in three.
+U = E ⇒ −
+
U
=
i
in
one
dimension
and
2
2m
2m ∂x
2m
dt
dt
Clearly, we need to add something on which the differentiations can act: the correct choice is the
wavefunction.
This yields the “master equation” for quantum mechanics, called Schrödinger’s equation:
2
2
d 2ψ
dψ
dψ
U
i
ψ
−
+
=
in
one
dimension
and
−
∇ 2ψ + Uψ = i
in three.
2
2m dx
dt
2m
dt
Solving this for ψ will provide us with all the information about a nonrelativistic system that it
is possible to obtain.
Summary and wrap-up discussion
Here’s what you were working on:
• a classical analysis of the rate at which electrons “should be” ejected from a conductor which
is exposed to an intense light source.
• a classical analysis of the relationship between the maximum kinetic energy of electrons
ejected from a conductor and the intensity of the light source illuminating the conductor.
• photons! (photon energy and momentum)
• modeling the resolution of a simple magnifier
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•
•
analyzing the “Heisenberg’s Microscope” thought experiment
estimating atomic sizes using the Uncertainty Principle
We also discussed the master equation of quantum mechanics, known as Schrödinger’s equation.
That's it!
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Physics 212
Honors topics in electrodynamics
Notes on special relativity
Spring, 2003
George Gollin
University of Illinois at Urbana-Champaign
2003
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236
Notes on Special Relativity
George Gollin
University of Illinois
Contents
The speed of light is finite ............................................................................................................1
The speed of light is constant, regardless of relative motion of source and observer ..................4
Nonsimultaneity of events viewed from different reference frames ............................................7
Things on which we can agree when comparing notes with other observers.............................12
Time dilation...............................................................................................................................16
Doppler shifts..............................................................................................................................18
Lorentz contraction .....................................................................................................................22
Quantitative description of nonsimultaneity...............................................................................24
Some simple paradoxes and their resolution ..............................................................................27
Coordinates of an event, viewed from different reference frames..............................................28
Time dilation in terms of events .................................................................................................29
Lorentz contraction in terms of events .......................................................................................30
Lorentz transformations ..............................................................................................................31
Addition of relativistic velocities................................................................................................33
Causality .....................................................................................................................................35
The speed of light is finite
All the strange features of relativity are consequences of the fact that the speed of light is exactly
2.99792458 × 108 meters per second. It doesn't matter whether the source of light moves with
respect to the measuring apparatus: any device measuring c (the speed of light) will obtain,
within experimental error, this value. Nothing else works this way: sound travels at fixed speed
with respect to the air, for example. The value 2.99792458 × 108 meters per second really is
exact: it serves as a definition, along with some time standard, of the length of one meter. The
speed of light is very close to 1 foot per nanosecond, and I'll use that as a convenient
approximation periodically.
Let's start by discussing how moving objects look to a single, stationary observer. Because the
speed of light is finite, an object moves between the time light rays leave it and the time they
arrive at an observer. For example, here's a diagram of what happens when a light bulb,
traveling at 0.2 c, is seen by an observer:
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v= 0.2c
true position
2 feet
apparent position
10 feet
.
Things are more complicated when the light comes from an extended source instead of a point
source. Imagine we put lights on the corners of a transparent rectangular block and observe it
from far away like this:
1 foot
Light from the back of the block takes about a nanosecond longer to reach the observer
than light from the front. If the block is in motion, the observer will see its back at an
earlier time from its front:
cube is moving up
.
As a result, the block looks like a parallelogram and not a rectangle:
.
Because the speed of the block is greater than the component of the light ray’s velocity along the
cube’s direction of motion, light from the far, back corner doesn’t travel through the transparent
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cube on its way to the observer. If the block had been opaque instead of transparent, light from
the front side would not have been visible while light from the back side would have been.
(Imagine a camera shutter.)
There's nothing exotic about this: the same effect would happen if you tried to locate airplanes in
formation at an air show by listening for their engine sounds.
Objects moving towards or away from an observer will also appear distorted. For example,
consider what happens when an observer looks at a square plate, as shown below:
2 feet
4.243'
4.123'
2 feet
4.000'
The center of the square is exactly 4 feet from the observer, while the centers of the sides and the
corners are 4.123 feet and 4.243 feet away, respectively. Light rays leaving the square's center
and corners simultaneously won't reach the observer simultaneously. If the square moves
towards the observer, he'll see the corners of the square at an apparent position that corresponds
to a time that's 0.243 nanoseconds earlier than the apparent position of the center of the square.
A diagram, assuming the square moves with speed close to c (not really accurate-- the square
looks curved in two dimensions):
apparent shape of the
square's surface
side view
0.243'
.
The outline of the square isn't "square" anymore, either:
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.
What does the square look like if it is moving away from the observer? What does a
transparent rectangular solid look like when it moves towards the observer at a speed
close to c?
Keep in mind that the unusual appearance of the objects we’ve been discussing just arises
from the finite speed of propagation of light. If several observers position themselves as shown
below, they’ll all agree that each was passed by the closest part of the square at the same time
(call it t=0).
In a sense, the objects just “look funny” but aren’t really distorted due to the finite
speed of light.
The speed of light is constant, regardless of relative motion of
source and observer
This is the tricky part in relativity. Let’s first consider how sound works, to show the
contrast with the way light propagates. Sound travels at a fixed speed with respect to still air.
The speed of sound is about 331 meters per second; let’s approximate this as 1000 feet per
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second or one foot per millisecond. Imagine that an observer on the ground measures the
propagation speed of the engine noise from a subsonic airplane:
1
2
100 feet
Even though the airplane is moving, it will take about 100 milliseconds for any
particular sound wave to travel between the two measuring stations labeled 1 and 2. The
speed of sound is independent of the speed of the airplane.
If we put the observer on a train, so that he’s moving towards the source of the sound with speed
v, the time for a particular sound wave to travel from the first sensor to the second will be
reduced.
1
2
v
From practical experience, we know that snowballs don't work the same way sound
does. A snowball thrown from a car hits a pedestrian a lot harder if it's thrown in the
direction the car's traveling than if it's thrown in the opposite direction.
Light has the (unusual??) property that any measurements made of its propagation speed always
yield the same answer, regardless of relative motion between light source and observer. Each of
the following observers will measure the same transit speed between the two sensors:
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v=0.9c
laser beam from rocket
100 feet
1
2
(Here we see the source moving towards a fixed observer.)
v=0.9c
laser beam from rocket
100 feet
1
2
(Here we see the source moving away from a fixed observer.)
light ray from a distant star
1
2
v = 0.8c with
respect to star
(Here we see the observer as moving towards a fixed source.)
This is not what we’re used to-- either in the way sound travels or in the way snowballs thrown
from cars behave. You can read about some experimental tests of this in a variety of textbooks.
Here's another, based on every-day things at Fermilab.
We know that many different kinds of unstable elementary particles can be produced in violent
collisions between high energy protons and target nuclei. The lightest of the particles made of
quarks and antiquarks which can be produced this way are call pi mesons, or pions. There are
three different kinds of pions: π+ and π- have identical masses of 139.57 MeV/c2 while the
lighter π0 has mass 134.96 MeV/c2. (The proton mass is 938.28 MeV/c2.) It is known that
nearly 100% of the time a π0 will decay into a pair of energetic photons (the quantum
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mechanical version of electromagnetic waves-- light). Each gamma ray photon from a pion
which decays at rest in the lab carries away 67.48 MeV of energy and travels travel with speed c
(which isn't surprising).
A source of energetic pions at Fermilab shoots π0's into my experiment, which measures the
kinematic properties of their decays. These pions are traveling close to the speed of light-useful π0's have values of v/c which range from 0.9996 to 0.9999993-- and their decay photons
can have energies as high as 100 GeV (1 GeV = 1000 MeV). A diagram comparing a typical π0
decay at rest with what we observe at Fermilab follows:
pion at rest
photons are
180° apart
pion with v/c = .999990
angle between the photons is about 1/2°
Even though the source of the photons is moving at high speed, we find that the speed
with which these photons travel is exactly the same as the speed with which the photons from
the stationary pion travel.
Nonsimultaneity of events viewed from different reference
frames
The consequences of the constancy of the speed of light are dramatic. Here's the first of them.
Consider the following hypothetical encounter between the rockets Nostromo and Exeter. Both
are capable of traveling at speeds approaching c, and each carries a pair of periscopes. A spark
jumps between the Nostromo’s and Exeter’s electrode as the two ships pass each other.
Observers on each ship, equidistant from the electrode, see the light from the spark. Let’s view
the creation of the spark from the perspective of passengers on the Nostromo. The spark is
created at the electrodes when the Nostromo’s clocks read t = 0, as shown in the figure below.
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N
Nostromo
spark gap
periscopes
?
periscopes
Exeter
?
E
As light from the sparks travels towards the Nostromo’s observer, the Exeter continues to
move to the left. Here’s a “motion picture” of what happens, from the perspective of
passengers on the Nostromo. Let’s assume that the Exeter is traveling with v = 0.5c.
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N
Nostromo clocks read
t=0 as the Exeter
passes.
Nostromo
Light from spark
leaves spark gap.
?
Exeter
?
E
Nostromo crew thinks
Exeter is moving to
the left with v = 0.5c
N
Nostromo
Nostromo clocks read
t=100ns.
light
pulse
Light from spark has
traveled 100 feet,
Exeter has traveled 50
feet. Light reaches aft
Exeter periscope.
?
?
Exeter
?
E
N
Nostromo
Nostromo clocks read
t=200ns.
Light from spark has
traveled 200 feet,
Exeter has traveled
100 feet. Light has
already reached both
Nostromo periscopes
and aft Exeter
periscope.
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light
pulse
?
?
Exeter
?
E
.10
245
As you can see, the Nostromo’s observers see the spark simultaneously (light travels one
foot per nanosecond, and the periscopes are equidistant from the spark gap). However,
the Nostromo crew members think the Exeter’s aft observer sees the spark before the forward
observer sees the spark.
Let’s shift to the Exeter’s frame of reference now. As far as the Exeter’s crew is concerned, they
are happily sitting around eating dehydrated food when the Nostromo glides past them, moving to
the right with v = 0.5c. The Exeter’s periscopes are equidistant from the spark electrode, and
will necessarily see the produced spark simultaneously, as far as the Exeter's crew is concerned.
This is in disagreement with the conclusion of the crew of the Nostromo. Here is a "movie" of
what things look like from the perspective of the Exeter’s crew. (There are certain inaccuracies
in the drawing concerning the relative lengths of the ships that we'll neglect for the time being.)
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Exeter clocks read t=0
as the Nostromo
passes.
Light from spark
leaves spark gap.
N
?
?
?
Exeter
Exeter crew thinks
Nostromo is moving to
the right with v = 0.5c
Nostromo
?
E
N
?
?
?
Nostromo
light
pulse
Light is about to reach
aft Nostromo
periscope.
Exeter
E
N
?
?
?
Nostromo
light
pulse
Light has already
reached both Exeter
periscopes and aft
Nostromo periscope.
Exeter
E
Things are rather different in the Exeter's reference frame.
So-- events which are simultaneous in one frame of reference are not necessarily
simultaneous in another frame of reference!! Curious!
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In a ship's rest frame, its clocks read the same value when the spark is produced (since they're
synchronized) and also read the same value when the light from the spark arrives at the
periscope. (The light travels the same distance from the spark gap to the forward periscope as it
does to the aft periscope.)
Imagine that each clock stops (breaks??) when the light reaches it. In a ship's rest frame, its own
clocks will read the same value when they break. An observer in any frame of reference will
agree with the ship's crew that the broken clocks read the same (stopped) time.
A ship's crew will conclude that the aft clock on the other ship breaks before the forward clock
on the other ship. However, once both its clocks are broken, they will be seen to read the same
time. Consequently, as seen from the rest frame of one ship, the aft clock on the other ship
reaches this time before the forward clock on the other ship.
The crew of one ship, looking at the clocks on the other ship (which is seen to be moving) will
conclude that the clocks on the other ship are not synchronized. In particular, the aft clock is set
ahead (reads a later time) relative to the forward clock.
Both crews will agree that the clocks at the spark gap (the center clocks in the pictures) read zero
when the spark was created. (Imagine that these clocks break when current flows through the
spark electrodes.)
Here's how the clocks will look in the rest frame of the Exeter:
N
Nostromo
Exeter
E
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Things on which we can agree when comparing notes with other
observers
The disagreement between the Exeter’s and Nostromo’s crews' observations about the
(non)simultaneity of the observations of the spark comes about because of the fact that the
observations take place at different points in space. Events which take place at the same point in
space and at the same time will be seen by both crews as happening simultaneously. Here are
some examples of pairs of events which work out this way:
(Nostromo’s electrode passes Exeter’s electrode)
.and.
(Nostromo’s center clock reads t=0)
(Nostromo’s rear clock breaks)
.and.
(light from spark arrives at Nostromo's rear periscope)
(Exeter’s front clock breaks)
.and.
(light from spark arrives at Exeter's forward periscope)
(I'm assuming that all clocks are tiny, and arbitrarily close to the electrodes or periscopes, etc.
etc.)
It’s easy to imagine how this works: imagine that one event (light arriving) causes some sort of
explosion which breaks the clock immediately. All observers will agree on the value read by the
(stopped) clock.
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Here is another example of a “reference frame-invariant” occurrence. I’ve underlined the
statement with which all observers will agree.
Left ruler is stationary, right ruler is
moving. Rods at ends of right ruler pass
through hoops at ends of left ruler when the
rulers collide.
Left and right rulers are moving towards each
other with equal speed. Rods at ends of right
ruler pass through hoops at ends of left ruler
when the rulers collide.
Note that this means that the two rulers will appear to have the same length, whether viewed
from a frame in which one is at rest or in which both are moving (perpendicular to their lengths).
Here’s another example: let’s say the Nostromo carries a pair of parallel mirrors. A crewmember
sets a light beam bouncing between the mirrors.
Nostromo, as seen by a crewmember floating outside the rocket.
Light ray leaves
bottom mirror
N
Nostromo
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Light ray
bounces off top
mirror
N
Nostromo
Light ray
bounces off
bottom mirror
N
Nostromo
The light beam bounces from mirror to mirror.
Nostromo, as seen from the Exeter.
Light ray leaves
bottom mirror, moving
up and to the right
N
Nostromo
N
Nostromo
Light ray bounces off
top mirror, now
moving down and to
the right.
Light ray bounces off
bottom mirror, moving
up and to the right
N
Nostromo
The light beam bounces from mirror to mirror as the Nostromo moves to the right.
Time dilation
Let’s examine the light-bouncing-off-a-mirror case in more detail. For convenience, let's assume
that the mirrors are three feet apart. The roundtrip pathlength between the bottom and top
mirrors is six feet; it takes a light beam about six nanoseconds to make the trip, as measured by a
crewmember of the Nostromo. The path taken by the light is longer, when viewed by
crewmembers of the Exeter who see the Nostromo as moving quickly to the right. Imagine that
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251
the relative velocity between the Nostromo and Exeter is 0.8c. Here's how the light beam seems
to move:
3 feet
Nostromo speed is 0.8c
An important point: an observer in either rocket who measures the speed with which the light
beam in Nostromo’s clock travels will always find that it travels 2.997...× 108 meters/second.
Let's calculate the round trip time (assuming c = 1 foot per nanosecond) for the moving mirrors.
Let's say that the time for a half-tick is ∆t', so that the round trip takes 2∆t' (note the prime here),
while the time for a half-tick of a stationary clock is ∆t (no prime):
In time ∆t', the mirrors travel ∆x ' = v∆t ' .
In this time, the light ray travels L ' = (∆x ') 2 + (∆y ') 2 = (v∆t ') 2 + (3) 2 .
Since c = 1ft/nsec in all frames, L' = c∆t' so... c∆t ' = (v∆t ') 2 + (3) 2 .
It's true that c∆t = 3 feet (no primes here: in the frame in which the clock is at rest), so
we can rewrite c∆t ' = (v∆t ') 2 + (c∆t ) 2 . Solve for ∆t' in terms of ∆t:
(c∆t ') 2 − (v∆t ') 2 = (c∆t ) 2 or ∆t ' =
∆t
1 − v2 / c2
.
Observers think the moving clock takes longer to tick by a factor of 1/ 1 − v 2 / c 2 than do
observers at rest with respect to the clock. The clock seems to be running slowly
compared to a "normal" clock. When v=0.8c, 1/ 1 − v 2 / c 2 = 5 / 3 .
The Nostromo's and Exeter's crews will not agree about the time required for Nostromo's "light
clock" to tick once. You can imagine how this analysis works if the light clock had been on the
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252
Exeter instead of the Nostromo: The Nostromo's crew would think the Exeter's light clock was
ticking too slowly. There's complete symmetry between the Exeter and Nostromo. Each crew
sees the other crew's rocket moving at high speed. In general, a light clock which requires time
t for each tick (as viewed by observers stationary with respect to the clock) will require a time
t ' = t 1 − v2 / c2
for each tick when viewed from a frame in which the clock seems to be moving with
speed v. Here's a plot of t'/t as a function of v/c:
14
Time dilation factor vs. v/c
1 1 − v2 c2
12
10
8
6
4
2
v/ c
0
0
0.2
0.4
0.6
0.8
1
So-- moving light clocks seem to run slow!
What about other moving clocks, for example mechanical or biological clocks? They all have to
seem to run slowly! Here's a plausibility argument for this. Let's say we stretch the distance
between the mirrors in the Nostromo's light clock so that it takes one second, as determined by
the Nostromo's crew, for the clock to tick. Imagine that one crewmember, whose pulse rate is 60
beats per minute, covers the bottom mirror with a black cloth, removing the cloth briefly each
time his/her heart beats. The light ray bouncing back and forth will always arrive at the bottom
mirror when it is uncovered, and the "clock" will continue to tick. A crewmember of the Exeter
will see the Nostromo’s crewmember uncover the mirror in time with his/her heartbeat, even
though the time between bounces will be longer than one second. The Exeter’s crewmember
will conclude that the Nostromo’s crewmember’s pulse has slowed down by the same factor as
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253
the rate of ticking of the light clock. All clocks, whether mechanical, biological, or electrical
must slow down by the same factor.
This “time dilation” effect is entirely consistent with observations of changes in average
lifetimes of unstable particles which are moving at high speed. Here’s an example, based on
what we see at Fermilab.
K mesons are unstable particles which can be produced in collisions between protons and target
nuclei at particle accelerators like Fermilab (which is in Batavia, about an hour’s drive west of
Chicago.) A K meson weighs about half as much as a proton. There are two kinds of neutral K
mesons and two kinds of charged K mesons; a “long-lived” neutral K, or KL, has an average
lifetime of 51.83 nanoseconds. KL‘s can decay into various combinations of lighter particles
such as pions, electrons, neutrinos, and muons. Fermilab shoots a beam of 1011 800 GeV
protons per second into a beryllium rod located 128 meters “upstream” of my experiment.
About 106 KL per second leave the target and travel towards the experiment. The K mesons are
traveling close to the speed of light; typical values for their time dilation factors range from 50 to
300. If time dilation didn’t slow down the KL‘s internal clocks, the average number surviving
after a time interval T would be e −T 51.83 . Traveling at one foot per nanosecond, only 0.03%
of the kaons (or about 300 per second) would survive to reach my experiment. In reality, more
than 90% of the kaons manage to travel the 128 meters without decaying in flight.
Doppler shifts
What would happen if we tried to make a “sound clock” which worked by bouncing the sound of
a click between two reflectors mounted on a railroad flat car? Would we get the same sort of
time dilation phenomena? (Of course we wouldn’t.) Let’s do the analysis. In the picture below,
the reflectors are moving about half the speed of sound:
331m/sec
stationary reflectors
click!
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331m/sec
moving reflectors
click!
.19
254
Because the speed of sound is 331m/sec with respect to still air, it will take longer for the
sound of the click to arrive at the upper reflector. Everyone will agree that the sound clock
moving with respect to still air is ticking slowly.
This time dilation effect will make Doppler shifts for relativistic motion work somewhat
differently from Doppler shifts for nonrelativistic systems. Remember about Doppler shifts?
Here’s how they work for sounds coming from nonrelativistic sources. Imagine a train emits
periodic clicks, as in the following diagram:
1
2
3
4
5
331 m/sec
1 2 34 5
The wavefronts from the clicks travel away from their points of creation at 331 m/sec.
A listener in front of the train will hear a shorter period between clicks than a listener
behind the train. Let’s say the train is moving at half the speed of sound, emitting 331
clicks per second, and that the first click is emitted at x=0. Here’s a table describing what
observers at x = +10 and x = -10 hear:
click
1
2
3
4
5
(x,t) for emission
of click
x=0, t=0
x=0.5, t=.003
x=1, t=.006
x=1.5, t=.009
x=2, t=.012
distance to
x = +10
observer
10m
9.5m
9m
8.5m
8m
arrival time at
+10 observer
.0302
.0317
.0332
.0347
.0363
distance to
x = -10
observer
10m
10.5m
11m
11.5m
12m
arrival time at
-10 observer
.0302
.0347
.0393
.0438
.0483
Because the source of the clicks is moving, the frequency with which the clicks arrive is shifted,
due to the different distance traveled by each click. For a stationary observer listening to a
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255
source which moves a distance D between clicks, the extra distance traveled by successive clicks
is -Dcos(θ):
D
θ
extra distance is
Dcos(θ)
observer
If the source emits clicks separated by time t, it will move a distance D = vt between
clicks. The difference in the transit time of successive clicks will be -Dcos(θ) / c where
c is the speed of sound. As a result, the ratio of the apparent time interval between clicks
to the “true” time interval will be
(t - Dcos(θ)/c) / t = (t - vtcos(θ)/c) / t = 1 - vcos(θ)/c.
If you prefer to deal in frequencies instead of periods, you can use the fact that frequency is the
same thing as 1/period to conclude that
fapparent / fsource = c / (c - vcos(θ)).
The same thing holds true for sinusoidal waves-- we’d just be describing the time between peaks,
instead of time between clicks.
There’s one other effect that has to be included when considering relativistic motion-- time
dilation. Imagine the Nostromo transmits flashes of light, evenly spaced by a time interval t as
measured on the Nostromo’s clocks. An observer on the Exeter will measure a different time
interval between the flashes due to the same effect as described above for sound, but with an
additional complication: the Nostromo’s clocks seem to be running slowly by a factor
(1 - v2/c2)1/2. Naturally, c now represents the speed of light. The time interval between flashes
is increased, or the frequency decreased, by this amount. As a result, the apparent frequency of
flashes seen by the Exeter is
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256
f apparent = f source × 1 − v 2 c 2 ×
c
.
c − v cos(θ )
Some algebra reveals that this is equivalent to
c−v
c+v
for the case that the source is moving directly away from the observer and
c+v
f apparent = fsource ×
c−v
for the case that the source is moving directly towards the observer.
f apparent = fsource ×
Notice that many of the algebraic expressions describing the effects associated with motion at
high speeds misbehave when v = c. The fact that any observer measuring the speed of a light
ray will determine it to be 2.997...× 108 m/sec makes it impossible for any massive object to
travel faster than the speed of light. Here’s an illustration of why this is the case. Imagine that a
nasty spaceship carrying a large bomb is constructed, as shown in the diagram below. The bomb
will detonate if the attached photosensor is struck by light from the rocket’s internal laser. The
laser shoots its beam upwards through a tube with black interior walls, from the perspective of
the crew of the rocket.
hollow tube with
black walls
Dr. Death Lasers, Inc.
Bombs 'r'
Us
Someone traveling on the rocket ship will get blown up a few nanoseconds after the laser is
turned on, as measured by the ships clocks. If the rocket can travel faster than the light beam
from the laser, when both are viewed from some other reference frame, the laser will never be
able to detonate the bomb: the top of the tube will have moved horizontally farther than the laser
beam, even if the beam travels horizontally, and the beam will be absorbed by the walls of the
black tube. However, in the rest frame of the spaceship, the bomb certainly will go off. We can’t
reconcile these two descriptions of events: either the ship blows up on its way to Alpha
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257
Centauri, or it doesn’t. Later, we’ll see why it takes an infinite amount of energy to boost an
object at rest to the speed of light.
Lorentz contraction
Things we know:
1. A moving clock seems to run slow.
2. Two clocks synchronized in their rest frame appear to be out-of-synch when viewed from
another frame. In particular, the front clock of a pair of moving clocks seems to be set back
while the rear clock seems to be set ahead.
What happens to (apparent) lengths of relativistic objects?
Careful: you must state exactly what you mean to do when you say you'll measure something's
length.
One way to measure the length of a moving object:
N
v
Nostrom
Solaris
see which clocks are adjacent to the "moving ship's" nose and tail when t = 0 on all your clocks;
call the distance between these two clocks the length of the moving object.
Another way: use one clock and the fact that distance = speed × time; the moving ship is known
to travel with speed v.
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258
N
v
Nostromo
Solaris
start
finish
Solaris
From the Nostromo's perspective:
N
Nostromo
v
...
...
Solaris
First method (two clocks): Solaris' clocks are not synchronized so their measurement is not in
agreement with the Nostromo's.
Second method (one clock): Solaris' single clock runs slow by a factor of 1 − v 2 c 2 so its
measurement is "inaccurate" due to this: it measures a length which is shorter than the
Nostromo's "rest length" by this same factor.
Both methods still have to agree, although they will produce a value which is smaller than the
length determined by the Nostromo's crew.
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259
Quantitative description of nonsimultaneity
Let's calculate the amount of missynchronization in a pair of moving clocks (which are
synchronized in their rest frame). The rest length of the ship is Lship.
1
(in ship's rest frame)
v
v
A
B
2
(in ship's rest frame)
v
A
v
B
3
(in ship's rest frame)
v
A
v
B
Clocks A and B are synchronized in their rest frame. Their spacing is chosen so that in their rest
frame each reads zero when passed by the corresponding end of the space ship. Let's call their
separation in their rest frame LAB.
We will do this:
1.
calculate the distance from clock A to the tail of the ship in picture 1, in the ship's rest
frame
2.
calculate how long it took in the ship's rest frame for clock A to travel this distance
(this is how much time has passed on the ship between pictures 1 and 2)
3.
calculate the change in the reading on clock A between picture 1 and picture 2.
Since we know clock A reads zero in picture 2, this will let us calculate what it read in picture 1.
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260
1.
The clocks are positioned so that their rest-frame separation LAB matches the apparent
length of the ship. In the clocks' rest frame, the ship is Lorentz contracted so
LAB = Lship 1 − v 2 / c 2 .
In the rest frame of the ship, the separation between the (moving) clocks is Lorentz
contracted. Call this L'AB: we have
2
L ' AB = LAB 1 − v 2 / c 2 = Lship  1 − v 2 / c 2  = Lship 1 − v 2 / c 2 .


(
)
As a result, the remaining distance to be traveled by clock A is
Lship − L ' AB =
Lship v 2
c2
.
An illustration:
1
(in ship's rest
f
)
A
v
B
v
L'AB
Lship - LAB
'
Lship
2.
→2
=
In the ship's rest frame, the clock takes an amount of time t1ship
Lship − L ' AB
v
=
Lship v
c2
to travel this distance.
3.
The amount of time which passes on the face of clock A between pictures 1 and 2 is
reduced, because of time dilation:
t
1→2
AB
=t
1→2
ship
1− v / c
2
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2
{L
=
ship
}
1 − v2 / c2 v
c
2
=
LAB v
.
c2
.26
261
LAB v
in picture 1. Recall that LAB is
c2
the separation between the clocks in their rest frame, and that clock B read zero in picture 1. As
a result, the amount of "missynchronization" between two clocks separated by a distance ∆x in
their rest frame, when viewed from a frame in which the clicks are in motion, is v∆x/c2. The
front clock reads an earlier time than the rear clock.
Here's a summary of what we know about the effects of relative motion on distance and time:
Since clock A reads zero in picture 2, it must have read −
x=-L
x=0
t=0
x=+L
t=0
t = -vL/c2
L
t=0
t=0
t = +vL/c2
v
t = +vL/c2
t=0
t = -vL/c2
v
From the perspective of the TOP ruler
The top ruler is stationary with respect to us. The middle ruler is moving to our left, the lowest
ruler is moving to our right. In each ruler's rest frame, the separation between clocks is L and the
three clocks are synchronized.
1. Effect on distance between objects:
The separation between adjacent clocks seems reduced. Clocks on the moving ruler will seem to
be a distance apart given by
L × 1 − v2 c2 .
2. Effect on passage of time on a single clock:
The rate of passage of time on moving clocks seems slowed. If a clock on the stationary ruler
advances by an interval T, a clock on the moving ruler will advance by an amount of time
Physics 212, Spring 2003.
Notes on special relativity
.27
262
T × 1 − v2 c2 .
3. Effect on synchronization of clocks:
Moving clocks which are synchronized in their rest frame will not appear synchronized when
viewed from another frame. In general, a clock that is "in front" will seem to be set to an earlier
time than a clock that is "in back". In the example with the rulers illustrated above, a clock at
x = L (as measured in its rest frame) will read t = − Lv c 2 when the moving clock at x = 0 reads
t = 0. Note that v < 0 corresponds to motion to the left.
Some simple paradoxes and their resolution
I. Relativistic train/tunnel
Train: rest length 1000 feet, travels with v/c = 0.8.
Tunnel: rest length = 800 feet.
Stationmasters try to slam entrance and exit doors to the tunnel simultaneously while the train is
inside, trapping it in the tunnel.
IC
Illinois Central
Illinois Central
Illinois Central
1000 foot rest length
door
door
800 foot rest length
a. Why does this seem like a paradox?
b. What happens? Describe from both the train frame and the tunnel frame.
II. Clock accelerators
A "clock accelerator" consists of a vacuum pipe and a series of accelerating structures, as shown
in the diagram. The vacuum pipe contains two rows of clocks; the upper row is moving to the
Physics 212, Spring 2003.
Notes on special relativity
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263
right at speed v = 0.8c while the lower row is stationary with respect to the accelerating
structures. The accelerating structures (and stationary clocks) are 1000 feet apart. All clocks in
a given rest frame are synchronized. The rightmost moving clock reads 0 as it passes the
stationary clock, which also reads 0. Just as the stationary clocks read 0, all the accelerating
structures fire, boosting the lower row of clocks into the same frame of reference as the upper
row of moving clocks. Because the acceleration takes place in an extremely short period of time
(let's say a picosecond), none of the clocks have moved an appreciable distance before the
acceleration ends. In addition, the readings on the faces of all clocks are virtually unchanged
compared with their pre-acceleration values. As a result, the apparent spacing between clocks in
the (now moving) lower row will STILL be 1000 feet, and all these clocks will STILL read 0.
How is this possible? What happened to all that Lorentz contraction stuff we've been discussing?
?
?
Beam pipe
accelerating structures
1000 feet
Coordinates of an event, viewed from different reference frames
An event is something that happens at one point in space, and one instant in time.
Examples:
collision of clocks on board space ships
Exeter forward clock reads 0 as the clock passes the Nostromo's aft clock
Not an event:
Exeter crew measures the length of the Nostromo (takes a finite time interval or finite distance
interval )
Why is this event business useful? Time dilation, Lorentz contractions, etc. etc. can be described
in terms of their effects on the separation between events when one shifts reference frames.
These are special cases where either the space or time interval between events is zero in ONE
frame of reference. For the general case where ∆x ≠ 0 and ∆t ≠ 0, describing measurements in
terms of space-time intervals between events will make it easier for us to calculate correctly.
Physics 212, Spring 2003.
Notes on special relativity
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264
Time dilation in terms of events
event 1: Nostromo nose and Solaris clock coincide and Solaris clock reads 0.
event 2: Nostromo tail and Solaris clock coincide.
Notice that these events happen in the same place, according to Solaris ...
v
Event 1
Event 2
N
Nostromo
Solaris
start
finish
Solaris
...but occur in different places, according to the Nostromo:
N
Nostromo
Event 1
v
Solaris
Physics 212, Spring 2003.
Notes on special relativity
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265
N
Nostromo
Event 2
Solaris
If the time interval between two events in the frame in which the events occur at the same place
is T, the time interval between the events in a different frame (in which the events occurred at
different locations) is larger, namely T
1 − v2 c2 .
Lorentz contraction in terms of events
Event 1: Nostromo's tail passes by Solaris' forward clock.
Event 2: Nostromo's nose passes by Solaris' aft clock.
From the perspective of the Solaris, the events occur at the same time...
N
v
event 1
Nostromo
event 2
Solaris
...but from the perspective of the Nostromo, they don't.
Physics 212, Spring 2003.
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266
If the distance interval between two events in the frame in which the events occur at the same
time is L, the distance interval between the events in a different frame (in which the events
occurred at different times) is larger, namely L
1 − v2 c2 .
Why bother?
•Avoid sloppy thinking this way: describe all measurements in terms of intervals between events
(back of rocket coasts past one part of my tape measure, front of rocket coasts past another part
of my tape measure).
•Also: gives us a natural language with which to describe things which happen neither at the
same time nor at the same place in our reference frame, when viewed from other reference
frames.
Lorentz transformations
Let's work up a description of this: two events are separated by distance x, and time t in one
frame; separated by a distance x' and a time t' in another frame.
Imagine we are at rest with respect to a string of clocks as shown below. A second string glides
past the first, moving with speed v. Two clocks in each string are attached to spark electrodes; a
spark is created as the corresponding electrodes align. The left electrode in the (to us) stationary
string is at x=0, the right electrode at x=L. In the rest frame of the moving string of clocks the
left and right electrodes in this string of clocks are at x'=0 and x'=L'. In general, L and L' can be
different. (For example, the electrodes are placed 11 clocks apart on the lower string, but 10
clocks apart on the upper string.)
Event 1: the left clocks' spark electrodes touch and a spark is created.
x'=L', t'=?
x'=0, t'=0
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
v
Event 1
x=0, t=0
x=L, t=0
Coordinates of event 1 from the perspective of the lower string: x=0, t=0.
Coordinates of event 1 from the perspective of the upper string: x'=0, t'=0.
Physics 212, Spring 2003.
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267
From the perspective of the lower clock string, the upper string's right clock is at
x = L′ 1 − v 2 c 2 (note that I am using x, not x' here), and it reads t ′ = − vL′ c 2 (note that I am
using t', L', not t, L here).
Moving clocks look to be out-of-synch, and moving objects look shorter, right?
Event 2: the right clocks' spark electrodes touch and a spark is created.
x'=0, t'=?
x'=L', t'=??
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
Event 2
x=0, t=T
v
x=L, t=T
Coordinates of event 2 from the perspective of the lower string: x=L, t=T.
Coordinates of event 2 from the perspective of the upper string: x'=L', t'=??.
Let's sort out what the upper string's right clock is reading. We know it was reading -vL'/c2
when event 1 occurred. We know that a time interval T has passed on the lower string's clocks.
We know the upper string's right clock is ticking more slowly than the lower string's clocks. As a
result, the right clock must be reading t ′ = − vL′ c 2 + T 1 − v 2 c 2 .
Do some work to replace L' with quantities known to the bottom clock string observers:
Distance between bottom clocks is the same as the sum of the Lorentz-contracted rest-frame
separation between top clocks and the distance the top clock string moves in time T. In
L − vT
= L′ .
equations: L = L′ 1 − v 2 c 2 + vT or
1 − v2 c2
Substitute this into the previous equation to get
 L − vT  2
T − vL c 2
after some algebra.
t ′ = −v 
c + T 1 − v2 c2 =

1 − v2 c2
 1 − v 2 c 2 
Collecting what we know, we can conclude the following about the space/time interval between
the events, when viewed from either frame:
Physics 212, Spring 2003.
Notes on special relativity
.33
268
interval
space
time
bottom string's frame
∆x = x2 − x1 = L
∆t = t2 − t1 = T
top string's frame
L − vT
∆x′ = x′2 − x1′ = L′ =
1 − v2 c2
∆t ′ = t2′ − t1′ =
T − vL c 2
1 − v2 c2
Slightly more compactly (and after some algebra to express ∆x, ∆t in terms of ∆x', ∆t'), we have
the Lorentz Transformations:
∆x′ =
∆x =
∆x − v∆t
1 − v2 c2
∆x′ + v∆t ′
1 − v2 c2
∆t ′ =
∆t =
∆t − v∆x c 2
1 − v2 c2
∆t ′ + v∆x′ c 2
1 − v2 c2
It takes a bit of thought to keep the sign straight in front of the term proportional to velocity.
Note that v refers to the velocity of the reference frame which measures "primed" quantities
when viewed from the reference frame which measures "unprimed" quantities.
Intervals along directions perpendicular to the relative velocity between the frames agree in both
frames: ∆y' = ∆y, ∆z' = ∆z.
A useful shorthand is to define γ ≡ 1 1 − v 2 c 2 and β ≡ v c . Using this, we can write
∆x′ = γ ( ∆x − β c∆t )
∆t ′ = γ ( ∆t − β∆x c )
∆x = γ ( ∆x′ + β c∆t ′ )
∆t = γ ( ∆t ′ + β∆x′ c )
Given the space-time interval between two events as measured in one frame, we can calculate
what observers in another frame will measure if we know the relative velocity of that frame with
respect to the other frame.
Addition of relativistic velocities
Let's use the Lorentz transformation formulas to derive a velocity addition formula. Here's the
sort of problem we'll need this for:
Imagine the Nostromo's shuttle craft leaves the Nostromo traveling at speed v, with respect to the
Nostromo.
Physics 212, Spring 2003.
Notes on special relativity
.34
269
Nostromo
v
An observer on the Solaris sees the Nostromo coasting to the right with velocity u. What is the
shuttle's apparent velocity according to this same observer?
u
Nostromo
?
Solaris
To use Lorentz transformations to sort this out, let's define two events like so:
event 1: Shuttle leaves Nostromo and clock at Nostromo's nose reads t' = 0.
event 2: Shuttle passes Nostromo crewmember on spacewalk who is at rest with respect to the
Nostromo, a distance x' to the right of the Nostromo. The crewmember's watch reads t' = x'/v.
The interval between the two events is ∆x' = x'-0 and ∆t' = x'/v - 0
From the perspective of the Solaris, the Nostromo rest frame is moving to the right with speed u,
so we can calculate the separation between the two events using the Lorentz transformations
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(with β = u/c here, not v/c): ∆x = γ ( ∆x′ + β c∆t ′ ) = γ ( x′ + β c x′ v ) and
∆t = γ ( ∆t ′ + β∆x′ c ) = γ ( x′ v + β x′ c ) .
The shuttle's speed, according to the Solaris, is ∆x/∆t:
(1 + u v ) = ( v + u ) .
∆x γ ( x′ + β c x′ v )
=
=
∆t γ ( x′ v + β x′ c ) 1 v + u c 2
1 + vu c 2
(
) (
)
In the limit that u → c and v → c, the apparent speed of the shuttle tends towards c, not 2c.
Again-- we find that c is the maximum velocity for any object.
Causality
Nonrelativistic version: Event 1 can only cause event 2 if event 1 happened before event 2.
Problem with this version: nothing can go faster than c.
Event 1: You pick up the phone at 9am on January 1, 3000AD and place a LONG distance call
while you are coasting through space on the starship Alpha.
Alpha
Omega
Event 1: x = 0, t = 9:00 am, Jan. 1, 3000AD. Alpha places call.
Event 2: The starship Omega is 24 trillion miles (about four light years) away from the Alpha,
but at rest with respect to the Alpha. Clocks on the two ships are synchronized. A phone rings on
the Omega at 9:00am, January 1, 3003AD, waking up the Omega's captain.
Alpha
Omega
Event 2: x = 24 trillion miles, t = 9:00 am, Jan. 1, 3003AD. Omega phone rings.
Is it possible that you woke up the Omega's captain? Of course not-- the signal from your
telephone can't travel faster than c; it'll take slightly more than 4 years for your phone call to
arrive at the Omega.
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Relativistic version of causality: event 1 can only cause event 2 if event 1 happened sufficiently
early with respect to event 2 so that a light beam leaving x1 at time t1 could arrive at x2 no later
than time t2: x2 − x1 < c t2 − t1 .
We expect: events which are "causally connected" when described using x, t coordinates
measured in one frame of reference will be causally connected when viewed from any other
frame of reference. Is this really true? (We can Lorentz-transform to check that
∆x < c∆t → ∆x' < c∆t'.)
How does the Alpha-Omega phone call look when viewed from a third ship, which sees both
ships in motion? Let's say the ships seem to be moving to the right with speed v equal to βc.
Use Lorentz transformations.
Event 1
Alpha
Omega
1
Solaris
cosmonauts tethered to Solaris
Event 2
Alpha
Omega
2
Solaris
You might think that the Solaris could see the events as causally connected if the Alpha-Omega
distance is Lorentz-contracted to less than three light years. Shouldn't a light beam be able to
make it from the Alpha to the Omega before the Omega's phone rings, in this frame of reference?
NO! It won't work. Non-simultaneity is at it again. Sure, The Omega's phone rings when its
clocks read 9am, Jan. 1, 3003AD, but the Omega's clock didn't read 9am, Jan. 1, 3000AD when
you placed the call from the Alpha.
Use Lorentz transformations to work out the ratio of the space and time intervals between the
events as seen from Solaris (primed coordinates are measured by the crew of the Solaris):
∆x′ = γ ( ∆x − β c∆t ) and ∆t ′ = γ ( ∆t − β∆x c ) . After some algebra, one finds that
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( c∆t′) − ( ∆x′ )
2
2
= ( c∆t ) − ( ∆x ) :
2
2
two events that are causally disconnected in one frame of reference will be causally disconnected
2
2
in ALL frames of reference. The quantity ∆τ 2 ≡ ( ∆t ) − ( ∆x c ) is an invariant under Lorentz
transformations: calculate it using values for the separation between two events as measured in
one frame of reference and you'll obtain exactly the same value that someone else using
measurements made in a different frame of reference will obtain.
∆τ2 is sometimes called the "invariant interval squared" or the "proper time squared" between
two events. Note that ∆τ2 can be negative (as it is in the case we've been discussing): don't be
mislead by the square.
Events whose separations yield a positive ∆τ2 can be causally connected; events with negative
∆τ2 cannot be causally connected. A pair of events with ∆τ2 = 0 can only be causally linked if
the first event used a light beam to make the second event take place.
Some comments on ∆τ2:
1. If ∆τ2 > 0:
• Event 2 always takes place after event 1, regardless of your frame of reference (t2 > t1
always).
• It is possible to find a frame of reference in which event 1 and event 2 occur at the same
point in space.
• The interval between the events is said to be timelike (|∆t| > |∆x/c|).
2. If ∆τ2 < 0:
• In some frames, event 2 takes place after event 1, in some frames event 2 takes place
before event 1. There is one particular frame of reference in which the two events
occur simultaneously.
• It is impossible to find a frame of reference in which event 1 and event 2 occur at the
same point in space.
• The interval between the events is said to be spacelike (|∆x/c| > |∆t|).
²²²²²²²²²²
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