Galilean relativity
This file contains a summary of all classes
covering Ch. 1-5 & lasers. I make these
available in one file for your convenience but I
strongly recommend to use the individual files
for studying. The individual files contain all the
concept tests with the answers at the end of
each pdf file and some optimization for static
viewing.
Reference frames
• Explore how things move
with respect to things that
are also moving
• Where are you? Where
are you going? Says
who?
Reference frames
A reference frame is a
set of coordinate axes
that never move with
respect to each other.
y
x
Think about laying out
a set of meter sticks in
a pattern like this one,
and never moving
them again relative to
each other.
Q: Where do I put the
origin, (0,0)?
A: Anywhere you like!
There is no “preferred
place.”
(0,0)
I could choose it to be
where I’m standing.
And it goes on forever
b/c you might want to
measure things far
away.
Reference frames
y
Now I locate an object
in my frame, at, say,
(x,y) = (3m,2m).
(3,2)
x
(0,0)
Reference frames
(3,2)
The blue ball is at
(3m,2m).
The red ball is at
(0m, -2m).
In principle, somebody
standing there yells out
the coordinate. If I go
myself to check, I’m
moving in this
reference frame!
Distance from red ball
to blue ball is
(0,-2)
2
2
(3m) + (4m) = 25m = 5m
Remark
Reference frames
In a different reference
frame, the blue ball
could be at (2m,0m)
and the red ball is at
(-1m,-4m). The balls
are still 5m apart!
(2,0)
(2,0)
We say this
distance is an
“invariant.”
Where I stand doesn’t
affect physical facts
like distance.
(-1,-4)
Compare two reference frames
(now in one-dimension only)
The distance
between two
objects that are not
moving in this
reference frame is
a constant.
(-1,-4)
Compare two reference frames
(now in one-dimension only)
Frame S has origin here.
x
... -3
-2
-1
0
1
2
... -3
-2
-1
x
3 ...
... -3
-2
-1
0
1
2
... -3
-2
-1
3 ...
x’
0
1
2
3 ...
Frame S’ has origin here, at x=3m
according to reference frame S.
(The frame S’ is drawn below S so
you can read both axes.)
Important conclusion
Observer in S finds the ball at x = 2 m
Observer in S’ finds the ball at x’ = -1 m
Q: Who is right?
a)
b)
c)
d)
Observer in S
Observer in S’
Both are right
Neither is right
Two observers in different reference frames can give
a different description of the same physical fact (in
this case, the location of the ball.) And they’re both
right!
This is relativity!
x’
0
1
2
3 ...
Observer in S measures ball at x = 2m.
Observer in S’ measures ball at x’ = -1m.
Reading Quiz
(closed book, no discussions)
An inertial reference frame is one:
a) that is not moving
b) that is not accelerating
c) in which objects are a fixed distance
apart
d) in which observers are a fixed distance apart
e) in which nothing is moving
Inertial reference frames
V
Inertial reference frames
TZD 1.3
Imagine a train car (it’s always a train!) moving on a
straight track with constant velocity with respect to the
ground. The train runs smoothly, so that you can’t tell
it’s moving by feeling the bumps on the track.
Would you expect the laws of Physics to be different
inside this train compared to the labs here at CU?
Inertial reference frames
Inertial reference frames
V
Now, you’re playing pool on the train. The balls
roll in straight lines on the table (assuming you put
no English on them). In other words, the usual
Newtonian law of inertia still holds. The frame as
a whole is not accelerating.
V
As I’m lining up my shot, the train slows and
approaches the station. I have not touched the
cue ball. What does it do?
This frame is no
a) Rolls to the front of the train
longer inertial!
b) Rolls to the back of the train
(Accelerated
c) Remains motionless
frame)
(Is this still an inertial reference frame?)
Comparing inertial frames
To cut a long story short:
An inertial reference frame is one
that is not accelerating.
x
... -3
-2
-1
0
1
2
... -3
-2
-1
3 ...
v
x’
0
1
2
3 ...
Here are two inertial reference frames, moving with
respect to one another.
According to S, S’ is moving to the right, with v = 1 m/s.
Comparing inertial frames
v
Important conclusion
Observer in S measures velocity of S’ to be +1 m/s
Observer in S’ measures velocity of S to be -1 m/s
Q: Who is right?
a)
b)
c)
d)
x
... -3
-2
-1
0
1
2
... -3
-2
-1
3 ...
x’
0
1
2
3 ...
Here are two inertial reference frames, moving with
respect to one another.
Observer in S
Observer in S’
Both are right
Neither is right
Two observers in different reference frames can give
a different description of the same physical fact (in
this case, the relative velocity of the ‘other’ reference
frame.) And they’re both right!
The two frames are moving relative to each other.
According to S’, S is moving to the left, with v = -1 m/s.
Comparing inertial frames
Which of these is an inertial reference
frame (in good approximation)?
A. My car without brakes rolling down a steep hill
B. A rocket being launched
C. A roller coaster going over the top of a hill
D. A sky diver falling at terminal speed
E. None of the above
... -3
-2
-1
0
1
2
3 ...
... -3
-2
-1
0
1
2
3 ...
v
At time t = 0, the two frames coincide. A ball is at rest in
frame S. Its position is
• x = 2 m in S
• x’ = 2 m in S’
Comparing inertial frames
... -3
-2
-1
0
1
2
... -3
-2
-1
v
3 ...
0
1
2
3 ...
Frame S’ is moving to the right (relative to S) at v=1m/s.
At time t = 3 sec, the position of the ball is
• x = 2 m in S
• x’ = -1 m in S’
Important conclusion
• Where something is depends on when you
check on it (and on the movement of your
own reference frame).
• Time and space are not independent
quantities; they are related by rel. velocity.
• Definition: An event is a measurement of
where something is and when it is there.
( x, y , z , t )
Comparing inertial frames
Galilean position transformation
If S’ is moving with speed v in the positive x direction
relative to S, then its coordinates in S’ are
... -3
-2
-1
0
1
2
... -3
-2
-1
v
3 ...
0
1
2
x′ = x − vt
y′ = y
z′ = z
t′ = t
3 ...
At time 0, the ball was at x = x’. At time t later, the ball is
still at x in S but where is it in S’ at the same time t?
a) x’ = x
b) x’ = x + vt
Note: In Galilean relativity, time is measured the
same in both reference frames; why wouldn’t it
be?
c) x’ = x-vt
Galilean velocity transformation
u
Galilean velocity transformation
x
... -3
-2
-1
0
1
2
... -3
-2
-1
v
3 ...
x’
0
1
2
If an object has velocity u in frame S, and if frame S’
is moving with velocity v relative to frame S, then the
position of object in S’ is:
x ' (t ) = x (t ) − vt
Velocity of the object is therefore:
u' =
u
3 ...
dx (t )
dx′(t ) d
= ( x (t ) − vt ) =
−v =u−v
dt
dt
dt
... -3
-2
-1
0
1
2
... -3
-2
-1
v
3 ...
0
1
2
3 ...
Same thing as before, but now the ball is moving in S,
too, with velocity u = –1 m/s.
Is the ball faster or slower, as measured in Frame S’?
A – faster
B – slower
C – same speed
Dynamics
Important conclusion
F
Two observers in different reference
frames can give a different description of
the same physical fact (in this case, the
velocity of the ball.) And they’re both right!
F
S
... -3
-2
-1
0
1
2
3 ...
F
v
S’
... -3
-2
-1
0
1
2
3 ...
Dynamics
Galilean relativity
In inertial frame S, we have (in x-direction, say)
F = ma
How about in inertial frame S’? Well,
F' = F
since you’re still applying the same forces, and
a' =
du
du ' d
= (u − v ) =
=a
dt
dt dt
The laws of
mechanics
(good old F = ma)
are the same in any
inertial frame of
reference.
no additional acceleration in an inertial frame.
Einstein’s
First Postulate of Relativity
But now we have a problem!!
From PHYS 1120 you might recall that Mr. Maxwell
told us the speed of light ‘c’ would be:
c=
If S is an inertial frame and if a second
frame S’ moves with constant velocity
relative to S, then S’ is also an inertial
frame.
1
ε 0 µ0
= 3.00 × 108 m / s
Now, Mr. Einstein tells us that ε0 and µ0 are
the same in any inertial frame of reference.
But Mr. Galileo just told us that c’ = c - v
What gives??
Peculiar light-waves
• A sound wave propagates through air, with a
velocity relative to the air (~330 m / sec)
• A water wave propagates through water, with a
velocity relative to the water (1..100 m / sec)
• “The wave” propagates through a crowd in a
stadium, with a velocity relative to the audience.
• An electromagnetic wave propagates through...
Answer (19th century physics): The “luminiferous ether.”
Ideas behind Einstein’s relativity
Is there an ether ?
(There where various other
motivations for special
relativity, but for simplicity we
will focus here on the quest
for detecting the ‘ether.’)
Michelson and Morley…
The ether
v
c
Suppose the earth moves through the fixed ether with speed v.
A light wave traveling at speed c with respect to the ether is
heading in the opposite direction. According to Galilean
relativity, what is the magnitude of the speed of the light wave
as viewed from the earth? (Further assume the earth is not
accelerating).
a) |c|
b) |c|+|v|
c) |c|-|v| d) |v|-|c| e) something else
…performed a famous*)
experiment that effectively
measured the speed of
light in different directions
with respect to the “ether
wind.”
*) some
Frame of reference
say, the most successful failure…
Ether in the laboratory frame
Observer on the sun:
v
‘Ether’
v
L
u'-v
v
How can we measure the speed v of the ether?
If the ether would be a river, we could measure the speed
of the water using a boat that travels at a known speed u’.
(u’ is the relative velocity between the boat and the water.)
Ether ‘viewed’ in the laboratory on the earth:
-v
-v
If the boat travels the distance L within the time t, then we
know v: L=(u’-v)t, therefore v = u’ – L/t
But: Very difficult with light! u’ = c
t ~ 10ns and v ~ 0.0001*c.
We would have to measure t with an absolute precision of
~0.0000000000001s and we have to know c very precisely!
Measuring only differences in c
-v
B
-v
u’+v
L
A u’-v
Michelson and Morley
Mirrors
L
-v
Detector
-v
L
L
Compare the round-trip times tA and tB for paths A and B.
This has the great benefit, that we do not have to
measure the absolute times tA and tB (which are only a
few ns) and we are less sensitive to uncertainties in the
speed of light.
Semi-transparent
mirror
“Interferometer”
Light source
The detector measures differences in the position of the maxima or minima
of the light-waves of each of the two beams. (Yes, light is a wave!)
Intermezzo: Interferometers
1881 Michelson invented a device now known as the
‘Michelson Interferometer.’ 1907 he received the Nobel
prize for it!
The Michelson interferometer
Mirrors
Light source
We will see it in action in the famous Michelson-Morley
experiment, which will lead us to the special relativity
theory.
So the interferometer had a huge impact!!
Such interferometers are nowadays widely used for
various precision measurements. State-of-the-art
visible-light interferometers achieve resolutions of
~100pm! (X-ray interferometers are ~1pm).
Semi-transparent
mirror
(100pm = 1Å = diameter of a Hydrogen atom.)
Detector
Electromagnetic waves
E-field (for a single color):
E(x,t) = E0 sin(ωt+2πx/λ +φ)
Light source
E
φ
E
λ
Free physics simulations!
http://phet.colorado.edu
λ = 2πc/ω, ω = 2πf = 2π/T
A0
x
Wavelength λ of visible light
is: λ ~ 350 nm … 750 nm.
Radio Waves.jar
Wave interference.jar
x
B
EM-Waves in an interferometer
Mirrors
L
Constructive interference
Esum(x,t) = ½ ·E0 sin(ωt+2πx/λ +φ) + ½ ·E0 sin(ωt+2πx/λ +φ) = ?
=
Light source
E0 sin(ωt+2πx/λ +φ) = Elight source(x,t)
L
+
Semi-transparent
mirror
Screen:
=
?
Unequal arm lengths
Destructive interference
Esum(x,t) = ½ ·E0 sin(ωt+2πx/λ +φ) + ½ ·E0 sin(ωt+2π(x+∆x)/λ +φ)
L
= ½ ·E0 sin(ωt+2πx/λ +φ) - ½ ·E0 sin(ωt+2πx/λ +φ) = 0
Light source
L+∆L/2
if ∆x = λ / 2:
sin(x+π) = - sin(x)
∆L/
2
+
∆L
=
?
Screen:
Moving mirror: What do you see?
Light source
Tilted mirror: What
do you see?
Fringes!
Light source
∆L
Screen
Intensity
λ/
2
Screen
∆L
Application: Flatness measurement
Michelson
interferometer
Ultrahigh power laser
(>3.5kW average and
250MW peak power)
Interference in daily life:
Do you want a bigger
‘interferometer’? There you go…
Gravitational wave detectors
Do you remember
this guy?
Interference
in ‘photonic
crystals’
Summary
The blue color originates from constructive interference of blue
light and destructive interference of all other colors.
Michelson interferometers allow us to measure tiny
displacements. Displacements of less than 100 nm are
made visible to the eye!
Interferometers find many applications in precision
metrology such as for displacement, distance and
stress measurements as well as flatness
measurements.
Interferometers have played an important role in
physics:
Michelson-Morley experiment
special relativity
Testing general relativity: Gravitational wave detection
Remember Michelson and Morley?
Mirrors
End of our little
“interferometer intermezzo”
L
-v
Detector
-v
L
Semi-transparent
mirror
“Interferometer”
Light source
The detector measures differences in the position of the maxima or minima
of the light-waves of each of the two beams. (Yes, light is a wave!)
t2 =
2L
⋅ β
c
(Homework!)
t1 =
L
L
2L
+
=
⋅ β , β = 1v 2
c −v
c+v
c
1− 2
c
∆t = t1 − t2 ≈
Last class: Michelson Morley
2
L v
⋅ ,
c c2
∆L ≈ L ⋅
2
v
c2
Experimental results
Over a period of about 50 years, the Michelson-Morley
experiment was repeated with growing levels of
sophistication. The overall result is a high level of confidence
that the wavelength shift is consistent with zero.
Michelson, 1881
Michelson & Morley 1887
Morley & Miller, 1902-04
Illingworth, 1927
Joos,1930
L (cm)
120
1100
3220
200
2100
Calc.
0.04
0.40
1.13
0.07
0.75
Meas.
0.02
0.01
0.015
0.0004
0.002
Shankland, et al., Rev. Mod. Phys. 27, 167 (1955)
Ratio
2
40
80
175
375
Michelson and Morley
They thought that the
experiment was a
complete failure because
no effect was found.
Yes, but...
Q: What if the ether
is “dragged along”
the surface of the
earth, like air flowing
around a tennis ball?
Michelson was awarded
the Nobel Prize in 1907!!
A: If so, this would
require a “viscosity”
of the ether, and
would require rewriting Maxwell’s
equations.
True result:
Speed of light is the
same in all directions!
Homework (part of your reading assignment): Work out
the math for this experiment (TDZ, Chapter 1.5)
There is no ether
Remark: Lots of effort tried to
save the idea of the ‘ether’, but
none held up.
Einstein’s
Second Postulate of Relativity
The speed of light is the same in
all inertial frames of reference.
Electromagnetic waves are special. A time-changing
electric field induces a magnetic field, and vice-versa.
A medium (“ether”) is not necessary.
There is no ether!
v
c
Suppose the earth moves through space with speed v.
A light wave traveling at speed c with respect to faraway stars
is heading in the opposite direction. According to Einstein’s
relativity, what is the magnitude of the speed of the light wave
as viewed from the earth? (Assume the earth is not
accelerating).
a) |c|
b) |c|+|v|
c) |c|-|v|
d) |v|-|c|
e) something else
This was still new in 1905 when Einstein
proposed it. Now it has been tested
experimentally many times.
Suppose you’re in a spaceship traveling through
the solar system at at a constant speed of one-half
impulse power, v = 1.5 X 108 m/s. You fire a
pulse of laser light out the front of your vessel.
(Speed of light = 3.0 X 108 m/s).
Q: How fast do you see the pulse leave your ship?
a)1.5 X 108 m/s b) 3.0 X 108 m/s
c) 4.5 X 108 m/s d) none of these
Suppose you’re in a spaceship traveling through
the solar system at at a constant speed of one-half
impulse power, v = 1.5 X 108 m/s. You fire a
pulse of laser light out the front of your vessel.
(Speed of light = 3.0 X 108 m/s).
Q: How fast does an inertial observer on Mars see
the pulse leave your ship?
Conclusion
The speed of light is the
same in all inertial frames
of reference.
Einstein, 1905
a)1.5 X 108 m/s b) 3.0 X 108 m/s
c) 4.5 X 108 m/s d) none of these
Now it’s time to talk about time!
• Measuring time in one
frame
• Synchronization of
clocks
• Measuring time in
different frames
Quiz on the reading
Recall ‘event’ (x, y, z, t)
Last Wednesday we have argued that to
describe a physical event, we must specify
both: where it is – say at (x, y, z) in some
inertial coordinate system – and what time
it occurs – say at time t according to some
clock. But what clock?
Time of an event
(No discussions during this one)
In a given reference frame, the time of an event is
given by…
a) The time the observer at the origin sees it.
b) The time that any observer anywhere in the
frame sees it.
c) The time according to the clock nearest the
event when it happens.
d) The time according to a properly synchronized
clock nearest the event when it happens.
Lightning strikes the top of Bear Mountain,
generating a clap of thunder. At what time did the
lightning strike?
A – At the instant you hear the thunder.
B – At the instant you see the lightning.
C – Very slightly before you see the lightning.
D – Very slightly before you hear the thunder.
E – Very slightly after you see the lightning.
‘Events’ in one reference frame
‘Events’ in one reference frame
An observer at (0,0)
has a clock; events
there are covered.
And there had better
be clocks
everywhere, so you
don’t miss any event.
An observer at
(3m,2m) had better
have a clock too, if
we want to know
about events there.
And, the two clocks
had better show the
same time.
Well, didn’t quite work…
Synchronizing clocks
... -3
-2
-1
0
1
2
3 ...
At the origin, at three
o’clock, the clock
sends out a light
signal to tell
everybody it’s three
o’clock.
... -3
-2
-1
0
1
2
At the origin, at three
o’clock, the clock
sends out a light
signal to tell
everybody it’s three
o’clock.
3 ...
Time passes as the
signal gets to the
clock at x = 3m.
Does this
scheme work?
When the signal
arrives, the clock at
x=3m is set to 3:00.
Synchronizing clocks
... -3
-2
-1
0
1
2
3 ...
At the origin, at three
o’clock, the clock
sends out a light
signal to tell
everybody it’s three
o’clock.
Time passes as the
signal gets to the
clock at x = 3m.
When the signal arrives, the
clock at x=3m is set to 3:00
plus the 10 ns delay.
Time passes as the
signal gets to the
clock at x = 3m.
If you do this, then the clock
at x = 3m is 10 ns slow,
because of the delay!
When the signal
arrives, the clock at
x=3m is set to 3:00.
Simultaneity in one frame
... -3
-2
-1
0
1
2
3 ...
Using this procedure, it is now possible to say that all the
clocks in a given inertial reference frame read the same time.
Even if we don’t go out there to check it ourselves.
Now I know when events really happen, even if I don’t find
out until later (due to finite speed of light).
Lucy
1
... -4
-3
-2
-1
0
Ricky
2
1
2
3
1
Lucy
4 ...
Two firecrackers sitting on the ground explode. This
time, Lucy is sitting twice as far from firecracker 1 as
from firecracker 2. She sees the explosions at the
same time. Which firecracker exploded first in Lucy’s
reference frame?
Two firecrackers explode. Lucy, halfway between the
firecrackers, sees them explode at the same time. Ricky
(same reference frame as Lucy) is next to firecracker 2.
According to Ricky, which firecracker explodes first?
A. Both explode at the same time
B. Firecracker 1 explodes first
C. Firecracker 2 explodes first
A. Both explode at the same time
B. Firecracker 1 explodes first
C. Firecracker 2 explodes first
Even though Ricky sees the flash from 1 after the one from 2,
she knows the local times at which each cracker went off.
Event 1: (x1, y1, z1, t1)
2
Event 2: (x2, y2, z2, t2), with t1 = t2
Simultaneity in two frames
Now we have time under control
in one frame!
v
How about if there are two frames
moving relative to each other?
... -3
-2
-1
0
... -3
-2
1
-1
2
0
3 ...
1
2
3 ...
A second frame has its own clocks, and moves
past me. What happens now?
L
R
L
R
v
Lucy
Lucy
Now Lucy is the middle of a railroad car, and sets off
a firecracker. (Boom goes the dynamite!) Light from
the explosion travels to both ends of the car. Which
end does it reach first according to Lucy?
a) both ends at once
b) the left end, L
c) the right end, R
These events are
simultaneous in
Lucy’s frame.
Sure! After the firecracker explodes, a spherical wave front
of light is emitted. (‘Spherical’, because the speed of light is
the same in all directions in any inertial frame of reference).
A little while later, it reaches both ends of the car.
Sometime later, Lucy finds out about it – but that’s a different
story. The synchronized clocks are all that matter.
L
R
v
L
R
v
Lucy
Ricky
Ethel
Lucy’s friends Ethel and Ricky are standing still next
to the tracks, watching the train move to the right.
According to Ethel and Ricky, which end of the train
car does the light reach first? (As before the
firecracker is still in the middle of the car.)
a) both ends at once
b) the left end, L
c) the right end, R
L
-2
-2
0
1
0
1
2
R
L
v
2
3 ...
Suppose Lucy’s firecracker explodes at the origin of
Ethel’s reference frame.
R
-1
-1
Ethel
In Ethel’s frame,
these events are not
simultaneous.
Lucy
... -3
... -3
3 ...
Ethel
The light spreads out in Ethel’s frame from the point
she saw it explode. Because the train car is
moving, the light in Ethel’s frame arrives at the left
end first.
An important conclusion
Given two spacetime events:
1) Light hits the right end of the train car
2) Light hits the left end of the train car
v
Lucy
... -3
-2
-1
0
1
2
3 ...
Ethel
Sometime later, in Ethel’s frame, the light catches
up to the right end of the train (assume the light is
going faster than the train).
Review: Lucy in the train plays with a
firecracker. Ethel watches from the track
L
R
v
Lucy
Lucy finds that the events are simultaneous.
Ethel (in a different reference frame) finds that they
are not simultaneous.
Ethel
And they’re both right!
They want to know what time the light reaches the left and
the right ends of the car (Two ‘events’ in each frame)
From last class: Timing of events depend
on the choice of the inertial frame!!
Lucy: in the train
L
Ethel: on the platform
LLLL L
R
x
0
RR RR R
vvvv v
0
Event L:
Event R:
(x=-3, t=3s)
(x=+3, t=3s)
Lucy says: ‘Simultaneous!’
x’
Event L’:
Event R’:
(x’=-2, t’=2s) (x’=+5, t’=4s)
Ethel says: ‘Not simultaneous!’
L
R
v
Lucy
... -3
-2
-1
0
1
2
3 ...
Ethel
Now suppose Lucy’s firecracker is just slightly
toward the right end of the train, so slightly that
Ethel still measures the light hitting the left end first.
According to Lucy, which end gets hit first?
a) both at the same time
b) the left end, L
c) the right end, R
An important conclusion
Given two spacetime events:
1) Light hits the right end of the train car
2) Light hits the left end of the train car
Lucy finds that the events are simultaneous.
Ethel (in a different reference frame) finds that they
are not simultaneous.
And they’re both right!
An important conclusion
• In Lucy’s frame:
– Firecracker explodes (event 1)
– Light gets to the right end of the train (event R)
– A little later, light gets to the left end (event L)
• In Ethel’s frame:
– Firecracker explodes (event 1)
– Light gets to the left end of the train (event L)
– A little later, light gets to the right end (event R)
And they’re both right!
An important conclusion
Not only can observers in two different inertial frames
disagree on whether two events are simultaneous, they
may not even agree which event came first.
Peep?
Time dilation
And that’s the relativity
of simultaneity.
Quiz on the reading
What does Proper Time refer to?
A – Time measured by an accurate clock that is at rest in
a specific inertial frame
B – Time measured by correctly synchronized clocks that
are at rest in a specific inertial frame
C – Time interval between any two events that take place
in the same inertial frame, measured by two correctly
synchronized clocks that are closest to these events.
D – Time interval between two events that occur at the
same place in a specific inertial frame measured by a
clock that is at rest in that same inertial frame.
E – Time measured by a Swiss watch.
Proper time (see TZD p. 18)
If two events have the same spatial coordinates in a
specific inertial frame, then the time between them
measured by a clock at rest in the same inertial frame is
the proper time.
Example: Any given clock never moves with respect to
itself. It keeps proper time in its own frame.
Any observer moving with respect to this clock sees it
run slow (i.e., time intervals are longer). This is time
dilation.
Same location
Mathematically:
Event 1: (x1,y1,z1,t1)
Event 2: (x1,y1,z1,t2)
Proper time:
∆t = t2 - t1
Speed of light
Comparing inertial frames
S
... -3
-2
-1
0
1
2
3 ...
x
x
An observer and a ball are at rest in reference frame S.
At t = 0, the observer in S flashes a light pulse to be
received at x = 3 m.
At ∆t = 10 ns, the light is received. Observer S
measures a distance ∆x = 3 m, so the speed of light in
frame S is
∆x 3m
u=
=
= 0.3m / ns
∆t 10ns
... -3
-2
-1
0
1
2
3 ...
... -3
-2
-1
0
1
2
3 ...
x’
S’ is moving with respect to S at v = 0.2 m/ns.
At t = 0, observer in S flashes a light pulse to be
received at x = 3 m.
Ten nanoseconds later
... -3
-2
-1
... -3
0
1
2
3 ...
-2
-1
0
1
v
2
3 ...
S’ is moving with respect to S at v = 0.2 m/ns.
At ∆t = 10 ns, the light is received. In Galilean realtivity,
how far does the observer in S’ think the light has
traveled?
a) 3 m
b) 2 m
c) 1 m
d) 0 m
v
Ten nanoseconds later
... -3
-2
-1
... -3
0
1
2
3 ...
-2
-1
0
1
v
2
3 ...
S’ is moving with respect to S at v = 0.2 m/ns.
At ∆t=10 ns, the light is received. In Galilean relativity,
the observer in S’ would therefore measure the speed of
light as
∆x′ 1m
u=
=
= 0.1m / ns
Uh-oh!
∆t ′ 10ns
More ‘evidence’ for time Dilation
Uh-oh!
If we are to believe Einstein’s second
postulate (and we do), then:
In frame S
In frame S’
Mirror
∆x
c=
∆t
c=
Lucy
h
v
∆x′
∆t ′
Conclusion: Since we accepted Einstein's second postulate of
relativity (‘c’ is the same in all inertial frames) and we found
that ∆x ≠ ∆x ' , we conclude that ∆t ≠ ∆t ′ . I.e., time passes
at different rates in the two frames of reference!!
More evidence for ‘Time Dilation’
Lucy measures the time interval: ∆t
(Not a big surprise!)
= 2h/c
More evidence for ‘Time Dilation’
Mirror
h
v
h
v · ∆t’/2
Ricky
Ethel
Ethel and Ricky measure the time interval:
Ethel
Ricky
Note: This experiment requires two observers.
∆t ' =
2h
γ,
c
γ =
1
1−
2
v
c2
But Lucy
measured
∆t = 2h/c !!
Time dilation in moving frames
Lucy measures: ∆t
Ethel and Ricky: ∆t’ = γ∆t, with
1
γ =
v2
1− 2
c
For Lucy time seems to run slower!
(Lucy is moving relative to Ethel and Ricky)
Intermezzo:
Spacetime diagrams
(very useful in SR!)
Spacetime Diagrams (1D in space)
Spacetime Diagrams (1D in space)
c ·t
In PHYS 1110:
In PHYS 2130:
v
x
x
x
x
∆x
v = ∆x/∆t
∆t
t
Spacetime Diagrams (1D in space)
t
Recall:
Lucy plays with a fire cracker in the train.
Ethel watches the scene from the track.
c· t
In PHYS 2130:
object moving with 0<v<c.
‘Worldline’ of the object
-2 -1
0
1
2
L
x
R
v
object moving with 0>v>-c
c·t
c·t
-2 -1
0
1
2
x
Lucy
object moving with v = -c.
x=0 at time t=0
object at rest
at x=1
-2 -1
0
1
2
x
Ethel
Example: Lucy in the train
ct
Example:
Ethel on the tracks
ct
Light reaches both
walls at the same time.
Light travels to both walls
L
R
x
Lucy concludes:
Light reaches both
sides at the same time
In Lucy’s frame: Walls are at rest
Ethel concludes:
Light reaches left
side first.
L
R
In Ethel’s frame: Walls are in motion
x
Frame S’ as viewed from S
S
... -3
-2
-1
0
1
2
3 ...
v=0.5c
S’
... -3
-2
-1
0
1
2
These angles
are equal
3 ...
ct
Frame S’ is moving to the right at v = 0.5c. The origins of
S and S’ coincide at t=t’=0. Which shows the world line of
the origin of S’ as viewed in S?
A
B
ct
C
ct
x
D
ct
x
This is the time axis
of the frame S’
ct’
x’
This is the space axis
of the frame S’
x
ct
x
x
Frame S’ as viewed from S
ct
ct’
x’
In S: (3,3)
In S’: (1.8,2)
x
Both frames are adequate for describing
events – but will give different spacetime
coordinates for these events, in general.
Example: Lucy in the train
ct
Light reaches both
walls at the same time.
Light travels to both walls
(End of Space time diagrams)
Now back to Time dilation!
L
R
x
Lucy concludes:
Light reaches both
sides at the same time
In Lucy’s frame: Walls are at rest
Example:
Ethel on the tracks
ct
Last class’ clicker question:
S
... -3
Ethel concludes:
Light reaches left
side first.
L
R
Frame S’ as viewed from S
ct
This is the time axis
of the frame S’
ct’
-1
0
1
2
v=0.5c
S’
... -3
-2
-1
0
1
2
3 ...
ct
Frame S’ as viewed from S
ct
ct’
x’
This is the space axis
of the frame S’
ct’
Frame S’ was moving to the right at v=0.5c.
The origins of S and S’ coincided at t=t’=0.
We found that the world line of the origin of
S’ as viewed in S was represented by the
blue line on the right:
x’
x
3 ...
x
In Ethel’s frame: Walls are in motion
These angles
are equal
-2
In S: (3,3)
In S’: (1.8,2)
x
Both frames are adequate for describing
events – but will give different spacetime
coordinates for these events, in general.
(End of Space time diagrams)
Now back to Time dilation!
x
What we found so far:
Time Dilation
Simultaneity of two events depends on the choice of
the reference frame
L
Time Dilation: Two observers (moving relative to each
other) can measure different durations between two
events.
R
v
Lucy
…-3 -2 -1
0
1
Ethel
2
v
h
3...
Ethel concludes:
Light hits left side first.
Lucy concludes:
Light hits both ends at
the same time.
Lucy
h
Ethel
Lucy measures:
∆t’ = 2h/c
Here: ∆t’ is the proper time
1
Ethel and Ricky:
γ =
2h
v2
∆t = γ /c , with
1− 2
c
L
R
v
Lucy
Are your clocks really
synchronized?
Ricky
... -3
-2
-1
0
1
2
3 ...
Ethel
Now Lucy and Ethel each have a set of clocks. Lucy’s
are synchronized in her frame (the train), while Ethel’s
are synchronized in her frame (the tracks).
How do the clocks of one frame read in the other
frame? (Pay attention, now!)
L
R
-2
-1
0
R RRR
v
Lucy
... -3
L L LL
1
2
3 ...
Ethel
At 3 o’clock in Ethel’s frame, two firecrackers go off to
announce the time. It so happens that these firecrackers
are at the left and right ends of the train, in Ethel’s frame.
Event 1: firecracker 1 explodes at 3:00
Event 2: firecracker 2 explodes at 3:00
v vvv
Lucy
Lucy
Lucy
Lucy
... -3
-2
-1
0
1
2
3 ...
Ethel
Sometime later, the wavefronts meet. The meeting point
is halfway between the firecrackers in Ethel’s frame, but is
somewhere in the left of the train car, in Lucy’s frame.
Event 3: two light pulses meet, shortly after 3:00.
The situation as seen by Lucy
L
R
Lucy
L
R
Ethel
Lucy
In Lucy’s frame, light left first from the right end of the
car. The light pulses both show clocks reading 3:00 in
Ethel’s frame. According to Lucy’s reference frame,
which of the following is true:
A) Ethel’s clock on the left reads a later time than
Ethel’s clock on the right.
B) Ethel’s clock on the right reads a later time than
Ethel’s clock on the left.
C) Both of Ethel’s clocks read the same time.
Ethel
Important conclusion
In Lucy’s frame:
L
Clocks in S’ (synchronized in S’)
moving to the left with respect to S
R
Lucy
v
S
Ethel
S’
This clock
reads a
little after
3:00 in S
If this clock
reads 3:00
in S, then:
This clock
reads even
a little later
in S
Clocks in S’
as seen by
observer in S
Even though the clocks in S’ are synchronized (in S’) the
observer in S sees each clock showing a different time!!
A little journey
v
Now let’s have some fun!
The Twin ‘Paradox’
Vicki
Carol
Carol and Vicki are identical twins. While Vicki stays
on Earth, Carol departs for the star Sirius, 8 lightyears away, traveling at a speed v = 0.8 c (Note γ =
5/3). According to Vicki, how long does the trip take?
(Assume Earth and Sirius are not moving relative to
each other.)
a) 6 years
b) 8 years
c) 10 years d) 16.67 years
A little journey
A little journey
v
Vicki
v
Vicki
Carol
Carol and Vicki are identical twins. While Vicki stays
on Earth, Carol departs for the star Sirius, 8 lightyears away, traveling at a speed v = 0.8 c (Note γ =
5/3). According to Carol, how long does the trip take?
(Assume Earth and Sirius are not moving relative to
each other.)
a) 6 years
b) 8 years
c) 10 years d) 16.67 years
A little journey
Why? Because Carol’s clock is present at both the events:
Carol is at Earth.
Carol is at Sirius.
So if ∆τ is Carol’s proper time between these events, and ∆t
is the time in the Earth-Sirius system, we have
∆τ =
∆t
γ
=
10 y
= 6y
5/3
Follow the proper time!
But wait....
From Carol’s point of view, it was Vicki who was moving.
So we should expect:
v
Vicki
Carol
Carol
Upon arriving at Sirius, Carol immediately turns around and
heads home at v = 0.8c. When she returns, she has aged
12 years, while Vicki has aged 20 years!
In Vicki’s frame (the Earth), Carol ages 12 years while Vicki
ages 20.
In Carol’s frame (the spacecraft), Vicki ages 12 years while
Carol ages 20.
And they can’t both be right!
Resolution: Carol has to turn around.
Vicki
Carol
Spacetime diagram
Carol and Vicky send out radio signals at the beginning of
every year (measured by their respective local clocks)
Vicky receives the last
six signals from Carol
within the last two
(earth) years.
Vicky receives the first
signal from Carol after
three (earth) years.
Length contraction
ct
{
Carol receives the first
signal from Vicky after
three (spaceship) years.
Vicky’s signals
Carol’s signals
x
(Consequence of time dilation and vice versa)
Quiz on the reading
Length of an object
Proper length of an object is the length of the
object measured…
(see TDZ p. 23)
A – …in its rest frame.
B – …in any inertial frame.
C – …in the inertial frame in which both ends of
the object have the same event coordinates.
D – …in the frame in which the object is not
rotating.
E – …by the speed of light.
... -3
-1
0
1
2
3 ...
This length, measured in
the stick’s rest frame, is its
proper length.
‘Proper length’
Proper length: Length of object measured at rest / object
measured in the frame where it is at rest (use a ruler)
-2
This stick is 3m long. I
measure both ends at
the same time in my
frame of reference.
“Same time” or not
doesn’t actually matter
here, because the stick
isn’t going anywhere.
Remember ‘proper time’
Proper time: Time interval ∆t = t2 – t1 between two events
measured in the frame, in which the two events occur at the
same spatial coordinate, i.e. time interval that can be
measured with one clock.
v
... -3
-2
-1
0
1
2
3 ...
Length of an object
Observer in S measures
the proper length L of the
blue object.
S
... -3
-2
Length of an object
-1
S’
0
0
1
2
3 ...
v
Another observer comes
whizzing by at speed v.
This observer measures
the length of the stick,
and keeps track of time.
Event 1 – Origin of S’ passes left end of stick.
S
... -3
-2
-1
0
1
S’
2
3 ...
v
0
Event 1 – Origin of S’ passes left end of stick.
Event 2 – Origin of S’ passes right end of stick.
A little math
S
S’
In frame S: (rest frame of the stick)
length of stick = L (this is the proper length)
time between events = ∆t
speed of frame S’ is v = L/∆t
A little math
v
In frame S’:
length of stick = L’ (this is what we’re looking for)
time between events = ∆t’
speed of frame S is v = L’/∆t’
Follow the proper time!
Q: a) ∆t = ∆t’
b) ∆t = γ∆t’
c) ∆t’ = γ∆t
Speeds are the same (both refer to the relative speed).
And so
| v |=
L′
L
L
=
=
∆t ′ ∆t γ∆t '
L′ =
L
γ
Length in stick’s rest frame
(proper length)
Length in moving frame
Length contraction is a consequence of time dilation (and
vice-versa).
Quiz on proper time/length
v
The Twin Paradox revisited
Vicki
Carol
Sirius
Carol travels from the Earth to Sirius. Which of the following
statements is correct? (Assume that Earth and Sirius are not
moving relative to each other)
A – Vicky measures proper time and proper length of the journey.
B – Carol measures proper time and proper length of the journey.
C – Vicky measures proper time and Carol measures proper
length of the journey.
D – Carol measures proper time and Vicky measures proper
length of the journey.
E – none of the above
Review: Proper length
Proper length: Length of object measured in the frame,
where it is at rest (use a ruler)
Review: Proper length
Proper time: Time interval ∆t = t2 – t1 between two events
measured in the frame, in which the two events occur at the
same spatial coordinate, i.e. time interval that can be
measured with one clock.
v
... -3
-2
-1
0
1
2
3 ...
Proper time & proper length
Length contraction vs. time dilation
Vicky
v
Vicki
Carol
Sirius
Now we know the following about this journey:
- Vicki measures the proper length: 8 light-years.
- Carol measures the proper time: 6 years.
- Both agree that Carol travels at a speed of v=0.8c
relative to the earth.
... -3
-2
-1
0
1
2
3 ...
Vicky measures:
Proper length: LVicky= 8 ly
Time: ∆tVicky = γ ∆tCarol = 10y
Carol
v
From Carol’s perspective:
Carol finds that she traveled only 6y*0.8c = 4.8 ly. But why
does she find herself at Sirius after 6 years??
Carol measures:
Length: LCarol= LVicky /γ = 4.8 ly
Proper time: ∆tCarol = 6 y
0
Length contraction!!
The Lorentz transformation
S
Lorentz transformation
(Relativistic version of Galileo transformation)
v
S’
x’
0
A stick is at rest in S’. Its endpoints are the events
(x,ct) = (0,0) and (x’,0) in S’. S’ is moving to the
right with respect to frame S.
Event 1 – left of stick passes origin of S. Its
coordinates are (0,0) in S and (0,0) in S’.
Lorentz transformation
An observer at rest in frame S sees a stick flying past him
with velocity v:
S
S
x
v
S’
0
The Lorentz transformation
x’
As viewed from S, the stick’s length is x’/γ. Time t passes.
According to S, where is the right end of the stick? (Assume
the left end of the stick was at the origin of S at time t=0.)
a) x = γvt
b) x = vt + x’/γ
c) x = -vt + x’/γ
d) x = vt – x’/γ
e) something else
v
S’
0
x’
x = vt + x’/γ . This relates the spatial coordinates of
an event in one frame to its coordinates in the other.
Algebra
x’ = γ(x-vt)
A note of caution:
Transformations
If S’ is moving with speed v in the positive x direction
relative to S, then the coordinates of the same event
in the two frames are related by:
Galilean transformation
(classical)
Lorentz transformation
(relativistic)
x′ = γ ( x − vt )
y′ = y
x′ = x − vt
y′ = y
z′ = z
S
-1
1
2...
0
1
x2
x1
v
S’
... -3
-2
-1
2
3 ...
An object moves from event A=(x1,t1) to event B =(x2,t2).
∆x
with: ∆x = x1 - x2
As seen from S, its speed is u =
∆t
∆t = t1 - t2
∆x′
with: ∆x’ = x’1 - x’2
As seen from S’, its speed is u ′ = ∆t ′
∆t’ = t’1 - t’2
Velocity transformation in 3D
y
S
S'
x
z
S'
(x',y',z',t')
v
x'
z'
Velocity transformation (1D)
∆x′
, where ∆x=x1 - x2, ∆x’=x’1 - x’2
∆t ′
Use Lorentz: x’ =. γ(x-vt)
1
..
∆x′
γ (∆x − v∆t )
∆t
×
u′ =
=
∆t ′ γ (∆t − (v / c 2 )∆x) 1
∆t
Galilean result
∆x ,
∆t
u′ =
u′ =
u −v
1 − uv / c 2
New in special relativity
Velocity transformation (3D)
The velocity u=(ux, uy, uz) measured in S is given by:
y'
(x,y,z,t)
z
u=
∆x
0
(x,y,z,t)
x
A
B
y'
S
Velocity transformation (1D)
-2
y
v
t ′ = γ (t − 2 x )
c
Note: This assumes (0,0,0,0) is the same event in both frames.
... -3
An observer in S would like to express an event (x,y,z,t) (in
his frame S) with the coordinates of the frame S', i.e. he
wants to find the corresponding event (x',y',z',t') in S'. The
frame S' is moving along the x-axes of the frame S with the
velocity v (measured relative to S) and we assume that the
origins of both frames overlap at the time t=0.
See homework #3
z′ = z
t′ = t
The way the Lorentz and Galileo transformations are presented
here (and in the textbook) assumes the following:
u
ux=∆x/∆t ,
(x',y',z',t')
uy=∆y/∆t ,
uz=∆z/∆t
v
x'
z'
In a more general case we want to transform a velocity u
(measured in frame S) to u’ in frame S’. Note that u can point
in any arbitrary direction, but v still points along the x-axes.
, where ∆x=x1-x2 …
To find the corresponding velocity components u’x, u’y, u’z in
the frame S’, which is moving along the x-axes in S with the
velocity v, we use again the Lorentz transformation:
x’1=γ(x1-vt1), and so on…
t’1=γ(t1-vx1/c2), and so on…
Algebra
Velocity transformation (3D)
(aka. “Velocity-Addition formula”)
u' x =
ux − v
1 − ux v / c 2
Some applications
uy
u' y =
γ (1 − u x v / c 2 )
u' z =
uz
γ (1 − u x v / c 2 )
Relativistic transformations
x′ = γ ( x − vt )
v u −v
x)
1c−2 uv / c 2
u−v
u' =
1 − uv / c 2
u=
t ′ = γu(′t =−
u′ + v
1 + u ′v / c 2
b) 0.5 c
u is what we were looking for!
(i.e. velocity measured in S)
y
Suppose a spacecraft travels at speed v=0.5c relative to
the Earth. It launches a missile at speed 0.5c relative to
the spacecraft in its direction of motion. How fast is the
missile moving relative to Earth? (Hint: Remember which
coordinates are the primed ones. And: Does your answer
make sense?)
a) 0.8 c
Velocity transformation:
Which coordinates are primed?
c) c
d) 0.25 c
y'
u
(x,y,z,t)
S
S'
x
Earth
(x',y',z',t')
z'
z
v
x'
Spacecraft
e) 0
Lucy
v
x′ = γ ( x − vt )
?
u′ =
u −v
1 − uv / c 2
u=
u′ + v
1 + u ′v / c 2
The “object” could be light, too!
Suppose a spacecraft travels at speed v=0.5c relative to
the Earth. It shoots a beam of light out in its direction of
motion. How fast is the light moving relative to the Earth?
(Get your answer using the formula).
a) 1.5c
b) 0.5 c
c) c
d) d
... -3
-2
-1
0
George
1
2
3 ...
t ′ = γ (t −
v
x)
c2
George has a set of synchronized clocks in reference
frame S, as shown. Lucy is moving to the right past
George, and has (naturally) her own set of synchronized
clocks. Lucy passes George at the event (0,0) in both
frames. An observer in George’s frame checks the clock
marked ‘?’. Compared to George’s clocks, this one reads
e) e
A) a slightly earlier time
B) a slightly later time
C) same time
Lucy
x′ = γ ( x − vt )
v
?
... -3
t ′ = γ (t −
-2
-1
0
1
George
2
v
x)
c2
3 ...
The event has coordinates (x = -3, t = 0) for George.
In Lucy’s frame, where the ? clock is, the time t’ is
t ′ = γ (0 −
v
3γ v
( −3)) = 2
c2
c
Spacetime interval
, a positive quantity.
‘?’ = slightly later time
Remember this? (from 1st class)
Remember Lucy?
The distance between
the blue and the red ball
is:
2
2
(3m) + (4m) = 25m = 5m
If the two balls are not
moving relative to each
other, we find that the
distance between them
is “invariant” under
Galileo transformations.
Remember Ethel?
c∆t’
h
Event 1 – firecracker explodes
Event 2 – light reaches detector
Distance between events is h
h = c∆t
Spacetime interval
Say we have two events: (x1,y1,z1,t1) and (x2,y2,z2,t2).
Define the spacetime interval (sort of the "distance")
between two events as:
h
(∆s )2 = (c∆t )2 − (∆x )2 − (∆y )2 − (∆z )2
∆x’
Event 1 – firecracker explodes
Event 2 – light reaches detector
Distance between events is c∆t’
But distance between x-coordinates is ∆x’
and: (c∆t’)2 = (∆x’)2 + h2
We can write
h 2 = (c∆t ′)2 − (∆x′)2
Lucy
And Lucy got
h 2 = (c∆t )2 − (∆x )2
since
∆x = 0
With:
∆x = x1 − x2
∆y = y1 − y2
Spacetime interval
∆z = z1 − z 2
∆t = t1 − t 2
The spacetime interval has the same value in all
reference frames! I.e. ∆s2 is “invariant” under Lorentz
transformations.
(Homework #3!)
Spacetime
ct
Here is an event in
spacetime.
Spacetime
ct
Here is an event in
spacetime.
Any light signal that
passes through this event
has the dashed world
lines. These identify the
‘light cone’ of this event.
x
The blue area is the future
on this event.
x
Spacetime
ct
Here is an event in
spacetime.
The yellow area is the
“elsewhere” of the
event. No physical
signal can travel from
the event to its
elsewhere!
A
x
The pink is its past.
Spacetime
Now we have two events
A and B as shown on the
left.
ct
B
A
x
The space-time interval
(∆s)2 of these two events
is:
a)Positive
b)Negative
c)zero
If (∆s)2 is negative in one frame of reference it is also
negative in any other inertial frame! ((∆s)2 is invariant under
Lorentz transformation).
Causality is fulfilled in SR.
Spacetime
Example: Wavefront of a flash
z
ct
D
C
B
A
x
(∆s)2 >0: Time-like events
(A – D)
(∆s)2 <0: Space-like events
(A – B)
(∆s)2 =0: Light-like events
(A – C)
t=0
t>0
y
x
Wavefront = Surface of a sphere with radius ct:
(ct)2 - x2 - y2 - z2 = 0
Spacetime interval for light-like event: (∆s)2 = 0
(∆s)2
is invariant under Lorentz transformation.
Einstein: 'c' is the same in all inertial systems.
Therefore: (ct')2 - x'2 - y'2 - z'2 = 0 in all inertial systems!
(Here we assumed that the origins of S and S' overlapped at t=0)
Momentum
The classical definition of the momentum p of a
particle with mass m is: p=mu.
This is the end of our excursion
to the relativistic spacetime.
In absence of external forces the total momentum is
conserved (Law of conservation of momentum):
n
∑p
i =1
Questions?
i
= const.
Due to the velocity addition formula, the definition p=mu is not
suitable to obtain conservation of momentum in special
relativity!!
Need new definition for relativistic momentum!
y
Review: Transformation rules and
conservation of momentum
Conservation of momentum is extremely
useful in classical physics.
For the new
definition of relativistic momentum we want:
y'
m
u2
u1
m
u'2
v = u1x
u1y
u'1
m
m u1x
S
S'
x
Classical:
x'
Relativistic:
u'x = ux – v
u' x =
u'y = uy
u' y =
ux − v
1 − ux v / c 2
uy
γ (1 − u x v / c 2 )
Relativistic momentum
Classical definition:
p=m
dr
dt
Say we measure the mass 'm' in its rest-frame ('proper
mass' or 'rest mass'). Since we measure 'm' it's restframe we agree on the same value for 'm' in all frames.
Assume we take the derivative with respect to the proper
time tproper , which has the same meaning in all frames.
Relativistic definition:
p≡m
1. At low velocities the new definition of p
should match the classical definition of
momentum.
2. We want that the total momentum (Σp) of
an isolated system of bodies is conserved
in all inertial frames.
Relativistic momentum
The time dilation formula implies that dt = γdtproper We can
therefore rewrite the definition of the relativistic
momentum as follows:
p≡γ m
dr
= γ mu
dt
dr
dt proper
This definition fulfills the conservation of momentum in SR!
To prove it you can apply the relativistic velocity transformation.
An important consequence of the Lorentz-factor γ is, that
no object can be accelerated past the speed of light.
L
Example:
Classical vs. Relativistic momentum
Relativistic momentum
An electron has a mass m ≈ 9·10-31kg. The table below
shows the classical and relativistic momentum of the
electron at various speeds (units are 10-22kg·m/s):
~52,000,000 mph
only ~1% change!
u
0.1c
p=m·u
classical
0.273
p=γm·u
relativistic
0.276
difference
[%]
1.1
0.5c
1.36
1.57
15.4
0.9c
0.99c
2.46
2.7
5.63
19.2
128.9
611.1
Particle A has half the mass but twice the speed of
particle B. If the particles’ momenta are pA and pB, then
a) pA > pB
b) pA = pB
c) pA < pB
Relativistic force
γ is bigger for the faster particle.
Example: Relativistic force
A charged particle (charge q) with mass m is at rest at x = 0
in a uniform electric field ℇ . Plot the velocity u of the
particle as a function of time t (assume the particle is
released at t = 0).
We can define the classical force using Newton's law:
F = ma
This is equivalent to:
F = dp/dt
Force acting on the particle: F = q·ℇ
Using the definition of the relativistic momentum we obtain a
suitable definition for a relativistic force:
Relativistic force F ≡ d(γ·m·u)/dt
Therefore:
dp d
= (γ m u ) , with γ =
dt dt
1
Or:
yields:
1−
u2
c2
yields: m·u = q·ℇ·t (1-u2/c2)0.5
Square both sides: m2·u2 = q2·ℇ2·t2(1-u2/c2)
a)
Bring u to the left: u2(m2c2 + q2·ℇ2·t2)= q2·ℇ2·t2·c2
Divide by term in bracket and take squareroot:
u
Classical
c
0
0
qEct
c
u
0
c)
( qEt ) 2 + ( mc ) 2
0
t
c
u
c
u
d)
E
u=
b)
E
γ =
Which graph best represents the total energy of a
particle (particle's mass m>0) as a function of its
velocity u, in special relativity?
E
Dividing by
1
q·ℇ · t = γ·m·u
Quiz: On the reading
for the velocity u.
γ·m·u = q·ℇ·t
q·ℇ dt = d(γ·m·u)
Integrating both sides: ∫ q·ℇ dt = ∫ d(γ·m·u)
u2
1− 2
c
Example: Relativistic force (cont.)
Now: Solve
q·ℇ = d(γ·m·u)/dt
E
F≡
p = γmu
B
A
0
c
u
0
Energy
Similar to the definition of the relativistic momentum we
want to find a definition for the energy E of an object that
fulfills the following:
1. At low velocity, the value E of the new definition should
match the classical definition.
2. The total energy (ΣE) of an isolated system of bodies
should be conserved in all inertial frames.
Kinetic energy
The work done by a force F to move a particle from
position 1 to 2 along a path s is defined by:
2
W12 = ∫ F ⋅ ds = K 2 − K1
1
K1,2 being the particle's kinetic energy at positions 1 and
2, respectively (true for frictionless system).
Using our prior definition for the relativistic force we can now
find the relativistic kinetic energy of the particle. (After some
'slightly involved' algebra.)
Relativistic kinetic energy
The relativistic kinetic energy K of a particle with a rest
mass m is:
Total energy
We rewrite the equation for the relativistic kinetic
energy and define the total energy of a particle as:
K = γmc2 - mc2 = (γ-1)mc2
E = γmc2 = K + mc2
Note: This is very different from the classical K= ½mv2 .
K = γmc2 - mc2 ≈ mc2 + ½ mc2v2/c2 - mc2 = ½ mv2
This definition of the relativistic mass-energy E fulfills the
condition of conservation of total energy.
(Not proven here, but we shall see several examples where
this proves to be correct.)
Quiz: Rest energy
Equivalence of Mass and Energy
For slow velocities the relativistic energy equation gives the
same value as the classical equation! Remember the
binomial approximation for γ:
γ ≈ 1+ ½v2/c2
E = γmc2 = K + mc2
In the particle's rest frame, its energy is its rest
energy, E0. What is the value of E0?
A:
B:
C:
D:
E:
0
c2
mc2
(γ-1)mc2
½ mc2
Note: This suggests an
equivalence of mass and energy!
m
v
-v
m
E2 = γmc2 = K + mc2
E1 = γmc2 = K + mc2
Total energy:
Etot = E1+E2 = 2K + 2mc2
Equivalence of Mass and Energy
v
m
-v
Example:
Rest energy of an object with 1kg
m
Conservation of the total energy requires that the final
energy Etot,final is the same
2 as the energy Etot, before
2
tot Therefore:
the collision.
E0 = mc2 = (1 kg)·(3·108 m/s )2 = 9·1016 J
E = γ2mc = 2K + 2mc
Etot,final =
Mc2 =
2K +
2mc2 =
Etot,initial
We find that the total mass M of the final system is bigger
than the sum of the masses of the two parts! M>2m.
Potential energy inside an object contributes to its mass!!!
How does nuclear power work?
Atomic cores are built from neutrons and protons. There
are very strong attractive forces between them. The
potential energy associated with the force keeping them
together in the core is called the binding energy EB.
We now know that the total rest energy of the particle
equals the sum of the rest energy of all constituents minus
the total binding energy EB:
Mc2 = Σ(mi c2) – EB
Or in terms of Mass per nucleon
9·1016 J = 2.5·1010 kWh = 2.9 GW · 1 year
This is a very large amount of energy! (Equivalent to the
yearly output of ~3 very large nuclear reactors.)
Enough to power all the homes in Colorado for a year!
Periodic table of elements
Example: Deuterium fusion
Isotopes of Hydrogen:
Example: Deuterium fusion
Relationship of Energy and momentum
Recall:
= γmc2
p = γmu
Total Energy: E
Momentum:
use:
Therefore: p2c2 = γ2m2u2c2 = γ2m2c4 · u2/c2
Isotope mass:
Deuterium: 2.01355321270 u
Helium 4: 4.00260325415 u
(1 u ≈ 1.66·10-27 kg)
u2 γ 2 − 1
= 2
c2
γ
p2c2 = γ2m2c4 – m2c4
=E2
This leads us the momentum-energy relation:
1kg of Deuterium yields ~0.994 kg of Helium 4.
Energy equivalent of 6 grams:
E0 = mc2 = (0.006 kg)·(3·108 m/s )2 = 5.4·1014 J
or:
Enough to power ~20,000 American households for 1 year!
Application: Massless particles
From the momentum-energy relation E2 = p2c2 + m2c4
we obtain for mass-less particles (i.e. m=0):
E = pc , (if m=0)
p=γmu and E=γmc2
Using E=pc leads to:
p/u =
u=c
E/c2
, (if m=0)
Massless particles travel at the speed of light!!
… no matter what their total energy is!!
E2 = (pc)2 + (mc2)2
E2 = (pc)2 + E02
Example:
Electron-positron annihilation
Positrons (e+, aka. antielectron) have exactly the same mass
as electrons (e-) but the opposite charge:
me+ = me-= 511 keV/c2 (1 eV ≈ 1.6·10-19J)
E1, p1
eBAM!
e+
E2, p2
At rest, an electron-positron pair has a total energy
E = 2 · 511 keV. Once they come close enough to each
other, they will annihilate one other and convert into two
photons.
Conservation
momentum:
. photons?
1 = -p2two
What can of
you
tell about pthose
Conservation of energy: E1+E2 = 2mc2 ,
E1 = E2 = 511 keV
Do neutrinos have a mass?
Do neutrinos have a mass? (cont.)
Neutrinos are elementary particles. They come in three
flavors: electron, muon, and tau neutrino (νe,νµ, ντ). The
standard model of particle physics predicted such particles.
The prediction said that they were mass-less.
Bruno Pontecorvo predicted the ‘neutrino oscillation,’ a
quantum mechanical phenomenon that allows the neutriono
to change from one flavor to another while traveling from the
sun to the earth.
The fusion reaction that takes place in the sun (H + H
He)
produces such νe. The standard solar model predicts the
number of νe coming from the sun.
All attempts to measure this number on earth revealed only
about one third of the number predicted by the standard
solar model.
Why would this imply that the neutrinos have a
mass?
Massless particles travel at the speed of light!
i.e. γ
∞, and therefore, the time seems to be standing
still in the frame of the neutrino:
∆tneutrino = ∆tproper = ∆tEarth / γ
In the HW: muon or pion experiments. The half-live time of the
muons/pions in the lab-frame is increased by the factor γ.
Summary SR
• Classical relativity
Galileo transformation
• Special relativity (consequence of 'c' is the same in all inertial
frames; remember Michelson-Morley experiment)
– Time dilation & Length contraction, events in
spacetime
Lorentz transformation
– Spacetime interval (invariant under LT)
– Relativistic forces, momentum and energy
– Lot's of applications (and lot's of firecrackers)
…
That’s it!
Now go home and work on your time machine…
… or on the warp-reactor.
Everything we have discussed to this point will be part of the first
mid-term exam (including reading assignments and homework.)
If you have questions ask as early as possible!!
Part 2: Quantum* Mechanics
Actually: don’t go home quite yet!
Part 2 of this course:
1. Basic properties of light (electromagnetic waves).
2. Photoelectric effect and how it shows light comes in
quantum units of energy. When is a wave not a wave?
(when it is a particle!)
2. Atomic spectra- quantized energy of electrons in atoms.
3. Bohr model of the atom. Where it works. Why it is wrong.
4. de Broglie idea- wave-particle duality of electrons etc.
When is a particle not a particle?
(when it’s a wave).
Courtesy of IBM
*We say something is quantized if it can occur only in certain discrete amounts.
5. Schrodinger Equation and quantum waves.
6. What they are, how to use.
7. Applications: chemistry, electronics, lasers, MRI, …
Properties of light
Interaction with matter
Electrostatic fields
http://phet.colorado.edu
Electric fields exert forces on charges
+
+
+
+
+
+
+
+
E
+
(e’s and p’s in atom)
E
F=qE
-
_
F=qE
Force = charge • electric field
F= qE
Light is an oscillating E(and B)-field. It interacts with matter by
exerting forces on the charges
– the electrons and protons in atoms.
Charges-and-fields.jar
Light is an oscillating E (and B)-field
Electromagnetic waves
http://phet.colorado.edu
• Oscillating electric and magnetic field
• Traveling at speed of light (c)
Snap shot of E-field in time:
At t=0
A little later in time
E
c
Emax
Electromagnetic
radiation
This symbolizes a local disturbance
of the electric field E(x,t)
x
The E-field is a function of position (x) and time (t):
E(x,t) = Emaxsin(ax+bt)
sin(ax-bt) , (here: b/a=c)
Radio Waves.jar
Remember this one?
Making sense of the Sine Wave
E-field (for a single color):
E(x,t) = E0 sin(ωt+2πx/λ +φ)
Light source
E
λ
λ = 2πc/ω, ω = 2πf = 2π/T
E0
φ
E
x
Wavelength λ of visible light
is: λ ~ 350 nm … 750 nm.
x
B
CQ: What does the curve tell you?
-For Water Waves?
-For Sound Wave?
-For E/M Waves?
Electromagnetic Spectrum
Snapshot of radio wave in air.
Spectrum: All EM waves. Complete range of wavelengths.
Length of vector represents strength of E-field
Orientation represents direction of E-field
What stuff is moving up and down in space as radio wave
passes?
a. electric field
b. electrons
c. air molecules
d. light ray
e. nothing
Frequency (f) =
# of times per second E-field at
point changes through complete
cycle as wave passes
λ
Red light
λ
Cosmic
rays
E. answer is nothing.
Electric field strength
increases and decreases
– E-field does not move up and down.
Electron
oscillates
with period
of Τ
Wavelength (λ) =
distance (∆x) until wave repeats
λ
Blue light
SHORT
LONG
Electromagnetic waves carry energy
λ
Emax=peak amplitude
c
Emax
X
How far away will the peak E-field be before the next peak is
generated at this spot?
b. c/Τ
c. Τ/c
e. none of the above
a. cΤ
d. sin(cΤ)
Intensity = Power = energy/time α (Eavg)2
area
area
ans. is a.
Distance = velocity * time
Distance = cΤ = c/f = 1 wavelength
so c/f = λ
α (amplitude of wave)2 α Emax2
Intensity only depends on the E-field amplitude but not
on the color (or frequency) of the light!
Maxwell’s Equations: Describes EM radiation
∫E •d A=
Qin
∫E •ds = −
ε0
dΦ m
dt
∫B•ds = µ I
∫B•d A= 0
E(x,t) = Emaxsin(ax-bt)
Light shines
on black tank
full of water.
How much
energy
absorbed?
0 through
Maxwell’s Equations: Describes EM radiation
∫E •d A=
+ ε 0 µ0
dΦ e
dt
Qin
ε0
∫B•d A= 0
In 3-D:
∫E •ds = −
∫B•ds = µ I
1 ∂2E
∇E= 2 2
c ∂t
2
dΦ m
dt
0 through
In 1-D:
+ ε 0 µ0
dΦ e
dt
∂2E 1 ∂2E
=
∂x 2 c 2 ∂t 2
B
Show that E(x,t) = Emaxcos(ax+bt)
E
is a solution (in 1-D)
with b/a=c.
How can we see that light
really behaves like a wave?
EM radiation is a wave
What is most definitive observation we can
make that tells us something is a wave?
Ans: Observe interference.
During 1600-1800s: lot’s of debate about what light really is.
After ~1876 (Maxwell): Light = EM radiation viewed as a
wave. How can it be tested?
Constructive interference:(peaks are lined up and valleys are
lined up)
c
What is most definitive observation we can make that
tells us something is a wave?
Observe interference!!
(Remember the Michelson interferometer?)
Two slit interference
EM radiation is a wave
What is most definitive observation we can
make that tells us something is a wave?
Ans: Observe interference.
Destructive interference: (peaks align with valleys
cancel each other)
c
E-fields
wave interfarence online
That's it about EM waves (for now)
wave-interference_en.jar
Review: How to generate light?
Stationary charges
constant E-field, no magnetic (B)-field
(We don’t see charges glow in the dark)
E
+
Charges moving at a constant velocity
B
Constant current through wire creates a B-field
but B-field is constant. (We don’t see DC.) I
Accelerating charges
changing E-field and changing B-field
B
(EM radiation
both E and B are oscillating)
We talked briefly about Maxwell equations
Questions?
E
The sun produces lots of light
Light might simply heat up the plate
X
#1
Why? How?
X
+ +
+
+
Surface of sun- very hot!
Whole bunch of free electrons
whizzing around like crazy. Equal
number of protons, but heavier so
moving slower, less EM waves
generated by protons.
From last class: Intensity α Emax2
Emax=peak amplitude
c
Emax
X
E(x,t) = Emaxsin(ax-bt)
Light shines
on black tank
full of water.
How much
energy
absorbed?
X
#2
#3
(Use E1max=E2max>E3max)
Which barrel will heat up the fastest?
a. 2>1>3
b. 1>2>3
c. 1=2>3
d. 1=3>2
e. 2>1=3 Intensity = power/area α Emax2
Does not depend on frequency/color!
Light is a wave
interference!
The definite check that light IS a wave
Observe interference!
Intensity = Power = energy/time α (Eavg)2
area
area
α (amplitude of wave)2 α Emax2
Intensity only depends on the E-field amplitude but not
on the color (or frequency) of the light!
So we found: light is a wave!
…and things made a lot of sense:
• EM waves can are described by Maxwell's theory
• double slit experiment and other diffraction phenomena
• explains grating spectrometers etc.
•…
The photoelectric effect (~1900)
The photoelectric effect is a
phenomenon in which electrons
are emitted from matter as a
consequence of their absorption
of energy from light.
But then there was this one, little experiment from Mr. Hertz
(~1887) that just couldn’t be explained with EM waves!
The effect was only observed with UV light, but not so with
red or IR light.
Effect is frequency dependent!?
Actually, some other problems started to surface, such the
lack of an accurate model for the black body radiation…
But Maxwell told us that the light intensity doesn’t depend on
frequency! (Intensity only depends on |E|2)
The photoelectric effect
The results from the photoelectric effect where inconsistent
with the classical view of EM theory. (Discovered 1887 by Hertz,
Explained by Einstein (using some of Plank's ideas) 1905. Nobel prize:
1921)
Experimental apparatus: PE effect
Metal surface
Glass cylinder
Einstein proposed:
"…the energy in a beam of light is not distributed
continuously through space, but consists of a finite
number of energy quanta, which are localized at points,
which cannot be subdivided, and which are absorbed
“…”
and emitted only as whole units." He took the energy of
these single units to be hf, as proposed earlier by Planck.
Vacuum
What could it be?
Experimental apparatus: PE effect
What happens?
A
Metal surface
B
Glass cylinder
2 ohms
-
10 V
+
Two metal plates in vacuum with a voltage between them.
Vacuum
Adjustable voltage
Current meter
What happens?
A
B
2 ohms
-
10 V
+
Two metal plates in vacuum with a voltage between them.
The potential difference between A and B is:
A) 0 V
B) 5 V
C) 10 V
D) infinite volts
How much current is flowing through the resistor?
A)
0A
B) 0.2 A
C)
5A
D) 10 A
E) infinite current
Potential difference between A and B is 10 V
Without light, no electrons can get across gap.
But if we put an electron close to the surface of plate A it 'feels'
the electric field between the two plates.
The electron accelerates towards the positive plate (B) and picks
up the energy = q(10V) = 1 electron charge x 10 V = 10 eV
Uniform E-field
between plates
A
Constant
force on
electron
constant
acceleration
0V
F=qE
E
+
+
+
+
+
- 10V +
0V
Current==0.1
0 AA
Current
10Volts
10V
B
A note about units of energy
Joules are good for macroscopic energy conversions.
But when talking about energy of single electrons,
Joules are inconvenient… (way too big!)
Define new energy unit: The electron-volt (eV)
1eV = kinetic energy gained by an electron when
accelerated through 1 volt of potential difference
How to put the e- close to plate A?
Shine light on the plate!!
Metal surface
A
Glass cylinder
Electrons get pulled
towards
Vacuum plate B by
F = E·q the electric field
1eV ≈ 1.6·10-19 J
e-
1V
+
+
+
+
0V
F
E
path
Adjustable voltage
+
1V
Current meter
Experimental apparatus: PE effect
Play with color and
intensity. Measure
current I. (I ~ #e-/s)
First we could argue that the light heats
up the plate
electrons pop-out
Measure the current!
BUT: # of electrons = constant
Here,
sec
electrons
So current is constant!
are
repelled
by neg.
electrode
Current
C
0 Voltage
Current
0 Voltage
D
Current
B
Current
Current
What is the current
vs. battery voltage?
0 Voltage
Measure the current!
What’s happening here?
Each electron that pops out is accelerated and
hits the plate on the right side.
Hot plate.
A few electrons get
enough energy to just
barely “splash” out.
A
+
B
+
+
+
+
0 Voltage
0
reverse V,
no electrons
flow.
not I = V / R !!
Battery Voltage
Vacuum tube diode. Works!
- early electronic device.
Vacuum tube diode
Swimming Pool Analogy
Current
If no water slops over side of pool, no flow. Little pump or big
pump, still no water current.
If electrons stuck inside metal plate, no current for little or big V.
pump
voltage
reverse V,
no electrons
flow.
What do you think will actually happen?
Optical power P
- frequency f
Now: Take out a piece of
paper and draw the
following graphs with
what you expect will
happen.)
Voltage U
?
Pool party: put bunch of energy into water, splash some out,
get flow through pump.
Put energy into metal by heating it very hot,
gives electrons energy, some “splash” out. Gives current.
Let's do the experiment!
Play with color and
intensity. Measure
current I. (I ~ #e-/s)
http://phet.colorado.edu
Current I
1. Current vs. Voltage with the lamp on (fixed color,
say UV light, and fixed intensity.)
2. Current vs. Frequency (color) at a fixed intensity
and voltage (right plate is on positive potential)
3. Current vs. Intensity for fixed color (right plate is at
fixed, positive voltage)
0
photoelectric online
Measure the current!
PE effect: A classical phenomena?
Electrons
NOT I=V/R !!
Battery Voltage
reverse V,
no electrons
flow.
Looks kind off what we have seen in the experiment…
Can we explain the PE effect with classical EM theory?
Assume we could treat light as a classical wave. Predict
the behavior (assume light puts energy into plate
heats plate up
get diode current voltage curve.)
Takes time to heat up the plate.
• Light on longer
heat more, more e / s
Current rises slowly over time.
• Color light would not matter, only intensity!
0
Voltage
Current
Current
Remember this?
photoelectric_en.jar
photoelectric_en.jar
I
I
B
A
0 Batt. V
I
0 Batt. V
D
C
0 Batt. V
I
• Color does matter! The velocity (and
number) of the electrons seems to increase
with frequency (fUV > fblue > fred)
• Positive voltages always yield the same
current (at fixed color and intensity).
• Large negative voltages make the current
go to zero.
• We never observe a negative current.
• We have discussed that if we heat the
metal, electrons could come out (slowly).
Questions?
Which graph represents low and high intensity
curves best for a fixed color of the light?
I
What did we observe so far?
0 Batt. V
F
0 Batt. V
photoelectric_en.jar
Remember definition of 'eV'
Define electron-volt (eV):
1eV = kinetic energy gained by an electron when
accelerated through 1 volt of potential difference
path
E
I
The lowest negative voltage required to stop the current
multiplied by the electron charge qe corresponds to the kinetic
energy of the fastest electrons! This lowest voltage is called
the stopping potential.
photoelectric_en.jar
HIGH intensity
LOW intensity
e’s
I
I
Fewer electrons pop out off metal
Current decreases.
Current proportional to light intensity.
Same kinetic energy
(KE). So same
“stopping potential”.
0
Battery Voltage
voltage to turn around
most energetic electron:
“stopping potential”
I
F
e’s
-U
-
0
photoelectric_en.jar
Predict what happens to
the initial KE of the
electrons as the frequency
of light changes? (Light
intensity is constant)
Battery Voltage
e’s
I
Predict shape
of the graph
Initial KE
0V
0
Frequency of light
0
Frequency
Initial KE
0
D
Correct answer is D.
Frequency
0
There is a minimum frequency
below which the light could not
kick out electrons…
even if we wait a long time
e’s
I
Frequency
E. something different
Summary of Phot. Electric experiment results.
(play with sim to check and thoroughly understand!)
http://phet.colorado.edu/simulations/photoelectric/photoelectric.jnlp
1. Current linearly proportional to intensity.
2. Current appears with no delay.
3. Electrons only emitted if frequency of light exceeds a
threshold. (same as “if wavelength short enough”).
4. Maximum energy that electrons come off increases linearly
with frequency (=c/wavelength).
(Max. energy = -stopping potential)
5. Threshold frequency depends on type of metal.
As the frequency of light increases
(shorter λ!), the KE of electrons
being popped out increases.
(it is a linear relationship)
Initial KE
C
Frequency
Initial KE
0
B
Initial KE
Initial KE
A
0
What about different metals?
(try sim)
photoelectric_en.jar
Frequency of light
Classical wave predictions vs. experimental observations
• Increase intensity
current increases.
Experiment matches with classical prediction
• Current vs. voltage: step close to zero Volts, then flat.
Flat part matches to classical pred., but experiment has 'tail'
of energetic electrons
Stopping potential, which depends
on color (not only intensity).
• Classical: Color of light does not matter, only intensity.
Experiment shows strong dependence on color
•Takes time to heat up ⇒ current initially low but increases
with time.
experiment: electrons come out immediately, no time delay
to heat up
how do these compare with classical wave predictions?
Einstein: Need “photon” picture of light to explain
observations:
-Light comes in chunks of energy (“particle-like”
We shall call those particles “photons”)
Is light a stream of particles?
Yes! Also….
E = hf
-a photon interacts only with single electron
- Photon energy depends on frequency of light, …
for lower frequencies, photon energy not enough
to free an electron (E = hf)
Ekin,max=hf - φ
“Work function” (I’ll explain later)
What did you think would happen?
Properties of photons
Optical power P
- frequency f
The energy of a photon is
The wavelength of a photon is
The momentum of a photon is
The mass of a photon is
E = hf
λ = c/f = hc/E
p = hf/c = E/c
m=0
Take out your paper again
on which you drew
following graphs.
Voltage U
Current I
1. Current vs. Voltage with the lamp on (fixed color of
light, say UV).
2. Current vs. Frequency (color) at a fixed voltage
(right plate is on positive potential)
3. Current vs. Intensity for fixed color (right plate is at
positive voltage)
h ≈ 6.626 ·10-34 J·s: Plank constant
It sometimes is useful to define h = h/(2π)
The energy of a photon is then: E = hf = hω
That's what happened:
high intensity
low intensity
U
0
A. There are more photons per second but each photon has
less energy.
B. There are more photons per second but each photon has
more energy.
C. There are fewer photons per second and each photon has
less energy.
D. There are fewer photons per second but each photon has
more energy.
E. Nothing happens to the photon number because light is a
wave.
F. I don’t do the reading assignments.
Intensity
or: Initial KE vs. f:
I
Initial KE
2. I vs. f:
0
The frequency of a beam of light is decreased but
the light’s intensity remains unchanged. Which of
the following is true?
I
I
1. Current vs. Voltage:
0
Photons
3. I vs. intensity:
Frequency
0
Frequency
Electromagnetic Waves and Photons
are describing the same thing!
Common confusion:
Recap: Intensity ∝ Emax2 but does not depend on frequency!
Need to have a model where these views/perspectives all fit together
and are consistent with each other!
X
vs.
X
Interference was definitive test that light is a wave!
“Why do higher frequency gamma rays carry more energy than
lower frequency radio waves, but frequency has nothing to do
with intensity, what gives?”
“But Tom said that energy of light depends on frequency (∝
∝ hf)!”
Do not confuse the energy carried by a beam of light
with the energy of a single quantum particle of light!
1. Light, Radio waves, X-rays are all electromagnetic waves!
2. The electromagnetic wave is made up from lots of photons.
3. Photons can be thought of as mini E/M wave segments
(each has a specific energy hf and a wavelength c/f )
Electromagnetic
Wave
E
Travels
Straight
Made up from
lots
photons
of photons
This picture can explain wave view and particle view.
“Particle” = little chunk of the electromagnetic wave.
Energy of photon (hf) is in its oscillating E and B fields.
(Sometimes it also helps to think of a photon as a tiny particle/energy packet).
When is it important to think
about particle aspect of light?
Only if your “detection system” is good enough to see
individual photons!
If you think of photons as particles you probably
automatically think of them as perfectly localized - like a
tiny billiard ball at a coordinate (x, y, z).
This is what get's you into trouble in QM!!
Examples where don’t need to think about particle behavior
• Billiard balls never produce a diffraction pattern
• Billiard balls have no wavelength/frequency
• Billiard balls cannot go through two slits at the same time
(photons can; electrons too! Will show later)
Bright Red Laser
Consistent descriptions:
Lots of light means …
• Big amplitude E/M wave
• Made from many photons
(mini E/M wave segments)
When a photon interacts with
something (e.g. an electron) all
the energy of its wave segment
ends up concentrated in one
spot (like a particle!)
Until a photon interacts with something (e.g. absorbed by an
electron), it is a wave. How does the wavelength of the photon
compare to the wavelength of the light in the red laser?
a. Photon has a smaller wavelength
b. Photon has same wavelength
c. Photon has a larger wavelength
d. Photons are points in space. They have no wavelength.
What if the intensity is really small?
Wimpy Laser beam: P = 10-19 W
Examples where important to think about particle behavior:
Photoelectric effect: Individual electrons popping out of metal
Lasers: Electrons in atoms transitioning between energy levels
Molecular bonds: Chemical bonds breaking
First photon
strikes here
(detection system isn’t good enough and wave behavior is easier):
Heating: Energy absorbed in microwave or by black asphalt.
Optics: Light bending through lenses or passing through slits
Laser beam: Treat it just like a beam of light… (Understanding
the working principle of a laser requires photon picture.)
Properties of photons
The energy of a photon is
The wavelength of a photon is
The momentum of a photon is
The mass of a photon is
E = hf
λ = c/f = hc/E
p = hf/c = E/c
m=0
h ≈ 6.626 ·10-34 J·s: Plank constant
It sometimes is useful to define h = h/(2π)
The energy of a photon is then: E = hf = hω
Where’s the photon?
Great question!
But I can't answer it…
It will strike the screen at an undetermined position!
3rd…
Next photon
strikes. here
..
screen
If the laser power is 10-19 W, we get only about one photon per
second in the beam! (assuming λ ≈ 780 nm)
Where's that one photon?
Well, we don't know until it strikes the screen.
Most photons seem to hit the screen where intensity is high.
However: We do observe that more photons strike at the
places where the intensity of the laser beam
is largest.
Important conclusion:
The probability to find a photon at a specific location in a
beam of light is proportional to the intensity of the beam
at that location.
Probability and randomness
Photon is 3-D-spread-out-little-chunk of an EM wave.
Gazillions of electrons in metal:
Which one will be kicked out?
Can’t tell, but photon uniformly
spread out so equal probability
everywhere.
What if shape of single photon wave looked like this?
Gazillion electrons
Which one will be kicked out?
Answer: Can’t tell, but probability
of photon collapse at particular
point (kicking out electron) related
to intensity of wave (Emax2)
How can light behave like a wave
(interference etc), but be made up of
little chunks (photons) that seem to
hit at random places?
Which is best answer, and why? (will
randomly ask for reasons)
If I shoot a photon through the two slits to hit the screen, it…
a. cannot hit in the middle, because block is in the way.
b. is completely random where it can hit. Has equal
chance of hitting anywhere on the screen.
c. must hit at the maximum of the interference pattern
d. has some chance of being anywhere, but on average
better chance of hitting at maximum of the interference
pattern
e. will hit anywhere it has a straight shot through either slit
quantum-wave-interference_en.jar
Probability of photon hitting is related to intensity at that location
(electric field strength)2~Intensity &
proportional to the probability of where photon will hit!
Two slit interference
standard electric field
representation of light field
Classical electric field wave
pattern describes probability
of where photons will be…
Higher intensity means higher
likelihood that photons will be
detected there.
If I shoot a photon through the two slits to hit the screen, it has some
chance of being detected anywhere on screen, but on average better
chance at being where interference pattern in brightest.
Double slit experiment at low intensities
Randomness in physics??!
A completely new concept in QM is that the outcome of a
measurement can often times not be predicted precisely.
We can only predict the probability of obtaining a
certain result!
Examples:
Where will a photon hit the screen?
Well, we don’t know, but the probability is largest where
the intensity of the light is largest ∝ (field amplitude)2
Where is the electron in a hydrogen atom?
Well, we don’t know, but the probability to find it is
largest at the location where the square of the matter
wave amplitude is largest. (Matter waves: see TZT chapter 6)
(Randomness is negligible for macroscopic objects but important on atomic scale!)
Let's revisit the PE effect one last time!
What actually happens in the metal
when a photon strikes?
What happens in the metal? Kicker analogy:
Photon is like a kicker in a pit…
Puts in energy. All concentrated
on one ball/electron.
Blue kicker has a fixed strength.
Red kicker (photon) kicks less
than blue one. Nothing gets out.
‘Photon’
Why do the emitted electrons have
different velocities/kinetic energies?
‘Electron’
electrons
metal
different ‘pit depths’
mgh = energy needed to
make it up hill and out.
mgh for highest electron
analogous to work function.
For photons:
KE = hf - Φ
h
What determines the work function ‘Φ’?
Different metals
Ball emerges with:
KE = kick energy - mgh
Fixed kick energy:
Top ones get out…
…bottom ones don’t.
PE effect: Apply Conservation of Energy
Energy in = Energy out
Energy of photon = energy needed to kick
Initial KE of electron
electron out of metal + as exits metal
h
platinum, hard to kick out
large work function ⇔ deep pit
PE effect: Apply Conservation of Energy
Electron Potential
Energy
Energy in = Energy out
Energy of photon = energy needed to kick + Initial KE of electron
electron out of metal
as exits metal
What happens if send in bunch of blue photons?
Ephoton
work function (Φ)
Outside
metal
Photon gives electron
“kick of energy”.
Inside
metal
All electrons have about equal chance of absorbing photon:
Max KE of electrons = photon energy - Φ
Min KE = 0
Some electrons, not enough energy to pop-out, energy into heat.
Electron Potential
Energy
sodium- easy to kick out
small work function ⇔ shallow pit
Loosely stuck electron, takes least energy to kick out
Φ
work function (Φ) = energy needed to kick
Outside
metal
Inside
metal
highest electron out of metal
Tightly stuck, needs more
energy to escape
Electrons over large range of energy have about equal
chance of absorbing photons.
Say you shine blue light on
Ephot
a metal plate
the metal
emits n electrons per sec.
Ephot
Electron potential
energy
h
work function Φ
Inside
metal
Now you change the
frequency to violet light
without changing the # of
photons per second.
What happens to the
number of electrons/sec.
coming out of the metal?
a. fewer electrons
b. same # of electrons but electrons are faster
c. more electrons
d. not enough information
Electrons over large range of energy have equal
chance of absorbing photons.
Typical energies
elect. potential
energy
photoelectric_en.jar
Photon Energies:
Each photon has: Energy = Planks constant * Frequency
(Energy in Joules)
Ephot
(Energy in eV)
E=hf=(6.626*10-34 J-s)*(f s-1)
E=hc/λ = (1.99*10-25 J-m)/(λ m)
work function Φ
Red Photon: 650 nm
metal
c. more electrons come out with violet
absorb blue light and have enough energy to leave
absorb blue light, but don’t come out
so the more energy the light has, the more electrons that come
out, until so much energy that every electron comes out.
(violet and ultraviolet would not be very different in this case)
E=hf= (4.14*10-15 eV-s)*(f s-1)
E= hc/λ = (1240 eV-nm)/(λ nm)
Ephoton = 1240 eV-nm = 1.91 eV
650 nm
Work functions of metals (in eV):
Aluminum
4.08 eV
Cesium
2.1
Lead
4.14
Potassium
2.3
Beryllium
5.0 eV
Cobalt
5.0
Magnesium
3.68
Platinum
6.35
Cadmium
4.07 eV
Copper
4.7
Mercury
4.5
Selenium
5.11
Calcium
Carbon
2.9
4.81
Gold
Iron
5.1
4.5
Nickel
Niobium
5.01
4.3
Silver
Sodium
4.73
2.28
Uranium
3.6
Zinc
4.3
Photomultiplier tubes: An application of photoelectric effect.
Most sensitive way to detect visible light, see single photons
(eye is incredibly good too; can see a few photons)
Application of the PE effect:
The photo multiplier tube (PMT)
Metal plate
big voltage
Electron
current
electron amplifier:
gives pulse of
current for each
photoelectron
1
2
3
4
5
Time (millisec)
What would be the best
choice of these materials to
detect visible light?
a. Platinum
Φ = 6.35 eV
b. Magnesium
= 3.68 eV
c. Nickel
= 5.01 eV
d. lead
= 4.14 eV
e. Sodium
= 2.28 eV
KE = photon energy-energy to get out
= hf – Φ
If λ is ½ then, f twice as big, Ephot150 = 2 hf300
V
Old KE300 = hf300 - energy to get out
New KEnew= 2hf300 - energy to get out
so KEnew is more than twice as big.
Energy
A photon at 300 nm kicks out an electron with a maximum
kinetic energy KE300. If the wavelength is halved to 150
nm and the photon hits an electron in the same metal what
would be the maximum possible kinetic energy KE150 of an
electron kicked out by this 150nm photon?
a. less than ½ KE300.
b. ½ KE300
c. = KE300
d. 2 x KE300
e. more than 2 x KE300
(remember hill/kicker analogy, draw pictures to reason out
answer, don’t just pick answer without careful reasoning)
hf150
hf300
KE300
KE300
KE300
V
Shine in light of 300 nm. The most energetic electrons come
out with kinetic energy, KE300. A voltage difference of 1.8 V is
required to stop these electrons. What is the work function Φ
for this plate? (e.g. the minimum amount of energy needed to
kick electron out of metal?)
a. 1.2 eV
b. 1.8 eV
c. 4.1 eV
d. 5.9 eV
e. none of the above
Remember: The energy of a photon
is hf (= hc/λ).
In units of eV you can write:
E= hc/λ = (1240 eV-nm)/(λnm)
(Your answer should be within ~0.1eV)
V
CQ: Shine in light of 300 nm, most energetic electrons come out with kinetic energy,
KE300. A voltage diff of 1.8 V is required to stop these electrons. What is the work
function Φ for this plate? (e.g. the minimum amount of energy needed to kick e out of
metal?)
Energy is conserved so:
a. 1.2 eV
b. 2.9 eV Ephot= energy need to exit (Φ) + electron’s left over energy
c. 6.4 eV
d. 11.3 eV so
Φ = Ephot – electron’s energy
e. none of
When electron stops, all of initial KE has been
the above
converted to electrostatic potential energy:
electron energy = q*∆V = e x 1.8V = 1.8 eV, and
Ephot = 1240 eV nm/300 nm = 4.1 eV.
So Φ = 4.1eV - 1.8 eV = 2.3 eV
Remember this picture?
This is the end of our discussion
about the PE effect
Questions?
Scanning tunneling microscope (STM)
Measure current
between tip and
sample
Electrons are wave packets too!
The probability to
find an electron that
is trapped inside this
ring of atoms is
highest at the place,
where the square of
the amplitude of the
electron wave
function is largest.
To all those students feeling confused now:
You should be bothered and ask questions!
QM can be kind of confusing in the beginning.
QM: Fundamental change in the way to think about physics:
Which slit did this photon go through?
a. left
b. right
c. both
d. neither
e. either left or right we just cannot know which one
Before (pre 1900, Physics I and II) -- everything could be
known exactly, if measured and calculated carefully enough.
Now-- physics behavior is fundamentally inexact.
Talks only about probability; Can only predict
probabilities for what will happen in a measurement.
Einstein didn’t accept this fact for long! And he certainly
was a pretty bright fellow! Today we fully accept the QM
description of small things (This was not the case at first!)
Photon before it goes through the slits
Which slit did this photon go through?
a. left
b. right
c. both!
d. neither
Photon as little segment of
wave moving towards slits
If one slit:
Get single slit pattern
(i.e. no interference)
But: that photon is part
of the two slit interference pattern,
the probability pattern of where it lands is described by
the 2 slit interference pattern, it must have gone through both
slits i. e. as a wave!
When it interacts with the screen it behaves particle-like!
Intensity of wave in various places, indicates probability
of finding the photon there if you looked at that moment.
quantum-wave-interference_en.jar
Photon after it went through the slits
Photon after it goes through the slits
When photon interacts with
an electron or atom, all
energy ends up in one
spot… Behaves like a
particle with energy = hc/λ
Photon is a wave…
It can interfere with itself.
Photon is a wave…
It can interfere with itself.
Intensity of wave in various
places indicates the probability of
the photon concentrating at that
spot if you had detector (e.g. a
bunch of atoms or a sheet of
metal)
Intensity of wave in various
places indicates probability of the
photon concentrating at that spot
if you had detector (e.g. a bunch
of atoms or a sheet of metal)
Which slit did this photon go through?
a. left
b. right
c. both
d. neither
e. either left or right we just cannot know which one
Which slit did this photon go through?
If one slit:
Get single slit pattern
(i.e. no interference)
Like this:
or this:
but not like this:
But: that photon is part
of the two slit interference pattern.
The probability pattern of where it lands is described by the 2 slit
interference pattern (the photon has to ‘know’ about both slits!)
It must have gone through both slits, i. e. as a wave!
(When it interacts with the screen it behaves particle-like!)
quantum-wave-interference_en.jar
Photon before it goes through the slits
Photon as little segment of
wave moving towards slits
Photon after it went through the slits
Still only one photon!
(But this photon has
a slightly complicated
wave function.)
Photon is a wave…
It can interfere with itself.
Intensity of wave in various places, indicates probability
of finding the photon there if you looked at that moment.
Photon after it went through the slits
When photon interacts with
an electron or atom, all
energy ends up in one
spot… Behaves like a
particle with energy = hc/λ
Intensity of wave in various
places indicates the probability of
finding the photon at that spot, if I
had detector there (e.g. a bunch
of atoms or a sheet of metal)
Probability of photon hitting is related to intensity at that location
(electric field strength)2 ~ Intensity &
proportional to the probability of where photon will hit!
standard electric field
representation of light field
Classical electric field wave
pattern describes probability
of where photons will be…
Higher intensity means higher
likelihood that photons will be
detected there.
If I shoot a photon through the two slits to hit the screen, it has some
chance of being detected anywhere on screen, but on average better
chance at being where interference pattern is brightest.
Two slit interference
Double slit experiment at low intensities
Randomness in physics??!
Remember this picture?
A completely new concept in QM is that the outcome of a
measurement can often times not be predicted precisely.
We can only predict the probability of obtaining a
certain result!
Examples:
Where will a photon hit the screen?
Well, we don’t know, but the probability is largest where
the intensity of the light is largest ∝ (field amplitude)2
Where is the electron in a hydrogen atom?
Well, we don’t know, but the probability to find it is
largest at the location where the square of the matter
wave amplitude is largest. (Matter waves: see TZT chapter 6)
(Randomness is negligible for macroscopic objects but important on atomic scale!)
Scanning tunneling microscope (STM)
Measure current
between tip and
sample
Electrons are wave packets too!
The probability to
find an electron that
is trapped inside this
ring of atoms is
highest at the place,
where the square of
the amplitude of the
electron wave
function is largest.
To all those students feeling confused now:
You should be bothered and ask questions!
QM can be kind of confusing in the beginning.
QM: Fundamental change in the way to think about physics:
Before (pre 1900, Physics I and II) -- everything could be
known exactly, if measured and calculated carefully enough.
Questions?
Now-- physics behavior is fundamentally inexact.
Talks only about probability; Can only predict
probabilities for what will happen in a measurement.
Einstein didn’t accept this fact for long! And he certainly
was a pretty bright fellow! Today we fully accept the QM
description of small things (This was not the case at first!)
What’s next?
Atoms and atomic spectra!
What happens if we bash atoms with
electrons?
In atomic discharge lamps, lots of electrons are given kinetic
energy (accelerated by a high voltage). When they bash into
atoms some of this kinetic energy is transferred to the atom
Atom get's excited!! (“Neon” lights, Mercury street lamps)
120V
Cathode (hot metal, so
Anode (positive potential)
electrons can come out)
Hydrogen
Na
Hydrogen
neon
neon
neon
Hydrogen
White light = whole spectrum.
Each type of atom produces
unique set of colors, called its
“spectrum”. None of the atoms
produces white light!
Na Na
Na
NaNa
Hg
Hg
Hg
Hg
Hg
Hold grating only by edges...oil from hands ruins grating!
Hold close to eye... See rainbow from lights. Turn grating
so rainbow is horizontal. (Rainbow appears quite a bit to
the side of the actual lamp.
What colors from white light?
What colors from neon?
What colors from hydrogen?
What from mercury?
What from sodium?
Hydrogen
Hydrogen
Hydrogen
Use a grating to look at the spectrum of
the discharge lamps
400
500
600
Wavelength (nanometers)
700
800
increasing
energy
What do the vertical arrows represent?
allowed energies
increasing
energy
allowed energies
ground state
ground state
a. arrows on right represent absorption of light,
arrows on left represent emission.
b. arrows on both left and right represent emission
c. arrows on both left and right represent absorption
d. arrows on left represent absorption of light,
arrows on right represent emission.
e. I don’t know. I have never seen such a diagram.
Announcements
d. arrows on left represent absorption of light,
arrows on right represent emission.
Atoms and Atomic spectra
• Reading for Friday: 5.8 – 5.10
• What are atoms made off? JELL–O?
• Interested in becoming an LA, or
simply interested in free food?
Go
to UMC, Room 235 tonight at 6pm.
• What happens when we hit atoms with various
stuff?
Early ideas about atoms
invisible!
• Atom - Greek “indivisible unit”
• The Blueberry-Muffin-Atom-Model (aka.
‘The Thomson model’): Uniform distribution
of positive charges with negative electrons
embedded in it.
But wait! We can get electrons from them
(scraping, chemical, or photoeffect) but no
positive charges. Hmmm?!...
-
• How do atoms interact with light?
Develop
model of how light interacts with and is
produced by individual atoms (Helps us learn
how atoms ‘work’)
Step 1: Shoot the JELL–O atom!
Have a heavy blob that seems like grape JELL–O, and you
have gun with rubber bullets. How to find out what the middle
of the blob is like?
Have a heavy blob that seems like grape JELL-O, and you
have gun with rubber bullets. How to find out what the middle
of the blob is like? Shoot bunch of bullets into it and see this.
Have a heavy blob that seems like grape JELL-O, and you
have gun with rubber bullets. How to find out what the middle
of the blob is like? Shoot bunch of bullets into it and see this.
What is the inside like?
Hard heavy core surrounded by JELLO
Only one thing is reflecting bullets, sending them straight
back so must be hard and heavy.
Essentially Rutherford experiment and conclusion (TZD 3.12).
Rutherford shot alpha particles = 2 protons, 2 neutrons
n
pn Positive charge
Bullets =
p
What is the inside like?
The Rutherford experiment
The Rutherford atom:
Tiny nucleus with protons and neutrons (~99.99% of mass)
Surrounded by large diffuse cloud of low mass electrons
10-10 m
Detector
n n
pnp p
p n
p
10-14 m
rutherford-scattering_en.jar
The ‘Rutherford atom’ is a much better model
than the ‘Thomson model’, but it is still far
from helpful.
A good model should:
• explain how stuff works
• predict the behavior of it
• be quantitative
• etc.
Need a better model!
Let’s see how atoms behave
when shot-at! (with
(with electrons)
electrons)
What happens if we bash atoms with
electrons?
Use a grating to look at the spectrum of
the discharge lamps
In atomic discharge lamps, lots of electrons are given kinetic
energy (accelerated by a high voltage). When they bash into
atoms some of this kinetic energy is transferred to the atom
Atom get's excited!! (“Neon” lights, Mercury street lamps)
- 120V +
Cathode (hot metal, so
electrons can come out)
Hold grating only by edges...oil from hands ruins grating!
Hold close to eye... See rainbow from lights. Turn grating
so rainbow is horizontal. (Rainbow appears quite a bit to
the side of the actual lamp.
Anode (positive potential)
Note: ‘Anode’ and ‘Cathode’ have different meanings in physics or
chemistry. Remember ‘Cathode Ray Tube’ (CRT): Electrons
leave the cathode (in physics).
Hydrogen
White light = whole spectrum.
Each type of atom produces
unique set of colors, called its
“spectrum”. None of the atoms
produces white light!
Na
Hydrogen
neon
neon
neon
Hydrogen
Each type of atom produces unique set of colors.
Discussion: What does this imply about electrons in atoms?
Implies that electrons only change between very specific energies.
Each time a photon is emitted an electron must be changing in
energy by that amount (releasing energy).
Only way for individual atoms to give off energy is as light.
Atoms are lazy - always want to go back to lowest energy state.
2. Excited atom ..electron
3. Electron
1. Fast electron
in atom goes to higher
jumps back to
hits atom
energy
Less KE low energy
Na Na
Na
NaNa
Hydrogen
Hydrogen
Hydrogen
Hg
Hg
Hg
Hg
Hg
What colors from white light?
What colors from neon?
What colors from hydrogen?
What from mercury?
What from sodium?
400
500
600
Wavelength (nanometers)
700
800
e
Higher
energy
e
e
e
e
e
Energy
Energy level diagrams represent energy levels the electron can
go to. Different height = different energy
e
~10ns
Excited
state
Ground
state
e
An single electron bashes into an atom in a
discharge lamp
d
10 V
D
e
For Hydrogen,
transitions to
ground state in
deep ultraviolet!
No light emitted with
colors in this region
because no energy
levels spaced with
this energy.
-2 eV
-3 eV
-6 eV
If atom fixed at this point in tube,
-10eV
list all the possible photon energies (colors) that you might see?
A. 1eV, 2eV, 3eV, 4eV, 7eV, 8eV
B. 4eV, 7eV, 8eV
C. 1eV, 3eV, 4eV
D. 4eV
E. Impossible to tell.
Electron energy = q∆V = e(Ed),
where E is the electric field = (battery V)/(total distance D),
and d is the distance it goes before a collision.
An single electron bashes into an atom in a
discharge lamp
An single electron bashes into an atom in a
discharge lamp
-2 eV
-3 eV
- 10 V +
d
- 10 V +
D
-6 eV
-6 eV
If atom fixed at this point in tube,
-10eV
list all the possible photon energies (colors) that you might see?
A. 1eV, 2eV, 3eV, 4eV, 7eV, 8eV
B. 4eV, 7eV, 8eV
C. 1eV, 3eV, 4eV
D. 4eV
E. Impossible to tell.
-10eV
If atom fixed at this point in tube,
list all the possible photon energies (colors) that you might see?
A. 1eV, 2eV, 3eV, 4eV, 7eV, 8eV
B. 4eV, 7eV, 8eV
C. 1eV, 3eV, 4eV
D. 4eV
E. Impossible to tell.
-2 eV
-3 eV
Electron energy = q∆V = e(Ed),
where E is the electric field = (battery V)/(total distance D),
and d is the distance it goes before a collision.
Remember the ‘Electron Volt’ (eV)?
Some handy relations:
Current flowing through a wire:
Define new energy unit: The electron-volt (eV)
1eV = kinetic energy gained/lost by an electron when
accelerated/decelerated by traversing an
area with 1 Volt potential difference (such as in
a plate capacitor)
1 Ampere = 1 Coulomb / Second
1 Watt = 1 Ampere · 1 Volt = 1 C/s · 1 V
1 Joule = 1 Watt · 1 Second = 1 C · 1 V
1eV ≈ 1.6 ·10-19 J
0V
F
E
path
1V
+
+
+
+
1C
0V
What we know so far:
Rutherford: Atoms have a tiny, but heavy core
surrounded by a cloud of electrons.
Discharge lamps: Atoms struck by fast
electrons emit light of distinct colors.
Energy levels: Electrons in atoms are found only
in discrete energy levels. When they jump
down to a lower level, a photon is emitted
carrying away the energy.
i.e: An electric potential of 1 V puts in one Joule
of energy into a charge of 1 C kinetic
energy increases by 1 Joule.
1V
+
+
+
+
An e- has a charge of 1.6·10-19C,
therefore, it gains a kinetic energy of
1eV=1.6·10-19J
Investigate energy levels in atoms
p
n nn
p p p
nn nn
p
120V
∆E
∆E=hf
e
Different Atoms: Different types of atoms have different level
structures (seen by the distinct set of colors they emit):
Neon: Strong red line; Sodium: strong yellow line …
energy levels of electron
stuck in atom
3
energy
2
of colliding
1
electron
G (ground)
If the colliding electrons have an energy between that of level 2
and level 3 when they hit the atom how many different colors
could be emitted by the atom?
a. no levels will be excited, and so no light will come out.
b. 1 color of light will come out
c. 2 colors of light will come out
d. 3 colors of light will come out
e. 4 colors come out.
ans. d. enough energy to excite level 2, then get 2⇒
⇒1 followed
by 1⇒
⇒G, but also can go 2⇒
⇒G.
A neon lamp emits a strong red line. Sodium emits a strong
yellow line. What accounts for this difference?
a. The electrons in the discharge lamp hit the neon atoms with
higher kinetic energy than the electrons hit the sodium atoms
b. The electrons in the discharge lamp hit the neon atoms with
less speed than the electrons hit the sodium atoms
c. The energy spacing between the electronic energy levels that
are responsible for these lines are smaller in the neon atom
than in the sodium atom.
d. The energy spacing between the electronic energy levels that
are responsible for these lines is larger in the neon atom than in
the sodium atom.
Applications of atomic spectroscopy
When an emission line appears very bright that
means:
1. Detecting what kind of atoms are in a material.
(excite by putting in discharge lamp or heating in flame
to see spectral lines)
a. Multiple photons are emitted for each single
electron transition
b. More electron transitions are occurring each
second
c. The electronic energy levels are farther apart
and thus the line appears brighter
d. a and b
e. b and c
Application:
Designing a better light
What is important?
(Well, we have to be able to see the light!)
What color(s) do you want?
( Choice of atom)
How do you excite them to desired level?
( Electron collisions)
How to get electrons with desired energy when
hit atoms? What determines energy of electrons?
2. Detecting what the sun and the stars are made of:
Look at the light from a star through a diffraction grating.
See what lines there are; Match up to atoms on earth.
telescope
star
diffraction
grating
3. Making much more efficient lights!
Incandescent light bulbs waste >90% of the electrical
energy that goes into them! (<10% efficient)
Streetlight discharge lamps (Na or Hg) ~80%
efficient. Fluorescent lights ~ 40-60% efficent.
Incandescent light (hot filament)
Temperature = 2500-3000K
Hot electrons jump
between many very
closely spaced levels
(solid metal). Produce all
colors. Mostly infrared at
temp of normal filament.
>90% is worthless
Infrared radiation (IR =
longer than ~700nm)
IR
P
λ
~10% of energy is
useful visible light
Discharge lamp
Energy levels in
isolated atom:
Only certain
wavelengths
emitted.
HV
Right atom, right pressure
and voltage, mostly visible light.
Streetlight discharge lamps
(Na or Hg) 80% efficient.
Florescent Lights. Discharge lamp, White= Red + green +blue
40-60% efficient (electrical power⇒visible light)
Florescent Lights.
Discharge lamp, but want to have it look white.
White = red + green +blue
40-60% efficient (electrical power ⇒ visible light)
Converting 180nm UV light into visible photons with “phosphor”.
How to do this?
phosphor
coating
Converting UV light into visible photons with “phosphor”.
Phosphor converts 180 nm UV to red+ green+blue.
Hg
180nm
6.9 eV energy per photon
633nm (red)
2 eV / photon
532nm (green)
2.3 eV / photon
475nm (blue)
2.6 eV / photon
180 nm far UV
Hg
}
6.9 eV
Hg
Hg
Hg
e
Hg
discharge lamp/flor. tube
120 V
real phosphors more than just 3
phosphor wastes 20-30% energy ⇒ heat
Summary of important Ideas
1) Electrons in atoms are found at specific energy levels
2) Different set of energy levels for different atoms
3) one photon emitted per electron jump down between energy
levels. Photon color determined by energy difference.
4) If electron not bound to an atom: Can have any energy. (For
instance free electrons in the PE effect.)
Hydrogen
Lithium
Energy
Electron energy levels in 2
different atoms: Levels have
different spacing (explains
unique colors for each type of
atom.
(not to scale)
Atoms with more than one
electron … lower levels filled.
Now we know about the energy
levels in atoms. But how can we
calculate/predict them?
Need a model
Step 1: Make precise, quantitative observations!
Step 2: Be creative & come up with a model.
Step 3: Put your model to the test.
energy of electron
in phosphor molecule
phosphor wastes 20-30% energy
⇒ heat
Quiz on the reading
(no talking and closed book)
The Rydberg formula describes
a. the colors of light emitted from all types of atoms.
b. the angular distribution of particles scattering off
of atoms that Rutherford observed in his famous
experiment.
c. the colors of light emitted by hydrogen atoms.
d. the energy distribution of cathode rays
e. the frequencies of light emitted by helium atoms.
Balmer series:
A closer look at the
spectrum of hydrogen
656.3 nm
410.3 486.1
434.0
Balmer (1885) noticed wavelengths followed a progression
λ=
91.19nm
1 1
−
22 n 2
where n = 3,4,5, 6, ….
As n gets larger, what happens to wavelengths of
emitted light?
λ gets smaller and smaller, but it approaches a limit.
Balmer series:
A closer look at the
spectrum of hydrogen
Hydrogen atom – Rydberg formula
Does generalizing Balmer’s formula work?
Yes!
It correctly predicts additional lines in HYDROGEN.
Rydberg’s general formula
656.3 nm
410.3 486.1
434.0
Balmer (1885) noticed wavelengths followed a progression
So this gets smaller
λ=
Balmer correctly
predicted yet
undiscovered
spectral lines.
91.19nm
where n = 3,4,5,6, ….
1 1
− 2
2
2 n gets smaller as n increase
gets larger as n increase,
but no larger than 1/4
λlimit = 4 * 91.19nm = 364.7nm
Predicts λ of n m transition:
n
(n>m)
m
(m=1,2,3..)
m=1, n=2
λ gets smaller and smaller, but it approaches a limit
Hydrogen atom – Lyman Series
Rydberg’s formula
λ=
91.19nm
1
1
− 2
2
m n
Hydrogen
energy
levels
Balmer-Rydberg formula
λ=
0eV
Can Rydberg’s formula
tell us what ground
state energy is?
(n>m)
0eV
91.19nm
1
1
− 2
2
m n
0
0eV
Einitial − E final =
m
Hydrogen energy levels
Look at energy for a transition
between n=infinity and m=1
Predicts λ of n m transition:
n
Hydrogen energy levels
91.19nm
λ=
1
1
− 2
2
2
m2 n
(m=1,2,3..)
m=1
-?? eV
Balmer/Rydberg had a mathematical formula to describe
hydrogen spectrum, but no mechanism for why it worked!
Why does it work?
Hydrogen energy levels
− E final =
hc
λ
=
hc  1
1 
 2− 2
91.19nm  m
n 
1
hc
91.19nm m 2
Em = −13.6eV
-?? eV
1
m2
The Balmer/Rydberg formula is a
mathematical representation of an
empirical observation.
It doesn’t explain anything, really.
Balmer’s formula
656.3 nm
410.3 486.1
434.0
λ=
How can we calculate the energy levels
in the hydrogen atom?
91.19nm
1
1
− 2
2
m n
where m=1,2,3
and where n = m+1, m+2
A semi-classical explanation of the
atomic spectra (Bohr model)
m=1, n=2
Rutherford shot alpha particles at atoms and he figured
out that a tiny, positive, hard core is surrounded by
negative charge very far away from the core.
• One possible model:
Atom is like a solar system:
electrons circling the nucleus
like planets circling the sun…
• The problem is that accelerating
electrons should radiate light
and spiral into the nucleus:
A. Well, they do, but very, very slowly.
B. Because planets obey quantum mechanics,
not classical mechanics.
C. Because planets obey classical mechanics,
not quantum mechanics.
D. Because gravitational forces work differently
than electrical forces and there is no such a
thing as gravitational radiation.
E. Because planets are much bigger than
electrons.
*Elapsed time: ~10-11 seconds
Electrostatic potential energy
Nucleus
-
Higher
Energy
Potential energy of the electron in hydrogen
D
F
+
Electron
++
++
Why don't planets spiral into the
sun?
-
Energy
levels
Force on electron is less, but Potential Energy is higher!
Electrons at higher energy levels are further from the nucleus!
Potential energy of a single electron in an atom
PE of an electron at distance D from the proton is
PE = −
ke 2
D
Coulomb’s constant
D
D
∞
∞
potential
energy
0 distance from proton
kqelect q prot
r2
dr
D
D
1
dr
ke 2
=−
PE = kqelect q prot ∫ 2 = kqelect q prot
r
r∞
D
∞
-e e
(k = 1/( 4πε0 ): Coulomb force const.)
(for hydrogen)
How can we calculate the energy
levels in hydrogen?
λ=
, ke2 = 1.440eV·nm
PE = -ke(Ze)
+ +
+
D
+ (For Z protons)
E
r
We define electron’s PE as 0 when far, far away from the proton!
Electron's PE = -work done by electric field from r1=∞…r2=D
∫ F • dr = ∫
When an electron moves to location further away from the
nucleus its energy increases because energy is required to
separate positive and negative charges, and there is an
increase in the electrostatic potential energy of the electron.
-
91.19nm
1
1
−
m2 n2
Step 1: Make precise, quantitative observations!
Step 2: Be creative & come up with a model.
How to avoid the Ka-Boom?
*Elapsed time: ~10-11 seconds
Bohr Model
Bohr's approach:
• When Bohr saw Balmer’s formula, he came up
with a new model that would predict it and
'solve' the problem of electrons spiraling into
the nucleus.
• The Bohr model has some problems, but it's
still useful.
• Why doesn’t the electron fall into the nucleus?
#1: Treat the mechanics classical (electron spinning around
a proton):
– According to classical physics, It should!
– According to Bohr, It just doesn’t.
– Modern QM will give a satisfying answer, but you’ll
have to wait till next week.
- Newton's laws assumed to be valid
- Coulomb forces provide centripetal acceleration.
#2: Bohr's hypothesis (Bohr had no proof for this; he just
assumed it – leads to correct results!):
- The angular momentum of the electrons is quantized in
multiples of ћ.
- The lowest angular momentum is ћ.
Original paper: Niels Bohr: On the Constitution of Atoms and Molecules,
Philosophical Magazine, Series 6, Volume 26, p. 1-25, July 1913.)
Bohr Model. # 1: Classical mechanics
The centripetal acceleration
a = v2 / r is provided by the coulomb
force F = k·e2/r2.
Bohr Model. #2: Quantized angular momentum
F=k e2/r2
v
(k = 1/( 4πε0 ): Coulomb force const.)
Newton's second law
mv2/r = k·e2/r2
or: mv2 = k·e2/r
The electron's kinetic energy is KE = ½ m v2
The electron's potential energy is PE = - ke2/r
E= KE + PE = -½ ke2/r = ½ PE
+
E
Bohr Model. Results
h2
= 52.9 pm , rB: Bohr radius
ke 2m
En = E R / n 2 , with E R =
F=k e2/r2
v
rn = rB n 2 , with rB =
h2
= 52.9 pm
ke 2m
En = E R / n 2 , with E R =
, rB: Bohr radius
m( ke 2 ) 2
= 13.6 eV , ER: Rydberg
2
Energy
2h
Only discrete energy levels possible.
Electrons hop down towards lowest level, giving off photons
during the jumps. Atoms are stable in lowest level.
0 distance from proton
2 2
m( ke )
= 13.6 eV , ER: Rydberg
Energy
2h 2
The Bohr model not only predicts a reasonable atomic radius
rB, but it also predicts the energy levels in hydrogen to 4 digits
accuracy!
Possible photon energies:
Bohr assumed that the angular
momentum of the electron could only
have the quantized values of:
L= nћ
And therefore: mvr = nћ, (n=1,2,3…)
or: v = nћ/(mr)
Substituting this into mv2 = k·e2/r leads to:
Therefore: If we know r, we know E and v, etc…
r = rB n 2 , with rB =
ћ = h/2π
1
 1
Eγ = En − Em = E R  2 − 2 
n 
m
(n > m)
The Bohr model 'explains' the Rydberg formula!!
potential
energy
Bohr couldn't explain why the angular
momentum is quantized but his
model lead to the Rydberg-Balmer
Lmin = h
formula, which matched to the
experimental observations very well!
He also predicted atomic radii reasonably well and
was able to calculate the Rydberg constant.
Successes of Bohr Model
Shortcomings of the Bohr model:
• 'Explains' source of Balmer formula and predicts
empirical constant R (Rydberg constant) from
fundamental constants: R=1/91.2 nm=mk2e4/(4πch3)
Explains why R is different for different single
electron atoms (called hydrogen-like ions).
• Predicts approximate size of hydrogen atom
• Explains (sort of) why atoms emit discrete spectral
lines
• Explains (sort of) why electron doesn’t spiral into
nucleus
• Why is angular momentum quantized yet
Newton’s laws still work?
• Why don’t electrons radiate when they are
in fixed orbitals yet Coulomb’s law still
works?
• No way to know a priori which rules to
keep and which to throw out…
• Can't explain shapes of molecular orbits
and how bonds work
• Can’t explain doublet spectral lines
Questions?
Which of the following principles of classical
physics is violated in the Bohr model?
A.
B.
C.
D.
E.
Opposite charges attract with a force inversely
proportional to the square of the distance between
them.
The force on an object is equal to its mass times its
acceleration.
Accelerating charges radiate energy.
Particles always have a well-defined position and
momentum.
All of the above.
Ideas for how to resolve these
problems?
Matter waves: TZD Chapter 6
Note that both A & B are used in derivation of Bohr model.
What’s so special about LASER light?
A) It doesn't diffract when it goes through two slits.
B) It can be collimated very well.
C) The photons in the laser beam travel a little bit
faster because they all go the same direction.
D) Laser light is pure quantum light, and therefore,
cannot be described with classical EM theory.
E) Laser light is purely classical light, and therefore,
it is incompatible with the photon picture.
What’s so special about LASER light?
Remember the word 'coherence'?
(We used it occasionally in context with
interferometers, diffraction and with photons.)
There are two types of coherence:
- Spatial coherence
- Temporal coherence
LASERs produce light with excellent spatial
and temporal coherence.
Spatial coherence
Temporal coherence
Collimating lens
Flashlight
(incoherent)
Flashlight
(incoherent)
'Mach-Zehnder'
interferometer
c ·τ 1
Laser
(coherent)
Laserpointer
(coherent)
c ·τ 2
For very stable lasers: (τ1- τ2)max ~ 1 second.
(This corresponds to 300’000 km path difference!)
Remember this one (from class 3):
Let's make a coherent light pulp!
Bad: Throw away most of
the light and individual
photons still have random
phases φ
E-field (for a single color):
E(x,t) = A0 sin(ωt+(x ·ω)/c +φ)
LASER
ω = 2πf
E
λ
A0
φ
x
Temporal coherence: ω and φ are very well
defined constants (i.e. time independent).
How can we make identical photons?
Aperture
Color filter (passes only
certain colors, ω)
Remember: Coherent light requires that ω and φ are constants.
How light interacts with atoms
e
out
Clone them!
2
1
spontaneous
emission of light:
Excited atom emits
one photon.
in
2
e
1
absorption of
light: Atom absorbs
one photon
in
e
2
out
1
stimulated
emission:
clone the photon
-- A. Einstein
Surprising fact. Chance of stimulated
emission of excited atom EXACTLY the
same as chance of absorption by lower
state atom. Critical fact for making a laser.
Laser: Stimulated emission to clone photon many times (~1020/s)
Light Amplification by Stimulated Emission of Radiation
Spontaneous emission
Stimulated emission
Random phase
Random direction
Similar energy (as absorbed photon)
Legend:
Photon
Identical phase
Identical energy
Identical momentum
Legend:
Atom in ground state
Atom in excited state
e
Photon
Atom in ground state
e
Atom in excited state
e
e
Chance of stimulated emission of
excited atom is EXACTLY the same
as chance of absorption by ground
state atom.
Glass tube below contains 9 atoms. Some are excited some
not excited (as shown). Light enters the tube on the left:
b. less come out right
3 excited atoms can emit photons,
6 ground state atoms can absorb. Absorption wins.
For the condition above: what do you expect?
a. More photons will come out (on the right) than go in.
b. Fewer photons will come out (on the right) than go in.
c. Same number as go in,
d. None will come out.
To increase the number of photons when going through the atoms,
more atoms need to be in the upper energy level than in the lower.
Need a “Population inversion”
(This is the hard part of making laser, b/c atoms jump down so quickly.)
Think about statistics / probabilities
Ok, if we have 'population inversion' we get 'gain'.
But where are these photons coming from?
out
}
Nupper > Nlower, more cloned than eaten.
From here!
Some fraction of the photons are 'recycled' through the
amplifier (feedback!). The rest is used as the laser's output.
Nupper < Nlower, more eaten than cloned.
This is done with an 'optical resonator/cavity'
Optical resonator
Semiconductor lasers are tiny!
Partially-silvered mirror
Mirror
out
Continuously supply energy to the atoms to
maintain population inversion.
But others can be fairly ‘involved’
…or very involved…
All the energy of this
laser is focused into this!
(Trigger Nuclear fusion.)
How to get population inversion?
No population inversion in 2 level atom!
e
out
1
spontaneous
emission of light:
Excited atom emits
one photon.
e
A) Use photons with hf < ∆E
B) Use photons with hf = ∆E
2
C) Use photons with hf > ∆E
D) Use very strong lamp with hf ≈ ∆E.
E) Will never get population inversion in this system.
2
e
1
absorption of
light: Atom absorbs
one photon
in
e
2
out
1
stimulated
emission:
clone the photon
Equal probability
∆E
excited
in
e
not excited
Population inversion means: More atoms are in the excited
state than in the ground state.
As soon as we have the same number of atoms in the excited
state as in the ground state, the probability of creating an
excited atom is same (or smaller, when considering spontaneous
emission) as the probability of having stimulated emission!
Can never reach population inversion in 2-level atom!
Need at least 3 energy levels!
Now, let's play!
Use a second color of light to create population inversion
2
2
t2
t1
1
1
t3t1<>t2t2
3
G
2
G
t1 < t2
t2
1
“Pumping” process produces population
inversion
t 1< t 2
G
http://phet.colorado.edu/simulations/lasers/lasers.jnlp
Various 'flavors' of Lasers
Summary
Mirror
Gas lasers
Half-silvered mirror
Dye lasers (liquid)
out
Chemical lasers
Solid state lasers
Fiber lasers
LASERs need:
• Population inversion
Gain
• Optical feedback (optical resonator)
Diode lasers
Coherent light
Gas dynamic lasers
…
Practically unlimited numbers of
applications for lasers
Just to name a few:
Medicine: Surgery (no bleeding, noncontact
eye)
Diagnostics (TP-Microscopy, tomography)
Machining: Tight focus allows very high intensity.
(100 W cut through hardened steel like butter)
Science: Huge variety of applications
(Ultra precise spectroscopy / light matter interaction...)
Commercial: DVD/CD, range finding, leveling
telecommunication…
Sharks with frickin' laser beams attached to their heads…