PHY1033C–Fall 2002 Peter J Hirschfeld Notes
Lectures 14-17 Oct. 15,17,22,24
Special Relativity
Viewed Episodes 41-43 of Mechanical Universe Film Series on Relativity
Maxwell’s equations of electromagnetism (1860) showed light was an electromagnetic wave moving with speed c = 3 × 108 m/s.
Relative to what? The speed of sound in air is 330 m/s, in solids it is several
thousand m/s – relative to the medium in which it is traveling. What is the
medium for light? Usual 19th century answer: the aether. Term was coined
by Aristotle or his predecessors as name for mysterious medium filling all space
(ancients didn’t like concept of vacuum!).
speed of wave
perceived by plane
c-v
speed of wave
perceived by plane
v+c
speed of plane
relative to air
v
sound wave moving
toward plane from
back, speed c with
respect to air
sound wave moving
toward plane from
front, speed c with
respect to air
Figure 1: The sound wave moving toward the plane is moving faster relative to
the plane than the sound wave coming from behind
Properties of aether according to 19th century physicists:
• stiff (c is a big number, and media where waves propagate fast are stiff in
our experience–compare sound in air with sound in metals)
• nonviscous (aether doesn’t seem to slow planets down much as they whiz
through this stuff, so it must be pretty thin)
1
hmm... don’t know about too many substances with these two properties. But
ok, we said it was mysterious.
Further problem (Einstein): results of an experiment on electromagnetism
would seem to depend on c, which, since it is the speed of a wave through a
medium, is different in different reference frames moving with respect to that
medium. Example: see figure 1
If we are doing an electromagnetism experiment in the plane, should we
replace c in our equations with c + v? Result of our experiment would then
depend on the velocity of our laboratory. Note this is definitely not the case
for Newton’s laws of mechanics: In mechanics we insist on Galileo’s principle of
Frame 1
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Frame 2
v
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Figure 2: Measuring the period of the pendulum gives the same answer in either
frame 1 or 2
relativity: the laws of mechanics are the same in every reference frame moving
with a uniform velocity (see Fig. 2). You can’t tell if you are at rest or moving
uniformly. This bothered Einstein: if it is true for mechanics, why not electromagnetism? In the meantime, Albert Michelson and Edward Morley set out to
mirror
v = vel. of aether wind
u=resultant
velocity
c = speed of light
rel. to aether
L
a)
b)
Figure 3: a) mirror clock configuration in its frame perp. to Earth’s motion; b)
resultant velocity ⃗u due to aether wind. From the geometry, u2 = c2 − v 2 .
measure the Earth’s motion relative to the aether. If the Earth moves through
2
the aether with velocity ⃗v , on Earth it is as if an aether “wind” is blowing over
us with velocity −⃗v . Let’s say we want to send light back and forth between 2
mirrors perpendicular to ⃗v (Figure 3 a), and light travels with speed c in the
medium: it’s like trying to fly your plane in a crosswind (or row your boat
directly across a moving river). If the aether wind is blowing by the clock, you
have to send light moving at speed c relative to the aether out at an angle so
that the aether wind blows it back on course to reach the other mirror. The
resultant velocity ⃗u is shown in Figure 3 b.
The time to go back and forth (one
√ “tick” of clock) is now 2 distance traveled
by the light / speed = 2 L/u = 2L/ c2 − v 2 .
In the parallel configuration (Figure 4), the time to travel to the right is
L
a)
b)
mirror
u
outbound:
v
c
u = c+v
inbound
v
c
Figure 4: a) mirror clock configuration in its frame parallel to Earth’s motion;
b) resultant velocity ⃗u due to aether wind on both inbound and outbound trips.
Resultant velocity ⃗u is different for two different parts of trip.
L/(c − v), while the time to travel back to the left is L/(c + v). The total time is
therefore L/(c − v) + L/(c + v) = 2Lc/(c2 − v 2 ). Compare the two answers–they
are different. This is the point: if light is just a wave traveling in a medium
like any other wave, we expect that the times for a pulse of light to go back
and forth should be different in the two configurations. Michelson and Morley
found, after careful measurement, that the time was the same. One obvious
conclusion was that the speed u was in fact c in both directions, that
there was no aether, and that the whole analysis above is wrong in
the case of light.
Einstein question to teacher: what would it look like if I could run fast
enough to catch up to a light beam? Teacher: don’t worry, it won’t be on the
test! Einstein’s own answer (1905): you can’t catch up, because light moves at
c relative to you regardless of how fast you run!1
1 Einstein himself claimed he was not aware of Michelson-Morley expt. when he wrote his
theory, and Michelson and Morley did not accept the “obvious” possibility that the aether
did not exist.
3
Postulates of Einstein’s theory of relativity:
• laws of physics are the same in all frames of reference in uniform
motion (“relativity”)
• speed of light in vacuum has same value for all observers
Consequences of Einstein theory:
• Time dilation
Imagine observing a clock moving with respect to you with speed v. Einstein says you observe this clock running slower than yours. That is, a
tick ∆t in the clock’s rest frame is shorter than the tick ∆t′ you observe
when you are looking at the clock in the the moving frame. Now any clock
is equivalent (it turns out) to a ”light clock” consisting of light bouncing
back and forth between two mirrors. Suppose we position the clock perpendicular to the direction of motion, and consider its motion to the right
for one “tick”, consisting of the light going back and forth once:
mirror
L
v∆ t
Figure 5: a) mirror clock moving for one tick
√
The distance traveled by the light pulse is 2 L2 + √
(v∆t′ /2)2 , so a tick
′
lasts a time equal to the distance/speed, or ∆t = 2 L2 + (v∆t′ /2)2 /c.
Squaring both sides, we get
c2 (∆t′ )2
4
= v 2 (∆t′ )2 + 4L2
⇒ (∆t′ )2 (1 − v 2 /c2 )
⇒ ∆t′
= 4L2 /c2 = (∆t)2
∆t
= √
( )2 = γ∆t
1 − vc
where I used the fact that in the rest frame a tick is just ∆t = 2L/c, and
defined a new quantity
1
γ=√
.
1 − v 2 /c2
This factor, which will always be > 1 since v < c always, appears all
over the place in the theory of relativity. If the velocity v is anything
reasonable, i.e. much less than c, γ is approximately 1, and we can forget
about relativity. But if v is a significant fraction of c, γ gets big. In this
case we can see that ∆t′ = γ∆t, i.e. “moving clocks run slower”. What
this really means is, if you observe a clock in another frame moving at a
uniform velocity with respect to yours, you measure it to run slower than
the equivalent clock at rest in your frame.
• Length contraction
With a little more work, the transformation of spatial coordinates can also
be worked out, but I will spare you the derivation. The result is that, if
something measures ∆x in length at rest in your frame, its length will be
shorter if you measure it as it goes by at some velocity v. Note that the
length in your frame gets divided by γ in this equation. This contraction
of objects (or really, space itself) is called the Fitzgerald contraction.
∆x′ = ∆x/γ
• Simultaneity
Here the basic idea is that the constancy of the speed of light in any
uniformly moving reference frame forces us to reexamine the idea of simultaneity of events. Think of an event as something happening at a
particular point in space and in time, e.g. a light flash at (x, t). The information about other types of events will travel to our eyes at the speed
of light, so this is the same. We looked at an example with a light flash
going off in the middle of a spaceship. The observer in the spaceship sees
the two events when the light hits the front and back of the spaceship as
simultaneous. On the other hand, an observer seeing the spaceship fly by
sees the light moving with respect to him at the same speed, c. Therefore
by the time the light gets to the back of the spaceship, the ship has moved
forward a bit so the light needs more time to reach the front. The observer
in the spaceship and the observer on the ground do not agree that the two
events are simultaneous.
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• Velocity addition
According to the basic ideas of relativity à la Galileo, if I am standing on
a railroad car moving to the right with speed 0.6c, and I throw a baseball
to the right with speed 0.6c relative to the car, the ball travels at speed 1.2
c relative to the ground. No problem, except Einstein says it can’t be true:
nothing can go faster than light. There must be a new way in Einsteinian
relativity to add velocities. When you work through the algebra, you get
the following result for the above example:
V =
v1 + v2
(
),
1 + v1c2v2
where V is the total velocity of the ball relative to the ground, v1 is the
velocity of the ball relative to the car, and v2 is the velocity of the car
relative to the ground, for example.2 If you plug in the numbers, you will
see again that the denominator changes the answer from what you expect
when the velocities v1 and v2 get to be an appreciable fraction of c. And
you can check that the formula never allows V to be larger than c. For
the numbers above, the ball moves relative to the ground at V = 0.88c.
• Relativistic momentum and energy
We saw in the film that problems with momentum conservation arise if
we try to stick to the Newtonian definition, p = mv in relativity. These
are fixed if we define the momentum of any object p = γm0 v. I attached
a “0” subscript to the mass to explicitly indicate that it is the mass of the
object as determined in its rest frame. Sometimes the mass in a moving
frame will be written as m = γm0 , so we say that the mass of a moving
object depends on its velocity. The kinetic energy changes in a similar way,
giving the famous Einstein formula
m0 c2
E = mc2 = m0 γc2 = √
.
1 − v 2 /c2
Note when v = 0, the energy is not zero! This is because Einstein found
that mass is equivalent to energy, or, in the old language, mass can actually
be converted to pure energy, e.g. in a nuclear reaction where you mass
the constituents before and after and find that mass has been lost. Where
did it go? It turned into energy usually in the form of radiation equal to
the missing mass times c2 .
The old, Newtonian result is recovered at low speeds by checking that an
approximation to the Einstein formula holds:
1
E ≈ m0 c2 + m0 v 2 ,
2
2 Note
the text has a misprint on p. 703, with the wrong sign in the denominator
6
which is the rest energy plus the usual kinetic energy. In the old mechanics
problems we did in the early part of the course, the mass is never lost, so
we could forget the constant first term. So as we found before, Newtonian
physics is found from the Einstein theory of relativity when velocities
are small compared to c. This is important to understand. In a sense
it is true that Einstein proved Newton “wrong”; in another, important
sense, Newton’s ideas were simply incorporated into Einsteins’. For what
Newton was interested in explaining, his ideas are essentially correct.
Problems:
Problem 1
In a laboratory experiment a muon is observed to travel d = 800 m before
disintegrating. A graduate student looks up the life time τ of a muon (τ =
2 × 10−6 sec) and concludes that its speed was
v=
800m
= 4 × 108 m/sec
2 × 10−6 sec
faster than light!!! Identify the error of the poor student, and find the actual
speed of the muon.
Solution: the number looked up in the table for τ , the lifetime of the muon,
is the average time the particle exists after being created, before decaying into
something else, in its own rest frame. As observed in the laboratory, the muon’s
“clock” runs slower, so it lives longer, as we saw in the film. Quantitatively, the
lifetime of the muon as viewed by the student is , τ ′ = τ γ, so the true velocity
as observed in lab frame is the distance L divided by time τ ′ :
v=
L
L
=
′
τ
γτ
We now need to solve for v. First divide both sides by c and write out the
definition of γ:
( )√
( v )2
v
L
=
1−
c
c
c
Square both sides, solve for v/c, and find
√
v = c/ 1 + τ 2 c2 /L2
When you plug the numbers in, this gives v = 2.4 × 108 m/s, less than c, of
course.
7
Problem 2
At rest a Rolls-Royce is found to be twice as long as a Cooper Mini. However,
as the Rolls-Royce is overtaking the Mini they are observed to have the same
length. The Mini is observed to be going at half the speed of light 2c (!) How
fast is the Rolls-Royce going for the observer (keep your answer as a fraction of
c)?
Solution: let’s first express the length of each car as seen by the observer at
rest in terms of the Rolls’ rest length LR and the Rolls’ velocity vR :
√
( v )2
R
′
LR = LR /γR = LR · 1 −
c
√
(
)2
LR
c/2
L′M = LM /γM =
· 1−
2
c
√
Note
1−
(
c/2
c
)2
=
√
3/2. Now the problem says the two cars are observed
to have equal lengths, so
√
L′R
= LR ·
1−
( v )2
√
( )2
so we need to solve 1 − vcR =
R
c
√
= LR
3
= L′M
4
√
3
4
to find vR =
√
13c/4
Problem 3
As the outlaws escape in their getaway car, which goes 43 c, the cop fires a
bullet from the pursuit car, which only goes 12 c. The muzzle velocity (speed
relative to gun) of the bullet is 13 c. Does the bullet reach its target
(a) according to Galileo’s rule of addition of velocities;
(b) according to Einstein’s rule of addition of velocities?
Solution: Galileo would say that the velocity of the bullet relative to the
ground is v1 + v2 = c/2 + c/3 = 5c/6. This is plenty to hit the getaway car.
Einstein, however, would say
V =
v1 + v2
5c/6
5c
=
v1 v2 =
1 + c2
1 + 1/6
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