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Phys 15a
Solutions
Fall ’14
Problem Set #6
1. First solution: In B’s frame the situation looks like:
A
C
4c/5
B
v
We want the relativistic subtraction of v from 4c/5 to equal the relativistic subtraction of 0
from v (which is simply v), because these two results equal the speed at which C sees A and B
approach him. So
4c
4c
v−0
5 −v
5 −v
=
=⇒
= v.
(1)
4c
v·0
v
1 − c2
1 − 45 vc
1 − 52
c
Letting β ≡ v/c, we have
−β
= β =⇒
1 − 45 β
4
5
4
4
− β = β − β 2 =⇒ 2β 2 − 5β + 2 = 0 =⇒ (2β − 1)(β − 2) = 0. (2)
5
5
Since the β = 2 root represents a speed larger than c, the solution we want is
β=
1
2
=⇒ v =
c
2
(3)
Second solution: In C’s frame the situation looks like:
A
C
v
B
v
From the velocity addition formula, B sees A approaching him at speed (v + v)/(1 + v 2 /c2 ).
(This setup is one of the two basic scenarios where the velocity-addition formula is relevant, the
other being a ball on a train on the ground.) But the given information tells us that this relative
speed is 4c/5. So
2v
4c
=
2
2
1 + v /c
5
=⇒ 5
v
v2
=2+2 2
c
c
=⇒ 2β 2 − 5β + 2 = 0,
(4)
as in the first solution. Since the speed of B with respect to C is the same as the speed of C
with respect to B, the v here is indeed the desired speed.
2. This problem is basically a derivation of the relativistic Doppler effect. It is one of the few cases
where we’re actually concerned with the time is takes light to reach someone’s eye. Let’s work
with a general v and then set v = 4c/5 at the end.
(a) In A’s frame, B’s clock runs slow, so he travels for a time γT (as measured by A) before he
sends out the signal when his clock reads T . He is therefore a distance v(γT ) away from A
(as measured by A) at this point. The photon then takes a time vγT /c to get back to A.
So the total time in A’s frame is (with β ≡ v/c)
√
T (1 + β)
1+β
γvT
= γT (1 + β) = √
=T
.
(5)
γT +
2
c
1−β
1−β
Plugging in v = 4c/5 gives a time of 3T on A’s clock.
Limits: If β ≈ 0, then the time is ≈ T , which is correct (there is no time dilation, and
the photon takes essentially no time to reach A). If β → 1 (that is, if v → c), then T goes
to ∞, which makes sense (the time-dilation factor γ is huge).
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(b) In B’s frame, when he sends out the signal, A is a distance vT away, because A is moving at
speed v. The photon must close this gap at a relative speed of c − v (because A is receding
from B at speed v), which takes a time vT /(c − v). (Yes, it is legal to simply subtract the
speeds, because they are both measured with respect to B.) The total time in B’s frame
is therefore T + vT /(c − v) = cT /(c − v). But B sees A’s clock run slow by a factor γ, so
the time on A’s clock is only
√
√
1 ( cT ) T 1 − β 2
1+β
=
=T
,
(6)
γ c−v
1−β
1−β
in agreement with the result in part (a).
3. First look at things in A’s frame. B must have proper length γL = (5/3)L, because this is
length contracted down to L in A’s frame.
Now look at things in B’s frame. B has length (5/3)L (just the proper length), and A is length
contracted down to L/γ = (3/5)L. Assuming the trains are facing eastward, then as measured
by B, the distance that A must travel between the moment when the east ends (the front ends)
coincide and the moment when the west ends (the back ends) coincide is the difference in the
lengths of the trains (in B’s frame). So the situation looks like:
(A's frame)
(B's frame)
4c/5
B
B
L
(west ends)
A
(east ends)
5L/3
(at rest)
3L/5
(at rest)
L
A
A
4c/5
(end)
(start)
A must travel a distance (5/3)L − (3/5)L = (16/15)L. It does this at speed 4c/5, so the time
in B’s frame is (16L/15)/(4c/5) = 4L/3c .
Note: The general form of this answer is
(
)
(
)
(
(
))
1
L
γL
1
γL
v2
γLv
γL −
=
1− 2 =
1− 1− 2
= 2 .
v
γ
v
γ
v
c
c
(7)
Alternatively: We can use the rear-clock-ahead effect, applied to clocks on train B. Consider
the situation in A’s frame. As B flies by, B’s rear (west) clock reads (γL)v/c2 more than B’s
front (east) clock; remember that it is the proper length that appears in this expression. So if
B’s east clock reads 0 when the east ends match up, then B’s west clock reads γLv/c2 at the
same instant (in A’s frame) when the west ends match up (because we are told that these events
are simultaneous in A’s frame).
Now consider things in B’s frame. The east ends match up when B’s east clock reads 0, and the
west ends match up when B’s west clock reads γLv/c2 (these are frame-independent statements).
So the desired time difference is γLv/c2 .
Alternatively again: We can use the rear-clock-ahead effect, applied to clocks on train A.
Consider the situation in A’s frame. From the given information, clocks at the ends of A read
the same time when the pairs of ends match up. Call this reading 0.
Now consider things in B’s frame. When the east ends match up, B sees A’s east clock read
0 (a frame independent statement) and also the west clock (which is the front clock, as far as
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the motion of A is concerned) read −Lv/c2 . The west ends finally meet when this clock reads
0 (again a frame independent statement). Due to time dilation, it takes a time of γ(Lv/c2 ) for
this clock to advance to 0 in B’s frame. So B has to wait γLv/c2 between the ends matching
up.
4. (a) In the ground/tunnel frame the situation looks like this (the rectangle is the tunnel, the
line is the train, and dot is the person):
v
v
tunnel
(start)
3L/5
L
3L/5
(end)
The train has length L/γ = 4L/5 in the ground frame. The distance it travels between
the moment when its front end coincides with the near end of the tunnel and the moment
when its back end coincides with the far end of the tunnel equals the sum of the lengths of
the train and the tunnel, which is 4L/5 + L = 9L/5. It covers this distance at speed 3L/5,
so the time in the ground frame is (9L/5)/(3c/5) = 3L/c .
(b) In the ground frame, the person moves a distance L during this time, so her speed with
respect to the ground is L/(3L/c) = c/3 .
√
(c) A ground observer sees the person’s watch run slow by a factor γ1/3 = 3/(2 2), so the
√
time that elapses on the person’s watch is (3L/c)/γ1/3 = 2 2 L/c .
Remarks: You can also solve this problem by working in the person’s frame. In this frame, the
tunnel and the train must move with the same speed (in opposite directions), because they both
move the same distance in the same time (they basically just switch positions as the person stands
at rest). If they each move with speed v with respect to the person, then the relativistic addition
of v with itself must yield the given speed of 3c/5, because this is the relative speed of the train
and tunnel. It then quickly follows that v = c/3 (the same type of calculation as in the second
solution to the first problem above), and you can take it from there.
As a double check, the time and space intervals in the ground frame are ∆tg = 3L/c and ∆xg = L
(because the events happen at the ends of the train). And in the person’s frame they are ∆tp =
√
2 2L/c and ∆xp = 0 (because the events happen at the person, who isn’t moving, by definition).
The invariant interval c2 (∆t)2 − (∆x)2 is indeed invariant, being equal to 8L2 in both frames. You
can also check that the Lorentz transformations work out.
5. (a) Since γ4/5 = 5/3, the situation in the ground frame looks like (the trees are at rest):
4c/5
(start)
3L/5
2L
(end)
The stick must travel a distance 2L − 3L/5 = 7L/5 at speed 4c/5, so the time separation
between the events in the ground frame is ∆tg = (7L/5)/(4c/5) = 7L/4c . And for part
(b) we’ll need ∆xg . This is 2L because the events happen at the trees. (Note that ∆xg is
not equal to 7L/5.)
(b) Using ∆xg = 2L and ∆tg = 7L/4c, the Lorentz transformations give ∆xp and ∆tp in the
person’s frame as (the sign here is a “−” because the person sees the ground moving to the
left):
(
)
(
) 5
4c 7L
2L −
·
= L
∆xp = γ ∆xg − v∆tg =
3
5 4c
3
∆tp
(
) 5
= γ ∆tg − (v/c2 )∆xg =
3
(
)
7L 4c/5
L
− 2 · 2L =
4c
c
4c
(8)
(c) In the person’s frame, the distance between the trees is 2L/γ = 2L/(5/3) = 6L/5. So in
the person’s frame the situation looks like (the stick is at rest):
4c/5
4c/5
L
(start)
6L/5
L
(end)
The right tree must travel a distance 6L/5 − L = L/5 at speed 4c/5, so the time separation
between the events in the person’s frame is ∆tp = (L/5)/(4c/5) = L/4c , in agreement
with part (b). And the spatial separation between the events is ∆xp = L (also in agreement with part (b)), because the events happen at the ends of the stick, which has length
L in the person’s frame. (Note that ∆xp is not equal to L/5.)
(d) We have
c2 ∆t2g − ∆x2g
= c2 (7L/4c)2 − (2L)2 = −15L2 /16,
c2 ∆t2p − ∆x2p
= c2 (L/4c)2 − (L)2 = −15L2 /16.
(9)
These are equal, as desired.
Remark: Note that the invariant interval is negative. If the stick were instead moving slowly,
then the time would be large, so the invariant interval would be positive. Therefore, by continuity,
there must be some intermediate speed for which the invariant interval is zero (that is, for which
a photon can travel from one event to the other). As an exercise, you can show that this speed is
3c/5.
6. (a) In the train frame, the distance is simply ∆xt = L . And the time is ∆tt = L/(c/3) =
3L/c .
c/3
(train frame)
L
(b)
i. The velocity of the ball with respect to the ground is
Vg =
c
3
+ 4c
17c
5
1 4 = 19 .
1+ 3 · 5
(10)
4c/5
L/γ4/5
Vg
(ground frame)
(start)
(end)
The length of the train in the ground frame is L/γ4/5 = 3L/5. At time t the position
of the ball is (17c/19)t, and the position of the front of the train is 3L/5 + (4c/5)t.
These two positions are equal when
4c
3L
17c
t− t=
19
5
5
4
=⇒ ∆tg =
3L
5
17c
19
−
4c
5
=
19L
3c
(11)
In short, the ball closes the initial head start of 3L/5 that the front of the train had,
at a relative speed of 17c/19 − 4c/5, as viewed from the ground.
The distance the ball travels is ∆xg = Vg t = (17c/19)(19L/3c) = 17L/3 .
ii. From part (a), the space and time intervals in the train frame are ∆xt = L and ∆tt =
3L/c. The γ factor between the frames is γ4/5 = 5/3, so the Lorentz transformations
give the coordinates in the ground frame as (the sign here is a “+” because the ground
sees the train moving to the right):
∆xg
∆tg
(
( ))
4c 3L
17L
L+
=
,
5
c
3
(
4c )
L
5 3L
19L
= γ(∆tt + v∆xt /c2 ) =
+ 52
=
,
3
c
c
3c
5
= γ(∆xt + v∆tt ) =
3
(12)
in agreement with the above results.
√
(c) In the ball frame, the train has length L/γ1/3 = 2 2L/3. Therefore, the time it takes the
√
√
train to fly past the ball at speed c/3 is ∆tb = (2 2L/3)/(c/3) = 2 2L/c . And the
distance is ∆xb = 0 , of course, because the ball doesn’t move in the ball frame.
c/3
c/3
L/γ1/3
(end)
(ball frame)
(start)
(d) The values of c2 (∆t)2 − (∆x)2 in the three frames are:
Train frame:
c2 ∆t2t − ∆x2t = c2 (3L/c)2 − L2 = 8L2 .
Ground frame:
c2 ∆t2g − ∆x2g = c2 (19L/3c)2 − (17L/3)2 = 8L2 .
√
c2 ∆t2b − ∆x2b = c2 (2 2L/c)2 − (0)2 = 8L2 .
Ball frame:
These are all equal, as they should be.
Remark: Note that because of the “0” in the ball frame, the invariance of c2 (∆t)2 −(∆x)2 implies
that ∆t takes on the smallest possible value in the ball frame; the ∆t is larger in any other frame.
This is consistent with time dilation, because (as we’ll see below) the time in any other frame is
obtained by multiplying the time in the ball frame by γ, which is larger than 1.
(e) The relative speed
of the ball frame and the ground frame is 17c/19. Therefore, since
√
γ17/19 = 19/6 2, the times are indeed related by
( √ )
19L
19
2 2L
∆tg = γ∆tb
⇐⇒
= √
,
which is true.
(13)
3c
c
6 2
(f) The relative speed of the ball frame and the train frame is c/3. Therefore, since γ1/3 =
√
3/2 2, the times are indeed related by
( √ )
3
2 2L
3L
= √
,
which is true.
(14)
∆tt = γ∆tb
⇐⇒
c
c
2 2
(g) The relative speed of the train frame and the ground frame is 4c/5. Therefore, since
γ4/5 = 5/3, we see that the times are not related by a simple time-dilation factor, because
( )
5 3L
19L
̸=
.
(15)
∆tg ̸= γ∆tt
⇐⇒
3c
3
c
5
We don’t obtain an equality here because time dilation is legal to use only if the two events
happen(at the same place)in one of the frames. Mathematically, the Lorentz transformation
∆t = γ ∆t′ + (v/c2 )∆x′ leads to ∆t = γ∆t′ only if ∆x′ = 0. In this problem, the “ball
leaving back” and “ball hitting front” events happen at the same place in the ball frame,
but in neither the train frame nor the ground frame. Equivalently, neither the train frame
nor the ground frame is any more special than the other, as far as these two events are
concerned. So if someone insisted on trying to use time dilation, he would have a hard time
deciding which side of the equation the γ should go on. When used properly, the γ goes on
the side of the equation associated with the frame in which the two events happen at the
same place. This frame is “special” (at least with regard to the two events).
(h) If the ball leaves the back of the train when a clock there says zero, then we know from
part (a) that it hits the front when a clock there says 3L/c. (These are frame-independent
statements; everyone agrees that the ball leaves the back when a clock there says zero, and
it hits the front when a clock there says 3L/c.) When viewed from the ground, we must
incorporate the rear-clock-ahead effect, which tells us that the front clock starts the process
not at zero, but rather at −Lv/c2 = −4L/5c.
(ground frame)
4c/5
4c/5
___
19L
5c
__
- 4L
5c
0
(start)
3L
__
c
(end)
If we look at this one clock then we see that it advances by 3L/c − (−4L/5c) = 19L/5c.
(We could just as well look at the rear clock; it reads 3L/c + Lv/c2 = 19L/5c at the end.)
Applying time dilation to this one clock correctly gives
(
)
19L
5 19L
∆tg = γ∆tone clock on train
⇐⇒
=
,
which is true.
(16)
3c
3
5c
The error in the application of time dilation in part (g) is that we tried to apply it by
taking the difference in readings of two different clocks at the front and back of the train.
But all that time dilation says is, “If you look at a single clock flying by, you see it run
slow.”
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