# Time Dilation: A Worked Example

*Michael Fowler, UVa Physics.*

### “Moving Clocks Run Slow” *plus* “Moving Clocks Lose
Synchronization” *plus* “Length Contraction” leads to
consistency!

The object of this exercise is to show explicitly how it is possible for two
observers in inertial frames moving relative to each other at a relativistic
speed to each see the other’s clocks as running slow and as being
unsynchronized, and yet if they both look at the same clock at the same time
from the same place (which may be far from the clock), they will *agree*
on what time it shows!

Suppose that in Jack’s frame we have two synchronized clocks *C*_{1}
and *C*_{2} set 18 x 10^{8} meters apart (that’s
about a million miles, or 6 light-seconds). Jill’s spaceship,
carrying a clock *C'*,* * is traveling at 0.6*c*, that is
1.8 x 10^{8} meters per second, parallel to the line *C*_{1}*C*_{2},
passing close by each clock.

Suppose *C' * is synchronized with *C*_{1} as they
pass, so both read zero.

As measured by Jack the spaceship will take just 10 seconds to reach *C*_{2},
since the distance is 6 light seconds, and the ship is traveling at 0.6*c*.

What does clock *C'* (the clock on the ship) read as it passes *C _{2}*?

The time dilation factor

so *C**'*, Jill’s clock, will read 8 seconds.

Thus if both Jack and Jill are at *C _{2}*
as Jill and her clock

*C'*pass

*C*, both will agree that the clocks look like:

_{2}*How, then, can Jill claim that Jack’s clocks C*_{1},*
C*_{2}* are the ones that are running slow?*

To Jill, *C _{1}*,

*C*

_{2}*are*running slow, but remember they are

*not synchronized*. To Jill,

*C*is

_{1}*behind*

*C*by

_{2}_{}

Therefore, Jill will conclude that since *C*_{2} reads 10
seconds as she passes it, at that instant *C*_{1} must be
registering 6.4 seconds. Jill’s own clock reads 8 seconds at that
instant, *so she concludes that C*_{1}* is running slow by the
appropriate time dilation factor of 4/5*. This is how the change in
synchronization makes it possible for both Jack and Jill to see the
other’s clocks as running slow.

Of course, Jill’s assertion that as she passes Jack’s second
“ground” clock *C*_{2} the first “ground”
clock *C*_{1} must be registering 6.4 seconds is not completely
trivial to check! After all, that clock is now a million miles away!

Let us imagine, though, that both observers are equipped with Hubble-style telescopes attached to fast acting cameras, so reading a clock a million miles away is no trick.

To settle the argument, the two of them agree that as she passes the second
clock, Jack will be stationed at the second clock, and at the instant of her
passing they will both take telephoto digital snapshots of the faraway clock *C*_{1},
to see what time it reads.

Jack, of course, knows that *C*_{1} is 6 light seconds away,
and is synchronized with *C*_{2} which at that instant is reading
10 seconds, so his snapshot must show *C*_{1} to read 4 seconds. That
is, looking at *C*_{1} he sees it as it was six seconds ago.

What does Jill’s digital snapshot show? It
must be identical—two snapshots taken from the same place at the same
time must show the same thing! So, Jill* must also* gets a picture
of *C*_{1} reading 4 seconds.

*How can she reconcile a picture of the clock reading 4 seconds with
her assertion that at the instant she took the photograph the clock was
registering 6.4 seconds? *

The answer is that she can if she knows her relativity!

*First point: length contraction*. To Jill, the clock *C*_{1}
is actually only 4/5 x 18 x 10^{8} meters away (she sees the distance *C*_{1}*C*_{2}
to be Lorentz contracted!).

*Second point: The light didn’t even have to go that far!* In
her frame, the clock *C*_{1} is *moving away*, so the light
arriving when she’s at *C*_{2} must have left *C*_{1}
when it was closer—at distance *x *in the figure below.* *The
figure shows the light in her frame moving from the clock towards her at speed *c*,
while at the same time the clock itself is moving to the left at 0.6*c*.

It might be helpful to imagine yourself in her frame of reference, so you
are at rest, and to think of clocks *C*_{1} and *C*_{2}
as being at the front end and back end respectively of a train that is going
past you at speed 0.6*c*. Then, at the moment the back of the train
passes you, you take a picture (through your telescope, of course) of the clock
at the front of the train. Obviously, the light from the front clock that
enters your camera at that instant left the front clock some time ago. During
the time that light traveled towards you at speed *c*, the front of the
train itself was going in the opposite direction at speed 0.6*c*. But you
know the length of the train in your frame is 4/5 x 18 x 10^{8} meters,
so since at the instant you take the picture the back of the train is passing
you, the front of the train must be 4/5 x 18 x 10^{8} meters away. Now
that distance, 4/5 x 18 x 10^{8}, is the sum of the distance the light
entering your camera traveled plus the distance the train traveled in the same
time, that is, (1 + 0.6)/1 times the distance the light traveled.

So the image of the first ground clock she sees and records as she passes
the second ground clock must have been emitted when the first clock was a
distance* x* from her in her frame, where

Having established that the clock image she is
seeing as she takes the photograph left the clock when it was only 9 x 10^{8}
meters away, that is, 3 light seconds, she concludes that she is observing the
first ground clock as it was three seconds ago.

*Third point: time dilation*. The story so far: she has a photograph of
the first ground clock that shows it to be reading 4 seconds. She knows that
the light took three seconds to reach her. So, what can she conclude the
clock must actually be registering at the instant the photo was taken? If
you are tempted to say 7 seconds, you have forgotten that in her frame, the
clock is moving at 0.6*c* and hence *runs slow* by a factor 4/5.

Including the time dilation factor correctly, she concludes that in the 3 seconds that the light from the clock took to reach her, the clock itself will have ticked away 3 × 4/5 seconds, or 2.4 seconds.

Therefore, since the photograph shows the clock to read 4 seconds, and she finds the clock must have run a further 2.4 seconds, she deduces that at the instant she took the photograph the clock must actually have been registering 6.4 seconds, which is what she had claimed all along!

The key point of this lecture is that at first it seems impossible for two observers moving relative to each other to both maintain that the other one’s clocks run slow. However, by bringing in the other necessary consequences of the theory of relativity, the Lorentz contraction of lengths, and that clocks synchronized in one frame are out of synchronization in another by a precise amount that follows necessarily from the constancy of the speed of light, the whole picture becomes completely consistent!