How two clocks can each run slow in the other's reference frame

[jtbell]

People often have trouble visualizing how time dilation can be "mutual" in special relativity. That is, if you have two clocks moving with respect to each other, in each clock's rest frame the other clock runs slower. The following example illustrates how this can be possible without contradictions.

Imagine a set of space stations, labeled A through E. They are at rest with respect to each other and spaced equally along a line, $\Delta x = \sqrt{3} = 1.732$ light-years (ly) apart. In their rest frame, their clocks are synchronized: they run at the same rate, and in step with each other.

Also imagine a set of space ships, labeled A through E. They are traveling along the line connecting the space stations. In the stations' rest frame the ships have the same spacing as the stations, and are traveling at a speed of $v = \sqrt{3}/2 = 0.866$ light-years per year (ly/yr). Note the speed of light is c = 1 ly/yr, exactly. The ships carry clocks that are constructed identically to the stations' clocks.

In the stations' rest frame, at time 0.00 according to the stations' clocks, each ship passes the corresponding station simultaneously. At this instant, set each of the ships' clocks to 0.00 also. We assume that the ships pass the stations closely enough that there is no problem with this procedure, e.g. from time delays in sending signals between them.

In the stations' rest frame, each ship takes $\Delta x / v = \sqrt{3} / (\sqrt{3} / 2) = 2.00$ yr to travel the distance from one station to the next, so 2.00 yr elapses on each station's clock as ship A travels from station A to station B, etc.

Meanwhile, because of time dilation, exactly half as much time, 1.00 yr, elapses on each ship's clock. You should be able to verify this from the given speed and the time-dilation equation.

The attached diagram shows a series of snapshots showing the positions of the ships and stations, and the the readings on their clocks (the "flags" attached to each object), at 2.00-yr intervals in the stations' rest frame.

In the next post I will start to consider what this looks like in the ships' rest frame.

 

[jtbell]

In the ships' rest frame, the stations are moving in the opposite direction, so the spacing between them is contracted by 1/2 compared to the stations' rest frame, to $\sqrt{3}/2 = 0.866$ ly. Similarly, the spacing between the ships is "un-contracted" by a factor of 2 compared to the stations' rest frame, to $2 \sqrt{3} = 3.464$ ly.

The total distance between stations A and E now equals the spacing between ships A and B. We can construct the attached diagram, a series of snapshots which shows stations A through E passing ship A, one after the other. I have left all the clock readings blank for the moment. In the next post I will start to fill them in.

 

[jtbell]

Referring back to the diagram for the stations' rest frame, we see that when ship A passes station A, their clocks both read 0.00. This must also be true in any other reference frame. Therefore, in the diagram for the ships' rest frame, we can fill in time 0.00 on both clocks when station A passes ship A, in the first snapshot.

Similarly, in any reference frame, when ship A and station B pass each other, ship A's clock must read 1.00 yr and station B's clock must read 2.00 yr. Therefore, we can fill in the times for the two clocks at each of the "passing events" in the diagram for the ships' rest frame, by looking them up on the diagram for the stations' rest frame. The attached diagram shows the result, with the remaining times still left blank.

 

[jtbell]

The first "snapshot" in the diagram attached to the preceding post shows that although the stations' clocks are synchronized in their rest frame, they are not synchronized in the ships' rest frame. Likewise, although the ships' clocks are synchronized in the stations' rest frame, they are not synchronized in the ships' rest frame.

This is an aspect of relativity of simultaneity: two events at different locations that are simultaneous in one frame are not simultaneous in any other frame. For example, in the ships' rest frame, ship A and station A pass each other simultaneously with ship B passing station E, but those two events are not simultaneous in the stations' rest frame.

Even though the ships' clocks are not synchronized in their rest frame, they still all run at the same rate. Also, each ship's clock is 3.00 yr out of sync with its neighbors. So we can easily fill in the rest of the ships' clock readings in the ships' rest frame.

Similarly, since the clocks on stations A and E are 6.00 yr our of of sync, neighboring stations' clocks should be 1.50 yr out of sync. So we can fill in the rest of the stations' clock readings. The attached diagram shows the final result.

As we go from one snapshot to the next, each (stationary) ship's clock advances by 1.00 yr, and each (moving) station's clock advances by 0.50 yr. Referring back to the diagram for the stations' rest frame, as we go from one snapshot to the next, each (stationary) station's clock advances by 2.00 yr, and each (moving) ship's clock advances by 1.00 yr. So, in one frame the ships' clocks run slower, but in the other frame the stations' clocks run slower. But these two phenomena nevertheless fit together because of the differences in synchronization of the two sets of clocks in the two frames.

In general, to get a consistent description of situations like this, you need to take into account all three of the following relativistic phenomena:

• Length contraction
• Time dilation
• Relativity of simultaneity

 

[granpa]

to prove that each sees the other as time dilated you have assumed that each sees the other as length contracted.

 

[Doc Al]

to prove that each sees the other as time dilated you have assumed that each sees the other as length contracted.

He's not proving time dilation, he's illustrating the symmetry of time dilation: how both frames can observe the other's clock running slow.

 

[jtbell]

Yes, the point is to show that if you start with just length contraction and time dilation, you need to add relativity of simultaneity (or differing synchronization of clocks in different reference frames) in order to make things come out consistently and symmetrically in both frames.

Oops, I was going to mention relativity of simultaneity in connection with clock synchronization, but I left it out. I'll add it to the fourth post.

 

[jtbell]

I'm going to do one more step to make this example a bit tidier. So far, the stations' clocks have been synchronized in the stations' rest frame, but the ships' clocks are not synchronized in the ships' rest frame. From the point of view of people on the ships, it would be nicer if their clocks were synchronized in that frame.

Therefore, let's set ship B's clock back by 3.00 years, ship C's clock back by 6.00 years, etc. This makes all the ships' clocks read 0.00, in the first snapshot of the ships' rest frame; 1.00 yr in the second snapshot, etc.

Now, when station E's clock reads 6.00 yr and ship B's clock reads 0.00 yr when they pass. To make this true in the stations' rest frame, we have to make the same adjustments on the ships' clocks there, also.

The attached diagrams show the final results for both frames.

 

[Dale]

I find it easier to see in this https://www.physicsforums.com/showpost.php?p=1620834&postcount=13" of a Lorentz transform.

If you look closely you can see that the line x'=0 crosses the line t=2 before it crosses the line t'=2, so the white clocks tick slower than the black clocks in the unprimed frame. Similarly you can see that the line x=0 crosses the line t'=2 before it crosses the line t=2, so the black clocks tick slower than the white clocks in the primed frame.