# Synchronization of clocks and special relativity

As shown in the diagram, a long rod has ends A and B, a spaceship is at A. A, B and the spaceship all have clocks attached to them. The clocks are then synchronized to time $$t_0$$, and immediately The spaceship flies with velocity $$u$$ relative to the rod from A to B. Suppose relative to the rod, the spaceship takes $$t_1$$ to get from A to B. When the spaceship arrives at B, The time of the clock on B would now be $$t_0+t_1$$, and the clock on the spaceship would now be $$t_0+\frac{t_1}{\gamma}$$ Which means the time shown on the spaceship clock is less than the time shown on the rod clock at B.

But if we use the spaceship as a reference frame, with the rod moving with velocity $$u$$ to the left, Suppose relative to the spaceship, B end of the rod takes $$t_2$$ to reach the spaceship. The spaceship clock would show $$t_0+t_2,$$ while the rod clocks would show $$t_0+\frac{t_2}{\gamma}$$ Which means the time shown on the spaceship clock is more than the time shown on the rod clock at B.

When spaceship meets B, the first scenario shows the rod clock showing larger time value than the spaceship clock, while the the second scenario shows the opposite. Both cannot be true, because one must occur when B and spaceship meet. What is wrong?

This looks very much like the twin paradox, but the twin paradox can be solved by saying that the twin cannot turn around and go back to find the other twin, because turning around would involve acceleration and special relativity cannot be used. How can this paradox I proposed here be solved? thanks！

• This is tangential to your main question and I therefore didn't include it in my answer, but note that the idea that acceleration cannot be handled in special relativity is a bizarrely common misconception. One need only compute the proper time along the accelerating twin's worldline to see that it is shorter than the proper time along the stationary twin's worldline, and thus the former ages less than the latter. No need for hand-waving - you can calculate it precisely with very little work. Dec 6, 2021 at 13:27
• Do you mean that in the twin paradox, the accelerating twin ages less than the stationary twin? Dec 6, 2021 at 13:45
• $\uparrow$ Yes. Dec 6, 2021 at 13:58
• Here's an anim from Wikipedia which illustrates the relativity of simultaneity. Dec 6, 2021 at 23:28

The clocks are then synchronized to time $$t_0$$ [...]

Which clocks, and in which frame? You must always remember that synchronization of clocks which are sitting at different points in space is a frame-dependent property. I assume that you mean that clocks $$A$$ and $$B$$ are synchronized in the rest frame at $$A$$ (or $$B$$), and that the clock on the spaceship is synchronized with clock $$A$$ as it passes by. However, note that in this case the clock at $$B$$ will not be synchronized with the others in the rest frame of the ship.

When spaceship meets B, the first scenario shows the rod clock showing larger time value than the spaceship clock, while the the second scenario shows the opposite. Both cannot be true, because one must occur when B and spaceship meet.

The $$t_2$$ to which you refer in your first scenario is different from the $$t_2$$ in your second scenario. In the first case, according to the clock at $$B$$, the ship arrives at $$B$$ at time $$t_2 = L/u$$, where $$L$$ is the distance between $$A$$ and $$B$$ in the rest frame of the observer at $$B$$. The ship's clock reads $$T = t_2/\gamma = L/u\gamma$$, where $$\gamma = (1-u^2/c^2)^{-1/2}$$.

In the second scenario, an observer on the ship sees the rod length contracted, with apparent length $$L' = L/\gamma$$. Accordingly, point $$B$$ reaches the ship when the ship's clock reads $$T=L'/u = L/u\gamma$$, in agreement with the previous calculation.

Furthermore, in the second scenario the clock at $$B$$ was never synchronized with the clock on the ship. Let $$(x',t')$$ be the coordinates in the ship's frame and $$(x,t)$$ be the coordinates in the Earth frame centered at clock $$A$$. The event $$(x,t)=(x',t')=(0,0)$$ corresponds to the ship passing clock $$A$$. According to the Lorentz transform equations, $$t' = \gamma\left(t-\frac{ux}{c^2}\right)$$ When the ship passes clock $$A$$, then the ship's clock reads $$t'=0$$, and clock $$A$$ (which sits at $$x=0$$) also reads $$t=0$$. However, clock $$B$$ is sitting at $$x=L$$, which means that $$t - \frac{uL}{c^2} = 0 \implies t = \frac{uL}{c^2}$$ In other words, clock $$B$$ starts with an offset of $$uL/c^2$$ according to the observer on the ship.

With that in mind, the relativist riding on the ship would recognize that when she reaches clock $$B$$, then it would read

$$t_2 =T/\gamma + \frac{uL}{c^2}=\frac{L}{u\gamma^2} + \frac{uL}{c^2} = \frac{L}{u}\left(\frac{1}{\gamma^2} +\frac{u^2}{c^2}\right) = \frac{L}{u}$$ where the first term is the reading on the ship's clock divided by $$\gamma$$ (due to the time dilation) and the second term is the initial offset on clock $$B$$. This is fully in agreement with our previous calculations, as expected.

"However, note that in this case the clock at B will not be synchronized with the clock on the ship." I'm afraid I don't understand this; since A is synchronized with B, and the ship clock is synchronized with A, then why aren't the three clocks synchronized?

This is a fairly subtle point which deserves some elaboration. A reference frame can be understood as an imaginary lattice of measuring sticks with a clock sitting at each point. Events are labeled by the position $$x$$ where they occur and a time $$t$$ as read on the clock which sits at $$x$$. This is crucial - the time of an event is local.

In principle, the clocks sitting at each grid point have nothing to do with each other. To make a sensible reference frame, we must synchronize them in some way, e.g. via Einstein synchronization. However, we may now imagine that a second observer moving relative to us has their own reference frame. The key insight of special relativity is that if both we and the moving observer sychronize our respective system of clocks, then each will observe the other's clocks to be out of sync.

The specific relation between our frames is given by the Lorentz transformation equations. Let my coordinates be $$(x,t)$$ and the moving observer's coordinates be $$(x',t')$$; then $$x' = \gamma\big(x-ct\big)$$ $$t' = \gamma\big(t - \frac{vx}{c^2}\big)$$

Consider two events - the first occurs at $$x=0$$ and $$t=0$$, while the second occurs at $$x=L$$ and $$t=0$$. That is, the coordinates I assign to the first event are $$(0,0$$), and the coordinates I assign to the second event are $$(L,0)$$. The coordinates which the moving observer assigns to the first event are also $$(0,0)$$, but the coordinates which they assign to the second event are $$x' = \gamma L$$ $$t' = -\gamma \frac{vL}{c^2}$$

So while I would say that the two events are simultaneous (both occurring at $$t=0$$), the moving observer will say that they are not (one occurs at $$t'=0$$, and the other at $$t' = -\gamma vL/c^2$$).

In keeping with your original question, let the first event at $$(0,0)$$ be "the ship passes the clock at $$A$$ when clock $$A$$ reads 0," while the second event is "clock $$B$$ reads 0." In my frame (the rest frame at clock $$A$$), these two events are simultaneous; because clocks $$A$$ and $$B$$ read zero at the same time, I say they are synchronized. However, according to the moving observer, the event "clock $$B$$ reads 0" occurs at $$t'=-\gamma vL/c^2\neq 0$$, which means that the clock at $$B$$ is not in sync with the ship's clock in their frame.

• "However, note that in this case the clock at B will not be synchronized with the clock on the ship." I'm afraid I don't understand this; since A is synchronized with B, and the ship clock is synchronized with A, then why aren't the three clocks synchronized? Dec 6, 2021 at 13:48
• @LeeLaindingold For clocks at different locations, synchronization is frame-dependent. This is perhaps the key insight of special relativity. Two clocks which are synchronized in one frame are generally not synchronized in another, with the only exception being if they are sitting at the same point. In this case, when the ship passes point $A$, then the ship clock and clock $A$ can be unambiguously synchronized; however, whether these two clocks are synchronized with clock $B$ depends on your frame (i.e. they are synchronized in the rest frame of $A$, but not in the ship frame). Dec 6, 2021 at 13:57
• @LeeLaindingold I have added a substantial elaboration to the end of my answer. However, there are also many very good answers regarding the relativity of simultaneity and synchronization on PSE which flesh this issue out more, e.g. this one. Dec 6, 2021 at 14:21
• J. Murray: "[...] The key insight of special relativity is that if both we and the moving observer sychronize our respective system of clocks, then each will observe the other's clocks to be out of sync." -- Your statement uses the notion "sync" in contrast to "synchronize[d] (in the sense of Einstein's definition, i.e. referring to clocks at rest wrt. each other)". It would be less confusing to say that: "If clocks were synchronized (as defined, with all others of their respective system) then the readings of any two clocks that met each other in passing were (generally) different." Dec 7, 2021 at 6:54

What is wrong?

Almost always in special relativity when there is a problem it is from forgetting about the relativity of simultaneity.

In this case, in the ship's frame, when the ship passes next to clock A and sets its clock to 0 the clock B is not set to 0. Instead (in natural units) the time in the ship’s frame when B reads 0 is:

$$t_B=-\frac{v \ x_B}{\sqrt{1-v^2}}$$

If you combine this offset with the time dilation then you get the correct conclusion that both frames agree. The time shown on the spaceship clock is less than the time shown on the rod clock at B according to both frames.

• Shouldn't that be + vxB / sqrt(1-v²) instead? From the spaceship's point of view, the clock of B is running ahead of A. Dec 7, 2021 at 16:29
• @fishinear oops, you are right. The text and the math don’t match. I have updated the text to correctly state what was calculated
– Dale
Dec 7, 2021 at 17:42

Let's suppose the experiment starts at 12 O'clock according to the three clocks, and that the relative speeds are such that when the spaceship reaches the other end of the rod, the clock at B on the rod shows 12:11, while the clock on the spaceship shows 12:10.

So according to the clocks on the rod, 11 minutes have passed during the time it has taken for the spaceship to move between one end of the rod and the other, while the clock on the spaceship says that only 10 minutes have passed.

In the spaceship's frame, however, it was only 11:59 at the far end of the rod when the spaceship left A, and at that time in the spaceship's frame, the clock at B on the rod was showing 12:01. So from the point of view of the spaceship, the clock at B on the rod has taken 11 minutes to travel to meet the ship, while the clock on the rod at B has moved on only ten minutes.

You will see, therefore, that the appearance of time dilation is entirely reciprocal, and it arises in this case from the fact that the clock at B on the rod is out of synch with the local spaceship time.

More generally, time dilation arises because the planes of constant time in one reference frame are tilted compared to the planes of constant time in any other moving reference frame. This means that you can only synchronise the clocks between the two frames at a single point- the further you move away from that point, the more the clocks in the two frames become out of synch.

You can generate the effect of time dilation quite simply. Imagine walking at 1m/s down a corridor at which there are people at 10m intervals each holding a clock, and the clocks are set so that each clock down the corridor is 1s ahead of the clock before it. When you pass the first clock it is set at 12:00, as is your watch. After ten seconds you have walked 10m and reached the next clock. On your watch it says 12:00:10, but the clock says 12:00:11, because it is running a second ahead. You walk another 10m and meet the next clock. Now your watch says 12:00:20, while the clock you have just reached says 12:00:22, because it is set a second ahead of the previous one. After walking another 10m you meet another clock which says 12:00:33, while your watch now says 12:00:30.

You will see that your watch seems to be running slow when you compare its reading with the clocks you are passing, but in fact your watch and the clocks are running at exactly the same rate- the appearance of time dilation arises because the clocks are all set out of synch with each other.

Now, suppose that walking along behind you are 10m intervals are people wearing watches that have been set so that each is 1s ahead of the watch before it in the line. Imagine how these walking people with watches are viewed by the first of the stationary people holding clocks. At 12:00:00 you pass with 12:00:00 showing on your watch. Ten seconds later the next walker in line goes past with 12:00:11 showing on their watch. After another ten seconds another walker passes with 12:00:22 on their watch. Ten seconds later another walker goes past with 12:00:33 on their watch, while it is only 12:00:30 on the clock. The appearance of time dilation is thus entirely reciprocal- the stationary person sees that their clock is losing time compared with the watches that are passing, just as you find your watch to be losing time compared with the clocks you are passing. Both groups of people- those walking with the watches and those standing holding the clocks- think that their timepieces are losing time (ie time dilated) compared with the timepieces they are passing, even though all the timepieces are ticking at exactly the same rate. The effect of time dilation arises not because the clocks slow down when they move but because time in one frame is out of synch with time in the other.