- Summary
- In this section of his 1905 paper, Einstein sets out to demonstrate that clocks that are synchronized in one frame of reference are not synchronized in another frame.

I'm struggling to understand Section 2 of Einstein's 1905 paper, "The Electrodynamics of Moving Bodies." This is my source for the paper: http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

In section 1 Einstein states that by definition light travels at the same speed, c, in every direction. He then proposes a means of synchronizing clocks at a distance:

Let a ray of light start at the “A time” tA from A towards B, let it at the “B time” tB be reflected at B in the direction of A, and arrive again at A at the “A time” t'A. In accordance with definition the two clocks synchronize if tB − tA = t'A − tB.

In section 2 he introduces a "rigid rod" and places a clock at each end of the rod:

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be. These clocks are therefore “synchronous in the stationary system.”

He then places the rod in motion so as to demonstrate that the synchronization of the clocks depends on frame of reference:

We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in section 1 for the synchronization of two clocks. Let a ray of light depart from A at the time tA, let it be reflected at B at the time tB, and reach A again at the time t'A. Taking into consideration the principle of the constancy of the velocity of light we find that tB − tA = rAB/c − v and t'A − tB = rAB/c + v where rAB denotes the length of the moving rod—measured in the stationary system. Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.

First off I assume by "the two clocks" Einstein means the clocks placed at the two ends of the rod, A and B (not the clocks in the stationary system). Observers co-moving with the rod find the clocks out of sync. But if the clocks are placed at each end of the rod and therefore moving with the rod, and if the observers are also moving with the rod, then the observers will not think the rod or the clocks are moving. This is because, from your own point of view, you're never in motion. In fact this is Galileo's principle of relativity. So why would "observers moving with the moving rod" find that the two clocks are no longer synchronized? Perhaps the answer is that the two clocks are not actually attached to the rod. This way the observers co-moving with the rod approach the clock at B while receding from the clock at A. As the light ray travels from A to B, the velocity v of the observers makes it seem as if the light ray's speed is c - v. As the light ray travels back from B to A, the velocity of the observers makes it seem as if the light ray's speed is c + v. Of course, as we know from Lorentz, the speed of light is always measured at exactly c. Assuming Einstein is right and this is because the speed of light is a law of nature and therefore frame-invariant, the motion of the observers is causing them to dilate in time so that they see the clock at B displaying an earlier time than what a stationary observer would perceive.

At this point I think I've got it. But then I realize the time dilation of the observers will cause them to see exactly the same earlier time displayed on clock A. So the clocks are once again synchronized!

Can anyone explain what I'm missing here?

In section 1 Einstein states that by definition light travels at the same speed, c, in every direction. He then proposes a means of synchronizing clocks at a distance:

Let a ray of light start at the “A time” tA from A towards B, let it at the “B time” tB be reflected at B in the direction of A, and arrive again at A at the “A time” t'A. In accordance with definition the two clocks synchronize if tB − tA = t'A − tB.

In section 2 he introduces a "rigid rod" and places a clock at each end of the rod:

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be. These clocks are therefore “synchronous in the stationary system.”

He then places the rod in motion so as to demonstrate that the synchronization of the clocks depends on frame of reference:

We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in section 1 for the synchronization of two clocks. Let a ray of light depart from A at the time tA, let it be reflected at B at the time tB, and reach A again at the time t'A. Taking into consideration the principle of the constancy of the velocity of light we find that tB − tA = rAB/c − v and t'A − tB = rAB/c + v where rAB denotes the length of the moving rod—measured in the stationary system. Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.

First off I assume by "the two clocks" Einstein means the clocks placed at the two ends of the rod, A and B (not the clocks in the stationary system). Observers co-moving with the rod find the clocks out of sync. But if the clocks are placed at each end of the rod and therefore moving with the rod, and if the observers are also moving with the rod, then the observers will not think the rod or the clocks are moving. This is because, from your own point of view, you're never in motion. In fact this is Galileo's principle of relativity. So why would "observers moving with the moving rod" find that the two clocks are no longer synchronized? Perhaps the answer is that the two clocks are not actually attached to the rod. This way the observers co-moving with the rod approach the clock at B while receding from the clock at A. As the light ray travels from A to B, the velocity v of the observers makes it seem as if the light ray's speed is c - v. As the light ray travels back from B to A, the velocity of the observers makes it seem as if the light ray's speed is c + v. Of course, as we know from Lorentz, the speed of light is always measured at exactly c. Assuming Einstein is right and this is because the speed of light is a law of nature and therefore frame-invariant, the motion of the observers is causing them to dilate in time so that they see the clock at B displaying an earlier time than what a stationary observer would perceive.

At this point I think I've got it. But then I realize the time dilation of the observers will cause them to see exactly the same earlier time displayed on clock A. So the clocks are once again synchronized!

Can anyone explain what I'm missing here?