Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

On Einstein original paper "On the electrodynamics of moving bodies", on section 2 of the first part (Kinematics), the following thought experiment is described: a rod is imparted constant speed $v$ in the direction of the $xx$ axis (growing $x$). At both ends of the rod, A and B, there is a clock, both clocks are synchronous in the stationary system. Then, a ray of light is sent from A at time $t_A$, which is reflected at B at time $t_B$, and arrives at A at time $t'_A$. It is then stated that "taking into consideration the principle of the constancy of the velocity of light", we get: $$ t_B - t_A = \frac{r_{AB}}{c - v} \text{and } t'_A - t_B = \frac{r_{AB}}{c + v} $$ where $r_{AB}$ is the length of the moving rod, measured in the stationary system. Now, if besides the rod's length, both time intervals refer to time in the stationary system, so must speed. Given the previous assumption of its constancy, why are terms like $c +v$ and $c -v$ in the above equations? Furthermore, from my previous knowledge of special relativity, I would have thought the observer moving with (moving) rod would declare the clocks to be synchronous, while the stationary observer would declare the clocks to be not synchronous. Yet the article states exactly the opposite. Why is this so? I know I'm missing something here, I just can't figure out exactly what it is.

Thanks in advance for your help.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

At time $t_A$, the ray of light has to travel distance $r_{AB}$ to get to the other end. Since the other end is travelling at a velocity $v$, the ray of light has to travel an additional distance $v(t_B - t_A)$ giving a total distance $r_{AB} + v(t_B - t_A)$. Since the ray of light travels at velocity $c$, this distance must also equal $c(t_B - t_A)$ $$r_{AB} + v(t_B - t_A) = c(t_B - t_A)$$ Which on rearranging gives one of the expressions in the question, and similar arguments gives the other.

Usually, the clocks in the same frame are configured to be synchronous with one another. However, Einstein is showing that clocks that are synchronous with one another in one frame, aren't so in another. This is near the start of the paper where he hasn't derived the Lorentz transformations yet, but is concentrating on the physics. So at this stage, he's defined the moving clocks to be synchronous with those in the stationary frame, meaning they show the same time at the same location. For two clocks a distance apart, he's defined them to be synchronous with one another if $t_B - t_A = t'_A - t_B$ using the transmission and relflection of light as described in the question. And since in this case $t_B - t_A\neq t'_A - t_B$, moving clocks A and B are not synchronous with one another in the moving frame if synchronous in the stationary frame.

share|improve this answer
    
Thanks, that was most helpful. However, I still have a small itch. The calculation you show is how an observer in the stationary platform would proceed (i.e. taking into account the motion of the rod). Furthermore, Einstein states that $r_{AB}$ is "the length of the moving rod---measured in the stationary system". This would indicate that from the perspective of the stationary observer, the clocks at both ends of the rod (thus moving with it) would not be synchronous. Yet the article states this conclusion is drawn by observers moving with the rod. Where am I going astray? –  wmnorth Sep 18 '11 at 11:12
    
The observers in the moving frame are applying the synchronisation criterion on their clocks as a check. But their clocks at each location in space is set to be the same as a clock in the stationary frame at the same position. From their point of view, a clock in the stationary frame is moving, of course, but that doesn't prevent them from reading off the time, and setting their stationary clock to match it. Their clocks therefore synchronize with those of the stationary frame. –  John McVirgo Sep 18 '11 at 15:03
    
cont. The calculation is for the time shown on a clock in the stationary frame at the events of transmission and reflection of light performed by the observer in the moving frame and therefore what their clock reads. Remember the observer in the moving frame has used this time to synchronise his clocks to as defined by Einstein when he says in the passage "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". –  John McVirgo Sep 18 '11 at 15:12
add comment

When setting up this description, the author states:

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.”

It is a part of the initial conditions that the clocks are synchronous in the stationary frame. The asynchroneity of the clocks in the rod frame is thus assured by the principle of relativity.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.