0
$\begingroup$

Exact Statement of Question: Two clocks are positioned at the ends of a train of length L (as measured in its own frame). They are synchronized in the train frame. The train travels past you at speed V. It turns out that if you observe the clocks as simultaneous times in your frame, you will see the rear clock showing a higher reading than the front clock. By how much?

I can't get how photons get related to a clock showing advanced time?

http://www.people.fas.harvard.edu/~djmorin/chap11.pdf

See page XI-11 of the pdf in the link.

|cite|improve this question
$\endgroup$
1
$\begingroup$

It's not quite clear to me what you're looking for here. It's straightforward to use the Lorentz transformation to show that if the two clocks are synchronized in the train's coordinate system (the primed system), then the rearward clock is observed to be ahead of the forward clock in the platform's coordinate system (the unprimed system).

So, why would the author of the linked document use photons to illuminate (heh) this result?

Well, one way to synchronize spatially separated clocks at rest in one's frame of reference is by the exchange of light signals. This synchronization convention is known as Einstein synchronization and it guarantees that one-way speed of light (in vacuo) measurements will yield $c$.

Honestly, I think a better approach to showing that the rearward clock is ahead in the unprimed system is to observe the train's clocks synchronization procedure from the platform.

Imagine that the photon source is position precisely at the midpoint of the train car. The photon source simultaneously emits a forward traveling photon and a rearward traveling photon.

The forward and rearward clocks are synchronized in the primed system when they record identical arrival times for their respective photons (all observers agree on the time each clock reads when their respective photon arrives).

As observed from the platform however, the elapsed time for the rearward photon to propagate from the source to the rearward clock is less (since the rearward clock closes the distance) than the elapsed time for the forward photon to propagate to the forward clock (since forward clock opens the distance).

Thus, if both clocks record identical time of arrival for their respective photons, it can only be that the rearward clock is ahead of the forward clock in the unprimed system.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ But Alfred, as the question contains the line - '...if you observe the clocks as simultaneous times in your frame....', hence will it not be contradicting that we have placed the object right in the middle wrt to primed frame because that would break simultaneity in the unprimed frame and hence not allowing us to see the clocks simultaneously in unprimed frame. As far as I have got a sense of the question (after DAYS!! though) is that seeing clocks simultaneously doesn't necessarily imply seeing no time difference in them. The author has used the same thing. $\endgroup$ – Onkar Singh Nov 7 '18 at 5:44
0
$\begingroup$

See [1]. The last answer is mine. The question is practically the same.

[1] How to do calculation in relativity of simultaneity

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Elio, I edited in the direct link to your answer at the linked question since your answer isn't necessarily the last one (the order depends on the sort setting). $\endgroup$ – Alfred Centauri Nov 4 '18 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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