0
$\begingroup$

I was going through a paper on De-Synchronization of 2 Clocks in Special Theory of Relativity. The author shows up number of ways clocks de-synchronize relative to 2 frames. The one I am stuck is due to Poincaré:

First consider A Spaceship moving with velocity V w.r.t ground toward right. Take right side of yours(yes you) to positive. Call this frame S'. There is another observer which is at rest w.r.t ground. We call it S. The observer in S' has 2 clocks. He Synchronizes them by following procedure:

-First measure the distance between 2 clocks(call it $l_{0}$) -Now clock A throws a photon towards B and starts clock A -You have already given B a head start by amount $l_{0}/c$ so that in frame S', the clocks are perfectly Synchronized.

Now,the author shows that these clocks are desynchronized in S frame by following argument:

  • Since the clock A throws Photon towards B and simultaneously Starting itself, the time taken for photon to reach B is (in S frame is) $l_{0} / \gamma *(c-v)$.

Now the problem I am struck at is the next step: The clock A advances by(due to time dilation) $$l_{0}/(\gamma^2)(c-v)$$

The question is why $\gamma$ is divided here. I think(which is obviously wrong) that I am observing clock A. The time measured by Photon to reach B is measured by me. So,my measurements are proper!!. Thats why I should multiply $\gamma$ rather then to divide. Where I am wrong?

$\endgroup$
1
$\begingroup$

I think for such problems it's helpful to work on a spacetime diagram to figure out what quantity is "proper" and what isn't.

Just to state things in a way I find clearer: according to an observer at rest in frame $S$, the signal from clock $A$ takes $l_0/\gamma (c-v)$ to reach clock $B$, at which instant clock $B$ will read $l_0/c$ (assuming clock $A$ starts from $0$). Also according to an observer at rest in frame $S$, when the signal from clock $A$ reaches clock $B$, clock $A$ will have advanced by (or display a time of) $l_0/\gamma^2(c-v)$.

Notice that the $l_0/\gamma (c - v)$ is the time difference with respect to S between two events co-located with clock $A$. The first event is when the signal is emitted by $A$ according to S, and the second event occurs when the signal reaches $B$ according to S. These events occur at two different locations according to an observer at rest in $S$ (since clock $A$ is moving in this frame), so the time difference measured in $S$ is not the proper time. These events occur at the same location according to clock $A$, which is at rest in $S'$, so clock $A$ measures the proper time. Since this proper time is smaller by a factor of $\gamma$ than the time difference measured by an observer at rest in $S$, clock $A$ reads $l_0/\gamma^2 (c-v)$ when the signal reaches clock $B$ according to an observer at rest in $S$.

You might already know this, but the fact that the proper time is smaller by a factor $\gamma$ can be easily derived from the inverse Lorentz transformation of time as follows. Starting with $t = \gamma(t' + vx'/c^2)$, writing this equation for two events that take place at the same position x', and taking the difference, we find $$\Delta t' = \Delta t / \gamma$$ where $\Delta t'$ is the proper time.

$\endgroup$
  • $\begingroup$ Thanks... Good explanation $\endgroup$ – Abhi7731756 Jul 19 at 11:50

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.