0
$\begingroup$

So I am given two clocks A and B moving in $S'$ frame with a velocity $V$ relative to $S$ frame. The two clocks are separated by a distance $L$ and are synchronized in $S'$ frame. The objective is to find the time difference between two clocks as measured in $S$ frame.

My approach:

Using Lorentz transformation,

$t_A = \gamma ({t_A}' + {x_A}'\frac{V}{c^2})$

$t_B = \gamma ({t_B}' + {x_B}'\frac{V}{c^2})$

where $t_A$ and $t_B$ are times in clock $A$ and $B$ as measured in $S$ frame, and ${t_A}'$ and ${t_B}'$ are times in clock $A$ and $B$ as measured in $S'$ frame. But since the two clocks are synchronous in $S'$ frame, ${t_A}'$ and ${t_B}'$ are same.

Hence, subtracting, we get, $t_B - t_A = \gamma ({x_B}'-{x_A}')\frac{V}{c^2} = \frac{\gamma LV}{c^2}$

Since the quantity on the right hand side is positive, this is giving me $t_B > t_A$. The hint on my question says the clock B lags behind. Could someone help me understand what is going on? Also, is this a correct setup of the problem?

$\endgroup$
1
$\begingroup$

We are measuring the readings on the co-moving clocks A and B, at the same time, in frame S. So, it is $t_A=t_B$ (and not $t'_A=t'_B$). In fact, it is $t'_B-t'_A$ that has been asked to determine.

Remember that $t'$ is the reading on the moving clocks (in other words, clocks that are affixed to frame $S'$ ; in your case, two such clocks were A and B) and $t$ is the reading on the clocks affixed to frame S. So, they refer to readings on entirely different clocks.

The moving clocks are synchronized in frame $S'$ but the question is nudging you to see if they are synchronized with respect to $S$ as well.

$$t'=\gamma(t- {{xv} \over {c^2}})$$ $$\Rightarrow t'_B - t'_A = -\gamma{{(x_B-x_A)v}\over{c^2}} = -{{Lv}\over{c^2}}$$

The last equality uses length contraction of the distance L.

EDIT (Responses to comments are made here) :

@Ufomammut : The variable $t'$ simply refers to the reading on the clocks affixed to frame $S'$ (By "affixed", I mean that they are at rest in frame $S'$) $-$ Let's call them $S'$-clocks. Observers of frame $S$ and $S'$ can both observe the readings of the $S'$-clocks. The $S'$-clocks are synchronous only to the $S'$ observers and not to the $S$ observers.

One should be very clear what the quantities related by the Lorentz transformation actually are. I highly recommend the chapter "Electrodynamics and Relativity" on Introduction to Electrodynamics by Griffiths as I greatly benefited from it. Figure 12.18 of the chapter should be clear in your mind.

$\endgroup$
  • $\begingroup$ The critical thing that I am not being able to wrap my head around is this: In $S'$ frame, the two clocks are always synchronized. So shouldn't $t_A^{'} = t_B^{'}$? I think I understand $t_A^{'}$ as the reading made by an observer co-moving with $S'$ frame. $\endgroup$ – Ufomammut Sep 29 '19 at 21:32
  • $\begingroup$ @Ufomammut : Hi, check my edit. $\endgroup$ – Ajay Mohan Sep 30 '19 at 7:38
0
$\begingroup$

When you apply the Lorentz transformation be careful about the velocity sign. Moreover, the sequence of light clocks locations is important. That is, If $A$ is located at the left end of the rod with the length $L$, and $B$ is located at the right end, and the rod moves from left to right at, say, $+v$, you will encounter a different answer ($B$ lags in this case) for your question compared to when the rod moves from right to left at $-v$ ($A$ lags in this case). Therefore, I think you just made a trivial mistake in choosing the correct sign for the velocity. Indeed, if you use the following transformation, you would get the favorite result:

$$t_A = \gamma ({t_A}' - {x_A}'\frac{V}{c^2})$$

$$t_B = \gamma ({t_B}' - {x_B}'\frac{V}{c^2})$$

Then, you will get:

$$t_B - t_A = -\gamma ({x_B}'-{x_A}')\frac{V}{c^2} = -\frac{\gamma LV}{c^2}<0 \rightarrow t_B <t_A$$

PS: According to your question, it is better to use $L^\prime$ instead of $L$, if the length is initially measured in $S^\prime$.

$\endgroup$
0
$\begingroup$

What's going on?

Consider the leading moving clock and the trailing moving clock passing one of the stationary clocks.

The difference in readings on the two moving clocks must be greater than the time difference on the stationary clock between the two events, because 'they' see it running slow.

Stationary observers see both moving clocks running slow.

So stationary observers must see the trailing clock showing a later time than the leading clock.

As an aside, that means they also see the gap between the clocks foreshortened. Be clear what L is.

$\endgroup$

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.