1. Feb 27, 2012

### mananvpanchal

Hello,

Suppose, in moving train Observer O is at middle point M, and two clocks is situated at A(rear) and B(front). O has a clock.

I have confusion about time of A and B after synchronization.

I have two scenario, please, tell me which one is right.

(1) If we assume that one way speed of light is same for all.
So, there is no problem if frame is moving or not.

Suppose, light signal takes 2 sec to reach at end and 2 sec for coming back. Using the equation $t=t_1+(t_2-t_1)/2$ we can calculate $t=2$. So we send "2" signal at $t_o=0$.

$t_o=0, t_a=?, t_b=?$
$t_o=1, t_a=?, t_b=?$
$t_o=2, t_a=2, t_b=2$
$t_o=3, t_a=3, t_b=3$
$t_o=4, t_a=4, t_b=4$

(2) If we assume that one way speed of light is not same for all, then motion of frame matters. Suppose, light signal takes 1 sec to reach at A and 3 sec for coming back, 3 sec to reach at B and 1 sec for coming back. Using the same equation we can calculate $t=2$. So we send "2" signal at $t_o=0$.

$t_o=0, t_a=?, t_b=?$
$t_o=1, t_a=2, t_b=?$
$t_o=2, t_a=3, t_b=?$
$t_o=3, t_a=4, t_b=2$
$t_o=4, t_a=5, t_b=3$

Please, tell me which is right. If both wrong then please tell me how to synchronize clocks in moving frame.

Thanks

2. Feb 27, 2012

### zonde

Both are right. You just have one more simultaneity "plane" in addition to simultaneity of train clocks and it is different in cases (1) and (2). In case (1) this simultaneity "plane" is the same as for train clocks. In case (2) this simultaneity "plane" is skewed in respect to train clocks.

Basically it looks like you are asking how to determine preferred frame. But relativity does not provide any means how to do that. So either you assume it in arbitrary fashion or do without it.

And maybe this can help Relativity of simultaneity

3. Feb 27, 2012