- #36
Sagittarius A-Star
Science Advisor
- 1,302
- 978
No, the sum gives ##\frac{x'(c+v) +x'(c-v) }{c^2-v^2} = \frac{2cx'} {c^2-v^2}##.DmitryS said:In the relatively moving frame it is ##\frac{x'} {c-v}## forth and ##\frac{x'} {c+v}## back. The sum of these gives you ##\frac{2vx'} {c^2-v^2}##
I still don't understand your calculation, but I guess you want to calculate the time-difference between the two lightnings with reference to the train rest frame.
This train/embankment thought experiment contains an easier clock synchronization procedure than the original one from Einstein. Two clocks are regarded as synchronous in the reference frame, in which they are both at rest, if an observer in the middle of them receives simultaneously their light pulses, that were sent of at their same clock-times.
Based on this, I continue my "closing speed" calculation from posting #12:
This calculation was done with reference to the embankment frame. I continue with this reference frame:Sagittarius A-Star said:The light-pulse from the front needs the following time-interval to reach the observer in the center of the car:
##\overline{BM}/(c+v)##
The light-pulse from the back needs the following time-interval to reach the observer in the center of the car:
##\overline{AM}/(c-v)##
Because the distances ##\overline{AM}## and ##\overline{BM}## are equal, the observer in the center of the car sees the light-pulse from ##B## earlier than that from ##A##.
The time-difference for the observer in the center of the car is:
##\overline{AM}/(c-v) - \overline{BM}/(c+v) = \frac{L}{2} \frac{2v}{c^2-v^2} = \gamma^2 v\frac{L}{c^2}##
To convert this time-difference to the train frame, this needs to be devided by ##\gamma## because of time-dilation. Then compare to ##t'_A- t'_B## in my posting #29:
Sagittarius A-Star said:##x'_A = \gamma (x_A - vt_A) = -\gamma \frac{L}{2}##, ##\ \ \ \ \ \ t'_A= \gamma (t_A - vx_A/c^2) =\gamma \frac{L}{2}v/c^2##.
Transform event B coordinates:
##x'_B = \gamma (x_B - vt_B) = +\gamma \frac{L}{2}##, ##\ \ \ \ \ \ t'_B= \gamma (t_B - vx_B/c^2) =-\gamma \frac{L}{2}v/c^2##.