Suppose we have a train moving. When the origin of train's frame coincides with the origin of observers frame; the the time is set to zero. At that very instant, a photon is emitted from train towards the direction train is moving. After time t measured by observer at rest the photon will be at a distance ct and the train at a distance vt; but the photon is at a distance ct−vt from the train, now the time that the passenger on train measures for the photon to reach that point is the distance divided by the velocity of light ( which of course is c) so time measured by observer on train is t'=(ct−vt)/c=(1−v/c)t . where t is time measured by observer at rest. Where is the mistake in this reasoning?

basaltico00

basaltico00

Answered question

2022-09-22

Suppose we have a train moving. When the origin of train's frame coincides with the origin of observers frame; the the time is set to zero. At that very instant, a photon is emitted from train towards the direction train is moving. After time t measured by observer at rest the photon will be at a distance c t and the train at a distance v t; but the photon is at a distance c t v t from the train, now the time that the passenger on train measures for the photon to reach that point is the distance divided by the velocity of light ( which of course is c) so
time measured by observer on train is t = c t v t c = ( 1 v c ) t . where t is time measured by observer at rest.
Where is the mistake in this reasoning?

Answer & Explanation

Marvin Hughes

Marvin Hughes

Beginner2022-09-23Added 6 answers

Your mistake is assuming that the distance measured by the two observers will be the same. In special relativity there is both a length contraction and a time dilation. The observer on the train will not agree with the observer on the ground that the length was c t v t. He will think the clock on the ground that measured time t was running slowly. In fact each will think the other clock is running slowly by the same factor of 1 v 2 / c 2 .

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