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 $t'=\frac{ct-vt}{c}= (1-\frac{v}{c})t$ . where $t$ is time measured by observer at rest.

Where is the mistake in this reasoning?


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 $ct-vt$. 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 $\sqrt{ 1 - v^2/c^2 }$.

  • $\begingroup$ Thank you for your answer. But, Why wouldn't the observer on the train not agree that the distance between him and the photon is $ct-vt$? $\endgroup$
    – pranphy
    Oct 31 '12 at 11:33
  • $\begingroup$ Because t is the ground observers time which he sees as running slow. $\endgroup$
    – FrankH
    Oct 31 '12 at 12:31
  • 2
    $\begingroup$ @PrakashGautam, the events (1) photon at $ct$ and (2) train at $vt$, while simultaneous in the observer's frame, occur at different times in the train's frame (they are not simultaneous in the train's frame). Look up "relativity of simultaneity". $\endgroup$ Oct 31 '12 at 12:52

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