I was reading Einstein's book on Relativity. What is the contradiction between the principle of relativity(in the restricted sense) and the law of the transmission of light in vacuum that Einstein is talking about? I am unable to understand it. Any help is appreciated.


1 Answer 1


I'm not sure I get your question right, but from what I understand, I think it is about the speed of light. You see, if I stay on a train and I launch a ball in the direction the train is moving, the ball speed will be the sum of it's speed relative to me + the speed that the train is traveling. If I point a laser from the train, (suppose we are in vacuum and I die trying to do this) the "speed" of the light coming from laser is not the sum of the light + the speed of the train. Light, wherever you stay has the same speed. You can even "ride" stay on a beam of light and all light will pass you with the speed of light, or you can stay still and you will measure same speed for light. It sounds like sci.fi. but this is how world works.

  • $\begingroup$ How does this contradiction explain his conclusions?(that time is relative) $\endgroup$
    – user150093
    Oct 6, 2017 at 9:08
  • $\begingroup$ Well you have to asume something. It becomes easy with a lil bit of maths. en.wikipedia.org/wiki/Time_dilation $\endgroup$
    – Xsnac
    Oct 6, 2017 at 9:21
  • $\begingroup$ This is because if you're moving relative to something, every process that makes you You starts accuring more slowly. This is why light always moves at the same speed, no matter the speed of objects. $\endgroup$
    – MaDrung
    Oct 6, 2017 at 11:31
  • $\begingroup$ @Xsnac See if I understand it right. If I am travelling in a train and light beam is travelling parallel to me, then I will say that the velocity of light relative to me is c-(my velocity).But this is wrong.Light will travel with a speed 'c' relative to me despite of my speed.And this is the contradiction between Galilean Relativity and Law of Propagation of light in vacuum.Am I right? $\endgroup$
    – user150093
    Oct 6, 2017 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.