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Consider an observer B is moving with speed $0.8c$ relative to another observer A in standard-configuration (I think it is called that in english: that the system B is only moving along the $x$-axis of system A).

  • At what speed will A observe the light leave the system B in the direction in which B travels away from A? Is it just $0.2c$ or would you have to use the velocity-addition formula to calculate it?
  • And what about the light leaving system B in the opposite direction (towards A) seen from A relative to B? B would observe the speed as being the same in every direction (c) relative to itself, but would A see the light moving with the same speed relative to B in every directions?
  • If that is actually true, then the speed of light is not the same observed by A relative to A, so I guess they would observe different happenings? So how would one calculate the speed of light emitted by B relative to B as observed by A in different directions?

Remember I do not ask about the speed relative to any of the systems themselves observed inside the system as that would always be $c$. I hope you understand the question. I gladly accept mathematical explanations.

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All inertial observers (including A) will see light move at exactly $c$. So both A and B see the light move at $c$. But from A's frame, it appears that the light moves with a speed relative to B of $(c - 0.8c) = 0.2c$. Likewise the light in the other direction would be observed by A to be moving relative to B at $(c + 0.8c) = 1.8c$

As long as you are asking questions about relative speeds in a single frame (that of A in the example), we can use simple vector addition.

There is no problem with measured relative speeds in excess of $c$, because nothing physical is really moving at that rate relative to an inertial frame. As an example, two photons fired in opposite directions will have a distance between them that is measured to be increasing at the rate of $2c$.

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  • $\begingroup$ Then how can you explain B observing the light leaving B with speed c? I understand how both observers messure the same speed in front of B (because lenght contraction and time dilation), but how can one explain that both observers messure same speed behind B with relativity consequences? $\endgroup$ – Malthe Andersen Oct 6 '15 at 12:20
  • $\begingroup$ It's because B doesn't measure time the same way A does. If we imagine two objects an equal distance in front and behind B, then the light reaches those objects at a time that is simultaneous in B's frame, but is not simultaneous in A's frame. But both observe the speed of the light to be the same $\endgroup$ – BowlOfRed Oct 6 '15 at 14:53
  • $\begingroup$ See also: physics.stackexchange.com/questions/199891/… $\endgroup$ – BowlOfRed Oct 6 '15 at 17:16
  • $\begingroup$ I agree that the time dilation and length contraction is the explanation, but I can't seem to get a picture of what is happening in my head. Yes, it does explain, but how/why? $\endgroup$ – Malthe Andersen Oct 6 '15 at 21:49
  • $\begingroup$ In this video at 3 min. there is an easy video picturing and explaining what happans in front of B, but how does that explain what happens behind B? m.youtube.com/watch?v=ACUuFg9Y9dY $\endgroup$ – Malthe Andersen Oct 6 '15 at 21:50

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