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So I understand that special relativity is all about the frame of reference and there is a lot to do with time dilation and how space-time is warped at velocities near $c$. So my question is what about an observer on Earth, watching two ships accelerate, in opposite directions, away from the observer at say 1g. As they pass a velocity between the observer on Earth and a ship of .51 , cant the observer on earth measure both of them going at .51c? At which point, with respect to the observer, they are moving away from each other at 1.02c? Now special relativity says the two ships would be moving away from each other at $(v+u)/[1+(vu/c^2)]$, but how does that relate to the stationary observer?

Thanks for your help on this a co-worker and I were unable to find another reference.

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  • $\begingroup$ Spacetime is not "warped at velocities near $c$" for approximately 7,000 reasons, the first of which is that spacetime does not have a velocity. $\endgroup$ – WillO Mar 27 '16 at 4:34
  • $\begingroup$ As @AndreaDiBiagio notes, the limitation is only on two objects observing each other, not on a third party observer observing two objects. In that case, the two individual limits are c, so the total limit is 2*c. No contradiction since there's no physical object moving faster than c, just the measurement of a distance. $\endgroup$ – barrycarter Mar 28 '16 at 2:35
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What you are saying is right. Another situation in which something similar happens is when you consider two proton beams in the LHC, travelling at $99.999\%$ the speed of light collide, the distance between them decreases at $199.998\%$ of $c$ in the lab frame.

The key is that these observations are made in the lab frame. From the point of view of one of the ships or the particle beams, in that frame, the speed of the other particle will be given by the formula you quoted.

So in the ship case: \begin{equation} \frac{u+v}{1+vu/c^2} = \frac{1.02c}{1+.51^2}=0.809c \end{equation} and the proton beam case: \begin{equation} \frac{u+v}{1+vu/c^2} = \frac{1.99998c}{1+.99999^2}= 0.9999999999 c \end{equation}

Basically that equation tells you: if you observe that there are two bodies, one moving at $u$ and the other at $v$ (either towards or away from each other), then the speed of either, in the frame of the other is given by tha formula.

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  • $\begingroup$ So to be clear. Two objects can actually move away from each other faster than c, but they can just never measure their velocity faster than c? Even though they are actually moving away from each other faster than c? I guess that just seems like a cop-out if they are actually moving faster than c then who cares if they can measure their velocities in that frame of reference? $\endgroup$ – Devbot920 Apr 1 '16 at 21:11
  • $\begingroup$ There is rarely "actually" in special relativity: statements about position, velocity, time and duration are all relative to the frame of measurement. But think about this: what happens if a ship shoots a laser towards the other ship? Speed of light is a constant in all frames of reference. What does the observer on earth see? What does the person in the ship see? $\endgroup$ – Andrea Apr 2 '16 at 10:33

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