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For instance: two protons are brought to collision at CERN'S LHC. An observer located at the center of the detector measures a reduction of the distance between both particles until the particles collide. From the reference frame of each particle, the relativistic formula for the addition of velocities yields a relative velocity smaller c. The observer at rest 'sees' both particles symmetrically approaching. So what is the speed of that reduction of distance between the protons, as seen by the observer?

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The speed in question is just under the double the speed of light. This however does not create any relativity contradiction. In relativity, the local speed of one object relative to another cannot exceed the speed of light. In your thought experiment, every relative speed is under the speed of light. The speed you are describing is not a relative speed, but two independent relative speeds. One is the speed of the first proton relative to you; the other is the speed of the second proton relative to you. You can combine these speeds the way you like and the way you are describing they combine to near double the speed of light. However, there is no such a concept in relativity, as the speed of one object relative to the other in the frame of the third (e.g. observer). "Relative" means one relative to the other in its frame, not in the frame of the third party. If you calculate the speed of one proton relative to the other proton, the result would be under the speed of light.

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  • $\begingroup$ But the observer at rest could measure the distance with a solid rod, which has it's center of mass at the supposed collision vertex. Due to symmetry reasons, he could see both ends of the rod simultaneously as he would see photons from both ends at the same time (keyword clock synchronization). So he would see that the length of the rod reduces itself with a rate > c? $\endgroup$ – harter kern Oct 21 '18 at 20:58
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    $\begingroup$ The length of the rod does not really reduce, unless you use a telescopic rod, which is two rods in one. So you'd measure two separate speeds anyway. You can simply add them, if you want, but this would not violate relativity, because its restriction is for a single relative speed. $\endgroup$ – safesphere Oct 21 '18 at 21:26
  • $\begingroup$ the reduction of the length of "rod" should have nothing to do with its mechanical properties nor should its state of matter. In this 'thought experiment' likewise each proton could emit photons in direction of the observer whereas both photons will reach the observer simultaneously, so you could substitute the solid rod with something equivalent, at least in this context. So does that mean, that the observer "sees" how the distance between the proton reduces with a speed >c ? $\endgroup$ – harter kern Oct 21 '18 at 21:38
  • $\begingroup$ Yes, as I stated in the first sentence of my answer, the speed as defined in your question is near double the speed of light. You cannot however measure it in a single measurement and for this reason it does not violate relativity. $\endgroup$ – safesphere Oct 21 '18 at 21:53
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    $\begingroup$ Consider a high energy proton coming to the Earth from the Sun near the speed of light. At the moment the proton passes by the Moon, what is its distance to the Earth? It is the same as the distance to the Moon regardless of the speed of the proton. The logical flaw here is in thinking that your hypothetical measuring rod is moving relative to you. It does not. It is your measurement tool in your frame of reference, so it is static relative to you. It is the proton that is moving passing by the marks in the rod. So the distance to the proton at the Moon is the same as the distance to the Moon. $\endgroup$ – safesphere Oct 22 '18 at 1:47

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