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The path integral formalism is used to get for example the propagator of particles. In this formalism we integrate over all mathematically possible paths (and weight them with the non-relativistic action) but many of them are relativistic or physically impossible because the velocity had to be larger than the speed of light. So why are we doing so? Why do we also count the paths where the particle must be viewed as relativistic but still weight it with the non-relativistic action? And why sum we over the paths where the particle must move faster then the speed of light?

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closed as unclear what you're asking by ACuriousMind, John Rennie, Sebastian Riese, Gert, user36790 May 1 '16 at 17:26

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    $\begingroup$ It would help, if you would add equation of the path integral you are referring to. Path integral formalism is not restricted to non-relativistic actions. You get out from the theory what you put in to the theory. $\endgroup$ – Mikael Kuisma Apr 30 '16 at 9:38
  • $\begingroup$ Related: physics.stackexchange.com/q/18835/2451 , physics.stackexchange.com/q/20823/2451 , physics.stackexchange.com/q/163622/2451 and links therein. $\endgroup$ – Qmechanic Apr 30 '16 at 9:43
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    $\begingroup$ The path integral is a formal integral over all paths in the mathematical sense. It is not an integral over "physical" paths, and the "reason" we integrate like that is simply that that is the correct integral to produce the quantities we want to compute, see any of the standard derivations of the path integral. I'm not sure what you're asking. $\endgroup$ – ACuriousMind Apr 30 '16 at 11:44
  • $\begingroup$ Ok, then the question is what interpretation has this integral over all paths (including the spacelike paths)? Or doesn't it has any interpretation? $\endgroup$ – Alpha001 Apr 30 '16 at 12:47