So, I am not asking is the square of four-momentum of a particle an invariant to Lorentz trasnformations, but rather,is it invariant in dynamic situations? It seems to me that this also has to hold. So, is four-momentum squared same before and after collision, not the total, but for one particle in that collision?
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$\begingroup$ four-momentum squared Is always c square $\endgroup$– EliCommented Mar 3, 2019 at 15:58
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$\begingroup$ Isnt it c squared times the mass squared with a minus in front? $\endgroup$– Žarko TomičićCommented Mar 3, 2019 at 15:59
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$\begingroup$ I don’t think so $P_{\mu }\cdot P^{\mu }=c^{2}$ $\endgroup$– EliCommented Mar 3, 2019 at 16:03
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1$\begingroup$ The sign depends on your convention for writing 4-vectors or for forming scalar products. Obviously the convention used by particle physicists is superior in every way to that used by cosmologists. Obviously. $\endgroup$– dmckee --- ex-moderator kittenCommented Mar 3, 2019 at 16:48
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$\begingroup$ I am not asking about the sign $\endgroup$– Žarko TomičićCommented Mar 3, 2019 at 16:55
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1 Answer
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As long as the particle has neither
- Changed kind (as happens in, for example, in charged-current weak scattering)
- Gotten excited (which can happen to atoms, nuclei, and hadrons; though in some cases this would be written as a change of type as in a proton turning into a Delta, for instance)
then the mass is the same.
But ... in a lot of ways what I wrote is a tautology. If a particle stays the same then it stays the same.
answered Mar 3, 2019 at 17:23
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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$\begingroup$ So square of four momentum is constant for a particle given that its mass stays the same? $\endgroup$ Commented Mar 3, 2019 at 17:28
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$\begingroup$ The square of the four momentum is the mass (up to uninteresting constants). Of course the change or don't change together. $\endgroup$ Commented Mar 3, 2019 at 17:31
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$\begingroup$ I am just surprised i was solving compton scattering and somewher squared four momentum of an electron only to find out its equal to its mass!! I was puzzled.... $\endgroup$ Commented Mar 3, 2019 at 17:34
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$\begingroup$ Ah. Either your resource has done you a disservice my not making that connection clear or you didn't notice the importance at first. But either way you have it now. $\endgroup$ Commented Mar 3, 2019 at 17:39