0
$\begingroup$

Two particles, moving at relativistic speeds in the x direction, are observed to have energies E1 and E2, and momenta p1 and p2 in frame A. Frame B moves at relativistic speed v relative to frame A, also in the x direction. The particle energies and momenta in frame B are E1′ , E2′ , p′1 and p′2.

How do I proof the following relationship?

enter image description here

$\endgroup$

closed as off-topic by AccidentalFourierTransform, John Rennie, user36790, Jon Custer, Kyle Kanos Dec 26 '16 at 19:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, John Rennie, Community, Jon Custer, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ You need some extra conditions here. How are the 4-vectors on each side related? $\endgroup$ – probably_someone Dec 25 '16 at 9:32
  • $\begingroup$ @probably_someone I've added some more information to the question $\endgroup$ – foxielmao Dec 25 '16 at 9:35
  • $\begingroup$ So does i run from 1 to 2 here? $\endgroup$ – probably_someone Dec 25 '16 at 9:36
  • $\begingroup$ @probably_someone yup, i run from 1 to 2 here $\endgroup$ – foxielmao Dec 25 '16 at 9:37
1
$\begingroup$

We know that Lorentz transformations preserve the length of the 4-momentum vector for each particle. In Minkowski space, this length is $E^2-p^2c^2$ up to sign convention, which is irrelevant here. Since this quantity is frame-independent (frame changes are Lorentz transformations) for each particle, the sum must also be frame-independent.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.