0
$\begingroup$

This question already has an answer here:

I have taken it granted that an action in the special relativity must be a lorentz scalar. However is there a fundamental reason for this requirement? I cannot think of a plausible reason for this question.

$\endgroup$

marked as duplicate by AccidentalFourierTransform, Qmechanic lagrangian-formalism Mar 29 '18 at 17:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Because in QFT you impose a Poincaré invariance and Poincaré includes Lorentz, this comes from the fact that physics must not depend from the frame you choose and so has to be rotation/translation(and boost for a Minkowskian metric) invariant, so your Lagrangian has to be what we call "Lorentz scalar" ie invariant under any Lorentz transfo.

$\endgroup$

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