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.
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1See these previous questions: https://physics.stackexchange.com/q/96009/ – enumaris Mar 29 '18 at 17:09
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1Possible duplicate of Does action really have to be Lorentz-invariant in SR? and/or Must the action be a Lorentz scalar?. – AccidentalFourierTransform Mar 29 '18 at 17:29
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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.
Giuseppe
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