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I think that the following statement is true, but can't seem to prove it.

Suppose we have a scalar field whose Lagrangian density$^1$ $\mathcal L(\phi, ~\partial_\mu\phi, ~X^\mu)$ is Lorentz invariant. By this I mean that $\mathcal L(\phi, ~\partial_\mu\phi, ~X^\mu)=\mathcal L(\phi, ~\Lambda_\mu^{\;\;\nu}~\partial_\nu\phi, ~\Lambda^\mu_{\;\;\nu}~X^\nu)$ for any arbitrary proper Lorentz transformation $\Lambda$. Then the Euler-Lagrange equations $$ {\partial\mathcal L\over\partial\phi} = \partial_\mu{\partial\mathcal L\over\partial(\partial_\mu\phi)} $$ are Lorentz covariant.

Can you help me prove it? Or is it false?

  1. The action is therefore $\int \mathcal L~~dt~dx~dy~dz$. Note that the space-time volume element $dt~dx~dy~dz$ is Lorentz invariant since $\det(\Lambda) = 1$ (proper transformations). So if $\mathcal L$ is Lorentz invariant too, then the action is also Lorentz invariant.
Atom
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  • This is a special case of this Phys.SE post. – Qmechanic Jul 15 '20 at 17:25
  • @Qmechanic You are a savior! Since you seem to have an enormous knowledge of such stuff, can you maybe recommend some of your favorite resources on this subject? – Atom Jul 15 '20 at 17:33
  • @Qmechanic, I am not familiar with the mathematical machinery that you use there... So can you at least just a definitive yes/no answer whether my proposition is correct? – Atom Jul 15 '20 at 17:41
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    Related: https://physics.stackexchange.com/questions/506259/what-if-the-lagrangian-mathscrl-a-lorentz-scalar-is-replaced-by-a-lorentz – SRS Jul 16 '20 at 01:34

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