# Lorentz invariance of Action or Lagrangian? [duplicate]

For what I know, the Action should be Lorentz-invariant in accordance to the axioms of SR. For example for a free particle the Action would be the integral of the proper time, which of course is invariant. Now my doubt is the following: does that mean that the Lagrangian should be Lorentz-invariant as well? In some texts I've read it is often stated that the Lagrangian should be a Lorentz scalar, but how does this assures the Action to be Lorentz-invariant too? For example, once one gets the Lorentz-invariant Lagrangian for the EM field $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$, why should its action:$$\int -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} d^4x$$ be Lorentz invariant too? Aren't we integrating a scalar over a specific set of coordinates which are not Lorentz-invariant?

• Does this answer your question? Lorentz invariance of the integration measure Mar 29, 2020 at 11:55
• The 4-dimensional volume-element $d^4x$ is invariant on Lorentz-transformations for a similar reason, why the 3-dimensional volume-element $d^3x$ is invariant on space rotations. Mar 29, 2020 at 12:03

It's a nice excercise to show that, under a Lorentz transformation $$\Lambda$$ the integral transforms are $$\int \mathrm{d}^4x\, f(x) \mapsto \int \mathrm{d}^4x' \,|\det\Lambda|\, f(\Lambda x')\,.$$ Since, by definition, Lorentz transformations have determinant $$1$$ or $$-1$$, the integral is invariant.
Action is constructed as Lorentz invariant dencity integrated over all Minkowski space. $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ is Lorentz invariant dencity.
$$d^4x$$ is Lorentz invariant measure of integration in rectangular coordinate system.
If you wanna to choose some another coordinate system (curvilinear coordinate system), in general you need consider another measure of integration $$d^4y \sqrt{-g}$$, $$g_{\mu\nu}$$ is Minkowski metric in specific coordinates, that one can choose.
$$g_{\mu\nu}(y) = \eta_{\rho\sigma}\frac{\partial x^{\rho}}{\partial y^{\mu}}\frac{\partial x^{\sigma}}{\partial y^{\nu}}$$