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What I will say is not exclusively true for the KG equation, but let's take it as a simple example. When proving the invariance of its action under a Lorentz transformation, it suffices to show that the Lagrangian density is covariant, because the absolute value of the Jacobian determinant is 1, and the integral giving the action is over the whole Minkowski spacetime, so we don't need to worry about the boundary of the region of integration. But the whole proof rests on the assumption that we are integrating over the whole spacetime manifold, thus over the whole time axis. But for ordinary particle action, we can take it to be from $t_1$ to $t_2$ whatever their values are. And if we did the same in field theory, we will have to start worrying about the change of the boundary of integration after we do a Lorentz transformation, and the usual proof doesn't hold. So, do I have a blind spot here that I cannot see? Am I mistaken in something? What's happening? I would appreciate your help very much!

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    $\begingroup$ What's wrong with the action for finite time intervals transforming under Lorentz boosts? For example, the Hamilton-Jacobi equation tells you that the variation of the action w.r.t. the ending time is equal to minus the Hamiltonian, and you know that the Hamiltonian transforms non-trivially under boosts. $\endgroup$
    – kaylimekay
    Dec 14 '20 at 5:48
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You need to transform the region of integration. It it's bounded by constant-time hyperplanes in the unprimed coordinate system, then it won't in general be bounded by constant-time hyperplanes in the primed coordinate system, but you can still integrate over it. If you restrict your boundaries to constant-time hyperplanes in all coordinate systems then you can't prove a Lorentz invariance result. It's no different from proving rotational invariance of integration over a cube that's aligned with the coordinate axes.

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  • $\begingroup$ I am aware of that, and that's my point exactly. If the boundary changes, then you will have for the two integrals giving the actions before and after the Lorentz transformation, the same integrand, the same measure, but not the same region of integration. Hence, the actions will not be the same and that means the action is not invariant! So, where is the problem? $\endgroup$
    – Arthur
    Oct 15 '20 at 5:34
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You are integrating over the same space-time region, but the integration is done in different coordinate systems (charts of the 4-manifold). The invariance is really that you get the same result from an arbitrary choice of observers. That, at least, is the "passive" picture.

The "active" picture is that physical law should be the same expression in any world (with any translation, rotation, or Lorentzian boosts). In that case, you don't have to worry about whether you are dealing with the same region in space-time. The physical law should have the same form, and that's all.

I found the passive picture more attractive, but either picture will give you the same prediction on experimental results.

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