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Lorentz invariance of the Klein-Gordon equation action

Physics Asked by Arthur on November 28, 2020

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 t1 to t2 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!

One Answer

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.

Answered by benrg on November 28, 2020

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