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For if I started by trying to make the Hamiltonian Lorentz invariant, I would have failed. Indeed, the Hamiltonian is part of a covariant tensor. But how do I know that the Lagrangian is not a part of such a tensor?

All of Lagrangian formulation, Hamiltonian formulation, and Poisson bracket formulation involve some partial derivative with respect to time at some point. None of them are manifestly Lorentz invariant.

We knew the system governed by Klein-Gordon equation is Lorentz invariant, but could I construct a non-Lorentz-invariant Lagrangian and derive a Lorentz invariant equation of motion out of that?

Qmechanic
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Shinjikun
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1 Answers1

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Going the way stated in the question's title is easy: The Euler-Lagrange condition is, inherently, a condition on the action -- the statement is that the classical path is the path for which the action takes a minimum value for the path. Since this is a statement about the value of the action, and the action is Lorentz-invariant, then this minimum value is left unchanged by a Lorentz transformation. Therefore, the equations of motion have to be lorentz covariant. None of this is automatically true in a Hamiltonian formalism, where you are explicitly doing a Legendre transform involving time, and ruining the manifest Lorentz invariance of the theory.

The question you ask in your conclusion is harder -- given a Lorentz-invariant equation of motion, is it possible to construct a non-Lorentz invariant Lagrangian? The trivial answer is "yes, you can add non-lorentz invariant boundary terms". But I don't know about the less trivial case.

Zo the Relativist
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