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?

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    $\begingroup$ Related: physics.stackexchange.com/q/51327/2451 , physics.stackexchange.com/q/144389/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Apr 20, 2016 at 5:53
  • $\begingroup$ The action is defined starting from the lagrangian, not from the hamiltonian. So, if you want to build an invariant set of equations - which are issued from the minimisation of the action - the lagrangian has to be automatically invariant. This is also the main reason of the usilisation of lagrangians in quantum field theory instead of hamiltonians as in non-relativistic quantum mechanics. $\endgroup$
    – Wellow
    Commented Apr 20, 2016 at 14:47

1 Answer 1


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.

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    $\begingroup$ Even the first part it is not so easy to me. A Lorentz transformation does change the stationary points. However, in view of the Lorentz invariance of the Lagrangian, the class of stationary points is invariant under a Lorentz transformation (but not each stationary point separately). A stationary point is completely determined by the boundary conditions. We are saying that, as the Lagrangian is Lorentz invariant, a stationary point determined by certain boundary conditions is transformed into another stationary point with correspondingly transformed boundary conditions. $\endgroup$ Commented Apr 20, 2016 at 14:33
  • $\begingroup$ @ValterMoretti: yes, I didn't say the path was the same, just that the value of the action is necessarily the same, so Lorentz transforms map solutions of the EOM to other solutions of the EOM. $\endgroup$ Commented Apr 20, 2016 at 15:33
  • $\begingroup$ But I should have said tha tthe EOM are lorentz covariant, not invariant. $\endgroup$ Commented Apr 20, 2016 at 17:07
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    $\begingroup$ OK, I misunderstood your answer, do not worry (+1) $\endgroup$ Commented Apr 20, 2016 at 18:35
  • $\begingroup$ @ValterMoretti: you improved my answer, irrespectively. :) $\endgroup$ Commented Apr 20, 2016 at 19:20

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