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If we have a Lagrangian $\mathcal L$ that depends on some scalar field $\phi$, we define the momentum as $\pi \doteqdot {\partial \mathcal L \over \partial \dot \phi}$. The Hamiltonian then is $\mathcal H \doteqdot \pi \dot \phi - \mathcal L$.

This rubs me the wrong way. It seems weird to be defining a quantity that puts special emphasis on time if you want a relativistic field theory. I would expect something like defining various momenta as $\pi^\mu \doteqdot {\partial \mathcal L \over \partial (\partial_\mu \phi)}$ and the Hamiltonian as $\mathcal H \doteqdot \pi^\mu \partial_\mu \phi - \mathcal L$, but I've never seen anyone use this.

I get that the Hamiltonian is the generator of time translations, but why should I care about it in the context of a relativistic field theory?

Qmechanic
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lobato
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  • You have hit the nail on the head - the Hamiltonian approach is not manifestly Lorentz invariant (though the theory turns out to be). But the alternative Hamiltonian you propose has no purpose - why should we care about it? As you say, the entire idea of the Hamiltonian is based around it generating the time translations. – ACuriousMind Nov 02 '14 at 17:23
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    It depends on what you want to do. At the end of the day you want to calculate something and you need the formalism that is the most suitable for that. The issue is then that while the theory may have all sorts of symmetries, the situation you are considering may not. E.g. the fact that the laws of physics are invariant under rotations does not mean that you can't drive your car in some direction. – Count Iblis Nov 02 '14 at 17:28
  • @ACuriousMind: Are there other parameters besides time which would make a better parameter for the Hamiltonian? For example, one imagines using the proper time or something like that. – DanielSank Nov 02 '14 at 17:29
  • I guess you would be more happy with the hamiltonian density defined via the energy-momentum tensor. There you have the other Lorentz indexes to form the full momentum density and, well, you know, the full energy-momentum tensor. – TwoBs Nov 02 '14 at 17:33
  • @DanielSank: I don't think so - the Hamiltonian lives on the phase space, and the phase space is the manifold of spatial positions & momenta. If you want manifest invariance, stay with the Lagrangian. – ACuriousMind Nov 02 '14 at 17:41
  • Possible duplicate: http://physics.stackexchange.com/q/38286/2451 – Qmechanic Nov 02 '14 at 18:57
  • @DanielSank, $\dot\phi =\partial_0 \phi$ is a derivative w.r.t proper time -- the proper time for an observer at rest in this frame. More invariantly consider an observer A with 4-velocity $u^\mu$ and take $\dot\phi = u^\mu \partial_\mu \phi$. Then $\dot\phi$ is the derivative of the field w.r.t. the proper time of A, and the Hamiltonian will generate translations in A's proper time. – Robin Ekman Nov 02 '14 at 19:13

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