I am working on vector fields in curved manifolds and arrive at the following question:

Why is it that we demand the Hamiltonian to generate time translations:

$$[i\mathcal{H}, A_\mu] = \partial_t A_\mu$$

instead of covariant time translations:

$$[i\mathcal{H}, A_\mu] = D_t A_\mu$$

For me the first definition seems very strange as we want the Hamiltonian to generate "real" time translations in which case the modes would be parallel transported won't they ?

Can you give a physical interpretation to both definitions perhaps ?


All right after a lot of thinking I came up with an answer myself

(feel free to give me some feedback ;) )

The crucial subtlety is that the covariant derivative of a vector $\vec{v}$ on a manifold with basis vectors $\vec{e}_\mu$ is defined such that $\partial_t \vec{v} = \partial_t (v^\mu \vec{e}_\mu) = \nabla_t(v^\mu)\vec{e_\mu}$. Indeed:

$$\partial_t(v^\mu \vec{e_\mu}) = \partial_t(v^\mu) \vec{e_\mu} + v^\mu \partial_t(\vec{e_\mu}) = \partial_t(v^\mu) + v^\mu \Gamma_{t\mu}^\kappa \vec{e_\kappa} = \nabla_t (v^\mu) \vec{e_\mu}$$

But this implies that partial time translation of the vector (as we are used to) equals covariant time translation of its indices ! Therefore the correct Hamilton equation is:

$$\partial_t \vec{a} = \{ \mathcal{H}, \vec{a} \} \leftrightarrow \nabla_t a_\mu = \{ \mathcal{H}, a_\mu \}$$

So it really depends on what object you are acting !

  • $\begingroup$ Well, I would rather say that the point is that you have done a 3+1 splitting and thus should not expect any covariance other than on the spatial hypersurface. The Hamiltonian is by definition the generator of the Lie group of time translations and the commutator then follows automatically. $\endgroup$ – Void May 8 '17 at 11:08
  • $\begingroup$ Yes indeed, but my concern was that I end up with the Hamiltonian as the generator of covariant time translations (of the indices) but as I pointed out above this is equivalent to normal time translation of the vector $\endgroup$ – gertian May 8 '17 at 11:50
  • $\begingroup$ I would also like to chime in with saying that although the question is too far removed from context for me, but in 3+1 spacetime splits, the "time derivative" is Lie derivative with respect to the timelike vector field you based the folitation upon. Which reduces to partial differentiation when evaluated in coordinate systems for which that vector field is the time-like coordinate vector field. $\endgroup$ – Bence Racskó May 8 '17 at 13:16
  • $\begingroup$ @Uldreth Indeed time translations should be characterized by the Lie derivative. However, this is not what happens in the Hamilton equations (as far as I known) I think that this may be interpretted as a condition for the Hamilton equations to make sense. i.e.: they only make sense when there is a time like coordinate vector field such that $\partial_t \sim \mathcal{L}_t$ $\endgroup$ – gertian May 8 '17 at 13:38

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