The Hamiltonian for a particle moving in a gravitational field can be taken as
$$\mathcal{H} = \frac12 \sum_{\mu,\nu=0}^3g^{\mu\nu}(x)p_\mu p_\nu\tag{1}$$
as long as the parametrization is affine. Given some 4-metric (say the Schwarzschild or Kerr metrics for definiteness), I would like to take the nonrelativistic limit of a slow particle in a weak gravitational field, and ideally arrive at the non-relativistic Hamiltonian
$$H = \frac{1}{2m} \sum_{i,j=1}^3 g^{ij} p_i p_j + V(x),\tag{2}$$
where $g_{ij}$ is just the Euclidean 3-metric written in the appropriate coordinates.
The problem is that the relationship between the two Hamiltonians isn't clear to me. They don't even have the same units, for starters. I think the fundamental issue is that the relativistic Hamiltonian includes $t$ as an independent degree of freedom, while in the non-relativistic case $t$ is just the parameter. I suppose that one would either have to do some kind of 3+1 decomposition of $\mathcal{H}$, or introduce a fictitious parameter $\lambda$ in $H$ as a sort of gauge invariance, and then the two Hamiltonians would be directly comparable and one could take the limit. Can this be done in general? Is this the right way to take the non-relativistic limit of $\mathcal{H}$?
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