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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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We start with the relativistic Hamiltonian Lagrangian$^1$

$$L_H~:=~ \sum_{\mu=0}^3p_{\mu} \dot{x}^{\mu} - \underbrace{\frac{e}{2}(p^2+(mc)^2)}_{\text{Hamiltonian}}, \qquad p^2~:=~\sum_{\mu,\nu=0}^3g^{\mu\nu}(x)~ p_{\mu}p_{\nu}~<~0, \qquad\dot{x}^{\mu} ~:=~\frac{dx^{\mu}}{d\tau}, \tag{A} $$ where $e$ is an einbein field, and $\tau$ is a world-line (WL) parameter (which does not have to be the proper time). OP's eq. (1) is the Hamiltonian (A) in the gauge $e=1$.

The derivation of OP's eq. (2) now goes as follows:

  • Assume that the metric tensor $g_{\mu\nu}$ has no spatio-temporal components $g_{0i}=0$.

  • Go to static gauge $x^0 =c\tau$.

  • For weak gravitational fields, it is well-known that we may identify the $00$-component of the metric $$-g_{00}~\approx~1+ \frac{2\phi}{c^2} \tag{B}$$ with the specific gravitational potential $\phi$, cf. e.g. this and this Phys.SE posts.

  • If we integrate out $p_0$ and $𝑒$ (i.e. eliminate them via their EL eqs.), we get $$\begin{align}\left. L_H\right|_{x^0=c\tau} \quad&\stackrel{p_0}{\longrightarrow}\quad {\bf p}\cdot \dot{\bf x}- \underbrace{\left(\frac{-g_{00}c^2}{2e} + \frac{e}{2}({\bf p}^2+(mc)^2)\right)}_{\text{Hamiltonian}}\cr \quad&\stackrel{e}{\longrightarrow}\quad {\bf p}\cdot \dot{\bf x} - \underbrace{c\sqrt{-g_{00}({\bf p}^2+(mc)^2)}}_{\text{Hamiltonian}}\end{align} \tag{C} .$$

  • The Hamiltonian becomes $$\text{Hamiltonian} ~\stackrel{(C)}{=}~c\sqrt{-g_{00}({\bf p}^2+(mc)^2)}~\stackrel{(B)}{=}~mc^2\sqrt{\left(1+ \frac{2\phi}{c^2}\right)\left(1+\frac{{\bf p}^2}{(mc)^2}\right)} $$ $$~=~ mc^2 + \frac{{\bf p}^2}{2m} + m \phi +{\cal O}(c^{-2}). \tag{D}$$

  • Eq. (D) is OP's eq. (2) apart from the rest energy $mc^2$, which can be ignored as it is a constant. $\Box$

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$^1$ See e.g. this Phys.SE post. We use the sign convention $(−,+,+,+)$.

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  • $\begingroup$ Hi, thanks for you answer! Two further questions: 1. What does it mean to integrate out a variable in this context? Replace it by its equation of motion? 2. And is it possible to do this process directly with the Hamiltonian, or does one have to pass through the Lagrangian? (Obviously it's not very hard to go from one to the other.) $\endgroup$ – Javier Mar 18 at 18:40
  • $\begingroup$ 1. Yes. I updated answer. 2. The derivation is clearest via a stationary action formulation. $\endgroup$ – Qmechanic Mar 18 at 18:54

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