Non-relativistic limit of Hamiltonian for a free particle in general relativity - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-20T05:18:38Z https://physics.stackexchange.com/feeds/question/466656 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/466656 3 Non-relativistic limit of Hamiltonian for a free particle in general relativity Javier https://physics.stackexchange.com/users/5788 2019-03-15T17:27:02Z 2019-03-18T18:54:18Z <p>The Hamiltonian for a particle moving in a gravitational field can be taken as</p> <p><span class="math-container">$$\mathcal{H} = \frac12 \sum_{\mu,\nu=0}^3g^{\mu\nu}(x)p_\mu p_\nu\tag{1}$$</span></p> <p>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</p> <p><span class="math-container">$$H = \frac{1}{2m} \sum_{i,j=1}^3 g^{ij} p_i p_j + V(x),\tag{2}$$</span></p> <p>where <span class="math-container">$g_{ij}$</span> is just the Euclidean 3-metric written in the appropriate coordinates. </p> <p>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 <span class="math-container">$t$</span> as an independent degree of freedom, while in the non-relativistic case <span class="math-container">$t$</span> is just the parameter. I suppose that one would either have to do some kind of 3+1 decomposition of <span class="math-container">$\mathcal{H}$</span>, or introduce a fictitious parameter <span class="math-container">$\lambda$</span> in <span class="math-container">$H$</span> 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 <span class="math-container">$\mathcal{H}$</span>?</p> https://physics.stackexchange.com/questions/466656/-/466999#466999 4 Answer by Qmechanic for Non-relativistic limit of Hamiltonian for a free particle in general relativity Qmechanic https://physics.stackexchange.com/users/2451 2019-03-17T14:20:38Z 2019-03-18T18:54:18Z <p>We start with the relativistic Hamiltonian Lagrangian<span class="math-container">$^1$</span> </p> <p><span class="math-container">$$L_H~:=~ \sum_{\mu=0}^3p_{\mu} \dot{x}^{\mu} - \underbrace{\frac{e}{2}(p^2+(mc)^2)}_{\text{Hamiltonian}}, \qquad p^2~:=~\sum_{\mu,\nu=0}^3g^{\mu\nu}(x)~ p_{\mu}p_{\nu}~&lt;~0, \qquad\dot{x}^{\mu} ~:=~\frac{dx^{\mu}}{d\tau}, \tag{A}$$</span> where <span class="math-container">$e$</span> is an einbein field, and <span class="math-container">$\tau$</span> 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 <span class="math-container">$e=1$</span>.</p> <p>The derivation of OP's eq. (2) now goes as follows:</p> <ul> <li><p>Assume that the metric tensor <span class="math-container">$g_{\mu\nu}$</span> has no spatio-temporal components <span class="math-container">$g_{0i}=0$</span>.</p></li> <li><p>Go to static gauge <span class="math-container">$x^0 =c\tau$</span>.</p></li> <li><p>For weak gravitational fields, it is well-known that we may identify the <span class="math-container">$00$</span>-component of the metric <span class="math-container">$$-g_{00}~\approx~1+ \frac{2\phi}{c^2} \tag{B}$$</span> with the specific gravitational potential <span class="math-container">$\phi$</span>, cf. e.g. <a href="https://physics.stackexchange.com/q/211930/2451">this</a> and <a href="https://physics.stackexchange.com/q/75006/2451">this</a> Phys.SE posts. </p></li> <li><p>If we integrate out <span class="math-container">$p_0$</span> and <span class="math-container">$𝑒$</span> (i.e. eliminate them via their EL eqs.), we get <span class="math-container">\begin{align}\left. L_H\right|_{x^0=c\tau} \quad&amp;\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&amp;\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} .</span></p></li> <li><p>The Hamiltonian becomes <span class="math-container">$$\text{Hamiltonian} ~\stackrel{(C)}{=}~c\sqrt{-g_{00}({\bf p}^2+(mc)^2)}~\stackrel{(B)}{=}~mc^2\sqrt{\left(1+ \frac{2\phi}{c^2}\right)\left(1+\frac{{\bf p}^2}{(mc)^2}\right)}$$</span> <span class="math-container">$$~=~ mc^2 + \frac{{\bf p}^2}{2m} + m \phi +{\cal O}(c^{-2}). \tag{D}$$</span></p></li> <li><p>Eq. (D) is OP's eq. (2) apart from the rest energy <span class="math-container">$mc^2$</span>, which can be ignored as it is a constant. <span class="math-container">$\Box$</span></p></li> </ul> <p>--</p> <p><span class="math-container">$^1$</span> See e.g. <a href="https://physics.stackexchange.com/q/90552/2451">this</a> Phys.SE post. We use the sign convention <span class="math-container">$(−,+,+,+)$</span>.</p>