2
$\begingroup$

The Lagrangian and Hamiltonian in Classical mechanics are given by $\mathcal{L} = T - V$ and $\mathcal{H}=T+V$ respectively. Usual notation for kinetic and potential energy is used.

But, in GR they are defined as $$\mathcal{L} = \frac{1}{2}g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu, \\ \mathcal{H}=\frac{1}{2}g^{\mu\nu}\dot{x}_\mu\dot{x}_\nu.$$

The Hamiltonian above is defined to be a "Super-Hamiltonian" according to MTW.

So, the Hamiltonian associated with a test particle in Schwarzschild geometry is then given by $$ \frac{1}{2} \left(A(r)^{-1} p_t^2 - B(r)^{-1}p_r^2-\frac{p_{\theta}^2}{r^2} - \frac{p_\phi^2}{r^2 \sin^2 \theta} \right). $$ This is only a kinetic energy term - - Where is the potential energy?

$\endgroup$
  • $\begingroup$ Which page in MTW? $\endgroup$ – Qmechanic Mar 7 '17 at 12:49
  • $\begingroup$ Not called GR? What is not called GR? It's Eq. 25.10 page 654. $\endgroup$ – Rumplestillskin Mar 7 '17 at 12:54
2
$\begingroup$

Writing a Hamiltonian down like that looks strange to me because if you look at those two expressions above they are identical: both scalars and in the end $\dot x_\mu \dot x^\mu/2$. In my edition of MTWs Gravitation they do not refer to a Hamiltonian in that form (at least I did not find one).

The typical way of deriving the Hamiltonian from a given Lagrangian is using canonical momenta/Legendre transformation: The canonical momentum conjugate to $x^\mu$ equals the momentum one-form of the particle \begin{equation}p_\mu=\frac{\partial\mathcal{L}}{\partial \dot x^\mu}=g_{\mu\nu}\dot x^\nu.\tag{1}\end{equation} The Hamiltonian is given by the Legendre transformation of the Lagrangian: $$\mathcal{H}=p_\mu \dot x^\mu-\mathcal{L}=\frac{1}{2}g^{\mu\nu}p_\mu p_\nu.\tag{2}$$

The Hamiltonian in form of eq. (2) is used by MTW in eq. (33.27c) and is called "super-Hamiltonian" in their book.

That being said using eq. (1) we may write (2) as $\mathcal{H}=\frac{1}{2}g^{\mu\nu}\dot x_\mu \dot x_\nu$ but this kind of makes our Legendre transformation pointless.

MTW use the prefix "super" and the related terms superspace and "super-Hamiltonian" to refer to different points and concepts of the related to the ADM formalism:

[Wikipedia, "Superspace", https://en.wikipedia.org/wiki/Superspace, 2017.03.07]: The word "superspace" is also used in a completely different and unrelated sense, in the book Gravitation by Misner, Thorne and Wheeler. There, it refers to the configuration space of general relativity, and, in particular, the view of gravitation as geometrodynamics, an interpretation of general relativity as a form of dynamical geometry. In modern terms, this particular idea of "superspace" is captured in one of several different formalisms used in solving the Einstein equations in a variety of settings, both theoretical and practical, such as in numerical simulations. This includes primarily the ADM formalism, as well as ideas surrounding the Hamilton–Jacobi–Einstein equation and the Wheeler–DeWitt equation.

The first time they use the term "super-Hamiltonian" is for eq. (21.12): $$\mathcal{H}(\pi_{ij},g_{mn})=0,$$ and they reference B. S. DeWitt, Phys. Rev. 160, 1113 (1967) and he presents the ADM Hamiltonian, related momenta and quantities in the 3+1 formalism.

$\endgroup$
1
$\begingroup$

OP is considering a massive relativistic point particle with mass $m=1$ in curved spacetime. It seems relevant to mention that MTW's "super-Lagrangian" and "super-Hamiltonian" can be seen as the $e=1$ gauge of the Lagrangian$^1$

$$ L~:=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2} -V , \qquad \dot{x}^2~:=~\sum_{\mu,\nu=0}^3 g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda}, \tag{1} $$

and the corresponding Hamiltonian Lagrangian

$$ L_H~:=~ \sum_{\mu=0}^3p_{\mu} \dot{X}^{\mu} - H, \qquad H~:=~ eT+V,$$ $$ T~:=~\frac{1}{2}(p^2+m^2), \qquad p^2~:=~ \sum_{\mu,\nu=0}^3g^{\mu\nu}(X)~p_{\mu} p_{\nu}, \tag{2} $$

see e.g. this, this & this Phys.SE post. Here $\lambda$ is the worldline parameter, $e$ is an auxiliary einbein field, and $H$ is the Hamiltonian.

The perhaps more familiar square root form appears by integrating out the einbein field $e>0$. Various possible gauge choices are given in my Phys.SE answer here for a free particle in Minkowski spacetime.

--

$^1$ In this answer we choose the Minkowski signature $(-,+,+,+)$ and put the speed of light $c=1$ equal to one.

$\endgroup$
0
$\begingroup$

I have been thinking about this "lack" of potential energy since I posted the question and I've came to the conclusion that the answer is to do with the equivalence principle in GR.

After all, we are concerned with an object in free fall in a gravitational field. As the equivalence principle states we cannot distinguish between a reference frame in free fall and a reference frame far-away from a gravitational field.

Hence, the "lack" of gravitational potential can be explained with the fact that the gravitational field cannot be detected and hence the only energy term that is present is given by the test particles kinetic energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.