# Einstein Hamiltonian

1. Jul 29, 2006

### eljose

If we have the Einstein Lagrangian...$$L= \sqrt (-g)R$$ my question is how do you get the Hamiltonian?..the approach by Wheeler-De Witt is to consider the line element:

$$ds^2 = N(t)dt^^2 + g_ij dx^i dz^ j$$ (Einstein sum convention) and then substitute it into the Lagrangian above and perform a Legendre transform in the form:

$$\pi_ij \dot g_ij -L$$ where "pi2 are the momenta.

2. Aug 22, 2006

### coalquay404

The approach you quote above is both (a) wrong, and (b) not due to Wheeler & De Witt. The original work on the Hamiltonian, or canonical, description of general relativity is called the ADM formalism (after Roger Arnowitt, Stanley Deser, and Charles Misner). Their paper is called "The dynamics of general relativity" and is available on gr-qc.

The essential content of their work is as follows. One supposes that one has a spacetime $$(\mathcal{M},g)$$ which represents a solution to Einstein's field equations. You can, of course, derive these field equations from the standard Einstein-Hilbert action

$$S = \int d^4x \sqrt{-g}R$$

where $$R$$ is the four-dimensional scalar curvature and $$\sqrt{-g}$$ is a volume element on the manifold. You then suppose that the spacetime can be "foliated" or sliced so that globally* it looks like the product $$\mathbb{R}\times\Sigma$$. You then perform a little bit of geometry to work out that the four-dimensional line element takes the form

$$ds^2 = -(N^2 - N_iN^i)dt^2 + 2N_i dx^idt + \gamma_{ij}dx^idx^j$$

where $$N$$ is the lapse function, $$N^i$$ is the shift vector, and $$\gamma_{ij}$$ is a three-dimensional metric on a spatial hypersurface in the spacetime. A little bit more geometry then allows you to show that the Einstein-Hilbert action can be rewritten in the following form:

$$S = \int dtd^3x\sqrt{\gamma}N(R- \mathrm{tr}K^2 + K_{ij}K^{ij})$$

where $$R$$ now stands for the three-dimensional scalar curvature of the hypersurface, $$K_{ij}$$ is the extrinsic curvature of the hypersurface in the spacetime, and $$\mathrm{tr}K=\gamma^{ij}K_{ij}$$. (I'm ignoring all boundary terms here.) Since the extrinsic curvature is defined in terms of the metric velocity $$\dot{\gamma}_{ij}$$ you can vary the action with respect to the metric velocity to obtain an expression for the canonical momenta

$$\pi^{ij} \equiv \frac{\delta\mathcal{L}}{\delta\dot{\gamma}_{ij}} = \sqrt{\gamma}(\gamma^{ij}\mathrm{tr}K - K^{ij})$$

You then simply write the Hamiltonian as you have suggested above. If you're interested, the best way to learn this is to look at section 2.8 of Hawking & Ellis, followed by chapter 21 of Misner, Thorne, and Wheeler.

*: I seem to recall reading on here recently that some people believe that this foliation has to hold only locally in order that the 3+1 formulation be well-posed. This is not true! You need a global foliation in order that the Cauchy problem is stable, something which becomes particularly important when you consider boundary contributions at spatial infinity.

Last edited: Aug 22, 2006