# How to derive the ADM form of this relativistic Hamiltonian?

The action for a general relativistic particle should be:

$$S=\int d\tau\int d^3x \sqrt{-g} g^{\mu\nu}\dfrac{dx^\mu}{d\tau} \dfrac{dx^\nu}{d\tau} \,\delta^3(x-y)$$

where $$g_{\mu\nu}$$ is the 4-metric. How to derive from this action the ADM form of the Hamiltonian?

The momenta should be $$\pi^\mu=\sqrt{-g} \dfrac{dx^\mu}{d\tau}$$, so after the Legendre transform I get something like this:

$$\int d\tau\int d^3x \dfrac{dx^\mu}{d\tau}\pi_\mu - \left[ \dfrac{\pi^\mu\pi_\mu}{\sqrt{-g}} \right]$$ I do not know how to precede from this point.

• Are you sure that this is the action you should be using? If $x$ is the particle's path, it cannot be integrated over, and also this looks neither like the Nambu-Goto nor the Polyakov action for a point particle. Nov 5 '20 at 9:16