Please don't take my abrupt comments as being rude; I certainly don't intend to be so and if I've come across that way, my apologies.
You claim that MTW say that the momentum is defined by:
\pi^{ij} = \frac{\delta L}{\delta g_{ij}}
You're correct when you claim that MTW say this is how the momentum is defined. However, the important point is that MTW are wrong! This is not as heretical as it may sound. Consider, for example, an ordinary classical system with generalised coordinates q^i[/tex], generalised velocities \dot{q}^i, and a Lagrangian L. In this trivial case we don't define the momentum conjugate to q^i as<br />
<br />
\pi_i = \frac{\partial L}{\partial q^i}<br />
<br />
This is the analogue of what you've written for the momentum in general relativity. The <i>correct</i> way to define the momentum is of course<br />
<br />
\pi_i = \frac{\partial L}{\partial\dot{q}^i}<br />
<br />
This is a trivial observation. The <i>correct</i> way to define the three-momentum in general relativity is then <br />
<br />
\pi^{ij} = \frac{\delta \mathcal{L}}{\delta\dot{g}_{ij}}<br />
<br />
where the Lagrangian density (and note that the correct thing to use is the Lagrangian <i>density</i>, not the Lagrangian itself due to the possible presence of surface terms) is understood to have been expressed in terms of the first and second fundamental forms of a hypersurface in the spacetime.<br />
<br />
If you work through the calculations you'll find that my correct way of defining the three-momentum gives<br />
<br />
\pi^{ij} = \sqrt{g}(g^{ij}K - K^{ij})<br />
<br />
On the other hand, if you use MTW's <i>incorrect</i> definition, you'll get something a whole lot more complicated (you'll get the euler-lagrange equations actually, as you could have noticed had you thought a little bit about the equation in MTW before transcribing it).
