# General Relativity, Energy-momentum tensor

1. Oct 13, 2009

### hjalte

1. The problem statement, all variables and given/known data
In Minkowski space, we are given a scalar field $\phi$ with action
$S= \int d\Omega (\frac{-1}{2}\phi^{,a}\phi_{,a} - \frac{1}{2}m^2\phi^2)$

We need to calculate the "translation-invariance" energy-momentum tensor:
$T^a_b = \frac{\partial \mathcal{L}}{\partial \phi_{,a}} \phi_{,b} - \mathcal{L}\delta^a_b$

and then rewrite the action in a generally covariant form, and calculate its "metric" energy-momentum tensor
$\frac{1}{2}\sqrt{-g}T_{ab} = -\frac{\delta(\sqrt{-g}\mathcal{L}}{\delta g^{ab}}$

2. Relevant equations
The action of matter (non-gravitational field) is given by
$S_m = \int \mathcal{L}\sqrt{-g}d\Omega$
Because we are in minkowski space $\sqrt{-g} = 1$, because g is the determinant of the metric tensor.

3. The attempt at a solution
The first part, where we have to calculate the translation-invariance metric tensor I have used, that the equation in the action integral, is the Lagrangian, and then we get
$T^a_b = \frac{\partial (\frac{-1}{2}\phi^{,c}\phi_{,c} - \frac{1}{2}m^2\phi^2)}{\partial \phi_{,a}} \phi_{,b} - \frac{-1}{2}\phi^{,c}\phi_{,c} - \frac{1}{2}m^2\phi^2\delta^a_b$
which give some more or less ugly result.

On the other part, where we have to calculate the metric energy-momentum tensor, I would say that, because the lagrangian does not depend on the metric, we get that
$\frac{\delta(\sqrt{-g} \mathcal{L})}{\delta g^{ab}} = 0$.

But I'm not sure if this is legal, or if I have to write the derivatives in the action, as the covariant derivatives, include the Christoffel symbols, and vary them with the metric tensor, and see what I get.