- #1
hjalte
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Homework Statement
In Minkowski space, we are given a scalar field [itex]\phi[/itex] with action
[itex] S= \int d\Omega (\frac{-1}{2}\phi^{,a}\phi_{,a} - \frac{1}{2}m^2\phi^2)[/itex]
We need to calculate the "translation-invariance" energy-momentum tensor:
[itex] T^a_b = \frac{\partial \mathcal{L}}{\partial \phi_{,a}} \phi_{,b} - \mathcal{L}\delta^a_b[/itex]
and then rewrite the action in a generally covariant form, and calculate its "metric" energy-momentum tensor
[itex] \frac{1}{2}\sqrt{-g}T_{ab} = -\frac{\delta(\sqrt{-g}\mathcal{L}}{\delta g^{ab}}[/itex]
Homework Equations
The action of matter (non-gravitational field) is given by
[itex] S_m = \int \mathcal{L}\sqrt{-g}d\Omega[/itex]
Because we are in minkowski space [itex]\sqrt{-g} = 1[/itex], because g is the determinant of the metric tensor.
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
[itex]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[/itex]
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
[itex] \frac{\delta(\sqrt{-g} \mathcal{L})}{\delta g^{ab}} = 0[/itex].
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.
Your help is much appreciated