General Relativity, Energy-momentum tensor

[hjalte]

Homework Statement

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
$\frac{1}{2}\sqrt{-g}T_{ab} = -\frac{\delta(\sqrt{-g}\mathcal{L}}{\delta g^{ab}}$

Homework 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.

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.



Your help is much appreciated

 

[Jhero]

! I would first like to clarify that the question is not entirely clear. It is not specified what the scalar field \phi represents or what the integration domain is (\Omega). Also, the notation used is not standard, so I will try to provide a general answer.

First, let's clarify the meaning of a "translation-invariance" energy-momentum tensor. This refers to the fact that the energy-momentum tensor is invariant under translations, meaning that it does not change when we shift the coordinates of the system. This is a fundamental property of physical laws and theories, and it is a consequence of the homogeneity of space.

Now, to calculate the energy-momentum tensor, we can use the definition given in the homework equations:
T^a_b = \frac{\partial \mathcal{L}}{\partial \phi_{,a}} \phi_{,b} - \mathcal{L}\delta^a_b

Here, \mathcal{L} is the Lagrangian density, which is a function of the fields and their derivatives. In this case, we have a scalar field \phi, so the Lagrangian density can be written as:
\mathcal{L} = \frac{-1}{2}\phi^{,a}\phi_{,a} - \frac{1}{2}m^2\phi^2

Now, we can calculate the derivatives with respect to \phi_{,a}:
\frac{\partial \mathcal{L}}{\partial \phi_{,a}} = \frac{-1}{2}\phi^{,a}

Substituting this into the expression for the energy-momentum tensor, we get:
T^a_b = \frac{-1}{2}\phi^{,a}\phi_{,b} - \frac{-1}{2}\phi^{,c}\phi_{,c}\delta^a_b - \frac{1}{2}m^2\phi^2\delta^a_b

Now, we can simplify this expression by using the fact that the scalar field is a function of the coordinates only, so its derivatives with respect to space and time are:
\phi^{,a} = \frac{\partial \phi}{\partial x^a} = \partial^a\phi

Substituting this into the energy-momentum tensor, we get:
T^a_b = \partial^a\phi