# I Derivation of E.M. Stress Energy Tensor

1. Aug 11, 2016

### Andrew Kim

From Carroll (2004)
It is possible to derive the Einstein Equations (with $c=1$) via functional variation of an action
$$S=\dfrac{S_H}{16\pi G}+S_M$$
where
$$S_H= \int \sqrt{-g}R_{\mu\nu}g^{\mu\nu}d^4 x$$
and $S_M$ is a corresponding action representing matter. We can decompose $\delta S_H$ into three subsequent actions, i.e.
$$\delta S_H=(\delta S)_1+(\delta S)_2+(\delta S)_3$$
$$(\delta S)_1=\int \sqrt{-g}\big(R_{\mu\nu}\big)\delta g^{\mu\nu} d^4 x$$
$$(\delta S)_2=\int R\delta\sqrt{-g}d^4 x = \int \sqrt{-g}\bigg(-\frac{1}{2}g_{\mu\nu}R\bigg)\delta g^{\mu\nu}d^4 x$$
$$(\delta S)_3=\int \sqrt{-g}g^{\mu\nu}\delta R_{\mu\nu}d^4 x$$
It turns out that $(\delta S)_3=0$, so we have
$$\delta S_H=\int\sqrt{-g}\bigg(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\bigg)\delta g^{\mu\nu}d^4 x$$
And therefore
$$\frac{\delta S_H}{\delta g^{\mu\nu}} = \sqrt{-g}\bigg(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\bigg)$$
Finding the extremal values of our original action yields
$$\frac{\delta S}{\delta g^{\mu\nu}}=\frac{\delta S_H}{\delta g^{\mu\nu}}+\frac{\delta S_{M}}{\delta g^{\mu\nu}}=0$$
or
$$\sqrt{-g}\bigg(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\bigg)=-16\pi G\frac{\delta S_{M}}{\delta g^{\mu\nu}}$$
It is at this point that we define
$$T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_{M}}{\delta g^{\mu\nu}}$$
and we find
$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi G T_{\mu\nu}$$
If we replace the matter action with an action for electromagnetism in GR, we have
$$T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_{EM}}{\delta g^{\mu\nu}}$$
$$S_{EM} = \int \sqrt{-g}\mathcal{L}_{EM}d^4 x$$
The value of $\mathcal{L}_{EM}$ that yields maxwell's equations when we use the Euler-Lagrange equations with respect to the fields $A_{\mu}$ is
$$\mathcal{L}_{EM} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+A_{\mu}J^{\mu}$$
We can find the electromagnetic stress energy if vary $S_{EM}$ with respect to the metric. However, I'm not sure how to conduct this functional variation. I know the fields in gravitational GR are $g^{\mu\nu}$ and the fields in relativistic EM are $A_{\mu}$. Should I vary with respect to both and only count the terms that have $\delta g^{\mu\nu}$, or should I treat $A_{\mu}$ as a set of constants like $J_{\mu}$?

2. Aug 12, 2016

### haushofer

Varying wrt the metric gives you the EM-tensor. So you should write the F-squared term in terms of the upper metric and do the variation. All the other fields are kept fixed. Varying with respect to the vector potential A gives you simply the equations of motion for this potential. You don't need the coupling of A to some current J; you only need that if there are other fields coupling to the electromagnetic field besides the metric (the coupling of the electromagnetic field to gravity is already described by the EM-tensor!) So you can put J=0.

3. Aug 12, 2016

### Andrew Kim

As I said, I'm reading from Carroll's textbook, and he discusses the idea of "coupling fields" without ever defining what fields are or what it means for fields to couple. Can you please briefly describe those concepts?

4. Aug 13, 2016

### dextercioby

I am afraid you are reading from the wrong book, and I mean the book which is not meant for such matters. Can you go to your closest univ. library and pick the 75 page booklet on the essentials of GR by PAM Dirac? You should have in 20 pages more than enough calculations and explanations.

5. Aug 13, 2016

### vanhees71

Dirac is always utmost clear in his writing. I marvel at his papers, which usually can just taken as they are also as a chapter in a textbook.

6. Aug 13, 2016

### haushofer

You can consult any textbook on field theory; i like Tomas Ortin's Gravity and Strings.