# Stress-energy tensor for electromagnetic field with interaction term

1. Dec 18, 2013

### maykot

First of all, I'm not sure if this thread belongs here or at the "Special & General Relativity" sub-forum, if I posted at the wrong place please move it.

1. The problem statement, all variables and given/known data
I encountered this problem working in my master's degree.
I need to find the stress-energy tensor of the following lagrangian:
$$\mathcal{L} = -\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta} - j^{\alpha}A_{\alpha}$$

Where: $F_{\alpha\beta}=\partial _{\alpha} A_{\beta} - \partial _{\beta} A_{\alpha}$

Also, I'm working in curved space-time with the (+---) sign convention.

2. Relevant equations
$$T^{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g_{\mu\nu}} = 2 \frac{\partial \mathcal{L}}{\partial g_{\mu\nu}}-g^{\mu\nu} \mathcal{L}$$
$$\frac{\partial g_{\alpha\beta}}{\partial g_{\mu\nu}} = \delta ^{\mu}_{\alpha} \delta ^{\nu} _{\beta}$$

3. The attempt at a solution
I raised the indices of all the tensors in the lagrangian ($F_{\alpha\beta}=g_{\alpha\rho} g_{\beta\sigma}F^{\rho\sigma}$, $A_{\alpha}=g_{\alpha\rho} A^{\rho}$), considered all of the contravariant tensors constant with respect to $g_{\mu\nu}$ and then used the definition of the SET above and the formula above for the derivatives of the metric, resulting in:
$$T^{\mu\nu} = \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} + g_{\alpha\beta} F^{\mu\alpha} F^{\beta\nu} + g^{\mu\nu}j_{\alpha} A^{\alpha} - 2j^{\mu} A^{\nu}$$

The problem is this is not the usual expression found in textbooks. Greiner, in his book Field Quantization, for instance, uses the Belinfante-Rosenfeld method to improve the canonical SET and finds:
$$T^{\mu\nu} = \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} + g_{\alpha\beta} F^{\mu\alpha} F^{\beta\nu} + g^{\mu\nu}j_{\alpha} A^{\alpha} - j^{\mu} A^{\nu}$$
The only difference being the absence of the factor two in the last term.

Besides that, I cannot help but feel uncomfortable with the procedure I used to get to the SET.
Take the last term for example, making the following change in the lagrangian: $j^{\alpha}A_{\alpha}=j_{\alpha}A^{\alpha}$, the last term in the SET goes from $-2j^{\mu} A^{\nu}$ to $-2j^{\nu} A^{\mu}$.
Since this could in principle be done for any pair of vector fields this would imply that: $A^{\mu} B^{\nu}=A^{\nu} B^{\mu}$ for any $A$ and $B$, which is not true.
I am not sure what is wrong with my calculations, if someone could help me pointing it out I would be very grateful.

2. Dec 19, 2013

### maykot

I traced the problem back to the expression $\frac{\partial g_{\alpha\beta}}{\partial g_{\mu\nu}} = \delta ^{\mu}_{\alpha} \delta ^{\nu} _{\beta}$. It is not symmetric under $\mu \to \nu$ nor $\alpha \to \beta$.
This is the expression I've always used to derive the energy momentum tensor, but I found an expression in this article http://www.sciencedirect.com/science/article/pii/S0003491603002033, page 365, equation 235.
$$\frac{\partial g^{\mu\nu}}{\partial g_{\kappa\lambda}} = -\frac{1}{2} (g^{\mu\kappa} g^{\nu\lambda}+g^{\mu\lambda} g^{\nu\kappa})$$
Using:
$$\frac{\partial \delta^{\mu}_{\nu}}{\partial g_{\kappa\lambda}} = \frac{\partial (g^{\mu\sigma}g_{\sigma\nu})}{\partial g_{\kappa\lambda}}=0$$
It implies that:
$$\frac{\partial g_{\alpha\beta}}{\partial g_{\mu\nu}} = \frac{1}{2} (\delta^{\mu}_{\alpha} \delta^{\nu}_{\beta} + \delta^{\mu}_{\beta} \delta^{\nu}_{\alpha})$$
Which is indeed symmetric under $\mu \to \nu$ and $\alpha \to \beta$ but has the strange property that:
$$\frac{\partial g_{12}}{\partial g_{12}} = \frac{1}{2}$$
And gives the following expression for the SET:
$$T^{\mu\nu} = \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} + g_{\alpha\beta} F^{\mu\alpha} F^{\beta\nu} + g^{\mu\nu}j_{\alpha} A^{\alpha} - (j^{\mu} A^{\nu} + j^{\nu} A^{\mu})$$
That is a bit better in relation to my original SET because it is at least explicitly symmetric, but still different from the expression from Greiner's book.
I'm getting more and more confused here.