# Electromagnetic Stress-Energy Tensor in curved space-time

I found on Wikipedia that the electromagnetic stress energy tensor in curved space-time with sign convention $$(-, +, +, +)$$ is

$$T_{\mu\nu} = -\frac{1}{\mu_0} \left ( F_{\mu \alpha} g^{\alpha \beta} F_{\beta \nu} - \frac{1}{4} g_{\mu \nu} F_{\sigma \alpha} g^{\alpha \beta} F_{\beta \rho} g^{\rho \sigma} \right ).$$

However, I need a reputable source for this equation. Does anyone know another source for this equation? All I could find was

$$T_{\mu\nu} = \frac{1}{\mu_{0}}( F^{\beta}{}_{\mu}F_{\beta\nu} - \frac{1}{4}g_{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}).$$

• The second equation you typed is the same as the first by definition of index raising. If you want to derive it from scratch, use the definition as the variation of the matter Lagrangian. – user195162 May 6 '19 at 2:58
• Pg. 177, A Relativist's Toolkit, Eric Poisson. – Avantgarde May 6 '19 at 18:54

## 2 Answers

You can derive this expression by your own from the Lagrangian $$\mathscr{L} = -\frac{1}{4}\sqrt{g} F_{\mu\nu}F^{\mu\nu}$$ where $$g$$ is the absolute value of determinant of $$g_{\mu\nu}$$. Find out how the action varies when the metric is varied. Then from the general definition $$S = \int{\text{d}^4x \sqrt{g}T^{\mu\nu}\delta g_{\mu\nu}}$$ The following relations might help you $$\delta F^{\mu\nu} = -F^{\mu\sigma}g^{\nu\lambda}\delta g_{\lambda\sigma}+F^{\nu\lambda}g^{\mu\sigma}\delta g_{\lambda\sigma}$$ $$\delta g=gg^{\lambda\sigma}\delta g_{\lambda\sigma}$$ Hope this helps.

Reminder: Because this is written in natural units you will not get the factor of $$\frac{1}{\mu_0}$$.

The metric tensor $$g_{\bullet\bullet}$$ is used for more than just most tensors are: it and its inverse $$g^{\bullet\bullet}$$ (defined so $$g_{\alpha\beta}g^{\beta\gamma}=\delta_\alpha^\gamma$$) are used for a canonical bijection between the vector and covector spaces.

So the vector $$\vec v$$ is considered to be equivalent to the covector $$(\vec v\cdot)$$ which takes vectors to scalars, and vice versa.

So by this bijection, tautologically, $$F^{\alpha\beta}=g^{\alpha\mu}~g^{\beta\nu}~F_{\mu\nu}.$$We could have chosen a different invertible tensor to make this correspondence, but we chose the metric because it is so ubiquitous.