# Confirming $\nabla_\mu T^{\mu\nu} = 0$ for the electromagnetic field tensor in curved spacetime

I'm trying to show that $$\nabla_\mu T^{\mu\nu} = 0$$ for the electromagnetic stress energy tensor which I derived to be,

$$T^{\mu\nu} = F^{\mu\beta}F^{\nu}_{\beta}-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$$ I have calculated $$\nabla_\mu T^{\mu\nu} = \nabla_\mu\left( F^{\mu\beta}F^{\nu}_{\beta}-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\right)$$

$$= (\nabla_\mu F^{\mu\beta})F^\nu_\beta+(\nabla_\mu F^\nu_\beta)F^{\mu\beta}-\frac{1}{4}g^{\mu\nu}(\nabla_\mu F_{\alpha\beta})F^{\alpha\beta}-\frac{1}{4}g^{\mu\nu}(\nabla_\mu F^{\alpha\beta})F_{\alpha\beta}$$ Now I know that the first term is zero since the equation of motion reads $$\nabla_\mu F^{\mu\nu} = 0$$ but I can't see how the rest of the terms are zero.

Important edit:

If we start from the equation of motion we have, $$\nabla_\mu F^{\mu\nu}=0$$. By multiplying both sides with $$g_{\mu\alpha}g_{\nu\beta}$$ we get $$g_{\mu\alpha}g_{\nu\beta}\nabla_\mu F^{\mu\nu}=\nabla_\mu F_{\alpha\beta} = 0$$ Does that imply $$\nabla_\mu F_{\alpha\beta} = 0$$?

Important edit response:

What I have written down is incorrect since you cannot multiply metrics with summed indices, the equation is invalid!

• The edit you just made is really another question. If you ask another question, I'll give you an answer there. – John Dumancic Nov 8 '19 at 15:45

## 1 Answer

We have $$\nabla_\mu T^{\mu\nu}=(\nabla_\mu F^\nu_{\:\,\beta})F^{\mu\beta}-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta}-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F^{\alpha\beta}\right) F_{\alpha\beta}\tag{1}$$ Now, note the following: \begin{align} -\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F^{\alpha\beta}\right) F_{\alpha\beta} & =-\frac{1}{4}g^{\mu\nu}g^{\alpha\gamma}g^{\beta\zeta}\left(\nabla_\mu F_{\gamma\zeta}\right)F_{\alpha\beta}\tag{2} \\ & = -\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F_{\gamma\zeta}\right)F^{\gamma\zeta} \\ & = -\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta} \end{align} So, now we have $$\nabla_\mu T^{\mu\nu}=g^{\nu\sigma}(\nabla_\mu F_{\sigma\beta})F^{\mu\beta}-\frac{1}{2}g^{\mu\nu}\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta}\tag{3}$$ I took the liberty of lowering an index in the first term. Multiply both sides by the metric \begin{align} g_{\nu\rho}\nabla_\mu T^{\mu\nu} &= \delta^\sigma_\rho\left(\nabla_\mu F_{\sigma\beta}\right)F^{\mu\beta}-\frac{1}{2}\delta^\mu_\rho\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta}\tag{4} \\ & = \left(\nabla_\mu F_{\rho\beta}\right)F^{\mu\beta}-\frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right)F^{\alpha\beta}\tag{5} \\ \end{align} Now, we now remember another of our equations of motion: $$\nabla_\mu F_{\nu\sigma}+\nabla_\nu F_{\sigma\mu} +\nabla_\sigma F_{\mu\nu}=0$$. Using this, \begin{align} g_{\nu\rho}\nabla_\mu T^{\mu\nu} &= -\left(\nabla_\rho F_{\beta\mu} + \nabla_\beta F_{\mu\rho} \right) F^{\mu\beta}-\frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right)F^{\alpha\beta}\tag{6} \\ & = -\left(\nabla_\beta F_{\rho\mu} \right) F^{\beta\mu} + \left(\nabla_\rho F_{\mu\beta}\right) F^{\mu\beta}-\frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right)F^{\alpha\beta}\tag{7} \end{align} Now, we rename indices (they are arbitrary labels; as long as we are consistent, we can call them what we want) and simplify: \begin{align} g_{\nu\rho}\nabla_\mu T^{\mu\nu} & = -\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} + \left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}-\frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right)F^{\alpha\beta}\tag{8} \\ & = -\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} + \frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}\tag{9} \\ & = -\left(\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} - \frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}\right) \tag{10} \end{align} So, we see that $$g_{\nu\rho}\nabla_\mu T^{\mu\nu}$$ is equal to both $$(5)$$ and $$(10)$$. Thus, $$\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} - \frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}=g_{\nu\rho}\nabla_\mu T^{\mu\nu}=-\left(\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} - \frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}\right)\tag{11}$$ $$2\left(\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} - \frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}\right)=0\tag{12}$$ $$\left(\nabla_\mu F_{\rho\beta} \right) F^{\mu\beta} - \frac{1}{2}\left(\nabla_\rho F_{\alpha\beta}\right) F^{\alpha\beta}=0\tag{13}$$ $$g_{\nu\rho}\nabla_\mu T^{\mu\nu}=0\tag{14}$$ Acting both sides with $$g^{\rho\omega}$$ and relabeling indices gives the desired result $$g^{\rho\omega}g_{\nu\rho}\nabla_\mu T^{\mu\nu}=\delta^\omega_\nu\nabla_\mu T^{\mu\nu}=\nabla_\mu T^{\mu\omega}=0\tag{15}$$ $$\nabla_\mu T^{\mu\nu}=0\tag{16}$$

• Aren't you missing a term in the first equation? $\nabla_\mu T^{\mu\nu}=(\nabla_\mu F^\nu_{~~~\beta})F^{\mu\beta}+F^\nu_{~~~\beta}(\nabla_\mu F^{\mu\beta})-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta}-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F^{\alpha\beta}\right) F_{\alpha\beta}$ – user137661 Nov 8 '19 at 4:23
• OP is working with zero charge density, so $\nabla_\mu F^{\mu\beta} = 0$, and the second term in your equation is zero. As OP pointed this out in the last sentence of the question, I elected to not include it, as it was already known that the term was zero. – John Dumancic Nov 8 '19 at 5:16
• Great answer! Could you assume (2) from the fact that the indices are summed over hence we can write them upstairs or downstairs as we want? – redhood Nov 8 '19 at 12:47
• Also, is there a more direct proof? It feels like we are subtracting the same equation and setting it equal to $\nabla_\mu T^{\mu\nu}$ instead of deriving it. – redhood Nov 8 '19 at 13:21
• Not sure on your first point; I always invoke the metric to make sure I do everything right. For your second point, what I did was prove that $g_{\nu\rho}\nabla_\mu T^{\mu\nu}$ was equal to a quantity and it's additive inverse. As the only quantity equal to it's own additive inverse is 0, $g_{\nu\rho}\nabla_\mu T^{\mu\nu} = \nabla_\mu T^{\mu\nu} =0$. For a direct proof, perhaps one would expand out the expression into components, but that would be both complex and tedious. – John Dumancic Nov 8 '19 at 15:17