4
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ The edit you just made is really another question. If you ask another question, I'll give you an answer there. $\endgroup$ – Tesseract Nov 8 '19 at 15:45
4
$\begingroup$

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 $$g^{\rho\omega}g_{\nu\rho}\nabla_\mu T^{\mu\nu}=\delta^\omega_\nu\nabla_\mu T^{\mu\nu}=\nabla_\mu T^{\mu\omega}=0\tag{15}$$ $$\nabla_\mu T^{\mu\nu}=0\tag{16}$$

$\endgroup$
  • $\begingroup$ Aren't you missing a term in the first equation? $\nabla_\mu T^{\mu\nu}=(\nabla_\mu F^\nu_{~~~\beta})F^{\mu\beta}+F^\nu_{~~~\beta}(\nabla_\mu F^{\mu\beta})-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta}-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F^{\alpha\beta}\right) F_{\alpha\beta}$ $\endgroup$ – user137661 Nov 8 '19 at 4:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Tesseract Nov 8 '19 at 5:16
  • $\begingroup$ 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? $\endgroup$ – fielder Nov 8 '19 at 12:47
  • $\begingroup$ 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. $\endgroup$ – fielder Nov 8 '19 at 13:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Tesseract Nov 8 '19 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.