# Energy-Momentum-Tensor of classical electrodynamics is conserved

I want to check if the energy momentum tensor of the classical electrodynamics with lagrangian \begin{align} L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \end{align} is conserved. The energy momentum tensor obtained from noethers theorem is \begin{align} T^{\alpha\beta} = -\frac{1}{4}\eta^{\alpha\beta}F_{\mu\nu}F^{\mu\nu}+F^{\alpha\sigma}\partial^\beta A_\sigma \end{align} It seems like an easy exercise, but I cant show, that indeed $$\partial_\alpha T^{\alpha\beta} = 0$$. My idea was to use the euler lagrange equation $$\partial_\mu F^{\mu\nu} = 0$$ which leads to \begin{align} \partial_\alpha T^{\alpha\beta} = F^{\alpha\sigma}\partial_\alpha\partial^\beta A_\sigma \end{align} but I cant see, why this should vanish.

• There is a mistake with your indices I think Commented Jan 19, 2023 at 12:33
• Thank you thats right I had a typing mistake which is now corrected. Commented Jan 19, 2023 at 12:34
• Perhaps you should look into the derivative of the first term of $T^{\alpha\beta}. Commented Jan 19, 2023 at 15:13 • You can see physics.stackexchange.com/questions/51778/… Commented Sep 5, 2023 at 7:30 ## 2 Answers You actually missed a serious term. You thought that $$\partial_\alpha \left(F_{\mu\nu}F^{\mu\nu}\right)=0$$. But actually it is not, because $$F_{\mu\nu}F^{\mu\nu}$$ represents a summation. You confused that with the notation in the indices! So here is the sketch of the proof below (also note, for the sake of easiness, it is better to write the expression of $$T^\alpha_\beta$$ instead of $$T_{\alpha\beta}$$ or $$T^{\alpha\beta}$$; let me do the most general proof, there you may put $$J^\mu=0$$. At first, from your Lagrangian, arrive at the following form of the stress energy tensor (and then follow the rest of the part accordingly): $$T^\alpha_\beta = \dfrac{1}{4\pi}\left(F^{\alpha\gamma}F_{\beta\gamma}-\dfrac{1}{4}\eta^{\alpha}_{\beta}F_{\gamma\delta}F^{\gamma\delta}\right)$$ alongwith two of the Maxwell's equation $$\partial_\mu F^{\mu\nu} = 4\pi J^\nu$$ $$\therefore \partial_\alpha T^\alpha_\beta = -\dfrac{1}{4\pi}\left[\dfrac{1}{4}\times 2~F_{\gamma\delta}\left(\partial_\alpha F^{\gamma\delta}\right) - F^{\alpha\gamma}\left(\partial_\gamma F_{\beta\gamma}\right) - F_{\beta\gamma}\left(\partial_\gamma F^{\alpha\gamma}\right)\right]$$ $$\Longrightarrow \partial_\alpha T^\alpha_\beta = -\dfrac{1}{4\pi}\left[\dfrac{1}{2} F^{\gamma\delta}\left(\partial_\gamma F_{\delta\alpha}\right) + \dfrac{1}{2} F^{\gamma\delta}\left(\partial_\delta F_{\alpha\gamma}\right) + \dfrac{1}{2} F^{\alpha\gamma}\left(\partial_\alpha F_{\beta\gamma}\right) + 4\pi F_{\beta\gamma}J^{\gamma}\right]$$ using the Bianchi identity, $$\partial_{[\alpha}F_{\beta\gamma]} = \partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta} = 0$$ . Now the first three terms in the last expression cancel each other (if unclear to you, you may check it!) and finally we obtain, $$\partial_\alpha T^{\alpha}_{\beta} = -F_{\beta\gamma}J^{\gamma}$$ ; This is the most general form of the desired conservation law of electromagnetic stress energy tensor (you can always manipulate this for $$T^{\alpha\beta}$$ or for $$T_{\alpha\beta}$$ by multiplying the metric tensor $$\eta_{\mu\nu}$$.). In vacuum, i.e. for $$J^\alpha=0$$, the conservation equation reduces to: $$\partial_\alpha F^{\alpha}_{\beta} = 0$$ . For more references, you may have a look at the famous as well as the legendary book by Thanu Padmanabhan, "Gravitation: Foundations and Frontiers". The entire proof, alongwith the related concepts are well described there. • The indices are not consistent in the expression of$\partial_{\alpha}{T_{\beta}}^{\alpha}\$ Commented Jan 24, 2023 at 8:23

\begin{align*} \partial_\mu T^{\mu\nu}_{\text{EM}} &= \partial_\mu \left(F^{\mu\lambda} F^\nu_{\ \lambda}\right) - \frac{1}{4} \eta^{\mu\nu} \partial_\mu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) \\ &= F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} + F^\nu_{\ \lambda} \partial_\mu F^{\mu\lambda} - \frac{1}{4} \eta^{\mu\nu} \partial_\mu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) \\ &= F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} + F^\nu_{\ \lambda} \partial_\mu F^{\mu\lambda} - \frac{1}{4} \partial^\nu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) \end{align*}

From Maxwell's Equations:

$$\partial_\mu F^{\mu\nu} = -J^\nu$$ We have: $$\partial_\mu T^{\mu\nu}_{\text{EM}} = -F^\nu_{\ \lambda} J^\lambda + F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} - \frac{1}{4} \partial^\nu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right)$$ Also we have: \begin{align*} \partial^\nu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) &= F^{\lambda\sigma} \partial^\nu F_{\lambda\sigma} + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= \eta^{\lambda\tau} \eta^{\rho\sigma} F_{\tau\rho} \partial^\nu F_{\lambda\sigma} + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= F_{\lambda\sigma} \partial^\nu \left(\eta^{\lambda\tau} \eta^{\rho\sigma} F_{\tau\rho}\right) + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= 2F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \end{align*} and the same: \begin{align} F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} &= \left(\eta^{\mu\alpha} \eta^{\beta\lambda} F_{\alpha \beta}\right) \left(\eta_{\mu\gamma} \partial^{\gamma}\right) \left(\eta_{\sigma\lambda} F^{\nu\sigma}\right) \\ &= \eta^{\mu\alpha} \eta^{\beta\lambda} \eta_{\mu\gamma} \eta_{\sigma\lambda} \left(F_{\alpha \beta} \partial^{\gamma} F^{\nu\sigma}\right) \\ &= \delta^\alpha_\gamma \delta^\beta_\sigma \left(F_{\alpha \beta} \partial^{\gamma} F^{\nu\sigma}\right) \\ &= F_{\gamma\sigma} \partial^{\gamma} F^{\nu\sigma} \\ &= F_{\mu\lambda} \partial^\mu F^{\nu\lambda} \end{align} so that: \begin{align} \partial_\mu T^{\mu\nu}_{\text{EM}} &= -F^\nu_{\ \lambda} J^\lambda + F_{\mu\lambda} \partial^\mu F^{\nu\lambda} - \frac{1}{2} F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= -F^\nu_{\ \lambda} J^\lambda + F_{\mu\lambda} \partial^\mu F^{\nu\lambda} - \frac{1}{2} F_{\lambda\mu} \partial^\nu F^{\lambda\mu} \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} F_{\mu\lambda} \partial^\mu F^{\nu\lambda} + \frac{1}{2} F_{\mu\lambda} \partial^\mu F^{\nu\lambda} - \frac{1}{2} F_{\lambda\mu} \partial^\nu F^{\lambda\mu} \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} \left(F_{\mu\lambda} \partial^\mu F^{\nu\lambda} + F_{\lambda\mu} \partial^\lambda F^{\nu\mu} - F_{\lambda\mu} \partial^\nu F^{\lambda\mu}\right) \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} \left(F_{\mu\lambda} \partial^\mu F^{\nu\lambda} + F_{\mu\lambda} \partial^\lambda F^{\mu\nu} + F_{\mu\lambda} \partial^\nu F^{\lambda\mu}\right) \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} F_{\mu\lambda} \left(\partial^\mu F^{\nu\lambda} + \partial^\lambda F^{\mu\nu} + \partial^\nu F^{\lambda\mu}\right) \end{align} Lastly we use Bianchi Identities: $$\partial^\mu F^{\nu\lambda} + \partial^\lambda F^{\mu\nu} + \partial^\nu F^{\lambda\mu} = 0 \Leftrightarrow \partial^{[\mu} F^{\nu\lambda]} = 0$$ we get: $$\partial_\mu T^{\mu\nu}_{\text{EM}} = -F^\nu_{\ \lambda} J^\lambda$$

I don't know if we can change the $$\lambda$$ to $$\mu$$ in the last equation, maybe we can because it is a dummy index.