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Question
When calculating the hamiltonian for the free Electromagnetic Field with Lagrangian density $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ Using Noether's theorem I found the answer to be $$T^{\mu\nu}=-F^{\mu\sigma}\partial^{\nu}A_{\sigma} - \eta^{\mu\nu}\mathcal{L}$$ However, this cannot be true because the energy-momentum tensor is symmetric but the expression that I have calculated is not symmetric. I've looked online and found that it should be $$T^{\mu\nu}=\eta_{\sigma\lambda}F^{\mu\sigma}F^{\lambda\nu}- \eta^{\mu\nu}\mathcal{L}$$

My Working
I began with Noether's theorem for the energy-momentum tensor $$T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\sigma)}\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ Now I substituted in the lagrangian for the free field which gives us $$T^{\mu\nu}=\frac{\partial}{\partial (\partial_\mu A_\sigma)}[-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}]\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ Factoring the constant outside of the derivitive leaves the expression as $$T^{\mu\nu}=-\frac{1}{4}\frac{\partial}{\partial (\partial_\mu A_\sigma)}[F_{\alpha\beta}F^{\alpha\beta}]\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ We can now use the chain rule to seperate the multiplied terms $$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}\frac{\partial F_{\alpha\beta}}{\partial (\partial_\mu A_\sigma)}\partial^\nu A_\sigma -\frac{1}{4}F_{\alpha\beta}\frac{\partial F^{\alpha\beta}}{\partial (\partial_\mu A_\sigma)}\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ Now if we substitute the definition for the $F_{\alpha\beta}$ tensor $$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})\partial^\nu A_\sigma -\frac{1}{4}F_{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha})\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ The second term in the expression can be lowered by the minkowski metric $$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})\partial^\nu A_\sigma -\frac{1}{4}\eta^{\alpha\alpha'}\eta^{\beta\beta'}F_{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial_{\alpha'}A_{\beta'}-\partial_{\beta'}A_{\alpha'})\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ To evaluate the derivitives we can look at what happens if we look at the expression $$\frac{\partial}{\partial x^\alpha} x^\beta$$ The derivitive should be $0$ for all indexes exept for when $\alpha=\beta$ which is the definition of $\delta_{\alpha}^{\beta}$. So this means that $$\frac{\partial}{\partial x^\alpha} x^\beta=\delta_{\alpha}^{\beta}$$ If we look at taking the derivitive with respect to a rank two tensor $$\frac{\partial}{\partial G^{\alpha\beta}} G^{\sigma\gamma}$$ This means that $\alpha = \sigma$ and $\beta = \gamma$ which can be written as $$\frac{\partial}{\partial G^{\alpha\beta}} G^{\sigma\gamma} = \delta_{\alpha}^{\sigma}\delta_{\beta}^{\gamma}$$ However we know that $\frac{\partial}{\partial (\partial_\mu A_\sigma)}$ is simpily the derivitive with respect to a rank 2 tensor so applying the results yeilds $$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma -\frac{1}{4}\eta^{\alpha\alpha'}\eta^{\beta\beta'}F_{\alpha\beta}(\delta_{\alpha'}^{\mu}\delta_{\beta'}^\sigma-\delta_{\beta'}^{\mu}\delta_{\alpha'}^\sigma)\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ We can raise the indexes of the Electromagnetic Field Strengrh tensor with the minkoiski metric $$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma -\frac{1}{4}F^{\alpha'\beta'}(\delta_{\alpha'}^{\mu}\delta_{\beta'}^\sigma-\delta_{\beta'}^{\mu}\delta_{\alpha'}^\sigma)\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ Since we are summing over $\alpha'$ and $\beta'$ we can replace them with $\alpha$ and $\beta$ $$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma -\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ Since we now have like terms we can combine them which gives $$T^{\mu\nu}=-\frac{1}{2}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ If we apply the kroneker delta we get that $$T^{\mu\nu}=-\frac{1}{2}(F^{\mu\sigma}-F^{\sigma\mu})\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$ Finaly since $F^{\mu\sigma}$ is anti symetric $$T^{\mu\nu}=-F^{\mu\sigma}\partial^\nu A_\sigma - \eta^{\mu\nu}\mathcal{L}$$

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    $\begingroup$ @NiharKarve Please don't close it I spent ages writing this out $\endgroup$ – Joshua Pasa Dec 28 '20 at 3:22
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    $\begingroup$ @NiharKarve The working out is just there for the background to show that I did research beforehand and that I gave enough detail so that the question wouldn't close $\endgroup$ – Joshua Pasa Dec 28 '20 at 3:24
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    $\begingroup$ Wikipedia: “Noether's theorem implies that there is a conserved current associated with translations through space and time. This is called the canonical stress–energy tensor. Generally, this is not symmetric...” So the fact that it is symmetric in EM is not to be expected! The Hilbert definition, on the other hand, is guaranteed to be symmetric. $\endgroup$ – G. Smith Dec 28 '20 at 4:35
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    $\begingroup$ Related: physics.stackexchange.com/q/37764 $\endgroup$ – Nihar Karve Dec 28 '20 at 4:40
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    $\begingroup$ The way this is symmetrised is usually by adding a term $C^{\mu\nu}$ to $T^{\mu\nu}$, such that it satisfies $\partial_\mu C^{\mu\nu} = 0$. You then pick $C^{\mu\nu}$ so as to make $T$ symmetric under $\mu \leftrightarrow \nu$ $\endgroup$ – Nihar Karve Dec 28 '20 at 4:47
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This is the Belinfante-Rosenfeld procedure (note that it is not necessary to invoke spin currents for the free electromagnetic theory). Just to spell things out, this involves modifying the canonical energy-momentum tensor by adding a divergenceless term (an alternative method is using the Hilbert definition, see this question for the relation between the two):

$$T^{\mu\nu}=-F^{\mu\sigma}\partial^{\nu}A_{\sigma} - \eta^{\mu\nu}\mathcal{L} +C^{\mu\nu},$$

where $\partial_\mu C^{\mu\nu}$ is a zero extremal variation of the Lagrangian, and thus preserves the conservation of $T_{\mu\nu}$. This does modify the $T^{0\nu}$ densities, but as G. Smith says:

But the point of Noether’s theorem is just to produce something that is conserved. What it produces isn’t necessarily the only thing that is conserved.

Anyway, if we make the ad hoc choice of $C^{\mu\nu}$ as $F^{\mu\sigma}\partial_\sigma A^\nu$, you can see that $$ \partial_\mu C^{\mu\nu}= (\partial_\mu F^{\mu\sigma})\partial_\sigma A^\nu+F^{\mu\sigma}\partial_\mu\partial_\sigma A^\nu=0, $$ since the first term is zero by the equations of motion, while the second term involves a contraction between a symmetric and antisymmetric indices, which can be shown to be zero. Hence the inclusion of this term modifies the energy-momentum tensor to be:

$$T^{\mu\nu}=F^{\mu\sigma}\partial_\sigma A^\nu-F^{\mu\sigma}\partial^{\nu}A_{\sigma} - \eta^{\mu\nu}\mathcal{L} \\ = \eta_{\sigma\lambda} (F^{\mu\sigma}\partial^{\lambda}A^{\nu}-F^{\mu\sigma}\partial^\nu A^\lambda) - \eta^{\mu\nu}\mathcal{L} \\ = \eta_{\sigma\lambda}F^{\mu\sigma}F^{\lambda\nu}- \eta^{\mu\nu}\mathcal{L} $$

which is exactly what you set out to show.

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  • $\begingroup$ Thank you for writing it out in full detail :) $\endgroup$ – Joshua Pasa Dec 28 '20 at 16:43
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    $\begingroup$ No problem! :-) $\endgroup$ – Nihar Karve Dec 28 '20 at 16:45
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The procedure, which makes the Noether tensor of EM symmetric, is known as Belinfante's method. You can find it in the standard textbooks, e.g. Landau's second volume, please google it, there are a lot of student notes discussing it, e,g. Blau's note.

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The standard way to fix this is to adopt the Belinfante-Rosenfeld tensor instead of the Noether tensor. This raises hairs. Why is the Noether tensor wrong? Not only the EM tensor must be symmetrized, also the angular momentum tensor had problems. Why is this Noether current also wrong? This problem must then be solved by ditching spin and writing it as an orbital momentum based on the the BR tensor. Clearly spin is an observable, as shown by the famous experiments of Beth in 1935. Why is there no expression for spin in the symmetrized theory?

In my paper on this, see https://arxiv.org/abs/physics/0106078, I argue that the problems stem from the Lagrangian itself and explore the alternative, the Fermi Lagrangian.

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  • $\begingroup$ Is this kind of like gauge fixing, but to make the stress-energy tensor symmetric? $\endgroup$ – Joshua Pasa Dec 28 '20 at 17:01
  • $\begingroup$ @JoshuaPasa It is not gauge fixing. A gauge transformation affects a solution of the equations of motion. This is a modification of the Lagrangian itself, so it affects the equations of motion and the conservation laws. Note that some might call it gauge fixing by extending the meaning of the term to modification of the Lagrangian. The effect of this Lagrangian is that there is no gauge invariance to begin with. $\endgroup$ – my2cts Dec 28 '20 at 18:14

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