The Lagrangian density of the EM field is given by $$ \mathcal{L} = \frac{1}{8\pi}\left(E^2-B^2\right) $$ Let $\vec{A}$,$\phi$ be such that $$ \vec{E} = -\frac{1}{c}\frac{\partial\vec{A}}{\partial t} - \nabla\phi $$ $$ \vec{B} = \nabla\times \vec{A} $$ Then $$ \mathcal{L} = \frac{1}{8\pi}\left(\frac{1}{c^2}\left(\partial_tA_i\right)^2+ (\partial_i\phi)^2 + \frac{2}{c}\left(\partial_tA_i\right)\left(\partial_i\phi\right) - \epsilon_{ijk}\partial_jA_k\epsilon_{ilm}\left(\partial_jA_k\right)\left(\partial_lA_m\right)\right) $$ From Noether's theorem, we have that $$ J^\mu_\nu = \frac{\partial\mathcal{L}}{\partial\left(\partial_\mu A_i\right)}\partial_\nu A_i + \frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi\right)}\partial_\nu\phi - \delta_{\mu\nu}\mathcal{L} $$ where $\partial_0=\partial_t,\partial_i = \partial_{x_i}$ The momentum density vector is then $J^0_\nu$, where $\nu=1,2,3$, and for the EM field it is $$ J^0_{\nu} = -\frac{1}{4\pi c}E_i\partial_\nu A_i $$ How can I reach the Poynting vector from this expression?
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$\begingroup$ The Noether current is the canonical energy-momentum: $J_{\nu}^{\mu} \equiv {\Theta^{\mu}}_{\nu}$ and it isn't unique since you can add a divergence to it without changing its local conservation. You just need to symmetrise the Noether current using the Belinfante-Rosenfeld procedure. $\endgroup$– ChamJan 14, 2019 at 14:24
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$\begingroup$ en.wikipedia.org/wiki/Stress–energy_tensor#Variant_definitions_of_stress–energy $\endgroup$– G. SmithJan 14, 2019 at 17:46
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$\begingroup$ Fix URL in @G.Smith's comment: Variant definitions of Stress Energy tensor $\endgroup$– NickDJan 4, 2022 at 19:11
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2 Answers
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The canonical energy momentum tensor extracted from the Noether construction is not the same as the symmetric Hilbert energy momentum tensor that is usually cited in E&M textbooks because the E&M field has spin. Thus you cannot get the Poynting vector by using Noether on her own.
answered Jan 14, 2019 at 13:22
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$\begingroup$ I should have added that the additional term you need is the Belinfante Rosenfeld correction. There is a wiki article on Belinfante–Rosenfeld stress energy tensor that will be useful. $\endgroup$ Jan 15, 2019 at 15:05
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Your derivation is entirely correct. However the Poynting vector is a hacked version of the Noether form. This hack is necessary because the Noether form is asymmetric and not gauge invariant. The alternative to hacking is to abandon gauge invariance, as I did successfully in my paper at https://arxiv.org/abs/physics/0106078.
