# Momentum density of the EM field - Classical field theory

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?

• 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. – Cham Jan 14 '19 at 14:24
• en.wikipedia.org/wiki/Stress–energy_tensor#Variant_definitions_of_stress–energy – G. Smith Jan 14 '19 at 17:46