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I am currently reading through a paper by Hughes and Ramamurthy (ref: https://arxiv.org/abs/1508.01205), which describes the electromagnetic response of a line-node semimetal by the action

$$S[A,B] = \frac{e}{16 \pi^2}\int d^4x \; \epsilon^{\mu \nu \rho \sigma} B_{\mu \nu} F_{\rho \sigma},$$ where $F_{\rho \sigma} = \partial_\sigma A_\rho - \partial_\rho A_\sigma$ is the usual electromagnetic field, and $B_{\mu \nu}$ is a two-form. At the top of the second page of this paper, the authors go on to say that from the form of the action, you can see that $B_{\mu\nu}$ is related to the magnetization and polarization as $$e B_{0 i} = 4 \pi^2 M_i$$ and $$e B_{ij} = 4 \pi^2 \epsilon^{ijk} P_k.$$

This statement is not obvious to me, how does one determine this relation between the magnetization/polarization and the two-form $B$?

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See here. For a more detailed source on the Lagrangian formulation of electrodynamics in matter, there is a wonderful book "Classical Field Theory" by David Soper.

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  • $\begingroup$ The Lagrangian that you linked to includes a term of the form $F_{\mu \nu} M^{\mu \nu}$, how does this apply to my question? $\endgroup$ – gene Jan 27 at 23:29
  • $\begingroup$ @gene Let $M^{\mu\nu}=\epsilon^{\mu\nu\rho\sigma}B_{\rho\sigma}$. $\endgroup$ – Richard Myers Jan 27 at 23:30

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