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The addition of

$$\mathcal{L}' = \epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} \propto \vec{E}\cdot\vec{B}$$

to the electromagnetic Lagrangian density leaves Maxwell's equations unchanged (shown here).

In Carroll's GR book, he appends the question: "Can you think of a deep reason for this?". What might this be? I imagine there is a physical argument other than simply "gauge invariance".

Qmechanic
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Aiden
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  • I think $L'$ is not a scalar under some discrete Lorentz transformation.　Since $L'$ contains the Levi-Civita symbol, it changes sign under some discrete Lorentz transformation, such as time reversal. CPT symmetry may be a keyword. Sorry, I don't know the deeper reason why Lagrangian must be scalar. – HEMMI Mar 29 '23 at 00:21
  • I think that the tha answer to "physical argument other than simply "gauge invariance" is that it fits existing data and is predictive of new. That is how successful theories of physics dominate. – anna v Mar 29 '23 at 03:45
  • @annav Any references how this term is "predictive of new" data? To my understanding, Maxwell's equations are invariant under this addition, so it should predict the same physics as Maxwell's theory. – Aiden Mar 29 '23 at 04:10
  • @Aiden anybody adding terms that do not affect the solutions of Maxwell equations (ME) must add them for some reason. From your GR , I assume general relativity, this term seems to be needed for some reason, so the argument is that an extra term for GR to work with electromagnetism does not destroy the success of ME fitting the data at low energies. – anna v Mar 29 '23 at 04:44
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    Just to spell out @Qmechanic's total-derivative argument:$$\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}=4\epsilon_{\mu\nu\rho\sigma}\partial^\mu A^\nu\partial^\rho A^\sigma=\partial^\mu(4\epsilon_{\mu\nu\rho\sigma}A^\nu\partial^\rho A^\sigma)-\underbrace{4A^\nu\epsilon_{\mu\nu\rho\sigma}\partial^\mu\partial^\rho A^\sigma}_0.$$This uses a symmetric-antisymmetric-contraction-is-$0$ argument. My answer to the linked question presented a different one, computing the term's derivative with respect to derivatives of the $A$-field. – J.G. Mar 29 '23 at 08:39

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As e.g. shown in OP's linked Phys.SE post the $F\wedge F$ term is a total divergence/topological term, and hence doesn't contribute to the Euler-Lagrange (EL) equations, i.e. Maxwell's equations, cf. e.g. this Phys.SE post.

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Qmechanic
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