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I've recently been interested in parity violating Lagrangians in general relativity.

One can obtain them using the totally antisymmetric tensor $\epsilon_{\alpha\beta\mu\nu}$. For instance the electromagnetic field Lagrangian,

$\mathcal{L}_{\rm EM} = -\sqrt{|g|}\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$,

which does not violate parity, has an analog that does change sign under a parity transformation

$\mathcal{L*}_{\rm EM} = -\sqrt{|g|}\frac{1}{4}\epsilon_{\alpha\beta\mu\nu}F^{\alpha\beta}F^{\mu\nu}$

These are just examples to illustrate what I am talking about.

My question is more general: it seems like any Lagrangian that violates parity by the inclusion of $\epsilon_{\alpha\beta\mu\nu}$ will change sign if one coordinate changes sign (a parity transform). This changes the "handedness" of the coordinate system, changing the sign of $\epsilon_{\alpha\beta\mu\nu}$. Since time and space in general relativity are really not separated (except time has a negative associated eigenvalue in $g_{\mu\nu}$), wouldn't any parity violating Lagrangian as above also be time-symmetry violating, since flipping the sign of the time coordinate changes the handedness of the 4-dimensional space-time, and the sign of $\epsilon_{\alpha\beta\mu\nu}$?

The only way this would not be true is if somehow the fact that time is associated with the negative eigenvalue in $g_{\mu\nu}$, $\epsilon_{\alpha\beta\mu\nu}$ does not change sign under time reversal, but I'm not seeing how that could be.

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    $\begingroup$ Is it common to use $\eta$ for the Levi-Civita symbol? If this is a convention from somewhere that I don't understand, that's fine, but otherwise, please use $\epsilon$, which is the only symbol I've ever seen for it. $\endgroup$ Apr 23, 2021 at 14:46
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    $\begingroup$ I changed $\eta$ to $\epsilon$. To be honest, I don't remember where I got that notation from, but I had seen in some papers, $\epsilon$ used for the non-tensor version with elements (1, -1, 0) and $\eta$ for the tensorial version $\eta_{\alpha\beta\mu\nu} = \sqrt{|g|}\epsilon_{\alpha\beta\mu\nu}$. In any case, you're right that $\epsilon$ is much more standard. $\endgroup$
    – juacala
    Apr 23, 2021 at 18:16
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    $\begingroup$ Found at least one reference that uses $\eta$: ui.adsabs.harvard.edu/abs/1980PhRvD..22.1915H/abstract (but again, not as standard) $\endgroup$
    – juacala
    Apr 23, 2021 at 18:33
  • $\begingroup$ Thanks, not a giant point, but it was a thing that caught my eye. $\endgroup$ Apr 23, 2021 at 22:10
  • $\begingroup$ maybe this question would of interest physics.stackexchange.com/questions/648705/… $\endgroup$ Jul 3, 2021 at 11:29

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I would say that this is really a question about particle physics and CPT, not a question about general relativity. GR is by definition a theory that has local Poincare invariance, and whose only geometrical apparatus is a dynamically determined metric $g$. Breaking parity symmetry in GR would mean altering the Einstein field equations in a way that breaks this invariance. That isn't what you're doing. You're using a certain toolkit of tensorial notation to write down a Lagrangian for a matter field. The structure of GR is such that it doesn't really make sense to talk about something like a global parity inversion. These things make sense locally, but GR is locally the same as SR, so your question is really a question about particle physics in SR.

Normally when we try to write down a Lagrangian density for a certain matter field, we restrict ourselves to a certain toolkit of tensor notation, which guarantees that the result will be a relativistic scalar. This avoids wasting our time with the infinitely many possible Lagrangians that won't produce a Lorentz-invariant theory. This toolkit is restricted. E.g., we don't get to assume there is "the" time coordinate, or use notation like $F^{0\mu}$.

When you throw in the Levi-Civita tensor, you are violating these rules. You're introducing some new piece of geometrical apparatus, which isn't part of GR. That's a choice, and it's not the only possible choice. There are other choices that would violate local Lorentz invariance in other ways. For instance, in the old steady-state cosmology theories, they introduced a preferred timelike field. In your case, the Levi-Civita tensor has its own particular characteristics. It can only be defined on an orientable manifold, and if so, it's determined everywhere up to normalization.

So what you're seeing is that the tool you've chosen allows you to write down Lagrangians of a certain type, which violate P and T. To get a more substantive statement about this kind of thing, one that isn't an arbitrary choice, you need the standard CPT theorem.

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    $\begingroup$ I have to disagree that the Levi-Civita tensor is foreign to general relativity. It exists for any orientable differential manifold. It's necessary to do any math with differential forms, which are very much a part of general relativity. Also, even if a scalar includes $\epsilon$, it is still a Lorentz scalar; $\epsilon$ transforms as a tensor. I believe both scalars I have written down above are invariant under any continuous coordinate transformation (rotation/boost); the second is just not so under a discrete parity transformation. $\endgroup$
    – juacala
    Apr 23, 2021 at 18:29
  • $\begingroup$ @juacala I'd also add that "past and future orientation" is absolutely a part of all sorts of considerations in General relativity, and if time inversion makes sense, then parity inversion should too. $\endgroup$ Apr 24, 2021 at 1:19

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