# Parity symmetry breaking implies time symmetry breaking in General Relativity?

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.

• 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. Apr 23, 2021 at 14:46
• 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. Apr 23, 2021 at 18:16
• Found at least one reference that uses $\eta$: ui.adsabs.harvard.edu/abs/1980PhRvD..22.1915H/abstract (but again, not as standard) Apr 23, 2021 at 18:33
• Thanks, not a giant point, but it was a thing that caught my eye. Apr 23, 2021 at 22:10
• maybe this question would of interest physics.stackexchange.com/questions/648705/… Jul 3, 2021 at 11:29

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 $$F^{0\mu}$$.
• 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. Apr 23, 2021 at 18:29