Physics Asked by juacala on May 2, 2021
I’ve recently been interested in parity violating Lagrangians in general relativity.
One can obtain them using the totally antisymmetric tensor $epsilon_{alphabetamunu}$. For instance the electromagnetic field Lagrangian,
$mathcal{L}_{rm EM} = -sqrt{|g|}frac{1}{4}F_{munu}F^{munu}$,
which does not violate parity, has an analog that does change sign under a parity transformation
$mathcal{L*}_{rm EM} = -sqrt{|g|}frac{1}{4}epsilon_{alphabetamunu}F^{alphabeta}F^{munu}$
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 $epsilon_{alphabetamunu}$ will change sign if one coordinate changes sign (a parity transform). This changes the "handedness" of the coordinate system, changing the sign of $epsilon_{alphabetamunu}$. Since time and space in general relativity are really not separated (except time has a negative associated eigenvalue in $g_{munu}$), 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 $epsilon_{alphabetamunu}$?
The only way this would not be true is if somehow the fact that time is associated with the negative eigenvalue in $g_{munu}$, $epsilon_{alphabetamunu}$ does not change sign under time reversal, but I’m not seeing how that could be.
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 $F^{0mu}$.
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.
Answered by user297146 on May 2, 2021
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