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In General Relativity is there a TE symmetry similar to CPT symmetry in the Standard Model ? It's pretty easy to understand that by flipping charge and parity you merely get a time reversed equivalent of your system, so flipping time as well would lead to an equivalent description. Similarly, since metric perturbations in GR are sourced by energy density, it seems to me that GR is invariant if we operate the transformation $(t,\rho) \rightarrow (-t,-\rho)$. Is this correct or am I missing something here ? Do the other $T_{\mu \nu}$ terms come into play in ways which I haven't considered here ? Is the symmetry actually $(g_{\mu \nu},T_{\mu \nu}) \rightarrow (-g_{\mu \nu},-T_{\mu \nu})$ ?

The reason I think this might work is if you make the energy density negative in the Friedmann equations expansion turns into contraction, then you time flip it and it turns back into expansion. Is this a ubiquitous behavior ? If this is indeed generally true can we say there is indeed an ET symmetry ?

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The structure of general relativity does not allow the theory itself to be classified in terms of global, discrete symmetries such as time-reversal. The Einstein field equations don't refer to a time coordinate; they're expressed tensorially, which means that they are completely independent of what coordinates you choose. Since there is no guarantee that you have a preferred time coordinate, there is no guarantee in general that you can define a global time-reversal operator. You can, for example, have spacetimes that aren't time-orientable.

This is similar to an issue that came up in your other question, which was about Lorentz symmetry. Operators like Lorentz boosts and time reversal are local, not global.

The Friedmann equations are not the field equations of GR, and they don't tell you anything about the symmetry properties of GR.

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Thanks again for the answer. I only brought up the Friedmann equations to explain the intuition behind my puzzlement. Simply put : how do we end up with what seems like a preferred direction in time ? I understand that you can't define a global time-reversal operator in principle, but I have trouble understanding how this leads to solutions where there seems to be a preferred direction in time. Does your answer mean I can choose either direction in time ? Or that it is indeed more accurate to say that space is expanding over time ? – ticster Jul 26 '14 at 16:13
@ticster: A solution to an equation can lack a symmetry possessed by the equation. Note that not all solutions to the Friedmann equations have a preferred time direction. Solutions that recollapse to a Big Crunch are symmetric. – Ben Crowell Jul 26 '14 at 17:12
"A solution to an equation can lack a symmetry possessed by the equation" I understand that perfectly well, for some reason though I can't seem to visualize how this emerges in the context of the Friedmann equations. Time to go back to the drawing board and re-derive them ! That should do the trick. Thanks for your help. – ticster Jul 26 '14 at 17:16
@ticster, an easy example is the "reversed" harmonic oscillator $\ddot{x}-\omega^2 x=0$, that has solutions of the form $x=A\cosh{\omega t}+B\sinh{\omega t}$. The $\cosh{\omega t}$ part is time-symmetric while the $\sinh{\omega t}$ is asymmetric. While the equation is perfectly invariant over time reflection it is the initial conditions which determine if the solution will possess this symmetry too. It is the exact same case for Friedman solutions, which are a family of solutions of a invariant equation, the initial conditions are matter content and extrinsic curvature. – cesaruliana Jul 26 '14 at 18:06
I should just add in light of Ben Crowell's excellent answer that of course in GR the issue of initial conditions is a bit more complicated as a consequence of absence of a preferred time coordinate, but in globally hyperbolic spacetimes, as is the case of FRW metrics, it is possible to formulate the Einstein Equations as a hyperbolic PDE with initial conditions – cesaruliana Jul 26 '14 at 18:12

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