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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@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