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In reading about various local solutions to Einstein's field equation it is easy to forget that they almost all assume a flat background spacetime (at least asymptotically). Considering this made me wonder about solutions in non-flat background spacetimes (analytical examples are few and far between).

For example if we consider some background spacetime that is some deformation of the three-sphere (an interesting topological choice) then one should be able to express any (symmetric) metric $g_{\mu\nu}$ in the form:

$$g_{\mu\nu}=\sum_{k,l,m}^{^{\infty}}g^{klm}Y_{\mu\nu}^{klm}$$

Where $Y_{\mu\nu}^{klm}$ are the tensor harmonics of the three-sphere (which are well-known being described for example in http://www.ccom.ucsd.edu/~lindblom/Publications/99_JMathPhys.S3Harmonics.pdf). The summation runs over k,l,m with the conditions that: $k\geq l\geq0$ and $l\geq m\geq-l$ . The coefficients are solved by the standard orthogonality relations:

$$g^{klm}=\frac{1}{R^{3}}\intop g^{\mu\nu}Y_{\mu\nu}^{klm}\sqrt{g}d^{3}x$$

It is important to note that these harmonics are not those of the unit-sphere (thus R is the radius of the three sphere). And so we have expanded the background metric out in terms of the so called hyper-spherical harmonics.

Now any tensor can be expanded out in terms of spherical harmonics (of $S^{3}$for our example), let us do the same with the Stress energy tensor $T_{\mu\nu}$

$$T_{\mu\nu}=\sum_{k,l,m}^{^{\infty}}T^{klm}Y_{\mu\nu}^{klm}$$

In general there are six classes of tensor harmonics and our summation will have to run over all of them (let us omit the indicitative notation for the sake of brevity). The Einstein tensor $G_{\mu\nu}$ can be considered as an operator on the metric (albeit a nonlinear one though there are coordinate systems where it resembles the D'alembertian or four-dimensional laplacian).

$$\hat{G}g_{\mu\nu}=T_{\mu\nu}$$

When we look at it in terms of spherical harmonics however:

$$\hat{G}\sum_{k,l,m}^{^{\infty}}g^{klm}Y_{\mu\nu}^{klm}=\sum_{k,l,m}^{^{\infty}}T^{klm}Y_{\mu\nu}^{klm}$$

We get stress energy tensor components acting like eigenvalues of the trace-reversed Ricci ( or rather Einstein) tensor acting upon the metric. Being orthogonal we might examine a single term of the series:

$$\hat{G}g^{klm}Y_{\mu\nu}^{klm}=T^{klm}Y_{\mu\nu}^{klm}$$

If we know the metric at some given time then we can solve for the Stress-Energy-Eigenvalues (ie energy momentum of the matter/fields when integrated). I've never seen the Einstein equation written this way, but I like the fact the large scale spacetime structure is taken into account unlike most treatments. It is also fairly straightforward to compute an arbitrary term of the series expansion.

There is of course also the similarity to the formalism of quantum theory as we have stress-energy/momentum as eigenvalues of a complete basis. Note I chose $S^{3}$ because I find it fascinating that the quantum numbers (or at least their classical analogs) appear with very similar constraints on their interrelatedness. My question: Is there anything wrong with solving the Einstein equation this way? one could choose harmonics of an arbitrary topology, but I (personal bias) like $S^{3}$ (for infinitely extended metrics we'll of course have the fourier transform type situation). Note that the radius R will in general be a function of time (to make things more fun)

PS: if one really wanted to "take it closer to home" we can note that there exist ladder operators for all arbitrary harmonics of the three sphere and hence we might represent all "deviations of matter momentum and energy" as latter operators acting upon the bare three sphere metric itself

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    $\begingroup$ You say almost all assume a flat background spacetime but you make this sound as if the solutions are perturbations of flat spacetime, which is not the case. The common solutions for stars assume the rest of the universe is a vacuum so we expect the solutions to go asymptotically to flat space at infinity because, well, that's the lowest energy vacuum solution. The FLRW metric is an example where we don't assume the universe is a vacuum, and it does not go asymptotically to flat space at infinity. $\endgroup$ – John Rennie Jul 17 '17 at 6:58
  • $\begingroup$ @JohnRennie to clarify, I didn't mean the solutions were non-analytic. I only meant that the assumption of a globally flat universe is quite an assumption. Our best measurements only place a minimum on the absolute value of the radius of curvature of our background spacetime. This is often expressed as "the universe is flat within 5% " but I find that misleading in that it is a bound on the radius of curvature. $\endgroup$ – R. Rankin Jul 17 '17 at 7:06
  • $\begingroup$ If you're assuming that the universe is a vacuum then the assumption that it's globally flat seems perfectly reasonable to me. Remember that these are idealised solutions e.g. the Schwarzschild metric is the solution for a universe containing just a single mass. You refer to experimental measurements of the curvature of our universe, but our universe is not a vacuum and geometries like Schwarzschild will only be approximations. There are solutions like the de Sitter-Schwarzschild that model masses that aren't alone in a vacuum. $\endgroup$ – John Rennie Jul 17 '17 at 7:10
  • $\begingroup$ There are several things unclear in the mathematical setup here: 1. Why are there indices on the spherical harmonics $Y^{klm}_{\mu\nu}$? If you are expanding each component $g_{\mu\nu}$ into harmonics, then these indices need to be on the $g^{klm}$, not the $Y$ (what would $Y_{\mu\nu}$ even be?) How do you even know that an expansion into the spherical harmonics is covariant, i.e. produces coefficients that transform properly? 2. How is spacetime supposed to be "a deformation of the 3-sphere"? Spacetime is four-dimensional (and the 4-sphere is not a Lorentz manifold). $\endgroup$ – ACuriousMind Jul 17 '17 at 9:39
  • $\begingroup$ Are you suggesting the spatial part of spacetime has the topology of a 3-sphere? 3. What do you mean by "background spacetime" for a solution to the EFE? Are you expanding the metric as a perturbation around a flat metric, i.e. working with linearized Einstein equations? 4. Since $G$ is not a linear operator, "eigenvalue" is not a very useful concept for it. What do you think this actually improves over the standard formulation? $\endgroup$ – ACuriousMind Jul 17 '17 at 9:40

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