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Is it true that the Einstein Field Equations have an infinite number of solutions when the pressure is zero?
Yes; but I would expect these to be described as "vacuum" solutions, not "zero pressure" solutions.All metrics with Weyl curvature and no Ricci curvature have vanishing SET
These are only pressureless in one coordinate chart; in other coordinate charts they are not. Or, for a more physical description, they are only pressureless to comoving observers; they are not pressureless to non-comoving observers. This is the kind of thing I was referring to in my previous post.there would be an infinite number of configurations of pressureless dust
But there is an invariant definition of a pressureless dust solution. See, for example, the criterion of contractions of the Einstein tensor given here:Yes; but I would expect these to be described as "vacuum" solutions, not "zero pressure" solutions.
These are only pressureless in one coordinate chart; in other coordinate charts they are not. Or, for a more physical description, they are only pressureless to comoving observers; they are not pressureless to non-comoving observers. This is the kind of thing I was referring to in my previous post.
I'm not disputing that the solution has an invariant definition. I'm just pointing out, for the OP's benefit, that "pressureless" only correctly describes that solution with respect to comoving observers. That's because the OP did not ask specifically about "pressureless dust" solutions defined as you say; he asked about "pressure zero" solutions, and he probably does not realize the limitations of that description.there is an invariant definition of a pressureless dust solution
Even more trivially, it is true for any differential equation for which a sufficient number of boundary conditions have not been specified.Seems trivially true. All metrics with Weyl curvature and no Ricci curvature have vanishing SET, thus vanishing pressure. In addition to this, there would be an infinite number of configurations of pressureless dust.
This case is not so obvious by that type of criteria, which is why I made a physical argument. One can sort of argue the SET is over constrained for a pressureless dust solution. You start with arbitrary symmetric tensor fields with 10 functional degrees of freedom. First, by coordinate invariance, they form equivalence classes leaving only 6. Then, vanishing divergence is 4 more conditions. But then 4 conditions need to be satisfied for a pressureless dust solution. Of course, vanishing divergence are differential conditions, which are weaker.Even more trivially, it is true for any differential equation for which a sufficient number of boundary conditions have not been specified.
I know that, but I am a little bothered by the counting argument I just gave. Is the flaw just that differential conditions are very weak?A cosmological solution for dust is not unique, you can choose the initial data in infinitely many ways. The Friedman equations are not overditermined.
See the earlier link I gave to Wikipedia.Hm, not sure. Are these independent? What are the 4 conditions for dust?
I guess what I am confused about is what 4 conditions need to be satisfied? The SET is ##T_{\mu\nu}=\rho u_\mu u_\nu##, why does this impose any restriction on the metric other than the EFE?See the earlier link I gave to Wikipedia.
I’m reasoning directly from the SET as a symmetric tensor field. 4 conditions are given on T itself.I guess what I am confused about is what 4 conditions need to be satisfied? The SET is ##T_{\mu\nu}=\rho u_\mu u_\nu##, why does this impose any restriction on the metric other than the EFE?
I was looking at the conditions involving G and R. But you are right, in terms of T there are only 3, but that still leads to over determination except for the idea that differential conditions are weaker.Oh, I see. But why do you say 4, the condition for no pressure should be 3 conditions, no?