What is the precise definition of a dust solution in general relativity? If the Einstein tensor of a metric has only the first diagonal term non-zero, it that sufficient for that solution to be called a dust solution?
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Yes, that would be sufficient (but not necessary). The stress-energy tensor of a perfect fluid takes the form $$T^{\mu\nu} = \left(\rho + \frac{p}{c^2}\right)U^{\mu}U^\nu + p g^{\mu\nu}$$ for a normalized, time-like vector field $U$ which we interpret as the fluid's 4-velocity field. Dust is the name of a perfect fluid which has $p=0$, and so the stress-energy tensor takes on the particularly simple form
$$T^{\mu \nu} = \rho U^\mu U^\nu$$
In the rest frame of the fluid (which may be different at different points, of course), $U = (c,\mathbf 0)$ and so $T^{\mu\nu} = \rho c^2 \delta^\mu_0 \delta^\nu_0$.
