# Cosmological Principle in Einstein Field Equations

In a homework, I've been asked to describe how The Cosmological Principle manifests in the Einstein Field Equations $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$ As I understand it, The Cosmological Principle states that at a large enough scale, the spatial distribution of matter in the Universe is homogeneous and isotropic, however I don't see how that principle holds for any metric or for any energy-momentum tensor. Wasn't the FLRW metric specifically constructed that way in order for it to be homogeneous and isotropic? If that's the case, why would any other metric be homogeneous and isotropic?

Maybe homogeneity could rise from the (strong) Equivalence Principle and from the general covariance of General Relativity. The strong Equivalence Principle states that, locally, our metric is $$\eta_{\mu \nu}$$ at any point, while general covariance means that the laws of Physics are independent from any coordinate system. This means that no matter where we stand on our manifold, there's no apparent curvature (locally) and everything should behave the same way. However, I don't like this line of reasoning since the Equivalence Principle is local while The Cosmological Principle holds at large enough scales.

If what I said before explains homogeneity, I don't see where isotropy could rise from, in fact I think isotropy is a particular property that depends on your metric and energy-momentum tensor of choice.

Ultimately, I think that The Cosmological Principle manifests itself if $$g_{\mu \nu}$$ is the FLRW metric and $$T_{\mu \nu}$$ is that of a perfect fluid, otherwise I don't see how homogeneity and isotropy are general properties of the Einstein Field Equations. I think I should clarify, I'm not looking for an answer to my homework, what I want to know is if my understanding of the concepts and topics in question is correct.

• My interpretation of "How does the cosmological principle manifest in the Einstein equations?" is more along the lines of "If we impose the cosmological principle as a requirement, what would that imply about the Einstein equations, or more specifically, $T$?" May 3, 2021 at 15:44
• But we also have to consider $T_{\mu \nu}$, don't we? I think there's no problem if said tensor is that of the vacuum or a perfect fluid, but generally speaking, it shouldn't be homogeneous and isotropic, should it? May 3, 2021 at 15:40
• @RMC777 Yes we need to consider both the curvature $G_{\mu\nu}$ and the stress energy $T_{\mu\nu}$. If we assume the cosmological principle, then as a consequence we can choose coordinates where the components of $T^{\mu}_{\ \ \nu}$ are ${\rm diag}(-\rho,p,p,p)$. Plugging this into Einstein's equations gives you the FLRW equations (so Einstein's equations are consistent with FLRW). If we don't assume the cosmological principle, then $T_{\mu\nu}$ is in general just some arbitrarily complicated tensor, and similarly the curvature and geometry have no reason not to be arbitrarily complicated. May 3, 2021 at 15:43
• @RMC777 I suspect that what they have in mind is that the cosmological principle is consistent with Einstein's equations. Eg: the cosmological principle implies you can find coordinates where $T^\mu_{\ \ \nu}={\rm diag}(-\rho(t),p(t),p(t),p(t))$ and ${\rm d}s^2=-{\rm d}t^2 + a^2(t) {\rm d}s_{3,\ {\rm max.\ sym.}}^2$ (where ${\rm d}s_{3,\ {\rm max.\ sym.}}^2$ is the metric for a maximally symmetric 3-dimensional space), and Einstein's equations permit geometries which are solutions to the equations with this kind of stress-energy. May 3, 2021 at 15:54