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If we apply the Friedmann Lemaître equation to the Universe, we find a critical density $\rho_c$ : if the actual density $\rho$ is under it, the Universe will continue to expand, if it is higher than it, it will end in a Big-crunch.

For what reason (mathematically or physically), can't we apply Friedmann-Lemaître to the Universe's volume associated to the Earth and conclude that this portion of space-time is shrinking because the density is higher than $\rho_c$ ?

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No, the Friedmann equations assume the cosmological principle: the universe at large scales is homogeneous and isotropic. The metric that describes Earth may be approximately isotropic (the same in every direction), but it is not homogeneous (the same in every place).

A better approximation to the space-time around Earth is actually the Schwarzschild solution, the one that also describes static black holes and any other spherically symmetric mass.

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  • $\begingroup$ Why it is not homogeneous ? The subvolume of the Earth, can be described as an homogeneous sphere of density $\rho$ isn't it ? $\endgroup$
    – Vincent
    Commented Oct 15, 2012 at 3:44
  • $\begingroup$ And as the opposite : if we cut a very large sphere in the Universe, if we suppose that the Universe is Homogeneous and Isotropic, we can suppose that this sphere will be a spherically symmetric mass ... that can be described with the Schwarzschild metric isn't it ? But with a Schwarzchild metric we will not be able to find that the sphere is expanding ? Where is my mistake ? $\endgroup$
    – Vincent
    Commented Oct 15, 2012 at 3:55
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    $\begingroup$ @Vincent The difference between a sphere of the universe (on the homogenous scale) and the sphere of the Earth is that the homogenous universe has reflective boundary conditions, not so with the Earth. The material is homogenous but the solution to GR equations is not. $\endgroup$
    – Alan Rominger
    Commented Oct 15, 2012 at 4:37

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