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Looking for solutions of the Friedmann equations

$$(\frac{\dot a}{a})^2+\frac{kc^2}{a^2} = \frac{8 \pi G \rho+\Lambda c^2}{3}, \tag{1}$$

$$\frac{\ddot a}{a} = \frac{-4 \pi G}{3} (\rho + \frac{3p}{c^2}) +\frac{\Lambda c^2}{3}. \tag{2}$$

There seems to be this possibility

with $\Lambda = 0$, $k=0$, constant $H$, and $a=e^{Ht}$, the equations reduce to

$$3H^2 = 8 \pi G \rho, \tag{3}$$

$$3H^2 = -4 \pi G(\rho + \frac{3p}{c^2}), \tag{4}$$

leading to the solution

$$\rho = \frac{3H^2}{8 \pi G}, \tag{5}$$

$$p = - \rho c^2. \tag{6}$$

A nice, simple solution with scaling symmetry and time symmetry. If the expansion happens to all length scales including the observer as in Cosmology - an expansion of all length scales then the solution is an apparently static universe (but with a redshift as described in the link) and a universe always at critical density.

But the question is, what is the best interpretation of $p = - \rho c^2$ in a universe with $\Lambda = 0$? One idea is that explosive events in the universe e.g. from the nuclei of galaxies provide the negative pressure - is there any other way to interpret this?

John Hunter
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  • In that answer, I mentioned that "Schrödinger in 1918 proposed that the Einstein’s static universe can be obtained from the Einstein field equations without a positive cosmological constant if a negative-pressure term was added to the stress-energy tensor on the right-hand side of the Einstein field equations! But this is controversial since this proposal is equivalent to introducing a cosmological constant at the level of the action of the theory."
    • – SG8 May 03 '21 at 11:23
  • And after that, I mentioned "After this proposal, Einstein stated that he, in this way, indeed discovered that the cosmological constant is able to account theoretically for the existence of a finite mean density in a static universe." So, the answer by @OON is quite reasonable (as OON wrote: a contribution to $T_{\mu\nu}$ that looks like an ideal fluid with $w=\frac{p}{\rho}=-1$). No need for emphasizing, they are completely (?) equivalent, as well understood by Einstein. – SG8 May 03 '21 at 11:28
  • In conclusion, we discussed a solution with $\Lambda =0$ and $k=0$ (spatially flat) and a negative-pressure component in $T^{\mu \nu}$, which is equivalent to a cosmological model with $\Lambda >0$ and $k=0$. This is exactly the (spatially) de Sitter model which is a special case of Lemaitre model. But, since this model assumes the matter radiation/matter densities are zero is not a realistic model. However, it's an interesting toy model. (I also upvoted the @OON answer for making it more clear) – SG8 May 03 '21 at 11:38
  • It would be, perhaps, more interesting to consider the case with $\Lambda=0$ but a non-zero $k$. Regarding this, we can have an static universe if we set $\rho+3p=0$ and $k=\frac{{8\pi\rho{a^2}}}{3}$. But, again, the problem is what is the negative-pressure component? Again, you will end up with a cosmological constant interpretation.
    • – SG8 May 03 '21 at 11:49
  • Regarding the first two (Friedmann) equations in your post: the case with $\Lambda=0$ and $k=0$ and a negative-pressure term could not be static since (I think) it cannot satisfies the static condition, i.e., $\ddot a= \dot a=0$. However the Hubble parameter is a constant but $a$ grows exponentially (I haven't read that link in your post yet. My time is very limited). But, the case with $\Lambda=0$ and $k>0$ and a negative-pressure could be static since it satisfies $\ddot a= \dot a=0$, as mentioned be Weinberg in his book (see my previous answer and the references therein, please)
    • – SG8 May 03 '21 at 11:57