# Another static solution of the Friedmann equations - interpretation of $p=-\rho c^2$

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?

• 1. 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
Commented May 3, 2021 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
Commented May 3, 2021 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
Commented May 3, 2021 at 11:38
• 2. 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
Commented May 3, 2021 at 11:49
• 3. 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
Commented May 3, 2021 at 11:57

The solution you consider is well-known de Sitter space. What you've got is the cosmological constant interpreted not as a separate $$\Lambda$$-term but as a contribution to $$T_{\mu\nu}$$ that looks like an ideal fluid with $$w=\frac{p}{\rho}=-1$$ (I'll use $$c=1$$). Note that the stress-energy tensor for the ideal fluid looks like, $$$$T_{\mu\nu}=(p+\rho)u_\mu u_\nu -p g_{\mu\nu}\underset{w=-1}{=} \rho g_{\mu\nu}$$$$ This means that there's no velocity field for such a fluid and if no other matter is present $$\nabla_\mu T^{\mu\nu}=0$$ enforces $$\rho=const$$. If you compare such a $$T_{\mu\nu}$$ with the $$\Lambda$$-term in the Einstein equation then you will see that they are completely equivalent.
• Thanks. The solution trys, by joining to the ideas in the link, to find a De Sitter universe that appears static, even without $\ddot a= \dot a=0$ and also has positive matter density. It's interesting that you mention that the cosmological constant can be interpreted as a contribution to $T_{\mu \nu}$ Commented May 3, 2021 at 13:53
• @ OON Is there any known interpretation of a contribution to $T_{\mu \nu}$ that has negative $\omega$? Not sure about this but here are examples: is negative pressure a collapse, or is it an explosion, or does antimatter have that property - is there anything at all within known physics that has negative $\omega$ - so we wouldn't need $\Lambda$ (or the unexplained dark energy)? Commented May 4, 2021 at 19:57
• ...p.s. or could a gravitational constant $G$ that increased with the increasing scale-factor of the universe have that property? Commented May 4, 2021 at 20:04