Does the Friedmann vacuum equation have a linear solution?

[jcap]

Does the Friedmann vacuum equation have a linear solution rather than an exponential one?

Using natural units one can write Friedmann's equation for the vacuum as
$$
\begin{eqnarray*}
\left(\frac{\dot a}{a}\right)^2 &=& \frac{8\pi G}{3}\rho_{vac}\\\tag{1}
&=& L^2 \left(\frac{\rho_0}{L^4}\right)
\end{eqnarray*}
$$
where I define the Planck length $L=(8\pi G \hbar / 3 c^3)^{1/2}$, $\hbar = c = 1$, and $\rho_0$ is a dimenionless constant.

Now let us interpret the Planck length $L$ to be the size of the smallest volume of space that can be described by general relativity.

But the Weyl postulate, together with cosmological observations, also imply that space is expanding.

Therefore we must have
$$L = a(t) L_0\tag{2}$$
where $L_0$ is the Planck length measured at the reference time $t_0$ where $a(t_0)=1$.

Inserting Eq.$(2)$ into Eq.$(1)$ we find
$$\left(\frac{\dot a}{a}\right)^2 = L_0^2 \left(\frac{\rho_0}{a^2L_0^4}\right)\tag{3}$$
where the Friedmann equation $(3)$ has been rescaled in terms of the Planck length $L_0$ measured at the reference time $t_0$.

Eq.$(3)$ has a linear solution
$$a(t) = \frac{t}{t_0}.$$
The scaled mass density $\rho(t)$ of the vacuum is not constant but rather given by
$$\rho(t) = \frac{\rho_0}{a^2 L_0^4} = \frac{1}{t^2 L_0^2}.$$

 

[Orodruin]

Therefore we must have

L=a(t)L0​

(2)(2)L=a(t)L0L = a(t) L_0\tag{2}
where L0L0L_0 is the Planck length measured at the reference time t0t0t_0 where a(t0)=1a(t0)=1a(t_0)=1.

I see no a priori reason to assume this. Even if $L$ is the "smallest volume" describable by GR (which in itself is not part of GR, just an assumption based on quantum considerations) this volume in itself does not need to expand, in fact, it should not expand as it is an abstract concept rather than a physical volume.

What you have effectively done is to replace the vacuum energy by a spatial curvature term so your universe is empty. You are just describing Minkowski space.