# Friedmann vacuum solution linear rather than exponential?

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

$$\rho(t) = \frac{\rho_0}{a^2 L_0^4} = \frac{1}{t^2 L_0^2}.$$

Depending on who you are asking, it's called

• empty Milne universe (Milne)
• coasting universe (Kolb)
• $$R_h = ct$$ universe (Melia)
• linear universe (in your case)

Everyone and her grandmother jumped on the bandwagon.

• Milne and Melia are not in the same category. Melia is a kook. See physics.stackexchange.com/a/73525/4552
– user4552
Apr 3 '19 at 18:08
• @Ben Crowell, The equation of state parameter w for the curvature turn in Milne's case and for the cosmic fluid in Melia's case are the same as w=-1/3. You could have different interpretations, but the cosmic evolution of the scale parameter according the Friedman equation is the same. Apr 3 '19 at 21:00