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Melia's $R_h=ct $ universe proposal suggests that our gravitational horizon $R_h(t)$ should be equal to $ct$ for all cosmic time $t$, not just its present value $t_0$:

"several unpalatable coincidences have emerged along with the successful confirmation of expected features (of ΛCDM). One of these is the observed equality of our gravitational horizon $R_h(t_0)$ with the distance $ct_0$ light has travelled since the big bang, in terms of the current age $t_0$ of the universe. This equality is very peculiar because it need not have occurred at all and, if it did, should only have happened once (right now) in the context of ΛCDM. In this paper, we propose an explanation for why this equality may actually be required by GR".

Can anyone clarify the definition of the gravitational horizon $R_h(t)$ vs. Hubble horizon $R_H(t) = \frac{c}{H(t)}$ (where $H(t)$ is the Hubble parameter)?

Upon some further reading of Melia's another paper, the following is my understanding:

The gravitational horizon $R_h(t)$ is defined as $$ R_h = \frac{2GM(R_h)}{c^2}, $$ where $M(R_h)$ is the mass enclosed within the sphere of radius $R_h(t)$.

For Friedmannian homogeneous universe, $M(R_h)$ is $$ M(R_h) = \frac{4}{3}\pi R_h^3\rho, $$ where $\rho$ is mass density (could be baryonic, radiation, dark matter, etc.). Thus we have $$ R_h = \frac{2G*(\frac{4}{3}\pi R_h^3\rho)}{c^2}, $$ or $$ \frac{c^2}{R_h^2} = \frac{8}{3}\pi G\rho. $$

Now we can find the relation between gravitational horizon $R_h(t)$ and Hubble horizon $R_H(t) = \frac{c}{H(t)}$.

According to the (first) Friedmann equation $$ \frac{c^2}{R_H^2} = H(t)^2 = (\frac{\dot{a}(t)}{a(t)})^2 = \frac{8}{3}\pi G\rho + \frac{c^2}{3}\Lambda - \frac{\kappa c^2}{R_0^2 a(t)^2}, $$ $$ = \frac{c^2}{R_h^2} + \frac{c^2}{3}\Lambda - \frac{\kappa c^2}{R_0^2 a(t)^2}, $$ the equality $$ R_h(t) = R_H(t) $$ is valid only if $$ \frac{c^2}{3}\Lambda - \frac{\kappa c^2}{R_0^2 a(t)^2} = 0, $$ where $\Lambda$ is cosmological constant and $\frac{\kappa}{R_0^2}$ is curvature of the universe ($\kappa = 0$ or $R_0 = \infty $ for flat universe).

Glorfindel
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MadMax
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2 Answers2

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The "gravitational horizon" is not a real horizon (event horizon or particle/causality horizon), just some kind of "apparent horizon". You may define it as in the Schwarzschild spacetime (I'm using $c = 1$ here): \begin{equation}\tag{1} D_h = 2 G M_h, \end{equation} where $M_h$ is any mass in your universe. You may also introduce the Hubble horizon (which isn't a real horizon neither, in the general case): \begin{equation}\tag{2} D_H = \frac{1}{H} = \frac{a}{\dot{a}}. \end{equation} Then the first Friedmann-Lemaître equation gives $D_H = D_h$ in the case of the flat space ($k = 0$) without a cosmological constant ($\Lambda = 0$): \begin{equation}\tag{3} \frac{\dot{a}^2}{a^2} = \frac{8 \pi G \rho}{3} \quad \Rightarrow \quad D_H = \frac{8 \pi G \rho D_H^3}{3} = 2 G M_H, \end{equation} where $M_H = 4 \pi \rho D_H^3 / 3$ is the mass enclosed inside the Hubble volume $V_H = 4 \pi D_H^3 / 3$, which have a physical sense only for the flat space ($k = 0$).

Thus, you see that $D_h = D_H$ if you use $M_h = M_H$ in equation (1). This doesn't imply $a(t) \propto t$. In the case of the dust universe : $\rho \propto a^{-3}$, you get $a(t) \propto t^{2/3}$.

Really, definition (1) is arbitrary and mostly useless.

Cham
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A detailed answer is provided at https://doi.org/10.1119/1.5045333 The short version is that an apparent (or gravitational) horizon in general relativity is a hypersurface that separates incoming from outgoing null geodesics. The apparent and event horizons are the same in static spacetimes, such as Schwarzschild, but not in those with time-dependent metric coefficients, such as most of the FLRW solutions.

Patty
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