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).

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    $\begingroup$ Possible duplicate of Fulvio Melia's linear Universe $\endgroup$
    – user4552
    Sep 28 '18 at 20:09
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    $\begingroup$ The newly edited version of the question asks just for the definition of "gravitational horizon." I don't recall ever hearing the term "gravitational horizon" in the normal physics literature. Googling on the phrase only turns up hits on Fulvio Melia's crank papers. If you want to know how he defines it, then you probably have to read his papers carefully and see if he defines it in some clear and meaningful way. Given all the detailed math in your question, it seems like you've done this. Since this is all kook stuff, it seems unlikely to me that others will clarify this any more. $\endgroup$
    – user4552
    Oct 3 '18 at 3:23
  • $\begingroup$ There's a very clear, pedagogical explanation in the paper "The Apparent (Gravitational) Horizon in Cosmology," which may be found at doi.org/10.1119/1.5045333 This doesn't look like kook stuff to me. Besides your declarative statements, Ben, do you have any formal, vetted evidence supporting your claims? $\endgroup$ Dec 9 '18 at 11:56
  • $\begingroup$ @PatrickHsu: Bilicki and Seikel, "We do not live in the R_h = c t universe," MNRAS 425 (2012) 1664, arxiv.org/abs/1206.5130 . Lewis, "Matter Matters: Unphysical Properties of the Rh = ct Universe," MNRAS, 2013, arxiv.org/abs/1304.1248 $\endgroup$
    – user4552
    Dec 9 '18 at 20:01
  • $\begingroup$ @BenCrowell, quote from Bilicki's paper "Our general conclusion is that the discussed model is strongly disfavoured by observations, especially at low redshifts (z < 0.5)." Do they effectively say that Melia's model is NOT strongly disfavored at high redshifts z > 0.5? $\endgroup$
    – MadMax
    Dec 10 '18 at 17:12

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.


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.


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