I would be interested in what people think of Fulvio Melia's argument for a linearly expanding Universe.

I realize that the experimental evidence seems to be pointing to an accelerating Universe but does Melia's argument have any merit?

The following is my own Newtonian argument for a linearly expanding universe (it is not Melia's argument).

enter image description here

Imagine that we are at point $P$ in a homogeneous, isotropic Universe. Point $P$ is on the surface of a sphere $S_1$ with mass $M_1$ and radius $R_1$. Therefore, in Newtonian terms, there is an acceleration $g_1$ at point $P$ directed towards the center of sphere $S_1$ with magnitude:

$$g_1=-\frac{G M_1}{R_1^2}.$$

But point P is simultaneously on the surface of a sphere $S_2$ with mass $M_2$ and radius $R_2$. Therefore one can construct another argument, independently of the previous one, that there is an acceleration $g_2$ at point $P$ directed towards the center of sphere $S_2$ with magnitude:

$$g_2=-\frac{G M_2}{R_2^2}.$$

However since the Universe is homogeneous and isotropic then, again in Newtonian terms, there should not be any net acceleration vector at any point.

How do we resolve this paradox?

Either the Universe is empty so that $g_1=g_2=M_1=M_2=0$ or General Relativity gives us another solution.

In GR pressure also has "weight" so that the gravitational acceleration $g$ at the surface of a uniform density sphere of mass $M$ and radius $R$ with density $\rho$ and internal pressure $p$ is given not by:

$$g=-\frac{G M}{R^2}=-\frac{4\pi G}{3}\rho R$$

but rather by:

$$g=-\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right) R.$$

Thus, in order that the net acceleration at any point $P$ is zero, we must require the acceleration $g$ due to any particular sphere touching point $P$ to be zero. Thus we must have the equation of state:

$$p=−\frac{\rho c^2}{3}.$$

When the above equation of state is substituted into the equivalent acceleration Friedmann equation (without cosmological constant) given by

$$\frac{\ddot a}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right),$$

with scale factor $a$, we find that $\ddot a=0$ so that $a(t)\sim t$ implying that the Universe can only be expanding linearly with time.

  • 2
    $\begingroup$ That link uses Birkhoff's Theorem and GR, which one needs to do because pure Newtonian gravity is undefined in an infinite homogeneous non-empty universe, as is the source of confusion here and here. As for the rest of that article, it sets off a number of warning bells. He seems to advocate an already falsified cosmology based on numerology, in addition to seemingly conflating the Hubble radius with a horizon. $\endgroup$ – user10851 Aug 8 '13 at 1:03
  • $\begingroup$ @ChrisWhite: In 1109.5189, he calls $R_h=(8\pi\rho/3)^{-1/2}$ the "gravitational radius," and cooks it up by throwing together some equations without any clear physical motivation. His $R_h$ equals the Hubble radius in the special case of spatial flatness; he states this on p. 4 of the popular-level article, but then immediately goes on, as you point out, to conflate it incorrectly with a horizon. $\endgroup$ – user4552 Aug 8 '13 at 1:50
  • $\begingroup$ This is closely related to this question (same person asking). $\endgroup$ – tfb Mar 14 '19 at 16:01

The article Melia 2012, linked to from the question, is a popularization. A more professional-level presentation of Melia's ideas is given here: http://arxiv.org/abs/1109.5189 . Neither of these appears to present the particular argument given in the question.

The question is in two parts.

(1) The first part of the question presents a paradox in Newtonian gravity. The resolution of the paradox is that solutions to the Poisson equation are only unique for certain specific types of boundary conditions, and those don't include the boundary conditions being discussed here. See this question for more discussion.

(2) The second part of the question includes some ingredients from GR but isn't a fully general-relativistic treatment. Its result is correct if stated as follows: cosmological expansion is linear if and only if the equation of state is $w=p/\rho=-1/3$, where $w$ represents an average over all sources including dark energy and the units are such that $c=1$. This follows directly from the Friedmann equations.

So this particular presentation isn't very compelling, since 1 is a mistake, and 2 is only of interest if 1 holds.

Melia 2011 points out two coincidences that hold only at the present time in our universe:

(a) Defining the quantity $R_h=(8\pi\rho/3)^{-1/2}$, we currently have $R_h\approx t$. (This is all in units with $G=c=1$.)

(b) The value of $w$ (averaged, as above, over all sources including dark energy) currently appears to be close to $-1/3$.

Melia appears to believe that these coincidences are a sign that Something is Terribly Wrong in the present state of cosmology, and that there must be some mysterious physical principle that makes $R_h=t$ identically for all times. This would require linear expansion and $w=-1/3$ for all times. This is pure kookery. As he himself admits in both Melia 2011 and Melia 2012, there are well-known and elementary physical reasons why $w$ must vary as the universe goes from radiation-dominated to matter-dominated to vacuum-dominated. Furthermore, there is empirical evidence that $w$ has varied this way, e.g., from big-bang nucleosynthesis.

There have been a couple of papers published that have debunked Melia's work: Lewis 2013 and Bilicki 2012.

Bilicki and Seikel, "We do not live in the R_h = c t universe," MNRAS 425 (2012) 1664, http://arxiv.org/abs/1206.5130

Lewis, "Matter Matters: Unphysical Properties of the Rh = ct Universe," MNRAS, 2013, http://arxiv.org/abs/1304.1248

Melia, "The Cosmic Spacetime," Australian Physics, May 2012, http://arxiv.org/abs/1205.2713

Melia and Shevchuk, "The R_h = ct Universe," 2011, MNRAS 419 (2012) 2579, http://arxiv.org/abs/1109.5189

| cite | improve this answer | |
  • 1
    $\begingroup$ Debates like this must be great for various researchers' careers - one can keep publishing more papers, like arxiv.org/abs/1309.6950 $\endgroup$ – user10851 Sep 28 '13 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.