Towards a physical interpretation of MOND's a0

In MOND, FUGE and Dark Matter Light, there’s a little play on words, based on a comment in Milgrom’s Scholarpedia article The MOND paradigm of modified dynamics:

For galaxy clusters, MOND reduces greatly the observed mass discrepancy: from a factor of ∼10, required by standard dynamics, to a factor of about 2. But, this systematically remnant discrepancy is yet to be accounted for.

In my post, I highlight that I consider “dark matter” to be more of a phenomenon related to the mass discrepancy, a placeholder if you like until such time as the mass discrepancy is explained.  One solution is an actual form of matter (cold dark matter) and another solution is the a₀ of MOND.  Milgrom seemed to be pointing to the possibility of a midway point, with a little cold dark matter (or missing baryons).

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I also, perhaps unadvisedly, said that Milgrom used a form of numeromancy to arrive at his value of a₀, the acceleration constant that is central (mathematically at least) to MOND.  I fiddled around – merely using the units (which could also be called a form of numeromancy) – and found that if a₀=c.H₀/2π, then we get a value of a₀ very close to what Milgrom calculated (~1.2×10^-10m/s²).

Now, according to Wikipedia, with no reference provided:

By fitting his law to rotation curve data, Milgrom found a₀ ≈ 1.2×10^-10 m/s² to be optimal.

According to Milgrom himself:

a₀ can be determined from several of the MOND laws in which it appears, as well as from more detailed analyses, such as of full rotation curves of galaxies. All of these give consistently a₀≈(1.2±0.2)×10⁻⁸cm s⁻².

And later:

Significantly perhaps, it’s measured value coincides with acceleration parameters of cosmological relevance, namely, a¯₀≡2πa₀≈cH₀≈c²(Λ/3)^1/2 (H₀ is the Hubble constant, and Λ the cosmological constant). This adds to several other mysterious coincidences that characterize the mass-discrepancy conundrum, and may provide an important clue to the origin of MOND.

So, it wasn’t quite numeromancy.  What I was really objecting to, in my own muddled way, was that there didn’t seem to be a physical meaning to a₀.  Sure, there’s an approximate numerical equivalency between a₀ and c.H₀/2π, but what does that mean?

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First off, there’s a problem tying anything to H₀ because of the Hubble tension which has the value of H₀ being 67.4±1.4 km/s/Mpc (CMB data), 67.36±0.54 km/s/Mpc or 67.66±0.42 km/s/Mpc (Planck 2018 data [the latter with BOA data added]), 73.04±1.4 km/s/Mpc (SH0ES data) and 78.3±3.4 km/s/Mpc (the most extreme of the quasar lensing measurements).  If we plug in these values, we would be saying that the value would lie in the range a₀≈1.04m/s² to a₀≈1.21m/s² (if calculated as above).  Milgrom’s value is right at the upper limit.

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In the FUGE model, the universe has been expanding by one unit of Planck length every unit of Planck time, and the mass-energy in it has been increasing by half a Planck mass per unit of Planck time.  This results in the density remaining critical throughout.

Critical density is given by the equation ρ_c = 3.H²/8πG.  This is the density of a (Schwarzschild) black hole with a radius of r=c/H (explained here).

That means that, in the FUGE model, if the universe has an age of approximately ꬱ.t_P=8×10⁶⁰ units of Planck time=13.77 billion years (explained here), then it has H₀=1/(ꬱ.t_P)=1/(13.77 billion years)=71km/s/Mpc (this is just saying that the Hubble value is the inverse of the age of the universe, which is related to how the universe expands), a radius of approximately ꬱ.l_P=8×10⁶⁰ units of Planck length=13.77 billion light year and a mass of approximately (ꬱ.m_P)/2=4×10⁶⁰ units of Planck mass=8.77×10⁵²kg (also explained here).

This gives us enough information to ask an odd question.  What is the gravity of the universe at its surface?  There are, of course, obvious objections to this question, which might be why it has not been asked before.  But let me work through it for the purposes of the exercise.

Gravity of the Earth is given by the radius of the Earth (more specifically the distance from the centre of the Earth’s mass at which we are considering, we can use sea level, 6,378km), the mass of the Earth in this equation (5.972×10²⁴kg) and the Gravitational Constant G (6.674×10^-11N.m²/kg²):

 g_E=Gm_E/r²=9.8m/s²

Using the same method, we could say that the “gravity of the universe” is:

g_U=Gm_E/r²=G.(ꬱ.m_P)/2/(ꬱ.l_P)²

We know that m_P=√(ħc/G), l_P=√(ħG/c³) and t_P=√(ħG/c⁵) and thus also that c=l_P/t_P.  So:

g_U=G.(ꬱ.√(ħc/G))/2/ꬱ²/(ħG/c³)=√(ħc/G)/(2.ꬱ.(ħ/c³))

Multiplying through by t_P/t_P:

g_U=√(ħc/G).√(ħG/c⁵)/(2.ꬱ.(ħ/c³).√(ħG/c⁵)

 =(ħ/c²)/(2.(ꬱ.t_P).(ħ/c³)=c/(2.(ꬱ.t_P))

And because, as mentioned above, H₀=1/(ꬱ.t_P):

g_U=c.H₀/2

Now this value is not what Milgrom and others arrived at but my question has to be, is there enough wriggle room in the mapping of the value of a₀ to rotation curve data to allow the π to be dropped?  There may be.  At his Scholarpedia entry, Milgrom has this chart:

Note that the selection of the a₀ value appears arbitrary and signifies the point "below which we are in the MOND regime".  If we consider instead the point above which we are are unequivocally in the Newtonian regime, a different line could be drawn:

This could easily equate to a₀=c.H₀/2.

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A TLDR for the above is this:

Consider the critical density of the universe, ρ_c = 3.H²/8πG, this is the same as the density of a (Schwarzschild) black hole with radius r=c/H.  Such a black hole has a mass of M=c³/2GH.  And the gravity at the radius of such a black hole is g=cH/2.  In the FUGE model, there is no inflation and on dark energy, so the radius of the universe would be c/H and a₀=cH/2 could therefore be the “implied gravity” on the “surface” of the universe.