a-cosmology-group/acg#80

So ΛCDM doesn’t work, we know that, but can your model solve these problems?

No, but I get sick by all those MOND advertising. So I shall throw up the following conjecture.

MOND is emergent from the Variable Mass Theory as has been implemented in UAC

Let’s start with a quick review of MOdified Newtonian Dynamics, especially the formulas for the deep-MOND regime ( $F=$ force, $m=$ test mass, acceleration $a \ll a_0$ ) :

$$ F_N = m \frac{a^2}{a_0} $$

Applying this to an object of mass $m$ in circular orbit around a point mass $M$ (a crude approximation for a star in the outer regions of a galaxy), we find:

$$ \frac{GMm}{r^2} = m\frac{(v^2/r)^2}{a_0} \quad \Longrightarrow \quad v^4 = GMa_0 $$

That is, the star’s rotation velocity $v$ is independent of $r$, its distance from the centre of the galaxy - the rotation curve is flat, as required. In comparison we have, with MOND = UAC:

$$ \mbox{observed } u \sim \sqrt{\mbox{distance to nucleus } r/R} \times \mbox{calculated } v $$

Where our previously dimensionless $0 \le r \le 1$ is replaced by the real world one, by multiplying it with the galaxy’s radius $R$. For the same crude approximation we see that the rotation curve is flat as required, but with a square instead of a fourth power and no free parameter $a_0$ :

$$ \frac{GMm}{r^2} = m\frac{v^2}{r} \quad \Longrightarrow \quad v^2 = \frac{GM}{r} $$

$$ u^2 = \frac{r}{R} v^2 \quad \Longrightarrow \quad u^2 = \frac{GM}{R} $$

Let us now calculate Mordehai Milgrom’s factor $a_0$ :

$$ \left(u^2\right)^2 (\mbox{UAC}) = v^4 (\mbox{MOND}) \quad \Longrightarrow \quad \left(\frac{GM}{R}\right)^2 = GMa_0 \quad \Longrightarrow \quad \boxed{ a_0 = \frac{GM}{R^2} } $$

So Milgrom’s constant - as it appears in UAC - is equal to the standard gravitational parameter divided by the radius squared of a galaxy.