# Deriving Faber-Jackson Relation from Milgrom's Law

I am reading Famaey & McGaugh 2011, a review paper on Modified Newtonian Dynamics. My question concerns bullet 3 in Sec 5.2, where the authors try to explain how Faber-Jackson Relation can be derived from Milgrom's Law.
They mention the equation of hydrostatic equilibrium for an isotropic isothermal system which I was having difficulty deriving, so I was led to this page. The derivation begins in Sec II of Page 2. I have one question here:

To write $$f_r=A\rho\bar{u}^2,$$ we require all the particles to have the same radial velocity $$u_r$$ so that $$\rho$$ can be taken out of the integral. Is this even realistic in the system under question?

Back to the paper, the equation is $$\frac{d(\sigma^2\rho)}{dr}=-\frac{\rho\sqrt{GMa_0}}{r}$$ which can be obtained by using Milgrom's Law and the hydrostatic equilibrium equation derived. This simplifies to $$\sigma^2\frac{d\rho}{dr}+2\sigma\rho\frac{d\sigma}{dr} =-\frac{\rho\sqrt{GMa_0}}{r}$$ $$\implies \sigma^2\frac{r}{\rho}\frac{d\rho}{dr}+2\sigma r\frac{d\sigma}{dr}=-\sqrt{GMa_0}$$ To end up with $$\sigma^4=\alpha^{-2}GMa_0$$ as stated in the paper, we require $$\frac{d\sigma}{dr}=0$$. Again, how realistic is this? The mean velocity is obviously zero because of the hydrostatic equilibrium assumption. As for the radial velocity I can't make any comment conclusively. Any help, or even references would be appreciated.