Misbehaving singular isothermal sphere potential

Question

The singular isothermal sphere (SIS) is a useful simple model often used in astrophysics. It has density profile:

$$\rho(r) = \frac{\rho_0 r_0^2}{r^2}$$

This is well known to have some quirks (infinite density at $r=0$, infinite mass as $r\rightarrow\infty$), but if one is willing to work around these two points the other properties of the SIS are supposed to be simple and well-behaved.

I'm running into an issue with the potential, though. Every textbook I've seen that discusses the SIS gives its potential as:

$$\Phi(r) = 4\pi G\rho_0r_0^2\ln(r) + {\rm constant}$$

At first glance this seems nice: simple $\ln(r)$ dependence, freedom to pick a constant. And if all I wanted to do was to calculate some relative potentials, this would be fine. But I've come across a paper that gives an absolute potential at a given radius. Typically I'd assume the author set $\lim_{r\rightarrow\infty}\Phi(r) = 0$, but I can't pick a finite constant to make that work.

One of the usual tricks with the isothermal sphere is to 'truncate' it at some radius to get around the infinite mass awkwardness. The author does state such a truncation radius, no harm in calling it $r_0$. Then I'm tempted to set the constant to something like $-GM(<r_0)/r_0$ and assume that potential is only valid for $r < r_0$. But then I'm left with the awkward logarithm of a dimensionful quantity: not good. I could also pick something of the form $$-\frac{AGM(<r_0)}{r_0}\ln(Br_0)$$ with $A$ and $B$ being dimensionless scalar constants, but then I have issues keeping $\Phi$ negative.

Anyway, does anyone know how to set that constant to get a well-behaved potential for a (truncated) SIS?

Answer 1

In order to obtain a mathematically correct expression for the potential of the singular isothermal sphere some care is needed. It is the analogous of the potential on an infinite, rectilinear charge distribution. The reference point for the potential cannot be fixed at the origin, and it cannot be fixed at infinity. What can (and should!) be done is to compute the potential as follows

$$\phi(r) - \phi(r_0) = -G\int_r^{r_0} M(x) x^{-2} \mathrm dx,$$

where $M(x)$ is the mass of the isothermal sphere contained in the sphere of radius $x$, centered at the origin. Of course, the value for $\phi(r_0)$ is fully arbitrary, the usual choice is $0$.

Answer 2

Returned to this and I think I have a working solution, although it doesn't yield much insight.

The paper linked in the question states (~halfway through Sec. 3.5) that the potential is sourced by a mass distribution distributed as $\rho\propto r^{-2}$ (a singular isothermal sphere), truncated at a radius of $1\,{\rm kpc}$ and with a total mass $10^8\,{\rm M}_\odot$. They give a normalization of the potential (they call it a perturbation $\Delta\Phi$ to the potential because this is being added to a background potential) $\Phi(500\,{\rm pc})=-700\,{\rm km}^2\,{\rm s}^{-2}$.

The form of the potential (at least within the truncation radius) should be $\Phi(r)=4\pi G\rho_0 r_0^2\ln(r)+C$. Simply substituting the given normalization immediately gives a value for $C$ of $-4\pi G\rho_0r_0^2\ln(500\,{\rm pc})-700\,{\rm km}^2\,{\rm s}^{-2}$; the logarithm of a dimensionful quantity is not a concern because substituting back into the potential gives:

$$\Phi(r)=4\pi G\rho_0r_0^2\ln\left(\frac{r}{500\,{\rm pc}}\right)-700\,{\rm km}^2\,{\rm s}^{-2}$$

This still leaves the constants $\rho_0r_0^2$ to be fixed, but this is straightforward from the given total mass and size. Integrating the density profile gives the mass enclosed within radius $r$, $M(<r)=4\pi\rho_0r_0^2 r$, so:

$$\rho_0r_0^2=\frac{10^8}{4\pi}\frac{\rm M_\odot}{\rm kpc}$$

If for some reason these are wanted independently (it's not really necessary), they can be interpreted as the density $\rho_0$ at some radius $r_0$. For instance, picking $r_0=1\,{\rm kpc}$ straightforwardly gives $\rho_0=\frac{10^8}{4\pi}\,{\rm M_\odot}\,{\rm kpc}^{-3}$.

The potential looks like (valid only for $r<1\,{\rm kpc}$):

I'm not fully satisfied with this because it doesn't really explain why the authors would have chosen this particular normalization, and they state it as though they've made some obvious choice, but it seems to work. And the mass distribution they used is unambiguously defined by the mass, truncation radius, and $\rho(r)\propto r^{-2}$.