Gravitational potential energy of an isothermal sphere

Question

I am trying to calculate the gravitational potential energy, W, defined as:

$$W = -\frac{1}{2}\int\rho(r)\Phi(r)\mathrm d^{3}r$$

for an isothermal sphere. I am given that the density profile varies with r as:

$$\rho \propto r^{-2}$$

and that $\Phi(r)$ is defined as:

$$\Phi(r)=\Phi_{0} + \eta\ln(r)$$

where $\eta$ is a constant. Given it is a spherical system I set up the following equation.

$$W = -\frac{1}{2}\int^{r_{0}}_{0}4\pi k(\Phi_{0} + \eta\ln(r))\mathrm dr$$

Where k is a constant of proportionality. I am not sure how to calculate this integral without ending up with infinities due to $\ln(0)$ values. I am also told that the sphere can be considered to be truncated at $r_{0}$, does this mean I can somehow ignore the $\ln(r)$ term in the integral?

Answer 1

The self-gravitational energy of the singular isothermal sphere diverges no matter the choice of the radius $r_0$ adopted to normalize the potential (you cannot take the logarithm of a lenght) and the adopted value of the potential at $r_0$, $\phi_0$. Notice that $r_0$ above is NOT a truncation radius! The integral diverges for the contribution of the potential at large radii. Basically, independently of the value of $\phi_0$, at large radii you are integrating $\ln(r/r_0)$, and this diverges as $r$ $\ln(r/r_0)$. A different case, with finite self gravitational energy, is obtained when the sphere is truncated at some truncation radius $r_t$ (of course, the sphere now is not isothermal: you cannot have a self-gravitating, truncated system of finite mass really isothermal. Finite mass means finite escape velocity, but in a maxwellian velocity distribution you have arbitrarily high velocities in its exponential tail). The resulting expression is very simple, it is a nice exercise.