\[
\begin{align}
W &= -4\pi G\int_0^\infty dr \, r\rho(r)M(r) \\
&= -G \int_0^\infty dr \, \frac{1}{r}\left(M \frac{dM}{dr}\right) \\
&= -\frac{G}{2}\int_0^\infty dr \, \frac{d}{dr}\left(\frac{M^2}{r}\right) - \frac{G}{2}\int_0^\infty dr \, \frac{M^2}{r^2}
\end{align}
\]
Since \(M\) is finite, the term \(M^2/r\) vanishes as \(r\to\infty\). Now, for \(r\to 0\), we have that \(M^2/r \sim r^5 \rho^2\) which implies that this term also vanishes at \(r=0\) provided that \(\rho(r)\) diverges more slowly than \(r^{-5/2}\). Thus, the first term in the last line above is zero and we obtain the desired result:
\[
W = -\frac{G}{2}\int_0^\infty dr \, \frac{M^2(r)}{r^2}
\]
Problem 2.2
For a spherical potential
\[
\begin{align}
W_{jk} &= - \int d^3\mathbf{x} \, \rho(r) x_j \frac{\partial \Phi}{\partial x_k} \\
&= - \int d^3\mathbf{x} \, \rho(r) r \left(\frac{d\Phi}{dr} \right)\frac{x_j x_k}{r^2} \\
&= -\int d^3\mathbf{x} \, \rho(r) r \left(\frac{GM(r)}{r^2} \right)\frac{x_j x_k}{r^2} \\
&= -G\int_0^\infty dr \, r \rho(r) M(r) \int d\Omega \, \frac{x_j x_k}{r^2} \\
&= -\frac{4\pi}{3} G \delta_{jk} \int_0^\infty dr \, r \rho(r) M(r) \\
&= \frac{1}{3} W \delta_{jk}
\end{align}
\]
One way to calculate the angular integral is to use Schur's lemma. Or we can represent the Cartesian coordinates in terms of spherical coordinates and perform the integral explicitly.
Problem 2.3
Using Gauss's theorem
Draw a cylindrical surface of radius \(R\) and height \(h\) aligned with the \(z\)-axis,
\[
2\frac{d\Phi}{dz} = 4\pi G \Sigma (\pi R^2) \implies \frac{d\Phi}{dz} = 2\pi G \Sigma,
\]
thus,
\[
\Phi(z) = 2\pi G \Sigma |z| + \mathrm{constant}.
\]
Using Poisson's equation
In cylindrical coordinates, Poisson's equation reads
In both cases, the Laplacian vanishes. Thus, the density is zero everywhere except at \(r=0\) and \(\theta = \pi/2\). Because of the absolute value, we conclude that a potential of the form \(\Phi(R, z) = v_c^2\ln(R + |z|) + \mathrm{const}\) introduces a \(\delta\)-function at \(\theta = \pi/2\), that is \(\rho(R, z) = \Sigma(R)\delta(z)\).
b. Disk density
We can integrate Poisson's equation across a thin pillbox of thickness \(2h\) around the plane \(z=0\).
\[
\begin{align}
R^2 + (a + |z|)^2 &= a^2\sinh^2u \sin^2v + (a + a \cosh u |\cos v|)^2 \\
&= a^2(\sinh^2u \sin^2v + 1 + 2\cosh u |\cos v| + \cosh^2u \cos^2v) \\
&= a^2(\cosh^2u + \cos^2v + 2\cosh u |\cos v|) \\
&= a^2(\cosh u + |\cos v|)^2
\end{align}
\]
The Kuzmin potential is then
\[
\begin{align}
\Phi_K &= -\frac{GM}{\sqrt{R^2 + (a + |z|)^2}} = -\frac{GM}{a(\cosh u + |\cos v|)} \\
&= -\frac{GM}{a}\frac{1}{\cosh u + |\cos v|}\frac{\cosh u - |\cos v|}{\cosh u - |\cos v|} \\
&= -\frac{GM}{a} \frac{\cosh u - |\cos v|}{\sinh^2u + \sin^2v}
\end{align}
\]
Problem 2.7
The an answer is No
Outside a bounded mass, the Newtonian potential is determined only by the multipole moments of the interior density. Saying
\[
\Phi(r) = -\frac{GM}{r} \quad\text{for } r > R
\]
is equivalent to saying that all exterior multipoles with \(l \geq 1\) vanish. This is only possible if the density is spherically symmetric inside \(r < R\). That does not force the interior density to be spherically symmetric, only that the integrated multipole moments are zero.
As an example consider a spherically symmetric mas \(M_0\) and add two thin concentric shells at radiia \(0 < b < a < R\), whose surface densities are given by
For a thin shell at radius \(r\), the exterior \(l = 2\) coefficient scales as \(r^2\) times the shell's \(Y_{20}\) coefficient. TThus the two sheel give equal and opposite contributions to the total \(l=2\) multipole moment for \(r > a\). The next exterior potential for \(r > R\) is exactly \(-GM/r\) but the interior density is not spherically symmetric.
Problem 2.8
For a density mass-to-light ratio \(\Upsilon\) we know that \(\rho(r) = \Upsilon j(r)\), and from (Problem 1.3)[./chapter01.3.md] the mass density is
\[
\begin{align}
\int_0^\infty dR\, I(R) &= 2\int_0^\infty dR \int_R^\infty dr \, \frac{r j(r)}{\sqrt{r^2 - R^2}} \\
&= 2\int_0^\infty dr \, r j(r) \int_0^r \frac{dR}{\sqrt{r^2 - R^2}} \\
&= \frac{2}{\Upsilon}\int_0^\infty dr \, r \rho(r) \cdot \frac{\pi}{2} \\
&= -\frac{\Phi(0)}{4G\Upsilon}
\end{align}
\]
Problem 2.10
a. Quadrupole expansion
Consider an axisymmetric density distribution \(\rho(R, z)\) with total mass \(M\). Since the density is axisymmetric, \(\rho_{lm} = \delta_{m0} \rho_{l0}\), and