Use the <relative energy> $\mathcal E=-E=\Psi-v^2/2$. At fixed $r$, write velocity-space spherical coordinates with polar angle $\alpha$ from the radial direction, so $L=rv\sin\alpha$. Then
$$
\rho=2\pi f_0r^{-2\beta}
\int_0^{\sqrt{2\Psi}}v^{2-2\beta}
(\Psi-v^2/2)^n,dv
\int_0^\pi\sin^{1-2\beta}\alpha\,d\alpha.
$$
The <beta function> integrals give
$$
\boxed{
\rho(r)=A r^{-2\beta}\Psi^{n+3/2-\beta}},
$$
where
$$
\boxed{
A=f_0,2^{3/2-\beta}\pi^{3/2}
\frac{\Gamma(1-\beta)\Gamma(n+1)}
{\Gamma(n+5/2-\beta)}}.
$$
Thus $\gamma=2\beta$ and $p=n+3/2-\beta$, with convergence for $\beta<1$ and $n>-1$.

Comparison with part a gives
$$
(\beta_1,n_1)=\left(\frac34,\frac54\right),
\qquad
(\beta_2,n_2)=\left(\frac12,2\right).
$$
Hence the model has the <constant-anisotropy distribution function>
$$
\boxed{
f(\mathcal E,L)=f_{01}L^{-3/2}\mathcal E^{5/4}
+f_{02}L^{-1}\mathcal E^2},
$$
where matching the two density coefficients gives
$$
\boxed{
f_{01}=\frac{3\,2^{5/4}\sqrt b}
{5\pi^{5/2}G^2M\Gamma(1/4)^2},
\qquad
f_{02}=\frac{3b}{4\pi^3G^3M^2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 320 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-320/2/b/solution