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

A <galactic distribution function> $f(\mathbf x,\mathbf v,t)$ is the stellar mass or number per six-dimensional <phase space> volume,
$$
dM=f(\mathbf x,\mathbf v,t)\,d^3x\,d^3v.
$$
Its velocity moments give the spatial density, mean velocity, and <velocity dispersion>; integrating those quantities along the line of sight and weighting by luminosity produces surface-brightness and line-of-sight-velocity observables. A model is compared with data only after the same projection, selection function, and instrumental convolution have been applied.

<Jeans theorem> states that every steady solution of the <Collisionless Boltzmann equation> depends on phase-space coordinates only through <integrals of motion>. Conversely, every nonnegative function of isolating integrals is a steady collisionless distribution function on the region where those integrals are defined.

For the stated <power law> in <relative energy>, isotropy gives
$$
\rho(\Psi)=4\pi F\int_0^{\sqrt{2\Psi}}
\left(\Psi-\frac{v^2}{2}\right)^{n-3/2}v^2\,dv.
$$
With the requested <change of variables> $v=\sqrt{2\Psi}\cos\theta$, the density becomes
$$
\rho=8\sqrt2\pi F\Psi^n
\int_0^{\pi/2}\sin^{2n-2}\theta\cos^2\theta\,d\theta
=2\sqrt2\pi^{3/2}F
\frac{\Gamma(n-1/2)}{\Gamma(n+1)}\Psi^n,
$$
where the last equality uses the <Beta function> and <Gamma function>. Thus
$$
\boxed{\rho\propto\Psi^n}
$$
for $n>1/2$.

The normalized second velocity moment is
$$
\overline{v^2}
=\frac{\int_0^{\sqrt{2\Psi}}v^4(\Psi-v^2/2)^{n-3/2}\,dv}
{\int_0^{\sqrt{2\Psi}}v^2(\Psi-v^2/2)^{n-3/2}\,dv}.
$$
The same substitution and the <Beta-function recurrence> give
$$
\overline{v^2}
=2\Psi\frac{B(5/2,n-1/2)}{B(3/2,n-1/2)}
=\boxed{\frac{3\Psi}{n+1}}.
$$
Consequently the one-dimensional isotropic <velocity dispersion> is $\sigma^2=\overline{v^2}/3=\Psi/(n+1)$, proving the required linear dependence on the <relative potential>.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 349 / 3 / Solution

{title2=$\mathcal E$}

Relative energy is the binding energy per unit mass measured from the chosen escape level, commonly $\mathcal E=\Psi-v^2/2$. Bound phase-space points have $\mathcal E>0$ in this convention.


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 Incoming links: Relative energy