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Anisotropic distribution functions

There are some expressions in section 5.6.4 that took me a while to figure out, I will write them here for my own reference and for anyone else who might find them useful.

Deriving Eq. 5.79

To derive Eq. (5.79) note that the velocity can be written in spherical coordinates as \(\mathbf{v} = (v_r, v_\theta, v_\phi)\), and we can further change \((v_\theta, v_\phi)\) to polar coordinates in the tangential velocity plane. Define the tangential speed as \(v_t = \sqrt{v_\theta^2 + v_\phi^2}\), and the angle \(\psi\) in the tangential plane so that \(v_\theta = v_t \cos\psi\) and \(v_\phi = v_t \sin\psi\). Then the volume element in velocity space is \(d^3\mathbf{v} = (dv_\theta dv_\phi) dv_r = (v_t\,dv_t\,d\psi)dv_r\).

\[ \begin{align} \rho(r) &= \int d^3\mathbf{v} \; f(r, \mathbf{v}) = \int_0^{2\pi} d\psi \int dv_r dv_t\; v_t \, f(r, v_r, v_t) \\ &= 2\pi \int dv_r dv_t\; v_t \, f(r, v_r, v_t). \end{align} \]

Since \(f = 0\) for \(\mathcal{E} \geq 0\), we need to consider only \(v_r^2 + v_t^2 \leq 2\Psi\). For a fixed \(v_t\), \(v_r\) ranges over \(\pm\sqrt{2\Psi - v_t^2}\), and \(v_t\) ranges from 0 to \(\sqrt{2\Psi}\). Using the symmetry \(f(\cdots, v_r) = f(\cdots, -v_r)\) to fold the \(v_r\) integral:

\[ \rho(r) = 4\pi \int_0^{\sqrt{2\Psi}} dv_t \int_0^{\sqrt{2\Psi - v_t^2}} dv_r \; v_t \, f(\mathcal{E},\, r v_t) \]

Now change the integration variable from \(v_r\) to \(\mathcal{E} = \Psi - \frac{1}{2}(v_r^2 + v_t^2)\) at fixed \(v_t\). Then \(d\mathcal{E} = -v_r\,dv_r\) and the limits are

  • \(v_r = 0 \;\Rightarrow\; \mathcal{E} = \Psi - v_t^2/2\)
  • \(v_r = \sqrt{2\Psi - v_t^2} \;\Rightarrow\; \mathcal{E} = 0\)

And the integral becomes

\[ \rho(r) = 4\pi \int_0^{\sqrt{2\Psi}} dv_t \int_0^{\Psi - v_t^2/2} d\mathcal{E} \; \frac{v_t \, f(\mathcal{E},\, r v_t)}{\sqrt{2(\Psi - \mathcal{E}) - v_t^2}} \]

Finally, swap the order of integration. The region is \(0 \leq v_t \leq \sqrt{2\Psi}\) and \(0 \leq \mathcal{E} \leq \Psi - v_t^2/2\), which is equivalent to \(0 \leq \mathcal{E} \leq \Psi\) and \(0 \leq v_t \leq \sqrt{2(\Psi - \mathcal{E})}\):

\[ \boxed{ \rho(r) = 4\pi \int_0^{\Psi} d\mathcal{E} \int_0^{\sqrt{2(\Psi - \mathcal{E})}} dv_t \; \frac{v_t \, f(\mathcal{E},\, r v_t)} {\sqrt{2(\Psi - \mathcal{E}) - v_t^2}} } \tag{5.79} \]

Deriving Eq. 5.81 from Eq. 5.79

To obtain \(\rho\,\sigma_r^2\) we need the second radial-velocity moment:

\[ \rho(r)\,\sigma_r^2(r) = \int d^3\mathbf{v}\; v_r^2 f(r, \mathbf{v}) \]

Repeating the same steps as above (integrate \(\psi\), fold \(v_r\), change variable to \(\mathcal{E}\)), the only difference is the extra \(v_r^2\) in the integrand. Since \(v_r^2\,dv_r = v_r\,d\mathcal{E}\), the \(1/v_r\) Jacobian and \(v_r^2\) moment combine into a single \(v_r = \sqrt{2(\Psi-\mathcal{E})-v_t^2}\) in the numerator (instead of the denominator as in 5.79). After swapping integration order:

\[ \boxed{ \rho(r)\,\sigma_r^2(r) = 4\pi \int_0^{\Psi} d\mathcal{E} \int_0^{\sqrt{2(\Psi - \mathcal{E})}} dv_t \; v_t \, f(\mathcal{E},\, r v_t)\, \sqrt{2(\Psi - \mathcal{E}) - v_t^2} } \tag{5.81} \]

Deriving Eq. 5.82 from Eq. 5.81

Equation 5.80 defines the augmented density as a function of \((\Psi, r)\) - identical in form to the density integral (5.79), but treating \(\Psi\) as a free variable rather than evaluating it at \(\Psi[r]\):

\[ \tilde{\rho}(\Psi, r) = 4\pi \int_0^{\Psi} d\mathcal{E} \int_0^{\sqrt{2(\Psi - \mathcal{E})}} dv_t \; \frac{v_t \, f(\mathcal{E},\, r v_t)} {\sqrt{2(\Psi - \mathcal{E}) - v_t^2}} \tag{5.80} \]

with the true density recovered as \(\rho(r) = \tilde{\rho}(\Psi[r], r)\).

Now consider Eq. 5.81

\[ \begin{align} \tilde{\rho}(\Psi, r)\,\sigma_r^2(r) &= 4\pi \int_0^{\Psi} d\mathcal{E} \int_0^{\sqrt{2(\Psi-\mathcal{E})}} dv_t \; v_t\, f(\mathcal{E}, rv_t)\, \sqrt{2(\Psi-\mathcal{E})-v_t^2} \\ &= 4\pi \int_0^{\Psi} d\mathcal{E} \int_0^{\sqrt{2(\Psi-\mathcal{E})}} dv_t \; v_t\, f(\mathcal{E}, rv_t)\, \int_{\mathcal{E} + v_t^2/2}^{\Psi} \frac{d\Psi'}{\sqrt{2(\Psi'-\mathcal{E})-v_t^2}} \\ &= 4\pi \int_0^{\Psi} d\Psi' \int_0^{\Psi'} d\mathcal{E}\int_0^{\sqrt{2(\Psi'-\mathcal{E})}} dv_t \; \frac{v_t\, f(\mathcal{E}, rv_t)}{\sqrt{2(\Psi'-\mathcal{E})-v_t^2}} \\ &\stackrel{5.80}{=} \int_0^{\Psi} d\Psi' \; \tilde{\rho}(\Psi', r) \end{align} \]

Where in the secont-to-last step we have re-arranged the domain

\[ \begin{align} 0 \leq \mathcal{E} \leq \Psi&, \quad 0 \leq v_t \leq \sqrt{2(\Psi-\mathcal{E})}, \quad \mathcal{E} + \frac{v_t^2}{2} \leq \Psi' \leq \Psi \\ \Rightarrow\quad& 0 \leq \Psi' \leq \Psi, \quad 0 \leq \mathcal{E} \leq \Psi', \quad 0 \leq v_t \leq \sqrt{2(\Psi'-\mathcal{E})}. \end{align} \]

Thus we have

\[ \tilde{\rho}(\Psi, r) = \frac{1}{\sigma_r^2(r)} \frac{d}{d\Psi}\tilde{\rho}(\Psi, r) \tag{5.82} \]