= Solution
The <pressure tensor> has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)\,d^3v$. With zero mean <velocity> and an isotropic <galactic distribution function>, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. Therefore
$$
p_{ab}=p\delta_{ab},\qquad p=\frac13\int f|\mathbf v|^2\,d^3v=n\langle v_a^2\rangle.
$$
Here $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor <divergence> is $(\operatorname{div}\mathbf p)_a=\partial_b(p\delta_{ab})=\partial_ap$. The time-independent, nonstreaming <Jeans equation> consequently reduces to
$$
\boxed{\nabla p=n\nabla\psi.}
$$
The <gravitational acceleration> convention is $\nabla\psi$: this $\psi$ differs by a sign from the usual <Newtonian gravitational potential> $\Phi$.
Let $z$ measure distance along the <line of sight>. <Spherical density projection> gives
$$
N(R)=\int_{-\infty}^{\infty}n(\sqrt{R^2+z^2})\,dz
=2\int_R^\infty\frac{n(r)r}{\sqrt{r^2-R^2}}\,dr.
$$
Because isotropy makes the local second <stellar velocity moment> along every <line of sight> equal to $p/n$, and because there is no streaming motion, the projected moment is
$$
N(R)\sigma^2(R)=\int_{-\infty}^{\infty}p(\sqrt{R^2+z^2})\,dz
=2\int_R^\infty\frac{p(r)r}{\sqrt{r^2-R^2}}\,dr.
$$
This is the forward relation for <isotropic stellar pressure deprojection>.
Both inversions follow from one <Abel transform> calculation. Write $F(u)=N(\sqrt u)$ and $a(v)=n(\sqrt v)$, so $F(u)=\int_u^\infty a(v)(v-u)^{-1/2}\,dv$. Compose the transforms, using <Tonelli's theorem> for nonnegative densities, or absolute convergence for signed profiles:
$$
\begin{aligned}
J(x)&=\int_x^\infty\frac{F(u)}{\sqrt{u-x}}\,du\\
&=\int_x^\infty a(v)\left[\int_x^v\frac{du}{\sqrt{(u-x)(v-u)}}\right]dv
=\pi\int_x^\infty a(v)\,dv.
\end{aligned}
$$
Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J'(x)=-\pi a(x)$. Setting $x=r^2$ gives the <spherical Abel deprojection>
$$
\boxed{n(r)=-\frac1{2\pi r}\frac d{dr}\int_{r^2}^\infty\frac{N(\sqrt u)}{\sqrt{u-r^2}}\,du.}
$$
Replacing $F(u)$ by $N(\sqrt u)\sigma^2(\sqrt u)$ and $a(v)$ by $p(\sqrt v)$ proves
$$
\boxed{p(r)=-\frac1{2\pi r}\frac d{dr}\int_{r^2}^\infty\frac{N(\sqrt u)\sigma^2(\sqrt u)}{\sqrt{u-r^2}}\,du.}
$$
These are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.
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