<p><strong>Background</strong>: If $f(\mathbf{x},\mathbf{v},t)$ is the distribution function of the stars in phase space and $n = \int f\,\mathrm{d}^3\mathbf{v}$ is the star density, then in the cylindrical coordinates $(R,\varphi,z)$, the $z$-component of the second of the <a href="https://en.wikipedia.org/wiki/Jeans_equations" rel="nofollow">Jeans equations</a> applied to an axisymmetric system is
$$\renewcommand{\expv}[1]{\langle{#1}\rangle}\partial_t(n\expv{v_z}) + \partial_R(n\expv{v_Rv_z}) + \partial_z(n\expv{v_z^2}) + \frac{n\expv{v_Rv_z}}{R} + n\partial_z\Phi = 0\text{.}$$
In a steady state, the first term vanishes, so if the positive and negative terms of the $z$-component of velocity terms balance, then so do the second and fourth, in which case:
$$\partial_z(n\expv{v_z^2}) = -n\partial_z\Phi\text{,}$$
which we can then integrate.</p>

<p>If the the star and mass distributions are axisymmetric and symmetric about the equatorial ($z = 0$) plane, then the potential has even symmetry in $z$, $\Phi(R,z) = \Phi(R,-z)$, and so does $n$. Thus their derivatives with respect to $z$ are odd functions, and $\Phi_{,z}(R,z) = -\Phi_{,z}(R,-z)$ implies
$$\frac{1}{n}\int_{-z}^z n\Phi_{,z'}\,\mathrm{d}z' = 0\text{,}$$
since the integrand has odd symmetry. Hence,
$$\frac{1}{n}\int_z^\infty n\Phi_{,z'}\,\mathrm{d}z' = \frac{1}{n}\int_{|z|}^\infty n\Phi_{,z'}\,\mathrm{d}z'\text{.}$$</p>

CC BY-SA 3.0
2014-12-06T23:56:42.553Z
2014-12-06T23:56:42.553Z
Stan Liou
<p>I'm solving the axisymmetric Jeans equations to determine the initial conditions for an exponential disk inside an NFW halo. The density profiles of the two components are 
\begin{equation}
\rho_d(R,z)=\rho_{0,d}\exp\left(-\frac{R}{R_0}\right)\exp\left(-\frac{|z|}{z_0}\right)
\end{equation}</p>

<p>\begin{equation}
\rho_h(r)=\frac{\rho_{0,h}}{r/r_s(1+r/r_s)^2}
\end{equation}</p>

<p>Due to the axial symmetry, for the meridional velocity dispersion we have
\begin{equation}
\sigma^2(R,z)=\frac{1}{\rho}\int_z^\infty \rho \frac{\partial \Phi}{\partial z'}dz'
\end{equation}</p>

<p>As the galaxy is symmetric about the disk plane ($z=0$), we would expect the distribution function at a distance $h$ above the disk plane to be identical to that at $z=-h$; in other words, if we flip the galaxy, the only difference will be a change in direction of particle motion, but not speed or dispersion.</p>

<p>Therefore, can't we replace the $z$ in the limit with $|z|$?</p>

CC BY-SA 3.0
2014-12-06T23:25:06.320Z
Is $z$ a function of $h$?
HDE 226868
<astrophysics><galactic-dynamics>
Exploiting symmetry in the axisymmetric Jeans equations
Robbie Fasano
# Exploiting symmetry in the axisymmetric Jeans equations

(Asked by: username_0 on 2014-12-06)

I'm solving the axisymmetric Jeans equations to determine the initial conditions for an exponential disk inside an NFW halo. The density profiles of the two components are
\begin{equation}
\rho\_d(R,z)=\rho\_{0,d}\exp\left(-\frac{R}{R\_0}\right)\exp\left(-\frac{|z|}{z\_0}\right)
\end{equation}

\begin{equation}
\rho\_h(r)=\frac{\rho\_{0,h}}{r/r\_s(1+r/r\_s)^2}
\end{equation}

Due to the axial symmetry, for the meridional velocity dispersion we have
\begin{equation}
\sigma^2(R,z)=\frac{1}{\rho}\int\_z^\infty \rho \frac{\partial \Phi}{\partial z'}dz'
\end{equation}

As the galaxy is symmetric about the disk plane ($z=0$), we would expect the distribution function at a distance $h$ above the disk plane to be identical to that at $z=-h$; in other words, if we flip the galaxy, the only difference will be a change in direction of particle motion, but not speed or dispersion.

Therefore, can't we replace the $z$ in the limit with $|z|$?

**Tags:** astrophysics, galactic-dynamics

> **Comments:**
> **username_1**: Is $z$ a function of $h$?  

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## 1 answers

### Answer by username_2 on 2014-12-06. Score: 0 | Accepted answer

**Background**: If $f(\mathbf{x},\mathbf{v},t)$ is the distribution function of the stars in phase space and $n = \int f\,\mathrm{d}^3\mathbf{v}$ is the star density, then in the cylindrical coordinates $(R,\varphi,z)$, the $z$-component of the second of the [Jeans equations](https://en.wikipedia.org/wiki/Jeans_equations) applied to an axisymmetric system is
$$\renewcommand{\expv}[1]{\langle{#1}\rangle}\partial\_t(n\expv{v\_z}) + \partial\_R(n\expv{v\_Rv\_z}) + \partial\_z(n\expv{v\_z^2}) + \frac{n\expv{v\_Rv\_z}}{R} + n\partial\_z\Phi = 0\text{.}$$
In a steady state, the first term vanishes, so if the positive and negative terms of the $z$-component of velocity terms balance, then so do the second and fourth, in which case:
$$\partial\_z(n\expv{v\_z^2}) = -n\partial\_z\Phi\text{,}$$
which we can then integrate.

If the the star and mass distributions are axisymmetric and symmetric about the equatorial ($z = 0$) plane, then the potential has even symmetry in $z$, $\Phi(R,z) = \Phi(R,-z)$, and so does $n$. Thus their derivatives with respect to $z$ are odd functions, and $\Phi\_{,z}(R,z) = -\Phi\_{,z}(R,-z)$ implies
$$\frac{1}{n}\int\_{-z}^z n\Phi\_{,z'}\,\mathrm{d}z' = 0\text{,}$$
since the integrand has odd symmetry. Hence,
$$\frac{1}{n}\int\_z^\infty n\Phi\_{,z'}\,\mathrm{d}z' = \frac{1}{n}\int\_{|z|}^\infty n\Phi\_{,z'}\,\mathrm{d}z'\text{.}$$


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