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 $|z|$?

  • $\begingroup$ Is $z$ a function of $h$? $\endgroup$ – HDE 226868 Dec 6 '14 at 23:25

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 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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