Density of a patch of an accretion disk

[ergospherical]

Homework Statement

A small patch of a thin accretion disk (around a point mass) at a radius where the angular velocity is $\Omega$ can be assumed to have an isothermal static atmosphere. Show that the density varies with $z$ (distance from the mid-plane) as\begin{align*}
\rho = \rho_0 \mathrm{exp}[-\gamma \Omega^2 z^2/(2c_s^2)]
\end{align*}($\gamma$ is adiabatic index)

Relevant Equations

Fluid equations

In the frame of the patch $-(1/\rho) \nabla p = - \nabla \phi$, and putting $\nabla p = (\partial p/\partial \rho) \nabla \rho = c_s^2 \nabla \rho$ and taking the $z$ component gives\begin{align*}
-\frac{c_s^2}{\rho} \frac{\partial \rho}{\partial z} = -c_s^2 \frac{\partial(\log{\rho})}{\partial z} = \frac{\partial \phi}{\partial z}
\end{align*}integrate:\begin{align*}
\rho = \rho_0 \mathrm{exp}[-\phi/c_s^2]
\end{align*}What is the form of the potential $\phi$? I thought $\phi = \phi_{\mathrm{rot}} + \phi_{\mathrm{grav}} = -\frac{1}{2} \Omega^2 r^2 + \phi_{\mathrm{grav}}$, but the centrifugal potential has no $z$ dependence and I don't see why the gravitational potential $\phi_{\mathrm{grav}}$ should depend on $\Omega$.

 

[hutchphd]

Isn't the gravitation $\Omega$ dependent by the assumption of a "static atmosphere"?

 

[TonyStewart]

use basic principles of hydrostatic equilibrium and consider the ideal gas law.

 

[TSny]

For an element $m$ of the accretion disk located on the central plane (z = 0), the gravitational attraction $F_G$ toward the central mass $M$ is balanced by the centrifugal force $F_C$ in the frame of $m$. However, for $z \neq 0$, the two forces no longer balance because of the tilt of $F_G$. For small $z$, the two forces produce a net downward force on $m$. Thus, equilibrium in the z-direction requires an additional upward force (caused by pressure variation in the z-direction).

 

[ergospherical]

Cheers! The gravitational potential at the point labelled in the diagram is $\phi = -GM(R^2 + z^2)^{-1/2}$, where $R$ is the cylindrical radial coordinate. Expanding to first order gives \begin{align*}
\phi = -\frac{GM}{R} + \frac{GM z^2}{2R^3} \implies \frac{\partial \phi}{\partial z} = -\frac{GMz}{R^3}
\end{align*}The circular speed of the disk $v_C(R) = \sqrt{GM/R} = \Omega R$ implies $\Omega^2 = GM/R^3$, so equivalently $\partial \phi / \partial z = -\Omega^2 z$. The momentum equation in the $z$ direction gives\begin{align*}
-c_s^2 \frac{\partial \log{\rho}}{\partial z} = \Omega^2 z \implies \rho = \rho_0 \mathrm{exp}[-\Omega^2 z^2 /(2c_s^2)]
\end{align*}Looks like I'm missing the adiabatic index $\gamma$?

 

[TSny]

Looks like I'm missing the adiabatic index $\gamma$?

Show that the ideal gas law can be written as $P = \large \frac{c_s^2}{\gamma} \rho$.

 

[ergospherical]

For an adiabatic gas I have $c_s^2 = (\partial p/\partial \rho) |_S$, and given the equation of state in the form $p = K\rho^{\gamma}$ that means $c_s^2 = \gamma p / \rho$. But starting from the hydrostatic equation\begin{align*}
\frac{1}{\rho} \frac{\partial p}{\partial z} = \frac{\partial \phi}{\partial z}
\end{align*}it looks like I can exchange\begin{align*}
\frac{\partial p}{\partial z} = \frac{\partial p}{\partial \rho} \frac{\partial \rho}{\partial z} = c_s^2 \frac{\partial \rho}{\partial z}
\end{align*}

 

[TSny]

For an adiabatic gas I have $c_s^2 = (\partial p/\partial \rho) |_S$, and given the equation of state in the form $p = K\rho^{\gamma}$ that means $c_s^2 = \gamma p / \rho$.

Ok. Sound vibrations are assume to be adiabatic. Thus, we use $(\partial p / \partial \rho) |_S$ when calculating $c_s^2$.

it looks like I can exchange\begin{align*}
\frac{\partial p}{\partial z} = \frac{\partial p}{\partial \rho} \frac{\partial \rho}{\partial z} = c_s^2 \frac{\partial \rho}{\partial z}
\end{align*}

The problem statement says to assume an isothermal static atmosphere. So, when considering how $P$ and $\rho$ vary with $z$, we would assume $T$ remains constant. So, $$\frac{\partial p}{\partial z} = \left(\frac{\partial p}{\partial \rho}\right)_T \frac{\partial \rho}{\partial z} $$

 

[ergospherical]

I've just noticed, the problem statement says that the adiabatic sound speed is $c_s^2$, i.e. $c_s^2 = \gamma p / \rho$, but the atmosphere is assumed isothermal - so $c_s^2|_{\mathrm{iso}} = p / \rho = c_s^2 / \gamma$, which clears it up. Thanks!