Density of a patch of an accretion disk

In summary, the potential ##\phi## at the point labelled in the diagram is ##-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##
  • #1
ergospherical
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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##.
 
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  • #2
Isn't the gravitation ##\Omega## dependent by the assumption of a "static atmosphere"?
 
  • #3
use basic principles of hydrostatic equilibrium and consider the ideal gas law.
 
  • #4
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).

1683487906907.png
 
  • #5
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##?
 
  • #6
ergospherical said:
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##.
 
  • #7
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*}
 
  • #8
ergospherical said:
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##.

ergospherical said:
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, $$\frac{\partial p}{\partial z} = \left(\frac{\partial p}{\partial \rho}\right)_T \frac{\partial \rho}{\partial z} $$
 
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  • #9
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!
 
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