- #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##.

-\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##.