• Hydrostatic Balance of an Adiabatic Fluid in a Background Potential
• Hydrostatic disks with varying surface densities

Hydrostatic Balance of an Adiabatic Fluid in a Background Potential

Consider an adiabatic fluid with an adiabatic index \(\gamma\). We want to compute the density structure \(\rho(z)\) in a potential \(\Phi(z)\). The gas pressure is \(P = K \rho^{\gamma}\).

First, note that \(c_s^2 = \gamma P /\rho_K = \gamma K \rho_K^{\gamma-1}\) or \(\gamma K = c_s^2 \rho_K^{1-\gamma}\)

The equation of vertical hydrostatic equilibrium is

\[\frac{dP}{dz} = -\rho(z) g(z)\] \[\gamma K \rho^{\gamma-1}\frac{d\rho}{dz} = -\rho(z) g(z)\] \[\gamma K \rho^{\gamma-2}d\rho = - g(z) dz\] \[\frac{1}{\gamma-1} \rho^{\gamma-1} + C = \frac{1}{\gamma K}\Phi(z)\] \[\rho^{\gamma-1} = (\gamma-1)\rho_K^{\gamma-1}\frac{\Phi(z)}{c_s^2} + C\] \[\rho = \left[(\gamma-1)\rho_K^{\gamma-1}\frac{\Phi(z)}{c_s^2} + C\right]^{\frac{1}{\gamma-1}}\] \[\rho = \rho_K\left[(\gamma-1)\frac{\Phi(z)}{c_s^2} + C\right]^{\frac{1}{\gamma-1}}\]

If we define the central midplane gas temperature at \(z=0\) to be such that the sound speed is \(c_s\), then \(\rho(z=0)=\rho_K\). Given the above, we then require

\[(\gamma-1)\frac{\Phi(z)}{c_s^2} + C = 1;\,z=0\]

We can arrange this if we define \(\Phi(z=0)=\Phi_0\) and write \(C = -(\gamma-1)\frac{\Phi_0}{c_s^2} + 1\)

and our density profile becomes

\[\rho(z) = \rho_K\left[(\gamma-1)\frac{[\Phi(z)-\Phi_0]}{c_s^2} + 1\right]^{\frac{1}{\gamma-1}}\]

A numerical integral sets \(\rho_K\) by demanding \(\rho_K = \Sigma/(2\int_0^{\infty}\left[(\gamma-1)\frac{[\Phi(z)-\Phi_0]}{c_s^2} + 1\right]^{\frac{1}{\gamma-1}} dz)\)

Hydrostatic disks with varying surface densities

So we can use the above to set the central density if the surface density is not a function of radius. But the surface density is declining with radius and the potential is varying as the radius increases.

Once we set \(\rho_K\) at \(z=0\), \(R=0\), we are stuck with that equation of state if the disk gas is all on the same adiabat.

Now, reconsider a different place in the disk. The central density in the midplane will be lower because of the surface density, but also lower if the potential has changed radially.

Clearly \(\rho(R\ne 0,z=0)\ne\rho_K\), so the constant of integration in determining the density by integrating the vertical force is not the same as at \(R=0\).

Instead we can write
\(\rho(R,z) = \rho_K\left[(\gamma-1)\frac{[\Phi(z)-\Phi_0]}{c_s^2} + D\right]^{\frac{1}{\gamma-1}}\)

Where \(D = \left(\frac{\rho_0(R)}{\rho_K}\right)^{\gamma-1}\)

such that \(\rho(R,z=0) = \rho_0(R)\). The density \(\rho_K\) where the sound speed is \(c_s\) is a constant, so the surface density constraint must be used to set \(\rho_0(R)\). This has to be done iteratively, but a first reasonable guess is \(\rho_0(R) \approx \rho_K \frac{\Sigma(R)}{\Sigma_0}\)

Note \(0<D<1\).