Hydrostatic Balance

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

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

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

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 $R=0$.

$\rho(R,z) = \rho_K\left[(\gamma-1)\frac{[\Phi(z)-\Phi_0]}{c_s^2} + D\right]^{\frac{1}{\gamma-1}}$
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 $% $.