Hydrostatic Equilibrium for rotating object

1) Hydrostatic Equilibrium for non-rotating object

Assumption

• point source gravity from the central supermassive black hole without other potential (halo, disk)
• gaseous self gravity is neglected
—> only SMBH gravity is balanced with pressure gradient
• adiabatic equation of state (EOS)

The hydrostatic equilibrium is

(1)
\begin{align} \frac{1}{\rho} \nabla P = - \nabla \Phi \end{align}

where the gravitational potential is $\Phi = - G M_{BH}/r$ and $M_{BH}$ is the black hole mass.

If we consider the adiabatic EOS;

(2)
\begin{align} P=K \rho^{\gamma} \end{align}

where $K=P_{c}/\rho_{c}^{\gamma}$, and $P_{c}, \rho_{c}$ are the arbitrary pressure and density in the central region, respectively.

Plugging the EOS and the potential equation into eq.(1),

(3)
\begin{align} \frac{K}{\rho} \frac{d \rho^{\gamma}}{dr} = - \frac{G M_{BH}}{r^{2}} \end{align}

Equating eq.(3), we can derive the density profile in the hydrostatic balance:

(4)
\begin{align} \rho(r) = \rho_{c} \left[ 1 + \frac{\gamma-1}{\gamma} \frac{\rho_{c}}{P_{c}} G M_{BH} \left( \frac{1}{r} - \frac{1}{r_{BH}} \right) \right]^{1/(\gamma-1)} \end{align}

where $r_{BH}$ is the schwarzschild radius. Since the inside of the bracket should be positive to get the real value of $\rho$,

(5)
\begin{align} 1- \frac{\gamma-1}{\gamma} \frac{\rho_{c}}{P_{c}} \frac{G M_{BH}}{r_{BH}} \geq 0 \\ \frac{P_{c}}{\rho_{c}} \geq \frac{\gamma-1}{\gamma} \frac{G M_{BH}}{r_{BH}} \end{align}

Assuming the ideal gas condition, the eq. (5) can be expressed with Temperature of the central region;

(6)
\begin{align} T_{c} \geq \frac{\gamma-1}{\gamma} \frac{G M_{BH}}{r_{BH}} \frac{\mu m_{H}}{k} \end{align}

where $\mu$ is mean molecular weight.

2) Hydrostatic Equilibrium for rotating object

Assumption

• point source gravity from the central supermassive black hole without other potential (halo, disk)
• gaseous self gravity is neglected
—> only SMBH gravity is balanced with pressure gradient
• adiabatic equation of state (EOS)
• solid body rotation with angular velocity, $\Omega$

In the rotating object, the hydrostatic equilibrium can be expressed as

(7)
\begin{align} \frac{1}{\rho}\nabla P = - \nabla \Phi_{eff}, \end{align}

where the effective potential is $\nabla \Phi_{eff} = \Phi + V$, and $V$ is the rotational energy, $V = -\frac{1}{2} ( \Omega r \sin{\theta} )^{2}$. Therefore, the eq.(7) can be rewritten as

(8)
\begin{align} \frac{1}{\rho}\nabla P = - \nabla \Phi + \frac{1}{2} \Omega^{2} \nabla (r \sin{\theta})^{2} \end{align}

Considering the point source gravitational potential, $\Phi=- G M_{BH}/r$, the eq. (7) can be solved to derive the density profile in the hydrostatic equilibrium for the solid body rotating object;

(9)
\begin{align} \rho(r,\theta) = \rho_{c,0} \left[ 1 + \frac{\gamma-1}{\gamma} \frac{\rho_{c,0}}{P_{c,0}} \left( \frac{G M_{BH}}{r} - \frac{G M_{BH}}{r_{BH}} + \frac{1}{2} r^{2} \Omega^{2} \sin^{2}{\theta} \right) \right]^{1/(\gamma-1)}, \end{align}

where $\rho_{c,0}, P_{c,0}$ are density and pressure at $(r,\theta) = (r_{BH},0)$, respectively. It is obvious that if $\Omega=0$, the eq. (9) is same with the eq. (4) for non-rotating case.