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

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

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

(6)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)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)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)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.