# Circum-Binary Barn-oulli

Since I am not going to upgrade yet, I will make these notes public.

# The non-viscous Bernoulli equation

The steady-state Euler equation reads $\left(\mathbf{v} \cdot \nabla \right)\mathbf{v} + \frac{1}{\rho}\nabla P + \nabla \Phi_G -\nu\left[\nabla^2\mathbf{v} + \frac{1}{3}\left(\nabla \cdot \mathbf{v} \right) \right] = 0$ where it is understood that the gravitational potential $$\Phi_G$$ and the coefficient of kinematic viscosity $$\nu$$ are time independent. Now use the identity $$\left(\mathbf{v} \cdot \nabla \right)\mathbf{v} = \frac{1}{2} \nabla \left( \mathbf{v} \cdot \mathbf{v}\right) - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right)$$, neglect viscosity for now, and integrate the momentum equation along a streamline from a reference point to the point of evaluation $\int{\mathbf{ds}\cdot \left[\nabla \left( \frac{1}{2}v^2\right) - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right) + \frac{1}{\rho}\nabla P + \nabla \Phi_G \right]} = 0.$ Since $$\mathbf{ds}$$ is the line element of a streamline, it is in the same direction as $$\mathbf{v}$$,so you get $\frac{1}{2}v^2 - \Phi_G +\int{\frac{dP}{\rho}} = \rm{cst}$ Which is Bernoulli’s equation for any steady, non-viscous flow. To evaluate the last term on the RHS let’s assume we are dealing with a monatomic ideal gas. Then $P = \left(\gamma - 1\right)\epsilon \rho = RT \rho.$ where R is the ideal gas constant. The first law of thermodynamics says that the change in internal energy $$\epsilon$$ plus the work done by the system is equal to the heat added, $dQ = d\epsilon + PdV.$ For an ideal gas \begin{aligned} d\epsilon &= \frac{R}{\gamma -1}dT \\ PdV &= -\left(\gamma -1\right)\epsilon \frac{d \rho}{\rho} = -RT\frac{d \rho}{\rho}\end{aligned} Then $dQ = \frac{R}{\gamma -1}dT - RT \frac{d \rho}{\rho}.$ We can rewrite $\frac{d\rho}{\rho} = \frac{1}{\rho} \frac{dP}{RT} - \frac{1}{\rho}\frac{P}{RT^2} dT$ so that \begin{aligned} dQ &= R\left( 1 + \frac{1}{\gamma-1}\right)dT - \frac{d P}{\rho} \\ &= \frac{\gamma R}{\gamma-1}dT - \frac{d P}{\rho}\end{aligned}

Let’s also introduce the enthalpy $$h$$ in the limit that the number of particles in the system is constant, $dh = TdS + VdP = TdS + \frac{dP}{\rho}$ and also the Gibbs free energy $$G$$ in the same particle conserving limit $dG = SdT + \frac{dP}{\rho}$ Now we assume two special cases, adiabatic and isothermal flows. First assume an adiabatic flow, $$dQ =0$$. Then $\int{\frac{dP}{\rho}} = \frac{\gamma }{\gamma -1} RT$ which we could have also seen from the expression for the enthalpy when $$dQ = TdS =0$$. For the isothermal case use the ideal gas equation $$P= RT \rho$$ and that $$T$$ is constant to write $$dP=RT d\rho$$, then $\int^{\rho}_{\rho_0}{\frac{dP}{\rho}} = RT \ln{\frac{\rho}{\rho_0}}.$ which we could have also seen from the expression for the Gibbs free energy with $$dT=0$$. We can also write out these expressions in terms of the isothermal or adiabatic sound speeds by noting that for the adiabatic case, $$dS=0$$ implies that $\frac{P}{\rho^{\gamma}} = cst$ then the adiabatic sound speed is $c^{\rm{ad}}_{s} = \sqrt{\frac{dP}{d\rho}} = \sqrt{\gamma \frac{P}{\rho}} = \sqrt{\gamma RT}$ The isothermal equation of state $$P=(c^{\rm{iso}}_{s})^2 \rho$$ gives us that $$c^{\rm{iso}}_{s} = \sqrt{RT}$$. So we can write $%\label{dP_ad} \int{\frac{dP}{\rho}} = \frac{(c^{\rm{ad}}_{s})^2 }{\gamma -1} = \frac{\gamma (c^{\rm{iso}}_{s})^2 }{\gamma -1} \quad \rm{Adiabatic} \ \rm{Flow}$ $%\label{dP_iso} \int{\frac{dP}{\rho}} = (c^{\rm{iso}}_{s})^2 \ln{\frac{\rho}{\rho_0}} \quad \rm{Isothermal} \ \rm{Flow}$

# Hydrostatic Balance

In a thin accretion disk around a point mass of mass $$M$$, hydrostatic balance gives $\frac{\partial P}{\partial z} = \rho \frac{GMz}{\left( r^2 + z^2 \right)^{3/2}}$ or when the disk scale height is much smaller than the disk radius, $$H \ll r$$, $\frac{P}{H} = \rho \frac{GMH}{r^{3}}.$ If we define the disk Mach number as the ratio of the isothermal sound speed to the keplerian orbital velocity $$v_K=\sqrt{GM/r}$$, then we have an expression for the Mach number in terms of disk sound speed for hydrostatic equilibrium in the vertical direction, $\mathcal{M} = \frac{H}{r} = \frac{v_K}{c^{\rm{iso}}_s}$ Now we can rewrite (\ref{dP_ad}) and (\ref{dP_iso}) in terms of the disk Mach number $\int{\frac{dP}{\rho}} = \frac{\gamma}{\gamma -1} \frac{v^2_K}{\mathcal{M}^2}\quad \rm{Adiabatic} \ \rm{Flow}$ $\int{\frac{dP}{\rho}} = \frac{v^2_K}{\mathcal{M}^2} \ln{\frac{\rho}{\rho_0}} \quad \rm{Isothermal} \ \rm{Flow}$