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

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 \[%\label{dP_ad} \int{\frac{dP}{\rho}} = \frac{(c^{\rm{ad}}_{s})^2 }{\gamma -1} = \frac{\gamma (c^{\rm{iso}}_{s})^2 }{\gamma -1} \quad \rm{Adiabatic} \ \rm{Flow}\] \[%\label{dP_iso} \int{\frac{dP}{\rho}} = (c^{\rm{iso}}_{s})^2 \ln{\frac{\rho}{\rho_0}} \quad \rm{Isothermal} \ \rm{Flow}\]

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 \[\int{\frac{dP}{\rho}} = \frac{\gamma}{\gamma -1} \frac{v^2_K}{\mathcal{M}^2}\quad \rm{Adiabatic} \ \rm{Flow}\] \[\int{\frac{dP}{\rho}} = \frac{v^2_K}{\mathcal{M}^2} \ln{\frac{\rho}{\rho_0}} \quad \rm{Isothermal} \ \rm{Flow}\]

## Share on Social Media