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( \cdot \nabla \right) + {\rho}\nabla P + \nabla \Phi_G -\nu\left[\nabla^2 + {3}\left(\nabla \cdot \right) \right] = 0 where it is understood that the gravitational potential ΦG and the coefficient of kinematic viscosity ν are time independent. Now use the identity $\left( \cdot \nabla \right) = {2} \nabla \left( \cdot \right) - \times \left( \nabla \times \right)$, neglect viscosity for now, and integrate the momentum equation along a streamline from a reference point to the point of evaluation \cdot \left[\nabla \left( {2}v^2\right) - \times \left( \nabla \times \right) + {\rho}\nabla P + \nabla \Phi_G \right]} = 0. Since DS is the line element of a streamline, it is in the same direction as V,so you get {2}v^2 - \Phi_G +{\rho}} = 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 ϵ plus the work done by the system is equal to the heat added, dQ = d\epsilon + PdV. For an ideal gas d\epsilon &= {\gamma -1}dT \\ PdV &= -\left(\gamma -1\right)\epsilon {\rho} = -RT{\rho} Then dQ = {\gamma -1}dT - RT {\rho}. We can rewrite {\rho} = {\rho} {RT} - {\rho}{RT^2} dT so that dQ &= R\left( 1 + {\gamma-1}\right)dT - {\rho} \\ &= {\gamma-1}dT - {\rho} Let’s also introduce the enthalpy h in the limit that the number of particles in the system is constant, dh = TdS + VdP = TdS + {\rho} and also the Gibbs free energy G in the same particle conserving limit dG = SdT + {\rho} Now we assume two special cases, adiabatic and isothermal flows. First assume an adiabatic flow, dQ = 0. Then {\rho}} = {\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 = RTρ and that T is constant to write dP = RTdρ, then \int^{\rho}_{\rho_0}{{\rho}} = RT {\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 {\rho^{\gamma}} = cst then the adiabatic sound speed is c^{}_{s} = {d\rho}} = {\rho}} = The isothermal equation of state $P=(c^{}_{s})^2 \rho$ gives us that $c^{}_{s} = $. So we can write % {\rho}} = }_{s})^2 }{\gamma -1} = }_{s})^2 }{\gamma -1} \quad \ % {\rho}} = (c^{}_{s})^2 {\rho_0}} \quad \ HYDROSTATIC BALANCE In a thin accretion disk around a point mass of mass M, hydrostatic balance gives {\partial z} = \rho {\left( r^2 + z^2 \right)^{3/2}} or when the disk scale height is much smaller than the disk radius, H ≪ r, {H} = \rho {r^{3}}. If we define the disk Mach number as the ratio of the isothermal sound speed to the keplerian orbital velocity $v_K=$, then we have an expression for the Mach number in terms of disk sound speed for hydrostatic equilibrium in the vertical direction, = {r} = {c^{}_s} Now we can rewrite ([dP_ad]) and ([dP_iso]) in terms of the disk Mach number {\rho}} = {\gamma -1} {^2}\quad \ {\rho}} = {^2} {\rho_0}} \quad \
Following Arnab Rai Choudhuri’s derivation in “The physics of fluids and plasmas” MICROSCOPIC DERIVATION Prelimenaries Let the distribution function $f(\bx, \bv, t)$ describe the phase space density of a distribution of N particles. If these particles move under a common potential and interactions between particles is negligible, then the dynamics of each particle can be described in terms of a Hamiltonian formulation which allows one to prove Liouville’s theorem for the distribution and arrive at the collisionless Boltzman’s equation {Dt} =\partial_t f + f + f= 0 which says that along the trajectory of a particle the distribution function is constant. In hydrodynamics however collisions are important. In the limit that there are no long-range forces (no EM fields) and that collisions between particles are binary (that is the volume of a particle is much less than the total volume of the space) one can compute the rate of particle collisions and construct a “collision” term for the right hand side of the above equation. The “Collisional” Boltzmann equation, in this special limit of binary, short range _reversable_ collisions, looks like, {Dt} =\partial_t f + f + f = d^3v_1 d\Omega} where primed (unprimed) quantities are those observed after (before) the collision. The subscripts 0 and 1 denote two different populations of particles with velocities $\bv_0$ and $\bv_1$. the nature of the collision is encoded in σ(Ω), the differential scattering cross section as a function of solid angle to which particles are scattered into. Now imagine a quantity χ which is conserved in such a binary interaction. Using the above notation, \chi_0 + \chi_1 = \chi'_0 + \chi'_1 Now multiply each side of ([EqCBEq]) by χ and integrate over the v₀. Starting with the RHS, d^3v_1 d\Omega} \ \chi_0 d^3v_0} Because the integral is over both the v₀ and the v₁ (hence they can be swapped and leave the integral the same) and because the collision is assumed reversible, the above integral can be re-written, {4} d^3v_1 d\Omega} \ d^3v_0}. but since χ is assumed to be conserved in binary interactions, the integral vanishes and we are left with f + }{m} f \right) d^3v } = 0 where I have dropped the 0 subscript and rewritten $$ via Newton’s 2nd Law. This can be re-written &\partial_t + \cdot - \chi \cdot \bv \ d^3v } \nonumber \\ & - {m} \chi \cdot \ d^3v } - {m} \cdot \ d^3v }= 0 where the term {m} \cdot \left( \chi f \right) \ d^3v } disappears via Gauss’s Law and the fact that f must vanish on a surface at infinity. Since each particle will have its own value of a quantity χ we now introduce the ensemble average \Cavg= }{} where the denominator is the number density of particles per unit volume, n. Now Eq. ([CsvMicro]) can be written in terms of average quantities: \partial_t\left(n \Cavg \right) + \cdot \left(n \left< \bv \chi \right> \right) - n \left< \bv \chi \right> - {m}\left< \cdot \chi \right> - {m}\left< \chi \cdot \right> = 0 This is the conservation equation which tells us how the volume density $n \Cavg$ (a macroscopic quantity) of any (microscopic) quantity χ, which is conserved in binary collisions, changes in time. In the macroscopic theory of hydrodynamics, quantities such as mass, momentum and energy are conserved in binary collisions and will give us our three usual conservation equations from ([Csv])! The Conservation Equations Mass Conservation To derive the continuity equation (mass conservation equation) choose our conserved quantity to be χ = m, assume all the particles have the same mass so that $\left<m\right> = m$, define the mass density by ρ = nm, and define an average flow velocity $\bu = \left<\bv \right>$. Then from ([Csv]) we get \partial_t \rho + \nabla \cdot \left( \rho \bu \right) = 0 where the last term of ([Csv]) is zero because we assume the force is independent of velocity. Momentum Conservation Choose $\chi = m \bv$ and substitute into ([Csv]) to get \partial_t \left( \rho \bu \right) + \nabla \cdot \left( \rho \left< \bv \bv \right> \right) - {m} \left(\cdot \right) = 0 where F ⋅ I = Fiδij = Fj and we define the second order tensor _{ij} &= \rho \left< \left( v_i - u_i\right) \left(v_j -u_j \right)\right> \nonumber \\ &= \rho \left< v_i v_j\right> -\rho u_i u_j Then rearranging we find, \partial_t \left( \rho \bu \right) + \nabla \cdot \left( \rho \bu \bu \right) + \nabla \cdot - {m} = 0 and we can derive later that = P - is the usual stress tensor with pressure P and viscous stress tensor σ. Internal Energy Conservation To derive an equation expressing the conservation of internal energy of the gas, we choose χ in equation ([Csv]) to be the translational kinetic energy of a particle which is conserved in binary collisions NOTE: THIS IS ONLY FOR A MONATOMIC IDEAL GAS WHERE ALL PARTICLES HAVE THE SAME MASS AND NO ROTATIONAL OR VIBRATORY DEGREES OF FREEDOM CAN SUCK UP TRANSLATIONAL ENERGY. So insert $\chi = {2}m| \bv-\bu |^2$ into ([Csv]). Starting with the first term {\partial t}\left(n \left<{2} m |\bv -\bu|^2 \right>\right) = \partial_t(\rho \epsilon) where we have defined the average internal energy per unit mass of the gas to be $\epsilon = {2} \left< |\bv - \bu|^2\right>$. The second term is a little trickier. We want a term that looks like the divergence of energy flux so write the second term as \nabla \cdot \left(n \left< \bv {2} m |\bv -\bu|^2 \right>\right) = \nabla \cdot \left(n \left< (\bv- \bu) {2} m |\bv -\bu|^2 \right>\right) + \nabla \cdot \left(n \bu \left< {2} m |\bv -\bu|^2 \right>\right) where we can pull the $\bu$ out of the average in the last term because $\bu$ is the average flow velocity. The first term looks like the average of the internal energy of a particle relative to the average flow times the velocity relative to the average flow. So the first term is an average energy flux which we denote as = {2} \rho \left< (\bv - \bu) |\bv -\bu|^2 \right>. Then our second term in ([Csv]) becomes \nabla \cdot \left(n \left< \bv {2} m |\bv -\bu|^2 \right>\right) = \nabla \cdot Q + \nabla \cdot \left( \rho \epsilon \bu\right) Let’s write the third term out in component form -n \left(\left< v^i {\partial x^i} {2} m (v_j -u_j)(v^j - u^j) \right>\right) = -{2} \rho \left( \left< v^i {\partial x^i} (v_j v^j) + v^i {\partial x^i} (u_j u^j) - 2 v^i {\partial x^i} (u^j v_j)\right>\right) The first divergence vanishes because the velocities are not coordinate dependent (the average velocities are however). The fact also allows us to write the last term as -2 v^i {\partial x^i} (u_j v^j) = -2 v^i v_j {\partial x^i} u^j Applying the divergence product rule to the second term, v^i {\partial x^i} (u_j u^j) = 2 u_j\left( u^i {\partial x^i} u^j + \epsilon^{jkl}u_k ^{ \ mn} {\partial x^m} u_n \right) = 2 u_j u^i {\partial x^i} u^j And we have -n \left(\left< v^i {\partial x^i} {2} m (v_j -u_j)(v^j - u^j) \right>\right) &= -\rho \left< u_j u^i {\partial x^i} u^j - v^i v_j {\partial x^i} u^j \right> \nonumber \\ % & = -\rho \left(u_j u^i {\partial x^i} u^j - \left< v^i v_j \right> {\partial x^i} u^j \right) Recalling the definition of 𝒯 and defining the symmetric tensor = {2} \left(\partial_i u_j + \partial_j u_i \right) we finally arrive at a full expression for the third term -n \left(\left< v^i {\partial x^i} {2} m (v_j -u_j)(v^j - u^j) \right>\right) = -\rho \left(u_j u^i {\partial x^i} u^j - \left< v^i v_j \right> {\partial x^i} u^j \right)=_{ij} \Lambda^{ij} The fourth term in ([Csv]) vanishes because the particle velocities are not coordinate dependent and the fifth term vanishes because we again assume any external force to be velocity independent. Finally our equation for conservation of internal energy is \partial_t\left(\rho \epsilon \right) + \nabla \cdot \left( \rho \epsilon \bu \right) + \nabla \cdot + \Lambda = 0 The Total Energy Equation To find the equation for conservation of total energy $E = {2} \rho u^2 + \rho \epsilon$, we need to use the above hydrodynamic equations to write out the time derivative of E. First start with \partial_t\left( {2} \rho u^2\right) &= {2} u^2 \partial_t \rho + \rho \bu \cdot \partial_t \bu \nonumber \\ &= {2} u^2 \nabla \cdot \left( \rho \bu \right) - \nabla \cdot \left( \rho \bu \bu + \right)\cdot \bu + {m} \cdot \bu \nonumber \\ % &= {2} u^2 \nabla \cdot \left( \rho \bu \right) - \bu\cdot \bu \nabla \cdot \left( \rho \bu \right) - \rho \bu \cdot (\nabla\cdot \bu) \bu - \left(\nabla \cdot \right)\cdot \bu + {m} \cdot \bu \nonumber \\ % &= -{2} u^2 \nabla \cdot \left( \rho \bu \right) - \rho \bu \cdot (\nabla\cdot {2} u^2) - \left(\nabla \cdot \right)\cdot \bu + {m} \cdot \bu \nonumber \\ % &= - \nabla \cdot \left({2}\rho u^2 \bu \right) - \left(\nabla \cdot \right)\cdot \bu + {m} \cdot \bu \nonumber \\ % &= - \nabla \cdot \left[\left( {2}\rho u^2 \right) \bu \right] - \nabla P \cdot \bu + (\nabla \cdot \sigma) \cdot \bu + {m} \cdot \bu where we used the continuity equation to replace the first term and the momentum equation to replace the second. The above are the kinetic energy terms. We have also expanded 𝒯ij = Pδij − σijCombing this with the expression for ∂t(ρϵ) in ([IEeq])we find, \partial_t\left( {2} \rho u^2 + \rho \epsilon \right) &= - \nabla \cdot \left[ \left( {2} \rho u^2 + \rho \epsilon \right) \bu \right] - \nabla P \cdot \bu + (\nabla \cdot \sigma) \cdot \bu + {m} \cdot \bu - \nabla \cdot - P \nabla \cdot \bu + \sigma \cdot \nabla u \nonumber \\ % &= - \nabla \cdot \left[ \left( {2} \rho u^2 + \rho \epsilon +P \right) \bu \right] + {m} \cdot \bu - \nabla \cdot + (\nabla \cdot \sigma) \cdot \bu + \sigma \cdot \nabla \bu Where we have used that $\sigma^{ij} = {2}\sigma^{ij}\left(\nabla_i u_j +\nabla_j u_i \right) = \sigma \cdot \nabla \bu$. Now we can write $\bu \cdot (\nabla \cdot \sigma) + \sigma \cdot \nabla \bu = \nabla \cdot (\sigma \cdot \bu)$ and the full energy equation becomes \partial_t\left( {2} \rho u^2 + \rho \epsilon \right) + \nabla \cdot \left[ \left( {2} \rho u^2 + \rho \epsilon + P \right) \bu \right] = \rho\cdot \bu - \nabla \cdot + \nabla \cdot (\sigma \cdot \bu). Where I have replaced the force F by the force per unit particle mass $\bf$. This equation says that the change in total energy per unit volume in time plus the divergence of energy through that volume is equal to the source of kinetic energy due to external forces minus the heat energy leaving or enetring the volume via sources of energy flux, plus the kinetic and internal energy sources due to viscosity. Macroscopic Approach to Internal Energy Equation Instead of deriving the internal energy equation from the microscopic treatment above. We can start with the already macroscopic first law of thermodynamics, dQ = dU + P dV where LTE is assumed and dQ is the heat added to the gas and U is the internal energy of the gas. We focus on a small element of gas with mass δm. Dividing by δm and differentiating w/r/t time, we have {dt} &= {dt} + P {dt} \left({\rho}\right) \nonumber \\ &= {dt} - {\rho^2} {dt} where we are moving along with a fluid element so d/dt denotes a Lagrangian derivative {dt} = {\partial t} + \bv \cdot \nabla X. Then in terms of Eulerian derivatives and using the Lagrangian continuity equation we multiply by ρ to find \rho {dt} &= \rho { \partial t} + \rho \bv \cdot \nabla \epsilon + P \nabla \cdot \bv \nonumber \\ &= { \partial t} - \epsilon {\partial t} + \rho \bv \cdot \nabla \epsilon+ P \nabla \cdot \bv \nonumber \\ & = { \partial t} + \epsilon \nabla \cdot {\rho \bv}+ \rho \bv \cdot \nabla \epsilon+ P \nabla \cdot \bv \nonumber \\ & = { \partial t} + \nabla \cdot(\rho \epsilon \bv ) + P \nabla \cdot \bv The LHS of the above is the heating rate per unit volume. There are two types of heating we can consider, viscous and a general energy flux, we can write these as \rho {dt} = -\nabla \cdot + \sigma \cdot \nabla \bv where for a general flux of energy, the heating rate per unit volume must be ∇ ⋅ Q by Gauss’s Law. The viscous term is more subtle (See http://rheology.tripod.com/z07.13.pdf). Now the internal energy equation becomes & { \partial t} + \nabla \cdot(\rho \epsilon \bv ) + \nabla \cdot + P \nabla \cdot \bv - \sigma \cdot \nabla \bv = 0 \nonumber \\ & { \partial t} + \nabla \cdot(\rho \epsilon \bv ) + \nabla \cdot + \Lambda = 0 which is exactly ([IEeq]) and the derivation of ([EtotEqn]) now follows identically as before.