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Since I am not going ot to  upgrade yet, I will make these notes public.\\ \section{The non-viscous Bernoulli equation}  The steady-state Euler equation reads  \begin{equation}  \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  \end{equation}  where it is understoond understood  that the gravitational potential $\Phi_G$ and the coefficient of kinematic viscosity $\nu$ are time independent. Now use the idenity 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 \begin{equation}  \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.  \end{equation} 
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\begin{equation}  \int{\frac{dP}{\rho}} = \frac{\mathcal{M}^2}{v^2_K} \ln{\frac{\rho}{\rho_0}} \quad \rm{Isothermal} \ \rm{Flow}  \end{equation}