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\end{equation}  where it is understoond that gravitational potential $\Phi_G$ and the coeficcient of kinematic viscosity $\nu$ are time independent. Now use the idenity $\left(\mathbf{v} \cdot \nabla \right)\mathbf{v} = \frac{1}{2} \nabla \left( \mathbf{v} \cdot \mathbf{v}\right) - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right)$ neglect viscosity for now, and integrate the momentum equation along a streamline from a refernce point to the point of evaluation  \begin{equation}  \int{\mathbf{ds}\cdot \nabla \left[\nabla  \left( v^2\right) \frac{1}{2}v^2\right)  - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right) + \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}  you get   \begin{equation}