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\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 that the gravitational potential $\Phi_G$ and the coefficient 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 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}