# Necessary and sufficient condition for Bernoulli's theorem

For an ideal fluid, if the vorticity is $\vec{\omega}=\nabla \times \vec{v}$, then Euler's equations can be rewritten as: $$\rho \dot{v}_i = \rho \epsilon_{ijk} v_j \omega_k - \frac{1}{2} \rho \partial_i v^2 - \partial_i p$$ Any textbook will then tell you that if you have a steady flow with zero vorticity: $$\frac{1}{2} \rho \partial_i v^2 + \partial_i p = 0$$ which is a differential form of Bernoulli's theorem. However as it is obvious from the previous equation the necessary and sufficient condition for this equation to hold is not a steady flow with $\vec{\omega}=0$ but a steady flow with $\vec{v} \times \vec{\omega}=0$, which is a more general condition. I am wondering if one can give a nice geometric interpretation of this condition. In other worlds, what is the geometric interpretation of a vector field having $\vec{v} \times (\nabla \times \vec{v}) = 0$?

The quantity $\vec{v}\times(\vec{\nabla}\times\vec{v})$ you are interested in here is known as the Lamb vector in theoretical fluid dynamics. It's gone a bit out of fashion these days, but there's some useful material in the JFM paper by C. W. Hamman, J. C. Klewicki and R. M. Kirby, "On the Lamb vector divergence in Navier–Stokes flows". Perhaps more pertinently, your question is a near-duplicate of a question asked here in February with an answer that has some more detail also. A Google search on "Lamb vector" may turn up more material you find interesting.
The actual derivation of the Bernoulli equation comes from the vorticity form of the incompressible Navier-Stokes equation. In terms of vorticity, the Navier-Stokes equation take the form, $$\frac{\partial \vec{V}}{\partial t} + \vec{\omega} \times \vec{V} = -\nabla\left(\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k\right) + \nu \cdot \left(\nabla \times \vec{\omega}\right)$$ Now if we have steady flow, $\frac{\partial \vec{V}}{\partial t} = 0$, and if we further assume the flow is inviscid, then the equation reduces to, $$\vec{\omega} \times \vec{V} = -\nabla\left(\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k\right)$$ Obviously, if the flow is irrotational, namely $\vec{\omega} = \nabla \times \vec{V} = 0$, then we are left with, $$\nabla\left(\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k\right) = 0$$ or equivalently, $$\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k = \textrm{constant}$$ This is the most famous form of the Bernoulli equation, which requires steady, incompressible, inviscid, and irrotational flow. Also, an important note on this relation, because the flow is irrotational, the Bernoulli equation can be applied across streamlines. Now for the case you specified, for instance, what if the flow is rotational? Well, you have to consider the direction of the vector quantity $\vec{\omega} \times \vec{V}$. The resulting vector of $\vec{\omega} \times \vec{V}$ is orthogonal to the velocity and vorticity vector. Therefore, along a streamline the quantity $\vec{\omega} \times \vec{V} = 0$. Hence, the resulting equation becomes, $$\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k = \mathrm{constant\big|_{streamline}}$$ Therefore, the conclusion is that the Bernoulli equation can be applied across streamlines if we have steady, incompressible, inviscid, and irrotational flow ($\vec{\omega} = \nabla \times \vec{V} = 0$). However, if the flow is rotational ($\vec{\omega} = \nabla \times \vec{V} \neq 0$), we can only apply the Bernoulli equation along a streamline.
• What you say is mostly true, but is not what I am interested in at the moment. I am stating that even if we want to apply the Bernoulli theorem across streamlines, across the entrie fluid, one can do that with a condition that is weaker than zero vorticity. However, I have no geometric picture of what a flow with non-zero vorticity, but zero $\vec{v} \times \vec{\omega}$ is. – evilcman Nov 21 '16 at 9:23
• Also, generally vorticity is not orthogonal to velocity. The scalar product of the velocity field with the vorticity actually has a name: [en.wikipedia.org/wiki/Hydrodynamical_helicity However, this does not change your argument as you never made use of your statement "we must recognize that the vorticity vector is always orthogonal to the the velocity vector". You only need that $\vec{\omega} \times \vec{v}$ is orthogonal to $\vec{v}$ for your result to hold. – evilcman Nov 21 '16 at 9:29