# What is the physical basis of the $\rho \bf{u} \times \bf{\omega}$ force on a fluid vortex line?

In fluid dynamics, the force density on a vortex line is $\bf{f} = \rho \bf{u} \times \bf{\omega}$. In Faber, Fluid Dynamics for Physicists, ch. 4, this is "derived" by analogy with magnetostatics, but the analogy does not appear to scan (top of p. 129 if you're following along at home; Faber appears to be using Bernoulli's principle even though the flow in fig. 4.4 is not irrotational across the boundary. There are some other implied steps in the analogy I don't think are well-justified either, but that's an aside.) Can you derive this force from basic fluids principles, say from Euler's equations for incompressible flow?

I've tried writing $\rho \bf{u}\cdot\nabla\bf{u} = \nabla(\frac{1}{2}\rho u^2) - \rho \bf{u}\times\bf{\omega}$, but I can't find an argument to move the $\rho\bf{u}\times\bf{\omega}$ term to the force side of the equation.

• Have you considered replacing $\omega$ with it's equivalent $\nabla\times\mathbf u$? Some vector calculus thereafter might lead somewhere useful. – Kyle Kanos Aug 20 '14 at 2:10

Bernoulli's equation does not require that the flow be irrotational, just inviscid. Let's consider a vortex filament, and denote its surface by $S$ and volume by $V$.

Using the identity you mentioned above, Euler's equation can be written as:

$$\rho\frac{\partial \boldsymbol{u}}{\partial t} + \nabla\left(\frac{1}{2}\rho\boldsymbol{u}^2\right)-\rho(\boldsymbol{u}\times\boldsymbol{\omega}) = -\nabla p + \rho\boldsymbol{g}$$

where $\boldsymbol{g}$ denotes the body force per unit mass. If we assume that

• flow is steady, $\frac{\partial \boldsymbol{u}}{\partial t} = 0$
• no external forces act outside the filament

then we get:

$$\nabla\left[\frac{1}{2}\boldsymbol{u}^2 + p \right] = \boldsymbol{u} \times \boldsymbol{\omega} + \boldsymbol{g}$$

Note that this is differs from Bernoulli's equation in that we do not assume that the body force is conservative. Since $\boldsymbol{f} = 0$ outside the filament, we have that $\frac{1}{2}\boldsymbol{u}^2 + p$ is uniform on $S$ (though it may be different on different vortex lines).

Integrating both sides and applying the divergence theorem, we have:

$$\int_S \left(\frac{1}{2}\boldsymbol{u}^2 + p\right)\boldsymbol{\hat{n}}dS = \int_V \left(\boldsymbol{u} \times \boldsymbol{\omega} + \boldsymbol{g}\right)dV$$

The vortex lines that bounding this filament either form closed loops or extend to a free surface, so the integral of a uniform quantity (along the vortex lines) over this surface is zero. Using $\boldsymbol{f} = \rho\boldsymbol{g}$ as the body force density, we have

$$\int_V \boldsymbol{f} + \rho\left(\boldsymbol{u} \times \boldsymbol{\omega} \right) dV = 0$$

In the limit as the vortex filament becomes a vortex line

$$\boldsymbol{f} + \rho\boldsymbol{u} \times \boldsymbol{\omega} = 0$$

For a vortex line going through rectilinear motion, the sign between the external force and the vortex force can be negated by moving the frame of reference with the vortex.