In multivariate calculus classes you learn a theorem that says that "A vector field is the gradient of a potential function on a domain $D$ if and only if it's curl-free on $D$."

When I try to apply my intuition to this theorem, I get somewhat confused. In the context of a 2D fluid flow:

On the one hand, the velocity field of a vortex looks like it should have non-zero curl.

On the other hand, I also expect the velocity field of any fluid flow to have some kind of conservation property.

The 'curl-free vector field' theorem seems to imply that the velocity field of a general fluid flow can't be characterised by a potential function.

Does this mean flows with vortices are not energy conservative somehow?

Otherwise, there some other conserved quantity that we can use to characterise the velocity field of a fluid? For example, can we use a pointwise kinetic energy function to characterize a velocity field?

(Disclaimer: I haven't had much formal physics education so I apologize if I've abused and/or confused fundamental concepts.)


1 Answer 1


In short, there is no requirement that the velocity field of a fluid be given by the gradient of some scalar function, but there are special cases & approximations in which it is. Also note that, in general, any flow can be decomposed into a part that is the gradient of a scalar function, a part that is the gradient of another scalar field which satisfies the Laplace equation, and the curl of a vector field. This is usually called a Helmholtz decomposition.

When a flow is describable entirely through the gradient of a scalar function, we call that flow a potential flow and the scalar function a velocity potential. When a flow has a very high Reynolds number, it turns out that regions far away from boundaries behave like potential flows and can be described as such.

Vortices and boundary layers put a wrench in all this because they induce curls in the flow. They also dissipate energy, but that’s because most fluids dissipate energy when velocity gradients lateral to the velocity occur—it’s not mathematically a requirement that they burn up energy in all fluids.

To track energy in a fluid, you can take the Cauchy momentum equation for the fluid and dot product it with the velocity to get an equation for the power density of the fluid. For an incompressible Navier-Stokes fluid, this looks like:

$$\rho\left(\mathbf{u}\cdot\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u}\cdot\nabla\left[\frac{||\mathbf{u}||^2}{2}\right] + \mathbf{u}\cdot\nabla\times\mathbf{u}\times\mathbf{u}\right) = - \mathbf{u} \cdot \mu (\nabla \times \nabla \times \mathbf{u}) + \mathbf{u}\cdot\nabla p + \mathbf{u}\cdot\mathbf{f}$$

The main culprit of energy dissipation in an incompressible Navier-Stokes fluid is the fourth term, which is nonzero when the curl of the curl of a flow is nonzero, and comes directly out of the Stokes constitutive equation. Essentially every vortex and boundary layer cause this term to be nonzero, and we need not add any additional physics to capture that energy loss.

Hope this helps!

  • $\begingroup$ Thanks! When I look at graphs of computed solutions to Navier-Stokes with high Reynolds numbers, I often see 'shedded vortices' in the flows. I'm only passingly familiar with the actual equations and know that they represent conservation of mass, momentum and energy. You mentioned that vortices and boundary layers dissipate energy-- is this also 'baked into' the math of N-S? Or do we need to introduce other physics to capture the loss of energy from a vortex? $\endgroup$
    – overseas
    Mar 4, 2019 at 2:23
  • $\begingroup$ I'll expand on my post a bit to answer your question! =) $\endgroup$ Mar 4, 2019 at 2:25

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