Yesterday I saw vortex shed from a kayak paddle around 40 yards behind the person paddling so it had been rotating for over a minute. Why isn’t the energy in this vortex dissipated quickly, e.g. why does it remain so long?

  • $\begingroup$ As an interesting aside, when doing modeling of fluid flows, some methods leverage the fact that vortexes are so long lasting and predictable. They detect them, "pull them out" of the underlying grid, and treat them as an entity, with a fixed amount of "vortex energy" which is dissipated over time due to viscosity, or through interaction with another vortex. $\endgroup$ – Cort Ammon Sep 7 '19 at 17:45
  • $\begingroup$ @CortAmmon I'm interested in reading more about those methods -- do you recall the name so I can pull up some papers? $\endgroup$ – tpg2114 Sep 7 '19 at 17:57
  • $\begingroup$ Seems to me I left a comment, but it's not here now. $\endgroup$ – Mike Dunlavey Sep 7 '19 at 21:35

The smaller the fluid's viscosity, the longer the vortex will last.

Viscosity is a measure of friction effects within a fluid. As with other forms of friction, viscosity transfers energy from macroscopic motion into complicated microscopic molecular motion (thermal energy). Intuitively, this gradual transfer of energy from visible macroscopic motion into invisible thermal energy should eventually cause a vortex to dissipate.

(By the way, an ideal superfluid has zero viscosity. In an ideal superfluid, a vortex can last forever.)

The rest of this answer is an attempt to deduce a quantitative estimate of how long a vortex can last, using a simple model that treats water as an incompressible fluid. Disclaimer: I'm not an expert in fluid mechanics! To write this answer, I had to review some of the basics. For the general theory, I used these sources:

For numerical info about the viscosity of water, I used these sources:

The key equation

Let $\mathbf{u}$ denote the velocity of the fluid. This varies with the location $\mathbf{x}$ in the fluid, and it may change with time $t$. Let $C$ be any closed loop in the fluid, and consider the integral of $\mathbf{u}$ around the loop: $$ \Gamma = \int_C d\mathbf{x}\cdot \mathbf{u}. \tag{1} $$ The quantity $\Gamma$ is called the circulation around $C$. It is non-zero only if there is a net circulation (in the colloquial sense) of fluid around the loop. In particular, it is non-zero if we take $C$ to be a small circle centered on a vortex. The goal is to understand why $\Gamma$ might remain essentially constant for a long time, and what eventually causes it to dwindle to zero.

To answer this, we need an equation that describes the time-dependence of the velocity field $\mathbf{u}$. Assuming that the fluid is incompressible (which is a good enough approximation for this question), we can use the Navier–Stokes equation $$ \frac{d \mathbf{u}}{d t} = \nu\nabla^2 \mathbf{u} -\nabla h \tag{2} $$ where $$ \frac{d \mathbf{u}}{d t} := \frac{\partial \mathbf{u}}{\partial t} +(\mathbf{u}\cdot\nabla)\mathbf{u} \tag{3} $$ is the convective derivative of $\mathbf{u}$, which says how quickly the velocity $\mathbf{u}$ changes in time at a point that moves with the fluid. The coefficient $\nu$ is the kinematic viscosity of the fluid, and the $\nabla h$ term accounts for gravity. To relate (2) to (1), take the integral of (2) around the contour $C$. The integral of the $\nabla h$ term around a closed contour is zero because the integrand is a gradient, so we are left with $$ \frac{d }{d t}\Gamma = \nu\int_C d\mathbf{x}\cdot (\nabla^2 \mathbf{u}). \tag{4} $$ This is a special case of Kelvin's theorem. Equation (4) is what we want: it tells us how the circulation $\Gamma$ changes with time, with the understanding that the contour $C$ moves with the fluid. This last qualification is important, because it says that if we take $C$ to be a loop around the center of the vortex, then $C$ remains centered on the vortex as the vortex is carried along by whatever overall current might be present.

Viscosity and stability

The intuition mentioned at the top of this answer is consistent with equation (4), which says that if the viscosity $\nu$ were zero, then $\Gamma$ would remain constant in time (insofar as the other approximations implicit in equation (2) are valid). In other words, if the viscosity were zero, a vortex could last forever.

Determining the relative sign

The viscosity term in (4) should cause the strength of the vortex to decrease with time, so the sign of the viscosity term should be the opposite of the sign of $\Gamma$. Let's check this using a simple model of the velocity field around a vortex. Outside the core of the vortex, suppose that the velocity field has the form $$ u_j = \frac{\sum_k\Omega_{jk} x_k}{\mathbf{x}^2} \tag{5} $$ for some antisymmetric matrix $\Omega_{jk}$, where $u_j$ is the $j$th component of $\mathbf{u}$. The model (5) ensures that the velocity is tangent to the radius from the center (as appropriate for a vortex), and it also ensures that the velocity increases toward the center. Equation (5) implies $$ \nabla^2 u_j = -2D\frac{u_j}{\mathbf{x}^2} \tag{6} $$ where $D=3$ is the number of dimensions of space. (Usually $D=3$, but sometimes pretending $D=2$ is useful for building intuition.) The minus sign in equation (6) confirms that the relative signs in equation (4) are correct: the strength of the vortex will decrease with time, as expected.

Rough numerical estimate

Now that we have the key equation (4), let's plug in some numbers. Under ordinary conditions, the kinematic viscosity of water is roughly $$ \nu \approx 10^{-6} \ \frac{\text{m}^2}{\text{s}}. \tag{7} $$ Getting numbers for the other quantities in equation (4) will involve some guesswork, because we don't have any measurements of the vortex that was observed. Take the loop $C$ to be a circle around the vortex with a radius of $10$ centimeters, and suppose that at that radius, the water is circulating at a rate of $5$ cycles per second. (I'm just guessing.) Since we're only estimating, we can say it like this: $\Delta r\sim 10$ cm is the typical spatial scale of the vortex, and $\Delta t\sim 1/5$ second is the typical time scale.

The quantity $\Gamma$ is the circumference $|C|$ of the circle times the velocity of the water at that radius (I mean the component of the velocity tangent to the circle), so we can think of $d\Gamma/dt$ as $|C|$ times the rate at which the water's velocity is changing. Since we're just estimating, we can use the model (5) to get $$ \int_C d\mathbf{x}\cdot (\nabla^2 \mathbf{u}) \sim -\frac{12\pi}{\Delta t}. \tag{8} $$ Using these numbers in equation (4) gives the estimate $$ |C|\frac{d }{d t}|\mathbf{u}| \sim -60\pi \times 10^{-6} \ \frac{\text{m}^2}{\text{s}^2}. \tag{9} $$ Divide both sides by $|C|\sim 2\pi \Delta r\sim 0.5$ meter to get $$ \frac{d }{d t}|\mathbf{u}| \sim -120\pi\times 10^{-6} \ \frac{\text{m}}{\text{s}^2} \sim -2\ \frac{\text{cm/s}}{\text{minute}}. \tag{10} $$ This estimate says that the circulation decreases slowly, loosing only a couple of centimeters-per-second of speed every minute.

(We could improve this estimate by going back to the original equations to see that the curve $C$, which is carried along with the fluid, can grow with time. In other words, the vortex can expand, which we perceive visually as the vortex weakening and disappearing into the noise of the ambient waves.)

This is only a rough estimate, and I'm not an expert in fluid mechanics, but it seems to confirm that a well-formed vortex can last a surprisingly long time in open water.

  • $\begingroup$ Great stuff...thx for your response $\endgroup$ – dbrub2 Sep 7 '19 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.