I saw the latest video from Sixty Symbols Little Swimmers. At the end of the videos he says that we do not know how to calculate the movement of the little swimmers. He says(6:14-6:40 in video) that mathematicians derive formulas for small velocities and that they do not work for this experiment. This got me thinking.

Is the speed really that high? Have a look at visualization of the flow behind the swimmer, time in the video 4:32-4:47. The flow looks calm, there is no turbulence. Therefore the argument that the velocity is too high might not be the right one.

Stokes, Oseen, ... drag: I did the calculation for Stokes and Oseen drag and yes you get that the swimmer should not move. Even more, if you assume the drag force as any function of velocity, than the calculation says the swimmer should not move. Therefore deriving more accurate drag formulas for constantly moving sphere in liquid does not help.

Acceleration is too high, not the velocity: Next step in approximation is to derive drag force on single oscillating sphere in fluid. The question is, what is the drag force on the sphere which is moving with velocity $\mathbf{v} = \mathbf{ v}_0(1 + a \sin(\omega t))$?

Will this drag force predict movement of the swimmer? If not than it can be tested in the experiment by making the spring really long and no movement should be seen.

$ \newcommand{bm}{\mathbf} $ My calculations:

Center of mass of the whole system, its velocity and total mass are $x, v,m$

Position, velocity, radium, mass of $i$-th sphere are $ x_i, v_i, r_i, m_i$.

Since $\bm x$ is center of mass than \begin{align} m_1 x_1 + m_2 x_2 &= m x \\ m_1 + m_2 &= m \end{align} Spheres are oscillating with frequency $\omega$ $$ x_1 - x_2 = L(1+\alpha \sin(\omega t)) $$ $L$ is rest length of spring, $\alpha$ is relative extension of the spring.

Thus \begin{align} x_1 &= x + \frac{m_2 L}{m}(1+\alpha \sin(\omega t)) \\ x_2 &= x -\frac{m_1 L}{m}(1+\alpha \sin(\omega t)) \\ v_1 &= v + \frac{m_2\omega L}{m}\alpha \cos(\omega t) \\ v_2 &= v -\frac{m_1\omega L}{m}\alpha \cos(\omega t) \\ \end{align}

The force on the system is sum of forces on those two spheres \begin{align} F &= F_1 + F_2 \\ F_i &= f(v_i,r_i) \end{align} Force on the single sphere is a function $f$ of its velocity and its radius.

Stokes drag: $$ f(v,r) = 6\pi \mu r v $$ Thus $$ F_i = 6\pi \mu r_i \left( v + \frac{m_2\omega L}{m}\alpha \cos(\omega t) \right) $$ Averaging over period and assuming that the CoM velocity is constant over the period. $$ \hat F_i =\frac{\omega}{2\pi} \int_t^{t+\frac{2\pi}{\omega}} F_i = \frac{\omega}{2\pi} \int_t^{t+\frac{2\pi}{\omega}} 6\pi \mu r_i \left( v + \frac{m_2\omega L}{m}\alpha \cos(\omega s) \right) ds = 6\pi \mu r_i v $$

So the averaged force over the time period is the same as if no oscillation is happening.

In general Suppose that the swimmer is not moving, $v=0$. Again assume that the velocity is constant over period. Then averaged force over period is $$ \hat F_i = \frac{\omega}{2\pi} \int_t^{t+\frac{2\pi}{\omega}} F_i = \frac{\omega}{2\pi} \int_t^{t+\frac{2\pi}{\omega}} f(v_i,r_i) = \frac{\omega}{2\pi} \int_t^{t+\frac{2\pi}{\omega}} f\left( \frac{m_2\omega L}{m}\alpha \cos(\omega t),r_i\right) $$ But drag force always acts in opposite direction to the velocity and its magnitude does not depend on the direction. We have to get $$ \frac{\omega}{2\pi} \int_t^{t+\frac{2\pi}{\omega}} f\left( \frac{m_2\omega L}{m}\alpha \cos(\omega t),r_i\right) = 0 $$

Possible flaw in the calculation I assumed that the acceleration of the swimmer is much smaller than acceleration of individual spheres. Thus I could treat $v$ as constant in the integration.

  • $\begingroup$ I am interested to see your derivation for Stokes' and Oseen drag. As i understand it the fact they don't move is due to time reversibility in laminar flows, i.e. the same movement forwards at time $t$ and an equal movement in reverse at time $t+\Delta t$ yields a net zero displacement. Once inertial effects become important, the non-linearity of the equations come into play and this leads to alot of interesting behavior (e.g. turbulence). $\endgroup$
    – nluigi
    Dec 29, 2015 at 12:37
  • $\begingroup$ You may be interested in the Golestanian swimmer model, especially if you want to check your math, where three spheres are connected to make propulsion possible in low Reynolds numbers. Most of his papers are on arXiv (he has several on the topic, the oldest one I think is arxiv.org/abs/cond-mat/0402070), as I believe is the one from Sixty Symbols (arxiv.org/abs/1501.05143). $\endgroup$
    – alarge
    Dec 29, 2015 at 14:18
  • $\begingroup$ If I remember correctly that talk about a jet of fluid which propels the swimmer, this would imply some turbulence/inertial forces on the liquid. Have you tried to approximate the Reynolds numbers for each sphere in the video? Is quite possible that the laminar forces (Stokes flow) yields a netto force of zero, however any nonlinear contribution might not cancel. So even if the viscus forces are dominant, you can't neglect the inertial forces. $\endgroup$
    – fibonatic
    Dec 29, 2015 at 15:08
  • $\begingroup$ Based on the video it does not look like that this is happening at low Reynolds number, at all. They are clearly saying that they have to increase the oscillation frequency considerably and the swimmer seems to leave a trail of microscopic turbulence, if not even cavitation. $\endgroup$
    – CuriousOne
    Dec 29, 2015 at 19:29

1 Answer 1


I am not quite sure if the following is helpful, but low-Reynolds fluid dynamicists have a difficult time predicting irreversible phenomena mathematically because the equations they generate are linear approximations to the real nonlinear equations with reversible solutions, and these solutions always fail in some sense.

To demonstrate it, take the Navier-Stokes equations for an incompressible Newtonian fluid:

$$\frac{\partial(\rho \vec{u})}{\partial t} + \nabla\cdot[\rho\vec{u}\otimes \vec{u}] = -{\nabla p} + \mu{\nabla^2}\vec{u} + \vec{f}$$

The key assumption microhydrodynamicists make is that the convective acceleration term $\nabla\cdot[\rho\vec{u}\otimes \vec{u}]$ is much smaller than the viscous drag $\mu{\nabla^2}\vec{u}$, and can therefore be removed from the equation, usually justified by dimensional scaling arguments. If you do that, you get a nice linear equation for $\vec{u}$:

$$\frac{\partial(\rho \vec{u})}{\partial t} = -{\nabla p} + \mu{\nabla^2}\vec{u} + \vec{f}$$

Even though this equation is "easily" solvable, the results you get are usually bad to varying degrees. You can test it by checking whether or not

$$\frac{\nabla\cdot[\rho\vec{u}\otimes \vec{u}]}{\mu{\nabla^2}\vec{u}} \ll 1$$

everywhere in the solution you get from the linear equation, and it turns out it's always bigger than $1$ somewhere! Verify it for the solutions you got; I'll bet it's bigger than $1$ somewhere for those too. That's why you get all those paradoxes (Whitehead paradox, Stokes paradox, etc.) people like to cite all the time. Using the Oseen formulation helps alleviate the more straightforward errors, but the (very complicated) Oseen solutions will probably also fail the criteria above too for some set-up even though they resolve the more well-known paradoxes.

In short, the -typical- equations of microhydrodynamics usually generate results consistent with experiment but also fail dramatically to capture some features (see this, for example), and are still being investigated—by people like me!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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