Is drag force on an oscillating sphere an effective model for a swimmer? 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.
 A: 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!
