Try this:

Make a cup of tea, add sugar, and stir it.

Now add some milk, while it's still spinning.

What I noticed is that the milk appears to move around a lot quicker in the middle of the cup of tea, than the outside.

This seems to be the opposite of what happens on a vinyl, though it's the same as what happens with planets orbiting the sun.

What's the explanation here?

  • $\begingroup$ I think the comparison with planets is not the best apology. I think you could better compare it with a hurricane or tornado. $\endgroup$ – fibonatic May 10 '14 at 13:05
  • $\begingroup$ Related: physics.stackexchange.com/q/3244/2451 and links therein. $\endgroup$ – Qmechanic Aug 11 '14 at 7:58

Ultimately, the Navier-Stokes equations explain this :)

OK, that's not a useful answer: here's how they explain the phenomenon in some cases. Under steady state conditions for a fluid (inviscid, incompressible) that doesn't differ too much from a cup of tea, the Vorticity Transport Equation shows that the vorticity $\omega = \nabla \times \vec{v}$ (the curl of the velocity $\vec{v}$) tends to nought in the steady state, as discussed in more detail in this answer here.

So now we have a potential flow; since $\nabla\times\vec{v}\approx \vec{0}$ we can set $v = -\nabla \phi$ where $\phi$ is a potential function and, since the continuity equation for an incompressible, steady state flow behests that $\nabla\cdot\vec{v}=0$, we must have $\nabla^2\phi=0$. So we now search for an axisymmetric solution to Laplace's equation with concentric circles for flow lines. The complex potential for a 2D vortex is (See the Wikipedia page for "Potential Flow" ):

$$\Omega(z) = \frac{\Gamma}{2\,\pi\,i}\,\log z$$

where $z = x+i\,y$ is the 2D position in the flow and $\Gamma\in\mathbb{R}$ the circulation. The implied velocity field (as a complex number field) is:

$$V(z) = (\mathrm{d}_z \Omega)* = -\frac{\Gamma}{2\,\pi\,i\,z^*} = \frac{\Gamma}{2\,\pi\,r} i\,e^{i\,\theta}$$

where we now write the position vector in polar co-ordinates $(r,\theta)$ i.e. the velocity (in the direction of $i\,e^{i\,\theta}$ is at right angles to the position vector $r\,e^{i\,\theta}$, and the streamlines are concentric circles centred at the origin.

This of course goes a bit haywire at $r=0$ and of course the description breaks down for small values of $r$, but you can see it describes pretty well the phenomenon you see in your cup of tea.

So let's summarise the physics that is expressed in the above:

  1. The fluid evolves into a state where fluid particles have zero spin angular momentum and $\nabla\times\vec{v}=0$; they can of course have orbital angular momentum, as in a whirlpool;

  2. Given incompressibility, conservation of mass is expressed at steady state by $\nabla.\vec{v}=0$;

  3. These two conditions force a a potential flow: $\vec{v}=-\nabla\phi$ where $\phi$ is harmonic;

  4. Lastly, the problem's circular symmetry defines a unique harmonic function.


If I understand correctly, the cup itself doesn’t rotate. In normal conditions it implies that linear velocity of the liquid at the boundary (the cup’s wall) equals to 0 (search for “no-slip condition” in Web for explanations). This is the main reason why the liquid moves “quicker in the middle”.

Also, if you stir your tea especially vigorously, you can form a vortex where the liquid near its center will move really a lot quicker. Such vortex has not necessarily be located at the center of the cup.


Indeed, "This of course goes a bit haywire at r=0". In fact the whole argument is just a trick to hide the fact that infinite vorticity has been inserted at the axis - see vortex. By stirring the tea you have introduced vorticity and hiding it in a singularity doesn't help. Nevertheless, "In the absence of external forces, a vortex usually evolves fairly quickly toward the irrotational flow pattern, where the flow velocity v is inversely proportional to the distance r. For that reason, irrotational vortices are also called free vortices."

The next question is - why?


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