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I was studying the process of coalescence in emulsions. We considered $N$ bubbles of liquid 1 floating in liquid 2. The result we derived, is that if there are some dissipative forces (diffusion) the most stable configuration is for the $N$ bubbles to coalesce/fusion and become 1 bubble (to reduce surface tension).

So I tried to make the experiment at home to see how long would it take. I emulsified oil in water. After one day, it has not changed much.

I tried to estimate the time it would take using Stokes-Einstein equation: \begin{equation} D=\frac{k_{\rm B}T}{6\pi \eta r} \end{equation}
where $\eta$ is the viscosity of (olive) oil (?) and $r$ the radius of the bubbles (about 1 mm).

And then using the result of Brownian motion for 2D: $$\overline{x^2}=4Dt$$ I replaced $\overline{x^2}$ by the surface of my container, about $R=$ 4 cm of radius so $\overline{x^2}=\pi R^2$. Mixing both equations, I can derive the time it takes (for room temperature). But I get an enormous duration of $10^5$ years.

Is this the true value or I am doing something wrong?

Update: Day five, the smallest bubbles ($r<1$ mm) are gone but still $N>100$.

Update2: Day six. No coalescence. I stopped the experiment.

Update3: Obvious problem of leaving the experiment for so long is that the water starts to evaporate.

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    $\begingroup$ are you sure you want to use the Stokes-Einstein equation, since it is specifically for low Reynold's number? $\endgroup$ – Daddy Kropotkin Dec 18 '18 at 13:51
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    $\begingroup$ A few points : 1. olive oil is not pure, so don't expect to see a real phase separation. 2. In your calculation, you should replace $x$ not by $R$, but by the average distance $d$ between the bubbles. 3. Gravity plays also a role since oil is less dense. Personnally, I would rather write $t=(R\rho v /gv\Delta \rho)^{1/2}$ that womes out from $ma=F$ with $v$ the volume of the small bubbles and $\rho$ their density, and $\Delta\rho$ the difference of density with water. $\endgroup$ – J.A Dec 18 '18 at 14:01

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