When applying soap liquid on the inner surface of swim goggles, the surface tension of the water decreases and small droplets of water on the surface won't form, therefore the fog won't form.

In this article describing the influence of plasma on the liquid (https://www.researchgate.net/publication/273137819_Influence_of_plasma_on_surface_tension_of_hydrocarbons), the authors showed that after applying plasma, the liquid breaks down onto smaller droplets, and claimed that the surface tension decreased.

So there is a contradiction here. Breaking the liquid onto small droplets means a decrease or an increase of the surface tension?


2 Answers 2


the physics in these two cases is completely different.

To begin with, soap renders glass wettable, so that any water that condenses on it forms a smooth, continuous sheet that you can easily see through, instead of tiny hemispheres of water stuck to the glass that diffract light.

This has nothing to do with the effects reported in the researchgate article.


Surface tension decreases as droplet size decreases.1

You can follow the derivation in the source, but the gist of it is:

From Gibbs' equation2, we have that in a two-phase system, the surface tension $\sigma$ depends on the potential $\mu$ as:

$$ {\rm d}\sigma = -\Gamma {\rm d} \mu $$

where $\Gamma$ is the superficial density of matter at the boundary of the phases computed with respect to the surface tension3.

Following the derivation in 1, we get $$ \frac{1}{\sigma}\frac{{\rm d}\sigma}{{\rm d}r} = \frac{(2\delta/r^2)\left[1 + (\delta/r) + \frac{1}{3}(\delta/r)^2\right]}{1 + (2 \delta/r)\left[ 1 + (\delta/r) + \frac{1}{3}(\delta/r)^2 \right]} $$

Which can be integrated and simplified (ignoring higher order terms) to $$ \sigma \sim \frac{1}{1 + 2 \delta /r} $$

$\delta$ is constant over a wide range of droplet size, but varies appreciably when droplets become small enough ($r \sim \delta \sim 10^{-10} {\rm m}$)

As $r$ decreases, $\delta/r$ increases, and so $\sigma$ decreases.


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