I was wondering how the radius of a soap bubble depends on the way it is formed and tried to come up with a simple argument using dimensional analysis. I am not sure if that argument if correct, or how I could verify if it is correct. Hopefully somebody can point out if I missed something and teach me a thing or two about both dimensional analysis and soap bubbles.

Soap Bubble Setup

Let's say we are blowing air with density $\rho$ and velocity $v$ through a loop of radius $a$. Spanning the loop is a film of soapy water, with surface tension $\gamma$. From those four parameters, we can form the dimensionless quantity $$ \theta = \frac{\rho v^2 a}{\gamma} \, , $$ which I guess could be interpreted as a ratio of energy densities$^*$. The bubble radius $r$ should then be of the form $r \propto a f(\theta)$, with an unknown function $f$ we have to guess at. Considering that it should be impossible to form bubbles the limit $\gamma \to \infty$ makes is plausible to guess $f(\theta) = \theta$. This leads to $$ r \propto \frac{\rho v^2 a^2}{\gamma} \, , $$ with an unexpected dependence on the square of the loop size. As a corollary, we also get the pressure difference $$ \Delta p = \frac{\gamma}{r} \propto \frac{\gamma^2}{\rho v^2 a^2} \, , $$ which depends on the square of the surface tension.

How can I verify if I (1) identified the 'correct' parameters to describe the system and (2) made a reasonable guess at the function $f$?

$^*$Side remark: The ratio $\rho v^2 a / \gamma$ reminds me of a Reynolds number, only with a 'viscosity' of $\gamma / v$. Just coincidence?


1 Answer 1


The volume of air that enters per second is $v\pi a^2$ and the mass of air in the bubble is $v\pi a^2 \rho t$ after a time $t$.

From density is mass/volume

$$\rho = \frac{3v\pi a^2\rho t}{4 \pi r^3}$$

$$r^3 = \frac{3va^2 t}{4}\tag1$$

assuming the bubble is spherical.

If you want to know how big the bubble can grow before popping, you'd need to know the thickness of the soap film before the bubble was formed $t_1$, and the thickness when it pops, $t_2$.

Dimensional analysis is useful for checking these formulae

If the volumes are equal when the film is in the circle and in a sphere, just before popping

$$4 \pi r^2 t_2 = \pi a^2t_1$$

$$r=\frac{a}{2} \sqrt{\frac{t_1}{t_2}}\tag2$$

  • $\begingroup$ Thanks! I was more interested in how the bubble radius depends on the loop radius and air velocity, rather than how it depends on time. The question came up because I noticed that a loop of fixed $a$ can produce bubbles of varying sizes. In your answer, you assume that the density of air inside the bubble is the same as outside. I'm not convinced that this is the case, since the pressure is different. Also, to calculate the radius when the bubble pops, it should be possible to express $t_2$ as a function of $\gamma$ and $t_1$, right? If rewritten as such, maybe me relation for r is recovered? $\endgroup$
    – leapsheep
    Jan 6, 2022 at 17:22

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