# What is the physics behind a soap bubble?

A soap bubble is an extremely thin film of soapy water enclosing air that forms a hollow sphere with an iridescent surface.

What fluid dynamical process occurs during the popping of a soap bubble?

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Very slow-motion videos of (pierced) bubbles: one and two. – Keep these mind Jun 25 '13 at 13:44

A bubble, while it still exists, is balanced by three factors:
1) Surface Tension of the soapy water.
2) Internal Pressure applied by the air inside the bubble on the surface.
3) Atmospheric Pressure.

When any of these are imbalanced, one force is greater than the others and this causes the bubble to pop.

If you're talking about why do they burst in practice? Well the bubble's surface is made out of soap water. Soap is lighter than water so when you create a bubble, the water is pulled towards the bottom of the bubble - pushing the soap upwards - due to gravity. Surface tension of water is higher than that of soap, so the upper part of the bubble is getting weaker as more soap is concentrated at the top. Soon, the internal pressure will be enough to break the bubble, causing it to burst.

Another reason is that the water gets evaporated, leaving it with a more concentrated solution of soap which - as mentioned above - has a lower surface tension.

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A bubble always behaves on the principles of "Bubble dynamics". These are governed by the Rayleigh-Plesset equation. As outlined by @mikhailcazi;

A bubble, while it still exists, is balanced by three factors:

1) Surface Tension of the soapy water.

2) Internal Pressure applied by the air inside the bubble on the surface.

3) Atmospheric Pressure.

the 3 pressures are integral in the behaviour of the bubble. Yet, the bubble doesn't pop immediately when internal pressures increase (Brennen explains this excellently). There are phenomena such as bubble growth and collapse as well. You might look into that if you are studying the soap-bubble in detail, and wish to use the Lagrangian approach for the same. Some assumptions for the Rayleigh-Plesset equation which you need to remember:

1. Bubble is assumed to be spherical throughout its life
2. Mass transfer through the bubble wall is assumed to be negligible there are many others about heat transfer, but these will suffice for the simple soap bubble case you need.

What fluid dynamical process occurs during the popping of a soap bubble?

It is basically the pressure variations that occur inside the bubble due to the external (ambient) pressure variations. The surface tension force adjusts by the change in radius that occurs due to expansion/contraction of the bubble; and balances the pressures till the soap film becomes too thin.

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Good answer. Just a small comment: a bubble doesn't always behave according to the Rayleigh-Plesset equation. It will only do that if the main assumption of this equation holds, which is in the case that the bubble is spherical. For small, unconfined bubbles this is a fair assumption, but for larger bubbles or in case of nearby walls it will break down. – Michiel Jun 26 '13 at 6:29
@Michiel; yes, I must've forgotten to add the very important assumption in the Rayleigh-Plesset eqaution. I will add these too. – aditya kp Jun 28 '13 at 5:32

If you look to Slow-Motion you will find the popping of a soap bubble is a kind of chain reaction, i would guess that bubble system may not be able to reach a lower energy state by releasing energy into the enviroment. so a reaction results in a small energy release but making way for more energy releases as a chain process. then the bubble system will typically collapse.

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Question number 1: A soap bubble is black when it bursts, why?

Answer: When a soap bubble bursts, it looks black due to destructive interference. When a bubble bursts,

1. Its thickness becomes negligible, that is, thickness ≈ 0, and
2. The condition for destructive interference is satisfied. $$2nt = m\lambda,\tag1$$ Where $n$ is the refractive index of water, $m$ is the order of a dark fringe, $\lambda$ is the wavelength of light used and $t$ is the thickness of the films. Thus when $t = 0$ and $m = 0$ we get the first order dark fringe due to destructive interference. Hence a soap bubble behaves like a thin film and when it bursts, it looks black due to destructive interference.
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This is the beginning of an answer, but is woefully incomplete. – dmckee Mar 5 '14 at 3:05

## protected by Qmechanic♦Mar 4 '14 at 19:18

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