Bubbles in Contact

Two bubbles and a tap

Blow a small soap bubble A on one end of a tube and a large bubble B on the other. A closed tap separates them. The last lesson gives their pressures. With \(T = 0.03\,\text{N/m}\), bubble A of radius \(1\,\text{cm}\) has an excess pressure of \(4T/r = 12\,\text{Pa}\). Bubble B, of radius \(2\,\text{cm}\), has only \(6\,\text{Pa}\). Now the tap is opened.

Air flows from the higher pressure to the lower, from A into B. A shrinks and B grows. As A shrinks, its radius falls, so its pressure rises further: the flow speeds up instead of stopping. The small bubble runs away into the large one. Its pressure is largest, \(30\,\text{Pa}\) in the figure above, when it is a hemisphere over the mouth of the tube, of radius \(0.4\,\text{cm}\).

After that, what is left of A is a cap flatter than a hemisphere. Its radius of curvature grows again, and its pressure falls. The flow stops when the cap is curved exactly as the large bubble is, so that both pressures are equal. In the figure above this happens at a radius of about \(2.08\,\text{cm}\), with \(5.8\,\text{Pa}\) in both. The cap is then almost flat across the mouth of the tube.

The wall between two bubbles

Now let two soap bubbles of radii \(r_1\) and \(r_2\), with \(r_1 < r_2\), touch and stick together, as in Fig 4.1. The film between them is shared. The air in the small bubble A is at \(P_0 + 4T/r_1\), and the air in the large bubble B at \(P_0 + 4T/r_2\). A has the higher pressure, so the common wall is pushed into B. It bulges into the larger bubble.

The common wall is itself a soap film with two surfaces, of radius \(r\). Across it the pressure steps by \(4T/r\). That step must be the difference between the pressures in the two bubbles.

For bubbles of \(2\,\text{cm}\) and \(4\,\text{cm}\), the pressures are \(6\,\text{Pa}\) and \(3\,\text{Pa}\) above the outside air. The wall must carry the difference, \(3\,\text{Pa}\), and a film with \(4T/r = 3\,\text{Pa}\) has \(r = 4\,\text{cm}\). For two equal bubbles, \(1/r = 0\): the wall is flat. The more unequal the bubbles, the more the wall bulges.

Where the three films meet, the outer surfaces of A and B and the common wall, they pull with equal tension. Three equal pulls can balance only if they are \(120^\circ\) apart. So the three films always meet at \(120^\circ\), as Fig 4.1 shows. Look at soap froth, and every edge where films meet has three of them, at equal angles.

Two bubbles that merge

Now let two bubbles merge into one bubble of radius \(R\), with no change of temperature. The amount of air is the same before and after. At a steady temperature, the pressure times the volume of a fixed amount of air stays the same. This is Boyle's law, which the chapters on gases will derive. So \(P_1V_1 + P_2V_2 = PV\), adding the air of the two bubbles.

In a vacuum there is no outside air, so the pressure inside each bubble is just \(4T/r\). Then \(PV = \dfrac{4T}{r} \cdot \dfrac43\pi r^3 = \dfrac{16}{3}\pi T r^2\). Each \(PV\) is a fixed number times \(r^2\).

Bubbles of \(3\,\text{cm}\) and \(4\,\text{cm}\) merge in a vacuum into one of \(\sqrt{9 + 16} = 5\,\text{cm}\). The result has a neat meaning. \(R^2 = r_1^2 + r_2^2\) says that \(8\pi R^2 = 8\pi r_1^2 + 8\pi r_2^2\): the surface area, and so the surface energy, is the same before and after.

The froth on a fizzy drink

Back to the glass of soda. Its froth is a crowd of bubbles of many sizes. The gas in a small bubble is at a higher pressure than the gas in a large one. So gas slowly seeps through the thin films from the small bubbles into their larger neighbours, as it flowed through the tube. The small bubbles shrink and vanish, and the large ones grow. Where two bubbles touch and their wall breaks, they join into one, with less surface than the two had. Both ways, the froth loses surface and energy, and it coarsens.