# How does extra pressure in an air bubble balance the surface tension force of a liquid?

For an air bubble in waer, because of surface tension force $\sigma$, the pressure inside the bubble $p_i$ is greater than the pressure out side the bubble $p_o$. That is $$p_i = p_o + p_{extra},$$ where $p_\sigma$ is the extra pressure needed to balance the surface tension force. This does not make sense to me as I don't see how adding the extra term $p_{extra}$ balances the surface tension force. Say the bubble is a unit sphere and we consider the point $x=[1,0,0]$. I believe the surface tension force at this point $\sigma(x)$ acts into the liquid, so at this point it will have the form $\sigma(x) = [\alpha,0,0]$ where $\alpha > 0$. If we say tha that the extra pressure term $p_\sigma$ balances this we would have $p_{extra}(x) = \alpha$. As the forces are balanced at all points on the surface of the sphere, there is no movement of the bubble.

But here's the problem. The extra pressure acts in all directions! So it gives rise to a force say $F_1 = [-\alpha,0,0]$ that balances the surface tension, but it also gives rise to a force $F_2=[\alpha,0,0]$ that pushes in the opposite direction, i.e. the direction of the surface tension. So adding this extra pressure does not seem to have balanced the surface tension force.

What I am misinterpreting here? Maybe I have misinterpreted how the forces are acting? Everything I've searched for has just spoke in general terms of how extra pressure is needed inside the bubble to balance the surface tension, but I have not been able to find a precise explanation which gives an equation for how this balancing happens at a point.

• You are aware that the surface tension acts like a balloon membrane surrounding the bubble, right? If you actually had a balloon, wouldn't the pressure inside exceed the pressure outside? – Chet Miller Oct 16 '17 at 13:01