Unstable Balloons
Most recent answer: 1/29/2016
- Estherika
Israel
That’s a really interesting question. Here’s what I think the answer is.
The air will move around to minimize the total free energy of the air and the balloons, the same way a fluid will flow downhill as far as it can, minimizing gravitational energy. Let’s make some approximations to see whether this weird behavior follows from simple ingredients or requires some special properties of the balloons.
Let’s say that the total volume of air is pretty much fixed. Then the free energy changes come from the stretching of the rubber in the balloons. Let’s say that, like most springy things, the excess free energy is proportional to the square of how far they’re stretched. Now the volume goes as radius cubed, but the free energy goes as radius squared. So the excess free energy is proportional to the balloon volume to the 2/3 power. The rate of change of free energy with respect to volume increase (the slope or derivative) then is inversely proportional to volume to the 1/3 power. So the more-inflated balloon increases free energy less for a given volume of added air. So free energy goes down when air flows from the less inflated to the more inflated balloon. The process will stop when the less inflated balloon isn’t stretched any more.
Mike W.
(published on 12/08/2007)
Follow-Up #1: unstable pair of balloons
- sarah (age 37)
iran
This old answer was hard to find, but I think it's what you need.
Mike W.
(published on 07/26/2015)
Follow-Up #2: wrong ideas about balloons
- Anonymous
Seriously? You think air spontaneously flows from low to high pressure? The fundamental laws of thermodynamics are all wrong?
I suggest you try it with ordinary balloons. Maybe you'll change your mind.
Mike W.
(published on 01/26/2016)
Follow-Up #3: unstable balloons revisited
- Anonymous
Your clarification does an excellent job of clearing up the logic of the problem, including where you took a wrong turn. We're exactly with you up to "This means there is less hoop tension in the large balloon skin." It was that very point which we addressed indirectly in our original answer. As you say, our argument may have been too technical sounding, but you've provided a nice structure to discuss it further.
What's the relation between the pressure difference across the balloon skin and the radius of the balloon? As you say, as the radius gets bigger, the tension in the skin goes up. How does that tension get translated into a pressure difference? In equilibrium the force on any little patch is zero. So the force from the tension must cancel the force from the pressure difference. And here's the key point you missed. If the skin were simply flat, no amount of tension in it would create any force at right angles to the skin! The only reason there's a force in that direction is that the surface curves, so that neighboring patches pull in a little on the patch we're considering. The force depends on the product of the tension and the curvature. Yes, as the balloon gets bigger the tension goes up, but the curvature goes down. In terms of the radius, R, the curvature goes as 1/R and the tension goes about as R. So it turns out that this line of argument is not getting us a unique answer.
Here's our simpler argument:
1. In equilibrium in the room, there's some volume V of air inside the balloons, at room temperature.
2. So let's assume that the air has that fixed volume and figure out how to distribute that V to minimize the free energy.
3. At fixed V and T, the distribution of V has no effect on the air free energy.
4. So the distribution is determined by minimizing the elastic free energy of the balloons, under the fixed-total-V condition.
So that's what we did.
Mike W.
(published on 01/27/2016)
Follow-Up #4: figuring out how to teach
- Anonymous
Thanks for your constructive thoughts on how to convey these ideas. It ain't always easy! If I figure out a way to take your advice and rephrase our argument in terms of forces without getting something wrong, I'll do it.
You do raise an interesting question, that applies to all of physics, not just some special little category. "How, do soap bubbles or light beam know calculus?" The amazing thing is that nature seems to know calculus and much fancier branches of math. In fact, it doesn't seem to know anything else.
Meanwhile, I think that for this not-too-fancy experiment, the advice to try it yourself is the key!
Mike W.
(published on 01/28/2016)
Follow-Up #5: axioms and theorems
- Anonymous
Well, speaking of axioms, your fundamental one is false. You write "Nature works with the immediate (differential) action / reaction forces. " That's a specific type of local realist theory. The violations of the Bell Inequalities (see e.g. https://van.physics.illinois.edu/qa/listing.php?id=30737&t=bell-inequalities) prove that no such theory can describe our world.
But ok, let's confine our attention to the subset of phenomena in which a local-realist classical approximation works fairly well. All of standard intermediate mechanics (Hamiltonian and Lagrangian) is based on formal proofs of the equivalence of the global principles to the local description you mention.
In our balloon case, however, none of that is particularly relevant. We're asking what condition things settle down to, not what the detailed motions are on the way. The little forces you describe obey laws that are the same if you reverse the direction of time. Have you ever observed any phenomenon that would be the same backward in time? I doubt it, since that would mean all the light rays would be going back into the sun or the light bulb, etc. So if you want to ask the question "what do things settle down to look like" you absolutely need the time-irreversible entropy maximization principle, called the Second Law of Thermodynamics. It basically says that given a chance nature wanders with equal probability through all the available quantum states. It turns out that the vast majority of these states look similar- e.g. with balls at the bottom of cups and not rolling around the rim. So our predictions of what we'll see as things settle down are based on state-counting, not dynamical details. Dynamical details cannot tell you whether balls spontaneously jump up. State-counting does tells you that big ones don't, but molecule-size ones do.
Incidentally, even the tension in rubber-like polymers is primarily driven by configurational entropy maximization and not by energy minimization. (And energy minimization is itself just another manifestation of more general entropy maximization.) The curled up polymer has more choices of ways to curl up than the stretched-out one has of being stretched. That's why in equilibrium you find them curled up. So even the most mechanical-seeming aspect of this problem is in fact statistical.
Mike W.
(published on 01/29/2016)