# Balloon experiment - Classical Physics vs. Statistical Physics

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• Juanda
Juanda
TL;DR Summary
Helium leaks into space continuously. Big things don't. What is the threshold?
While reading a similar and deservedly closed post a contradiction came to my mind. The supposed contradiction is related to Statistical Physics where my understanding is only conceptual so correct me where I might be wrong.

I remember reading that lightweight gasses can escape Earth's gravitational field because, due to the statistical nature of reality, some particles at the top of the atmosphere may get enough kinetic energy from heat (random kinetic energy of nearby particles) to reach the escape velocity. If there's nothing around to collide with it (and there isn't because we said it's already at the top of the atmosphere) then it'll just keep going and leave the gravitational field from Earth. I'd love to add a graph showing how the chances of this event happening are related to temperature. I recall they exist but I can't find a proper source.

As far as I know, this doesn't happen to hydrogen because it's such a reactive element that it'll bond with heavier elements making that event impossible to happen (I guess you'd always say there's a small chance in statistical physics but it's near zero). Helium on the other hand, being the noble gas it is, only bonds with itself so it doesn't get to be heavy enough to be retained by Earth in the long term so we're constantly losing helium to space. Helium is also being created on Earth by some unrelated processes but that's not the point of the post.

So that's the introduction. Let's now go with the post. There are many details that make the real-world scenario complicated such as the influence of the sun for example in the form of day and night, heat and solar winds so let's try to simplify it by imagining a starless perfectly round and solid planet similar in mass to Earth with a heat source within so the crust is uniformly at 20ºC (68F). Its atmosphere is composed of only CO2 (something I presume is heavy enough to make a stable atmosphere).
From that, we'd be able to establish the density and temperature distribution of the atmosphere as a function of height. I couldn't in my tries but I believe we'd have all the information necessary. If there is any prerequisite missing to derive the density function feel free to add it.

Now let's take an infinitely rigid spherical balloon 1m in diameter and 1g in mass. Then, it'll keep raising until its density equalizes with the density of the CO2 outside which as we said is a function of height. Anyways, at least in classical mechanics, this is not a especially hard problem.
Is there a point in statistical mechanics where we'd have to worry about the balloon escaping Earth's gravity? Is it a matter of shrinking the balloon? If it were the mass and size of a molecule it'd be able to escape. If it's much bigger we know the chances are near zero and the behavior can be described with classical mechanics.
I believe there's a point where we could say something like: For a spheric rigid balloon X in volume and Y in mass there is a Z% chance that it will escape Earth's atmosphere within 1 day.
What do you think?

Juanda said:
I remember reading that lightweight gasses can escape Earth's gravitational field because, due to the statistical nature of reality, some particles at the top of the atmosphere may get enough kinetic energy from heat (random kinetic energy of nearby particles) to reach the escape velocity. If there's nothing around to collide with it (and there isn't because we said it's already at the top of the atmosphere) then it'll just keep going and leave the gravitational field from Earth. I'd love to add a graph showing how the chances of this event happening are related to temperature. I recall they exist but I can't find a proper source.
What you are looking for (to start) would be a graph of the distribution of kinetic energies for a given temperature. This is the Maxwell Boltzmann distribution.

There are some complexities. Temperature does not relate to kinetic energy alone. For an ideal gas, it is temperature per degree of freedom. Helium molecules have three degrees of freedom. One for each dimension of space -- x, y and z. The available kinetic energy is shared (on average) equally in the three dimensions. Diatomic hydrogen molecules have six degrees of freedom. One for each dimension of space -- x, y and z, two for the two ways that the molecule can rotate (about the long axis does not count) and one for vibration. Water vapor has 12 degrees of freedom. You get three for linear motion, three for rotation (there is no excluded axis this time) and 6(!) for the various vibrational modes.

Given a particular molecular weight and knowing the escape velocity at the edge of the atmosphere you can calculate the required escape energy.

Divide that escape energy by the three degrees of freedom. That is how much kinetic energy per degree of freedom that you need to escape.

Read from the graph the probability that any given molecule will have that much kinetic energy per degree of freedom. [Or calculate it from the formula].

Divide by two (or more) since it is 50/50 whether the velocity points down toward the rest of the atmosphere or out into space.

Figure out how many molecules of a specified type (hydrogen, helium, oxygen, nitrogen, argon, water vapor, CO2, etc) are in a region near the edge of the atmosphere where the mean free path is roughly five kilometers. This part is a handwave on my part. The atmosphere halves in pressure approximately every 5 km. If the mean free path is 5 km then you have a decent chance of making it all the way to infinity.

Multiply the number of molecules by the probability per molecule. This will give the number of molecules that are on an escape trajectory at this moment.

But what you really want is a rate, not a number. So you are going to have to divide by the time it takes to re-randomize the distribution. I am not certain of the right way to get this figure. However, the average mean free path (5 km as above) divided by the average particle velocity would seem to be a good figure of merit. Maybe after multiplying by a small constant.

I am no expert. But this is how I would attack the problem.

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russ_watters and Juanda
That seems reasonable and conceptually I think could follow the explanation.
However, I think my main concern is: At which balloon size is it necessary to drop the classical mechanics' approach and embrace the statistical one to better describe what will happen to the rigid balloon? I'm assuming it is always possible to solve it with statistical mechanics but the answers you get are 99.999% certain so you're better off with classical mechanics that are more straightforward.

Still, regarding your answer, I have a few questions. They might be very basic in terms of statistical mechanics so feel free to point me to a book if they're too long or tedious to explain online.
jbriggs444 said:
There are some complexities. Temperature does not relate to kinetic energy alone. For an ideal gas, it is temperature per degree of freedom. Helium molecules have three degrees of freedom. One for each dimension of space -- x, y and z. The available kinetic energy is shared (on average) equally in the three dimensions. Diatomic hydrogen molecules have six degrees of freedom. One for each dimension of space -- x, y and z, two for the two ways that the molecule can rotate (about the long axis does not count) and one for vibration. Water vapor has 12 degrees of freedom. You get three for linear motion, three for rotation (there is no excluded axis this time) and 6(!) for the various vibrational modes.
Since I'm considering this to be a rigid sphere, is it right to assume it's got 6 degrees of freedom (3 translations with the same energy share and 3 rotations with the same energy share)? In this case, I'd be only interested in the translation.

EDIT. Sorry, I clicked "SEND" before finishing my writing. I'll edit it.
jbriggs444 said:
Given a particular molecular weight and knowing the escape velocity at the edge of the atmosphere you can calculate the required escape energy.
Given the rigid sphere scenario, is density a valid equivalency for molecular weight?

jbriggs444 said:
Figure out how many molecules of a specified type (hydrogen, helium, oxygen, nitrogen, argon, water vapor, CO2, etc) are in a region near the edge of the atmosphere where the mean free path is roughly five kilometers. This part is a handwave on my part. The atmosphere halves in pressure approximately every 5 km. If the mean free path is 5 km then you have a decent chance of making it all the way to infinity.
How would you define the atmosphere edge? In classical mechanics, I believe it simply extends to infinity with density going lower and lower as the distance increases.

Juanda said:
Since I'm considering this to be a rigid sphere, is it right to assume it's got 6 degrees of freedom (3 translations with the same energy share and 3 rotations with the same energy share)? In this case, I'd be only interested in the translation.
If you are considering this as a rigid sphere then I would count either three degrees of freedom if it is slippery or six degrees of freedom if it has a rough surface. Three for translation, of course. Three for rotation if it is rough. If it is slippery then it can never experience a torque so rotation is irrelevant.

If it is perfectly rigid then there can be no vibrational modes.

However, now we have this huge rigid sphere being considered as if it were a single molecule. Its molecular weight is off the charts. A one gram balloon would have a molecular weight of ##6.02 \times 10^{23}##. Its escape energy would be correspondingly huge.

If you plotted the balloon's escape energy on a Maxwell Boltzmann graph, it would be ludicrously far out to the right.

Note that steering this thread to a re-discussion of rigid balloons rather than atmospheric outgassing is risking a thread closure and some points on your record.

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Juanda said:
How would you define the atmosphere edge? In classical mechanics, I believe it simply extends to infinity with density going lower and lower as the distance increases.
For purposes of the calculation that I proposed, it would be "where the mean free path of the molecule in question exceeds 5 km".

A quick trip to Google says that the mean free path is 10 cm at 100 km altitude. We want a factor of 50,000 larger. At a factor of two per 5 km, that would be another 80 km up. So I'm guessing 180 km.

If the balloon is rigid, only its volume and mass matter. Whether it is filled with helium or depleted uranium, its buoyancy will depend only upon its volume and mass.
The two questions are therefore completely distinct.
Non Sequitur

jbriggs444
Juanda said:
Given the rigid sphere scenario, is density a valid equivalency for molecular weight?
No.

Consider, for instance, a basketball at rest on the floor at half court. Let it sit there and equilibriate with the air. Technically its center of mass will take on an energy per degree of freedom based on the air temperature. The same energy per degree of freedom as one of the air molecules surrounding it. [Though I may not be grasping all of the details of Brownian motion].

Watch that basketball. How long before it ejects itself into orbit do you think?

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russ_watters and hutchphd
hutchphd said:
Whether it is filled with helium or depleted uranium
That went over like a lead balloon.
Or will if you wait enough half-lives.

dlgoff, hutchphd and jbriggs444
jbriggs444 said:
However, now we have this huge rigid sphere being considered as if it were a single molecule. Its molecular weight is off the charts. A one gram balloon would have a molecular weight of ##6.02 \times 10^{23}##. Its escape energy would be correspondingly huge.

If you plotted the balloon's escape energy on a Maxwell Boltzmann graph, it would be ludicrously far out to the right.

Note that steering this thread to a re-discussion of rigid balloons rather than atmospheric outgassing is risking a thread closure and some points on your record.
My point is not to re-open a closed thread but to understand the boundary between classical and statistical mechanics.
The balloon thread was the inspiration but what I'm interested in is how the problem changes as this rigid balloon decreases in size or mass. I think that's conceptually more understandable than "inflating a molecule" until statistical mechanics no longer apply to it. Or I guess a better description would be until statistical mechanics is not worth using because classical mechanics will give you the same results more easily.
As you pointed out, any real scenario involving rigid balloons/tanks I can imagine would be so far to the right in the graph that applying statistical mechanics to it would yield the same result as classical mechanics just with much more effort. Still, I thought it would be an interesting concept to pound about.

hutchphd said:
If the balloon is rigid, only its volume and mass matter. Whether it is filled with helium or depleted uranium, its buoyancy will depend only upon its volume and mass.
The two questions are therefore completely distinct.
Non Sequitur
It's true the content doesn't matter since the container is rigid. It just came from the previous reasoning.
I'm not completely sure about what two questions you're referring to. These?
Juanda said:
Now let's take an infinitely rigid spherical balloon 1m in diameter and 1g in mass. Then, it'll keep raising until its density equalizes with the density of the CO2 outside which as we said is a function of height. Anyways, at least in classical mechanics, this is not a especially hard problem.
Is there a point in statistical mechanics where we'd have to worry about the balloon escaping Earth's gravity? Is it a matter of shrinking the balloon? If it were the mass and size of a molecule it'd be able to escape. If it's much bigger we know the chances are near zero and the behavior can be described with classical mechanics.
I believe there's a point where we could say something like: For a spheric rigid balloon X in volume and Y in mass there is a Z% chance that it will escape Earth's atmosphere within 1 day.
What do you think?
What I was trying to get at is that if we shrink the balloon down in terms of mass (less energy necessary to reach escape velocity) and cross-section (less likely to hit anything) getting closer and closer to a light molecule such as helium its behavior should approach it too. By its approaching behavior, I mean that it could escape Earth's gravity.

jbriggs444 said:
No.

Consider, for instance, a basketball at rest on the floor at half court. Let it sit there and equilibriate with the air. Technically its center of mass will take on an energy per degree of freedom based on the air temperature. The same energy per degree of freedom as one of the air molecules surrounding it. [Though I may not be grasping all of the details of Brownian motion].

Watch that basketball. How long before it ejects itself into orbit do you think?
That I understand. With big enough objects this kind of thing happening is impossible. Just like how in macroscopic scales heat always flows from hot to cold.

I'm trying to understand where the boundary in size is if that makes sense to begin with.

Juanda said:
I mean that it could escape Earth's gravity.
You will find that once your balloon consists of more than a few molecules, the probability is infinitesimal.

Juanda
You will find that once your balloon consists of more than a few molecules, the probability is infinitesimal.
All right so that's what I was aiming at. Thanks!
That sets the boundary I was trying to understand. It's actually much smaller than I expected.
So the original for the original point: For a spheric rigid balloon X in volume and Y in mass there is a Z% chance that it will escape Earth's atmosphere within 1 day.
The "answer" would be that if the container reaches the size, mass and volume (X and Y) of just a few molecules the chances Z are practically zero.

Juanda said:
My point is not to re-open a closed thread but to understand the boundary between classical and statistical mechanics.
Maybe there's something I don't understand, but I was under the impression that statistical mechanics can very well be a part of classical mechanics. Wasn't the Maxwell Boltzmann distribution part of classical mechanics? Wasn't it developed prior to the development of relativity or quantum mechanics?

Mister T said:
Maybe there's something I don't understand, but I was under the impression that statistical mechanics can very well be a part of classical mechanics. Wasn't the Maxwell Boltzmann distribution part of classical mechanics? Wasn't it developed prior to the development of relativity or quantum mechanics?
I didn't realize that. I was missing the right word to describe the difference. So one is statistical mechanic. What's the name for "normal" Newton physics? Deterministic Newtonian physics perhaps?

Juanda said:
I didn't realize that. I was missing the right word to describe the difference.
I don't think that missing the right word is the problem so much as a lack of understanding. In classical statistical mechanics we are just applying the principles of newtonian physics to large collections of particles and focusing on the behavior of the collection rather than on the behavior of any one particle.

PeroK
Classical statistical physics does not introduce new fundamental laws but uses the fundamental laws of Newtonian mechanics in situations, where you don't have the complete information about a system, i.e., you don't know the precise initial state, which would be given a point in ##6N##-dimensional phase space. For typical particle numbers of everyday macroscopic bodies (in the order of ##N=10^{24}##) this is not possible, let alone to exactly solve the equations of motion of an interacting system of ##N## particles. Also this detail is not needed to describe the relevant macrosopic properties of this body. Usually a few macroscopic collective observables are sufficient, e.g., to describe how a ball moves it's enough to calculate the equations of motion of its center of mass and it's usually even sufficient to treat this as a "point particle".

To also understand some relevant properties of the motion of the molecules or atoms making up the body, you use statistical methods. Instead of describing the full ##N##-body phase-space distribution function, which is as well impossible to consider for ##10^{24}## particles, you just ask for, e.g., the single-particle distribution function.

Now Newton's equations of motion for the ##N##-body phase-space distribution function is the socalled Liouville equation, and integrating out all phase-space variables of this ##N##-body distribution leads to the one-particle distribution function, but the exact equation for its time evolution, derivable from the Liouville equation needs to know the two-body distribution function; for the calculation of this two-body distribution function you'd need the three-body distribution function and so on. That's the socalled BBKY hierarchy (named after its discoverers, Born, Bogoliubov, Kirk, Yvon). Of course at the end you'd need to know the complete ##N##-body distribution function, if you'd try to solve this set of equations.

Now comes the ingenious trick invented by Boltzmann! It's of course impossible to know the ##N##-body distribution functions are the entire tower of ##n##-body distribution functions for all ##n<N##. Now usually there's also a "separation of scales". The minute details of each single particle is rather quickly fluctuating due to a lot of collisions with other particles, while the macroscopically relevant motions, like the flow of a gas or liquid, are pretty slow. The same holds for the variations of spatial distributions or distributions of momenta. For the macroscopic description you can thus look at macroscopically small but microscopically large regions of phase space and average over all the many particles in this region. For these averages even the details of the exact two-body correlation function are pretty unimportant and on the average you can just assume that two randomly picked particles out of the macroscopically small, microscopically large phase-space cell are uncorrelated. Thus within this macroscopic cell you approximate the two-body distribution function by the product of the one-body distribution functions, i.e., you neglect any possible statistical correlations between two randomly picked particles. That's Boltzmann's molecular-chaos assumption ("Stoßzahlansatz").

This coarse-graining process cuts the BBGKY hierarchy, i.e., for the approximate calculation of the one-body functions you now only need the one-body distribution functions. The result is the Boltzmann transport equation with a collision term taking into account only 2->2 scattering processes. This already leads to very good descriptions of the macroscopic behavior of the gas. It leads to the H-theorem, according to which the entropy always increases, and the stationary long-time limit leads to the Maxwell-Boltzmann distribution for thermal equilibrium as the state of maximum entropy. From this you can derive equilibrium thermodynamics as well as hydrodynamics for situations close to thermal equilibrium.

hutchphd and Juanda

## What is the balloon experiment in the context of classical physics and statistical physics?

The balloon experiment typically involves observing the behavior of a gas inside a balloon under various conditions. In classical physics, the focus is on macroscopic properties like pressure, volume, and temperature, using the ideal gas law. In statistical physics, the experiment examines the microscopic behavior of gas molecules, using probability distributions and statistical ensembles to describe the system.

## How does the ideal gas law apply to the balloon experiment?

In classical physics, the ideal gas law (PV = nRT) relates the pressure (P), volume (V), and temperature (T) of the gas in the balloon, where n is the number of moles of gas and R is the universal gas constant. This law helps predict how the balloon will expand or contract when the temperature or pressure changes.

## What role does entropy play in the balloon experiment from a statistical physics perspective?

In statistical physics, entropy is a measure of the disorder or randomness of the gas molecules inside the balloon. It provides insight into the number of possible microstates that correspond to a given macrostate. As the gas expands or contracts, the entropy changes, reflecting the distribution of molecular energies and positions.

## How do fluctuations in gas molecules' behavior differ between classical and statistical physics?

In classical physics, gas behavior is often treated as smooth and continuous, with fluctuations being negligible. In contrast, statistical physics explicitly considers these fluctuations, analyzing the probability of deviations from average behavior. This approach helps explain phenomena like Brownian motion and provides a more detailed understanding of gas dynamics at the microscopic level.

## Can you explain the concept of equipartition of energy in the context of the balloon experiment?

Equipartition of energy is a principle from statistical physics stating that energy is distributed equally among all degrees of freedom of the gas molecules. For a gas in a balloon, this means each molecule's translational, rotational, and vibrational motions contribute equally to the total energy, influencing the temperature and pressure of the gas.

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