# Non-interacting gas in homogeneous gravitational field

I am getting really tripped up by this exercise: You're given a non interacting gas of particles each having a mass m in a homogeneous gravitational field in a set temperature T. You're told it is non interacting so it obeys the Maxwell distribution of speeds. Then you are asked to find the altitude $$z$$-dependence of pressure and density in thermodynamic equilibrium.

It even gives a hint, it says "consider two horizontal surfaces $$z1$$ and $$z2$$ and think about what thermodynamic equilibrium means for particles traveling from one surface to the other". This really trips me up because I am not sure what to do with this. Obviously in equilibrium the number of particles between the surfaces should stay the same, but what does this tell me really? I can't see where this is going. The best conclusion I can draw out of it is that the same number of particles crossing either surface going downwards is crossing it to go upwards. But so what?

Another thing I considered is that the energy of some particles won't be enough to go above a certain altitude. But again, so what? It doesn't tell me that much as far as I can tell.

Normally I would try to find the canonical partition function, use it to find the entropy etc. But given that it tells me it obeys the Maxwell-Boltzmann distribution and given the hint, there must be something else I am supposed to do. Any ideas?

• It is quite a common misconception that the Maxwell-Boltzmann distribution is valid only for non-interacting systems. Actually, for a classical system does not matter whether it is interacting or not, or if it is in a gas, liquid, or solid phase. The velocity distribution function in the canonical and grand-canonical ensembles can't be anything else. Jan 5 at 23:17

You could certainly use the full statistical physics approach, if you'd like. From this POV, you would consider adjacent horizontal layers to be separate systems in thermal and chemical equilibrium, and then note that the external potential serves as a chemical potential which varies between those layers.

However, the more down-to-earth treatment would be to basically repeat the standard derivation for the pressure variation in a liquid. One can consider the sum of the forces acting on a small cube of air at height $$z$$. The forces in question are supplied by gravity and the pressure at the various points in the system. Noting that the average molecular speed is constant, you can relate all of these forces directly to density, which (in the limit as the size of the cube goes to zero) becomes a differential equation for $$\rho(z)$$.

• What you're saying about the cube thing makes sense I guess, but the exercise talks about non-interacting particles, so how come I can use that? Jan 5 at 15:27
• @AndreasC That the particles are non-interacting means that there are no long-range forces of the type which would require including some kind of attractive or repulsive interparticle potential in the Hamiltonian. However, they must still be able to collide (e.g. with the walls of a cube), otherwise pressure would not be well-defined. Ideal collisions with each other are also necessary to drive the system to thermodynamic equilibrium. Jan 5 at 15:53
• Ι guess this does make sense... Ι'll try that... Jan 5 at 16:59

I am not familiar with the Maxwell distribution of speeds myself, because I am not so familiar with the mathematical tools needed (I am a high school senior), but here is my attempt at a partial answer:

Suppose the two layers are separated by a very small height $$dh$$. Say the average energy of a particle moving from the higher layer to the lower layer is $$E$$. When they arrive at the lower layer, they must have gained some gravitational potential energy, $$nmg\, dh$$, where $$g$$ is the field strength. As such, the particles arrive with average kinetic energy $$E + mg \,dh$$. At the same time, for the same amount of heat (random kinetic energy) to be transferred to the layer above, the particles must also arrive with average kinetic energy $$E + mg \,dh$$; thus it should follow that they leave the lower layer with an average kinetic energy $$E' = E + 2nmg\,dh$$.

Remembering that the temperature is constant throughout the whole system, the particles must follow the same Maxwell distribution of speeds (and Maxwell-Boltzmann distribution of kinetic energies, if that is something in the toolkit).

Suppose that $$E$$ is the average kinetic energy of the particles in the upper layer. You could force $$E'$$ to be the average kinetic energy of the particles with the Maxwell-Boltzmann distribution at temperature T by excluding some proportion of the particles, $$p$$, $$0 < p << 1$$ (I suppose this is a mathematical exercise in finding a cut-off area of low-energy particles, at energies above which the mean energy is $$E'$$ if the mean energy is $$E$$ over the entire M-B distribution). This implies that, for every particle in the upper layer, there must be $$\frac{1}{1-p} \approx 1 + p$$ particles in the lower layer.

The difference in the number of particles (a common ratio of $$\frac{1}{1-p}$$) could be extrapolated to find the difference in density between the two layers, as well as the difference in pressure (from $$P\,\,\alpha\,\,n$$, $$T, V\,constant$$).