Reference: 'Thermal walls in computer simulations', R. Tehver, Phys Rev E 1998,

<http://pre.aps.org/abstract/PRE/v57/i1/pR17_1> , which can also be downloaded from

<http://polymer.chph.ras.ru/asavin/teplopr/ttk98pre.pdf>

In the above paper, one finds that molecules emitted from a wall of temperature T have a

velocity distribution of

[1]

where v is the velocity normal to the wall. In our case, let us take a ground plane as a

thermal wall with temperature T1, have gravity present so that the normal component of

velocity reduces with altitude, and find how this distribution varies as a function of

altitude.

First we note that for a molecule of mass m which reaches a plane 2 at height h, its

kinetic energy will be reduced by the gravitational potential energy.

[2]

The velocity at height h for this particular molecule can then be found as

[3]

Now let us deal with the distribution of velocities. First note that the number of particles

which can be found within velocities va and vb at ground level is given by

[4]

This is not hard to solve, but for convenience can be solved via ‘The Integrator’, e.g.

<http://integrals.wolfram.com/index.jsp?expr=%28m%2F%28kT%29%29*x*exp%28-

0.5*m*x%5E2%2F%28kT%29%29&random=false>

This gives the result

[5]

Now let us assume that va is large enough so that all the particles reach height 2. We do

not know the equation for the distribution of velocities, or the temperature, at height 2,

but we do know that va will be reduced according to Eqn(3) to

[6]

and vb will be reduced to

[7]

and that the particle number will remain constant so that

[8]

We then write an equation for the unknown distribution at level 2 to get

[9]

From Eqns(6,7), we have a relation between va and vc, and between vb and vd, so solving

for va and vb, and substituting in, one gets

[10]

Fortunately, we already know a function that when integrated between its limits gives

such a dependence on v, it is of course 1. Hence we can relate 2 to 1 as

[11]

showing that the distribution at altitude 2 has the same characteristic temperature as the

distribution at altitude 1, the functional form of the distribution is the same, the only

difference being a reduction in amplitude. This reproduces the standard result of an

isothermal column of gas under gravity, with a density drop off according to the

barometric formula.

Note on assumptions:

By using the 1 distribution, one circumvents the complaint that the equilibrium volume

distribution in a gravitational field is assumed to be that of the MB form. The only

requirement in the above derivation is that there is thermal continuity between the gas

and the ground at the interface between the two, and this requirement is independent of

the presence or absence of a gravitational field.

The density drop off is of course due to those particles which leave ground level but do

not reach altitude 2. As these molecules have v<v1min, they do not affect the condition

ncd=nab which transfers molecules with va>v1min to altitude 2 (v1min being the minimum

velocity needed for a molecule leaving ground level to reach altitude 2).

Other assumptions are usual for an ideal gas: no collisions, no Van der Waals forces, etc.