Maxwell=s Distribution

In September 1859 James Clerk Maxwell came
before the British Association of Aberdeen and read a paper that he was to have
published the following year in the *Philosophical Magazine*. In that
speech Maxwell described how to deduce the distribution of speeds among the
particles comprising an ideal monatomic gas. He thus laid the foundation of
modern statistical thermodynamics (as distinct from classical thermodynamics,
which developed in parallel with the probabilistic version). Now I want to
follow Maxwell=s reasoning while
paraphrasing and elaborating upon what he said. Let Maxwell begin:

AIf
a great many equal spherical particles were in motion in a perfectly elastic
vessel, collisions would take place among the particles, and their velocities
would be altered at every collision; so that after a certain time the *vis
viva* will be divided among the particles according to some regular law, the
average number of particles whose velocity lies between certain limits being
ascertainable though the velocity of each particle changes at every collision.@

We have N identical spherical particles whose
aggregate carries a total kinetic energy (Maxwell=s
*vis viva* or living force) E and we put them into a container with
perfectly hard, perfectly elastic walls (think of a foot-thick steel wall
relative to a tennis ball). The average energy per particle stands in the ratio
e=E/N. Maxwell has tacitly assumed into the premises the proposition that the
particles have no internal modes of energy storage (such as vibration), so they
can only undergo perfectly elastic collisions with each other. After sufficient
time has elapsed since we put the gas into our container, the particles achieve
an equilibrium, in which for every particle that a collision kicks out of the
velocity range between **v** and
**v**+d**v** another particle gets kicked into that same velocity range.

Every particle has a velocity

(Eq=n 1)

in which equation **i**, **j**, and **k** represent the
unit vectors pointing in the positive x-, y-, and z-directions respectively.
When the system comes to equilibrium we expect to find a number of particles
that have an x-component of velocity in the range between v_{x} and v_{x}+dv_{x}
in accordance with

(Eq=n 2)

The function f(v_{x}), whose form we want to determine,
depends only upon v_{x} as a variable factor. No other properties of the
particles come into play in this case because they don=
t differ from particle to particle. We take Leibniz=s
principle of sufficient reason in the form AIt
takes a difference to make a difference@
and say that because we have no differences in the particle=s
properties we can have no differences in their behaviors. We interpret the
expression f(v_{x})dv_{x} as expressing the probability of any
given particle possessing a velocity in the range dv_{x} around the
value v_{x}.

We also know that within the ranges between v_{y}
and v_{y}+dv_{y} and between v_{z} and v_{z}+dv_{z}
we have particles in the amounts

(Eq=n 3)

and

(Eq=n 4)

In Equations 2, 3, and 4 f(v) represents the same function of the component velocities. If that statement did not stand true to physics, then the number of particles in a given component-velocity range would depend upon the orientation of our coordinate grid. But the distribution of the particles among the available velocities cannot depend upon any feature of our purely imaginary grid; only our description of that distribution can have such a dependence.

Of course, each particle has three components
in its velocity. Those components, as Maxwell noted, represent motions parallel
to three mutually perpendicular directions, so they do not affect each other. If
a collision changes the x-component of a particle=
s velocity, the y- and z-components remain unchanged. That fact means that we
can calculate the number of particles whose velocity components lie in the
ranges between v_{x} and v_{x}+dv_{x}, between v_{y}
and v_{y}+dv_{y}, and between v_{z} and v_{z}+dv_{z}
simply by multiplying together the probabilities in Equations 2, 3, and 4;

(Eq=n 5)

Maxwell got that result from standard probability theory applied to mutually independent outcomes.

Imagine an abstract space in which the points represent velocities. In this velocity space our gas would appear as a cloud more or less centered on the origin of the grid. At equilibrium that cloud will display a spherical symmetry about the origin, again because the distribution cannot show any effects due to our choice of orientation of the axes of the grid. If we calculate the density of the particles in velocity space about one of its points, the function that we use can only depend upon the magnitude of the velocity and not on how we divide it into components, so we must have, as Maxwell expressed it,

(Eq=n 6)

A function of the sum of components corresponds to the product of the functions of the components. That fact tells us that the function has the same relation to its argument that a number has to its logarithm, so Maxwell solved that equation with

(Eq=n 7)

and

(Eq=n 8)

He then noted that A cannot represent a positive number, because integration of the function would then yield an infinity that certainly would not describe the gas. So he set A=-γ, in which γ represents a positive number (actually he used the inverse of alpha squared rather than gamma, but I whither this reasoning leads, so I=m cheating a little here).

So now we have Equation 2 in the form

(Eq=n 9)

If we sum N(v_{x}) over the range
-4
<v_{x}<+4
, we must get N as the result, so we
must have

(Eq=n 10)

which necessitates that

(Eq=n 11)

Maxwell then drew four conclusions:

1. The number of particles with the
x-component of their velocities lying in the range between v_{x} and v_{x}+dv_{x}
comes out to

(Eq=n 12)

2. The number of particles with speeds in the
range between v and v+dv occupying, in velocity space, a spherical shell of area
4πv^{2},
thickness dv, and density NΦ,
which gives us

(Eq=n 13)

3. To calculate the average speed of the particles we multiply that equation by v, integrate over the range 0# v<4 , and divide the result by N. We get

(Eq=n 14)

4. To calculate the average of v^{2}
we multiply Equation 13 by v^{2}, integrate over the range 0#
v<4 , and divide the result by N. We get

(Eq=n 15)

And that=s as far as Maxwell took his derivation. But he could have gone further. If we multiply Equation 15 by one half the mass of one particle in the gas, we get a description of the average kinetic energy per particle;

(Eq=n 16)

But if we want to use kinetic energy, rather than velocities, in our calculations (and we do), we must make the substitution γ=mβ/2 in Equation 13 to get

(Eq=n 17)

Equation 16 then gives us

(Eq=n 18)

Let=s use that derivation to calculate the pressure that the gas exerts upon the walls of its container. We consider an area dA that lies parallel to the y-z plane and we examine the phenomenon of particles striking that area and bouncing off it; in particular, we want to determine the rate at which linear momentum gets transferred from the particles to the wall. Impinging upon the minuscule projections that give even the smoothest surface a texture, the particles of the gas transfer linear momentum to the wall in all three of the cardinal directions. In the y- and z-directions we have as many particles carrying a certain amount of linear momentum in one direction as we have carrying the same amount in the opposite direction, so the net transfer of linear momentum to the wall in those directions averages out to zero.

In the x-direction the particles put their
momenta into the wall as the wall stops them and put in an additional amount in
rebounding from the wall. If the gas and the wall share thermal equilibrium,
then for every gas particle that loses a certain amount of energy in striking
the wall (and thereby rebounding with less momentum than it had when it struck
the wall), another gas particle gains as much energy in its rebound. So all we
need to do to calculate the total momentum transferred to the wall is to
calculate the amount of momentum that the gas particles bring to the wall and
double it. If we have a particle density of n, then in an interval dt a
description of the number of particles with x-ward component of velocity between
v_{x} and v_{x}+dv_{x} striking the wall, filling a
volume dAv_{x}dt, comes from Equation 12 in the form

(Eq=n 19)

We multiply that equation by a general description of each
particle=s linear momentum in the
x-direction, mv_{x}, integrate the result over all possible values of v_{x}
from zero to infinity, double the integral, and get

(Eq=n 20)

in which Dividing that equation by dt converts it into a description of the force that the gas exerts upon the area dA and dividing that result by dA yields a description of the pressure that the gas exerts upon the walls of its container,

(Eq=n 21)

If we multiply that little equation by the volume of the container, we get

(Eq=n 22)

That equation tells us that if we can change the average energy of the particles in our gas, then either the pressure or the volume of the gas or both must change. That means that if we contrive a way to keep the volume in the gas unchanged, then any change in the pressure of the gas will reflect a proportional change in the average energy per particle. We can exploit that fact to measure the average energy of the particles in the gas.

Imagine that we have the gas in a non-expandable container that has a hole leading to a vacuum chamber in one wall. A piston, free to slide without allowing any of the gas to escape, fills the hole and a very stiff spring holds it in position. Through a series of levers we connect the piston to a pointer that moves along a graduated scale; the levers convert small movements of the piston into large movements of the pointer. If we have properly calibrated the device, the position of the pointer along the scale tells us the pressure of the gas. But the number that we read off the scale also correlates with the average energy of the particles in the gas. It doesn=t actually tell us the average energy per particle outright, so we give it a new name and call it the temperature of the gas.

We relate the temperature of a gas to the average energy per particle by way of

(Eq=n 23)

in which k represents an absolute constant that converts units of temperature (degrees Celsius or degrees Fahrenheit) into units of energy per particle (ergs or joules). Note that with reference to Equation 18 I have simply defined kT=1/β. The factor of three in Equation 23 represents the fact that each particle has three independent modes of storing kinetic energy, each mode corresponding to one of the axes along which the particle is free to move;

(Eq=n 24)

Thus we have, because no direction is preferred over any other,

(Eq=n 25)

Finally we want to calculate the number of particles in the gas that have energies lying in the range between E and E+dE. We simply take Equation 17 and replace beta with the reciprocal of kT and convert the velocities into the equivalent kinetic energies (remembering that dE=mvdv). We claim then that we have thus made dn(v)=dn(E). Now we need to prove and verify that claim.

By hypothesis we have

(Eq=n 26)

We begin by integrating that equation over all possible values of E, from zero to infinity, in accordance with

(Eq=n 27)

Making the substitution x=E/kT gives us the full integration of Equation 26 as equal to N, the number of particles in the gas, as we require.

To make certain that we have the correct formula in Equation 26, we now use it to calculate the average energy of the particles in the gas. We thus take

(Eq=n 28)

and integrate it over all possible values of E. In this case we must integrate by parts. Using the same substitution used above, we have

(Eq=n 29)

Thus we get the average energy per particle in the gas as

Eq=n 30)

as we should. Thus we have proven and verified our claim that Equation 26 gives a correct description of the distribution of energy over the particles in the gas. And that gives us a full description of the Maxwell Distribution also displaying the Boltzmann factor (the exponential), which plays an important role in the full statistical formulation of thermodynamics.

Appendix: kT=1/β

We have Equation 22 as pV=N/β, in which beta, in accordance with Equation 18, is inversely proportional to the average energy carried by the N particles comprising a gas under pressure p in a container of volume V. In Equation 23 I seem to have arbitrarily replaced the reciprocal of beta by the product kT. Mathematically that=s a legitimate substitution, but I want to show further that the T actually corresponds to our basic concept of temperature, of relative hotness and coldness.

We know that heat is a form of energy. Thus when an object is hot it contains more energy than it does when it is cold. If the object is a body of gas inside a container, then that comment also applies to the average energy of the particles comprising the gas: the hotter the gas, the more energy the particles carry on average. But Equation 30 relates average energy per particle to the parameter T, so T has a greater value for a hot object that it does for a cold object. Thus T represents the temperature of the body.

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