2Classical gases

II Statistical Physics

2.3 Maxwell distribution

We calculated somewhere the average energy of our gas, so we can calculate

the average energy of an atom in the gas. But that is just the average energy.

We might want to figure out the distribution of energy in the atoms of our gas.

Alternatively, what is the distribution of particle speed in a gas?

We can get that fairly straightforwardly from what we’ve got so far.

We ask what’s the probability of a given particle being in a region of phase

space of volume d

3

q d

3

p centered at (q, p). We know what this is. It is just

Ce

−βp

2

/2m

d

3

q d

3

p

for some normalization constant

C

, since the kinetic energy of a particle is

p

2

/

2

m

. Now suppose we don’t care about the position, and just want to know

about the momentum. So we integrate over

q

, and the probability that the

momentum is within d

3

p of p is

CV d

3

p e

−βp

2

/2m

.

Let’s say we are interested in velocity instead of momentum, so this is equal to

CV m

2

d

3

v e

−βmv

2

/2

.

Moreover, we are only interested in the speed, not the velocity itself. So we

introduce spherical polar coordinates (v, θ, φ) for v. Then we get

CV m

3

sin θ dθ dϕ v

2

dv e

−mv

2

/(2kT )

.

Now we don’t care about the direction, so we again integrate over all possible

values of θ and φ. Thus, the probability that the speed is within dv of v is

f(v) dv = Nv

2

e

−mv

2

/(2kT )

dv,

where we absorbed all our numbers into the constant

N

. Then

f

is the probability

density function of v. We can fix this constant N by normalizing:

Z

∞

0

f(v) dv = 1.

So this fixes

N = 4π

m

2πkT

1/2

.

This f(v) is called the Maxwell distribution.

We can try to see what

f

looks like. We see that for large

v

, it is exponentially

decreasing, and for small

v

, it is quadratic in

v

. We can plot it for a few

monoatomic ideal gases:

T

C

4

He

20

Ne

40

Ar

132

Xe

We see that the energy distribution shifts to the right as we increase the mass.

This is expected, because we know that the energy of the particle is always

1

2

kT

,

and so for lighter particles, we need to have higher energy.

We can sanity-check that the expected value of v is correct. We have

hv

2

i =

Z

∞

0

v

2

f(v) dv =

3kT

m

.

So we find that

hEi =

1

2

mhv

2

i =

3

2

kT.

This agrees with the equipartition of energy, as it must. But now we have an

actual distribution, we can compute other quantities like hv

4

i.

Note that in these derivations, we assumed we were dealing with a monoatomic

ideal gas, but it happens that in fact this holds for a much more general family

of gases. We will not go much into details.