Kinetic considerations

1. Goal of this section

They want to show two things:

Pressure from microscopic motion
Pressure of a gas can be written in terms of how fast the molecules move and how much momentum they carry.
Effusion rate
How fast molecules escape through a tiny hole in a container (effusion) can also be written in a very simple form.

And they want formulas that work no matter what statistics the particles obey (classical, Bose-Einstein, Fermi-Dirac).

2. A general formula for the pressure

They first derive pressure using statistical mechanics (partition function, etc.) and then rewrite it into a kinetic-theory form.

After some integration gymnastics (the long integrals at the top of page 153), they arrive at:

\[ P = \frac{1}{3} \frac{N}{V} \left\langle p\frac{d\varepsilon}{dp}\right\rangle \]

which they also write as

\[ P = \frac{1}{3} n \langle p u \rangle \tag{3} \]

where

\(N\) = total number of particles
\(V\) = volume
\(n = N/V\) = number density (particles per unit volume)
\(p\) = magnitude of the momentum of a particle
\(\varepsilon(p)\) = energy of a particle with momentum \(p\)
\(u = d\varepsilon/dp\) = speed of the particle
\(\langle \dots \rangle\) = average over all particles

Interpretation:

Pressure is one-third of (number of particles per volume) × (average of "momentum × speed").

The factor 1/3 comes from the fact that the gas is the same in all directions, so on average only one-third of the motion is along any particular axis (say the z-axis, perpendicular to the wall).

Special cases (eq. 4)

If the energy is a power of momentum:

\[ \varepsilon \propto p^s \]

then you can show

\[ P = \frac{s}{3} n \langle \varepsilon \rangle = \frac{s}{3}\frac{E}{V} \tag{4} \]

where \(E\) is total energy.

Non-relativistic: \(\varepsilon = p^2/2m\) → \(s = 2\)
⇒ \(P = \frac{2}{3} \frac{E}{V}\) (usual ideal gas result).
Ultra-relativistic: \(\varepsilon = pc\) → \(s = 1\)
⇒ \(P = \frac{1}{3} \frac{E}{V}\) (like photon gas / radiation).

So these fancy integrals just lead to very simple relations between pressure and energy.

3. Re-deriving \(P = \frac{1}{3}n\langle p u\rangle\) from collisions with a wall

Now they say: "Let's forget the partition function and derive pressure just from particles bouncing off the wall." That's the picture on page 154.

Step-by-step picture

Take a small patch of wall with area \(dA\), perpendicular to the z-axis.
Look at a short time \(dt\).
Which molecules will hit that patch during this time?
A molecule with z-component of velocity \(u_z\) will travel distance \(u_z dt\) towards the wall.
So the molecules that can hit the patch lie inside a skinny cylinder of:
- base: \(dA\)
- height: \(u_z dt\)
Number of molecules in that cylinder:

They define a speed distribution \(f(u)\):

\(n f(u)du\) = number of molecules per unit volume with speeds between \(u\) and \(u+du\).
Normalized so that \(\int f(u)du = 1\).

Accounting for directions and the velocity component \(u_z\), they argue that the number hitting is proportional to

\[ dA \cdot dt \cdot u_z \cdot n \cdot f(u) \cdot du \]

Momentum change per collision:
A molecule hits the wall with momentum component \(p_z\) and bounces back with \(-p_z\).
Change in momentum for that molecule: \(2 p_z\).
Total momentum transferred per unit area per unit time (i.e. pressure):
Multiply "number hitting" × "momentum change per particle".
Divide by \(dA \cdot dt\).
Integrate over all velocities pointing towards the wall.

This gives eq. (6):

\[ P = 2n \int p_z u_z f(u) \, d^3u \tag{6} \]

After some symmetry arguments (isotropy and evenness in \(u_z\)), this simplifies to

\[ P = n \int (p_z u_z) f(u) \, d^3u \tag{7} \]

Use isotropy (all directions equally likely):

Because the gas has no preferred direction, the average of \(p_z u_z\) is one-third of the average of \(p u\):

\[ \langle p_z u_z \rangle = \frac{1}{3}\langle p u\rangle \]

So

\[ P = \frac{1}{3} n \langle p u\rangle \tag{9} \]

Exactly the same result as the earlier "big integral" derivation. This shows the formula is purely kinetic and doesn't care if the gas is classical, Bose, or Fermi.

4. Effusion through a small hole

Now they apply the same idea to molecules escaping through a hole instead of hitting and bouncing off a wall (page 155).

What is effusion?

Effusion rate \(R\) = number of particles per unit area per unit time passing through a tiny hole.

Again:

Take a patch of wall with a small hole, area = 1 (for convenience).
Only molecules with positive \(u_z\) (moving towards the opening) can go through.
In time \(dt\), molecules in a cylinder of base area 1 and height \(u_z dt\) will make it through. So:

\[ R = n \int u_z f(u) \, d^3u \tag{10} \]

(Almost the same as before, but no factor 2 because they don't bounce back.)

Switch to spherical coordinates in velocity space and integrate over angles where \(u_z > 0\) (i.e., over half the sphere). This yields

\[ R = n\pi \int_0^\infty f(u) u^3 du \tag{12} \]

Use the normalization condition:

\[ \int_0^\infty f(u)(4\pi u^2)du = 1 \tag{5a} \]

From this, they massage the integral to show

\[ R = \frac{1}{4} n \langle u \rangle \tag{13} \]

where \(\langle u\rangle\) is the average speed of the particles.

Interpretation:

The flux of molecules through a small hole is just

\[ R = \frac{1}{4} \times \text{(number density)} \times \text{(average speed)}. \]

Again, the factor \(1/4\) comes from directional averaging: only half the molecules move towards the hole, and their average \(\cos\theta\) factor gives another 1/2.

5. Big picture summary

Pressure is the rate of momentum transfer to the wall per unit area.
That gives \(P = \frac{1}{3}n\langle p u\rangle\), or for usual gases \(P = \frac{2}{3}E/V\).
Effusion rate is the number of molecules crossing a unit area per unit time.
That gives \(R = \frac{1}{4}n\langle u\rangle\).
Both results are very general, not tied to Maxwell-Boltzmann specifically; they rely mainly on:
isotropy (no preferred direction),
definition of pressure / flux,
and the velocity distribution normalization.