Maxwell-Boltzmann Distribution: Stationary, Collisionless Solution to the Boltzmann Equation

In our previous lesson, we undertook the significant task of deriving the Boltzmann transport equation. We saw how, through a series of physically motivated approximations—most notably the Stosszahlansatz or molecular chaos assumption—we can describe the evolution of a dilute gas using the one-particle distribution function $f (r, p, t)$ . The resulting equation,

\frac{\partial f}{\partial t} + v \cdot \nabla_{r} f + \frac{F _{ext}}{m} \cdot \nabla_{p} f = (\frac{\partial f}{\partial t})_{coll}

masterfully balances the collision-free "streaming" of particles through phase space with the scrambling effects of inter-particle collisions.

Now that we have this powerful tool, our first objective is to find its most fundamental solution: the equilibrium distribution. In this lesson, we will demonstrate that the celebrated Maxwell-Boltzmann distribution is the stationary, collisionless solution to the Boltzmann equation. This is a crucial result, as it provides a dynamical foundation for the equilibrium states that form the bedrock of statistical mechanics, justifying their structure from the perspective of particle kinetics.

1. The Condition for Equilibrium: Detailed Balance

A system is in equilibrium when its macroscopic properties no longer change with time. For our description, this means the distribution function must become stationary, $\partial f / \partial t = 0$ . The Boltzmann equation then implies that the change due to streaming must be exactly cancelled by the change due to collisions.

However, a much stronger and more intuitive condition for equilibrium exists. The collision term, which we derived as

(\frac{\partial f}{\partial t})_{coll} = \int d^{3} p_{2} \int d Ω \frac{d σ}{d Ω} ∣ v_{1} - v_{2} ∣ [f (p_{1}^{'}) f (p_{2}^{'}) - f (p_{1}) f (p_{2})]

represents the net effect of all possible collisions. For this integral to vanish, it is sufficient that the integrand itself is zero for every possible collision process. This condition is known as the principle of detailed balance.

Detailed Balance: In equilibrium, the rate of any microscopic process is exactly equal to the rate of its reverse process.

For a binary collision ${p_{1}, p_{2}} \to {p_{1}^{'}, p_{2}^{'}}$ , detailed balance requires:

f_{eq} (p_{1}) f_{eq} (p_{2}) = f_{eq} (p_{1}^{'}) f_{eq} (p_{2}^{'})

This elegant statement asserts that the probability of two particles entering a collision with momenta $p_{1}$ and $p_{2}$ is the same as the probability of two particles entering the reverse collision with the final momenta $p_{1}^{'}$ and $p_{2}^{'}$ . When this holds, the gain and loss terms in the collision integral cancel perfectly for every collision channel, and $(\partial f / \partial t)_{coll} = 0$ .

2. From Detailed Balance to the Maxwell-Boltzmann Distribution

The detailed balance condition provides the key to finding the mathematical form of the equilibrium distribution. By taking the natural logarithm of both sides, we get a more revealing relationship:

ln f_{eq} (p_{1}) + ln f_{eq} (p_{2}) = ln f_{eq} (p_{1}^{'}) + ln f_{eq} (p_{2}^{'})

This equation tells us that the quantity $ln f_{eq} (p)$ , when summed over the colliding particles, is a collisional invariant.

What quantities are conserved in a binary elastic collision? From your background in classical mechanics, you know there are a few fundamental ones:

Mass: (trivial, as particles are not created or destroyed)
Momentum: $p_{1} + p_{2} = p_{1}^{'} + p_{2}^{'}$
Kinetic Energy: $\frac{p _{1}^{2}}{2 m} + \frac{p _{2}^{2}}{2 m} = \frac{p _{1}^{'2}}{2 m} + \frac{p _{2}^{'2}}{2 m}$

For $\sum_{i} ln f_{eq} (p_{i})$ to be a conserved quantity for any collision, $ln f_{eq} (p)$ must be a linear combination of the quantities that are additively conserved for a single particle. These are the particle's momentum components ( $p_{x}, p_{y}, p_{z}$ ) and its kinetic energy ( $p^{2} /2 m$ ). Therefore, the most general form for $ln f_{eq} (p)$ is:

ln f_{eq} (p) = A - β \frac{p ^{2}}{2 m} + C \cdot p

where $A$ , $β$ , and the vector $C$ are constants that do not depend on $p$ . Exponentiating both sides gives the form of the equilibrium distribution:

f_{eq} (p) = e^{A} exp (- β \frac{p ^{2}}{2 m} + C \cdot p)

By completing the square in the exponent, we can write this in a more recognizable form. Defining a constant drift velocity $u$ such that $C = β m u$ , the distribution becomes:

f_{eq} (p) = N exp (- \frac{β}{2 m} (p - m u)^{2})

Here, $N$ is a normalization constant. For a gas at rest ( $u = 0$ ), this simplifies to:

f_{eq} (p) = N exp (- \frac{β p ^{2}}{2 m})

This is precisely the Maxwell-Boltzmann distribution. The derivation is laid out clearly in David Tong's lecture notes.

2. Kinetic Theory

This resource formalizes the argument we just made, starting from the principle of detailed balance.

Please read Section 2.2.2, "Equilibrium and Detailed Balance" (pages 23-24). Focus on how the author uses the conservation laws (Eqs. 2.18 and 2.19) to justify the general form of \ln(f_{\text{eq}}) (Eq. 2.27), leading directly to the Maxwell-Boltzmann distribution (Eq. 2.28).

The Maxwell-Boltzmann distribution for O₂ gas at 300 K. The vertical axis represents the fraction of molecules per unit speed interval, and the horizontal axis is the molecular speed. The peak of the curve corresponds to the most probable speed, $v_p$.

This result is profound. Starting from the dynamics of two-body collisions and the assumption of molecular chaos, we have shown that the only distribution for which collisions have no net effect is the Maxwell-Boltzmann distribution. Any other distribution will be driven towards this form by the collision integral.

3. The Stationary Solution

We have established that the Maxwell-Boltzmann distribution, which we'll now denote $f_{M B}$ , makes the collision integral vanish. Now we must check under what conditions it is a true stationary solution of the full Boltzmann equation, meaning $\partial f / \partial t = 0$ . This requires the streaming term to also be zero:

v \cdot \nabla_{r} f_{M B} + \frac{F _{ext}}{m} \cdot \nabla_{p} f_{M B} = 0

Let's examine two important cases.

Case 1: No External Forces ( $F_{ext} = 0$ )

In this simple case, the condition for a stationary solution reduces to:

v \cdot \nabla_{r} f_{M B} = 0

The Maxwell-Boltzmann distribution depends on macroscopic parameters like particle density $n (r)$ and temperature $T (r)$ (where $β = 1/ (k_{B} T)$ ). If these parameters vary with position $r$ , then $f_{M B}$ will have a non-zero spatial gradient, $\nabla_{r} f_{M B} \neq = 0$ . In that situation, the streaming term will be non-zero, meaning that although collisions have ceased to change the form of the distribution locally, the distribution itself is being transported from one place to another. This describes phenomena like heat conduction or diffusion, which we will study later.

For a true, global equilibrium, the streaming term must vanish. This requires $\nabla_{r} f_{M B} = 0$ , which implies that the density $n$ and temperature $T$ must be spatially uniform.

Conclusion: In the absence of external forces, the only true stationary equilibrium solution to the Boltzmann equation is a spatially homogeneous Maxwell-Boltzmann distribution.

Case 2: Conservative External Force ( $F_{ext} = - \nabla_{r} U (r)$ )

Now consider a gas in a potential, such as a gravitational field. The total energy of a particle is $H (r, p) = p^{2} / (2 m) + U (r)$ . Let's propose that the stationary equilibrium distribution is given by the full Boltzmann factor for the single-particle Hamiltonian:

f_{eq} (r, p) = N e^{- β H (r, p)} = N exp (- β (\frac{p ^{2}}{2 m} + U (r)))

Let's verify that this makes the streaming term zero, assuming the temperature is uniform ( $β$ is constant). We compute the gradients:

$\nabla_{r} f_{eq} = f_{eq} (- β \nabla_{r} U) = f_{eq} (β F_{ext})$
$\nabla_{p} f_{eq} = f_{eq} (- β \nabla_{p} (p^{2} /2 m)) = f_{eq} (- β p / m) = - f_{eq} β v$

Substituting these into the streaming term:

v \cdot (f_{eq} β F_{ext}) + \frac{F _{ext}}{m} \cdot (- f_{eq} β v) = β f_{eq} (v \cdot F_{ext}) - β f_{eq} (F_{ext} \cdot v) = 0

The streaming term vanishes identically. Since the collision term also vanishes (as the momentum dependence is still Maxwellian), this distribution is indeed the correct stationary solution. A classic example is the barometric formula for an isothermal atmosphere, where the density decreases exponentially with height: $n (z) \propto e^{- m g z / k_{B} T}$ .

The following resource confirms this finding, showing that the collisionless equation is satisfied by a general time-independent Boltzmann distribution.

Contents

This document discusses the collisionless Boltzmann equation and its solutions.

Please read the short section 8.3.2, "Collisionless Boltzmann equation" (page 5). The author first writes down the collisionless equation (8.22) and then shows that a general function of the form f(\mathbf{r} - \mathbf{v}t, \mathbf{p}) is a solution. Importantly, they explicitly state that the time-independent Boltzmann distribution (Eq. 8.25) is a possible solution when forces are absent. Our derivation above extends this to include conservative forces.

Conclusion

In this lesson, we have connected the dynamics of kinetic theory to the static picture of equilibrium. We used the Boltzmann equation not to describe evolution, but to find the unique state towards which that evolution tends.

Key Takeaways:

The condition for equilibrium where the collision integral vanishes is known as detailed balance, which states that the rates of forward and reverse collision processes are equal: $f_{eq} f_{eq,2} = f_{eq}^{'} f_{eq,2}^{'}$ .
Requiring that $ln f_{eq}$ be a linear combination of collisional invariants (quantities conserved in a collision) uniquely determines the equilibrium momentum distribution to be the Maxwell-Boltzmann distribution.
A distribution is a stationary solution of the full Boltzmann equation only when the streaming term also vanishes.
In the absence of external forces, the stationary solution is a spatially homogeneous Maxwell-Boltzmann distribution.
In the presence of a conservative external force, the stationary solution is $f_{eq} \propto exp (- β H)$ , where $H$ is the single-particle Hamiltonian, incorporating the potential energy.

We have now established the destination of a relaxing gas. But how does it get there? In the next lesson, we will prove Boltzmann's H-theorem, which demonstrates mathematically that the system's evolution under the Boltzmann equation is always directed towards this unique equilibrium state, providing a microscopic origin for the second law of thermodynamics and the arrow of time.

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