Proving the simple form of a function from statistical mechanics

Question

I have discovered a pertinent solution to my problem in the article On the Kinetic Theory of Rarefied Gases by Harold Grad and the book Thermodynamik und Statistik by Arnold Sommerfeld, both of which present the same proof for the issue I was addressing. The proof assumes the continuity of the function $f$.

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Suppose we have a function $f_0:{\mathbb R}^3\rightarrow {\mathbb R}_+$ that satisfies the following property

\begin{equation} \begin{split} &\mathbf{v}_1^2 + \mathbf{v}_2^2 = \mathbf{v}_1'^2 + \mathbf{v}_2'^2\newline &\mathbf{v}_1 + \mathbf{v}_2 = \mathbf{v}_1' + \mathbf{v}_2' \end{split} \quad \Rightarrow \quad f_0(\mathbf{v}_1)f_0(\mathbf{v}_2) = f_0(\mathbf{v}_1')f_0(\mathbf{v}_2') \end{equation}

Question: Under what conditions, such as continuity, smoothness, or even analyticity as assumed by physicists, can $\log f_0$ be written as a linear combination of $v^2$, the three components of $v$, and an arbitrary constant?

$f_0$ originates from Boltzmann’s distribution of particles in the velocity space, which specifies the equilibrium state in the absence of external forces in classical statistical mechanics. The two equalities represent the conservation laws of kinetic energy and momentum, respectively, in a collision between two perfectly elastic spheres. I encountered $f_0$ in the book Mathematical Statistical Mechanics by Colin J. Thompson.

Here is the original text from the book

Taking logarithms of both sides of Equation 6.1 we have $$ \log f_0({\bf v_1})+ \log f_0({\bf v_2}) = \log f_0({\bf v_1’}) + \log f_0({\bf v_2’}) $$ which has the form of a conservation law. Since for spinless molecules (e.g., hard spheres) the only conserved quantities are energy and momentum (and constants), it follows that must be a linear combination of $v^2$ and the three components of $v$, plus an arbitrary constant, i.e., $$ \log f_0({\bf v})= \log A-B({\bf v}-{\bf v_0})^2 $$

I cannot see a rigorous proof provided to explain it, nor are any reference bibliographies given, despite the book being titled "Mathematical Statistical Mechanics"…

Answer 1

We can indeed prove this for reasonable functions, $\log f_0\in C^2$, say.

Let me write $F=\log f_0$. By replacing $F$ by $F(v)-C-d\cdot v$, we can also assume that $F(0),\nabla F(0)=0$.

If $a,v$ are orthogonal, then, by assumption, $$ F(v)+F(a)=F(v+a)+F(0)=F(v+a) , $$ and for small $a$, we have $F(v+a)\simeq F(v) +a\cdot\nabla F(v)$, $F(a)\simeq a\cdot\nabla F(0) =0$. This shows that $a\cdot \nabla F(v)=0$ for all $a$ with $a\cdot v=0$. In other words, $\nabla F(v)$ has the same direction as $v$. It follows that $F$ is radial since for a fixed sphere, the directional derivatives tangential to the sphere are zero. We have $F(v)=g(x^2+y^2+z^2)$, if $v=(x,y,z)$.

On the other hand, we can again decompose $v=xe_1+(ye_2+ze_3)$, and our basic assumption gives $F(v)=g(x^2)+g(y^2+z^2)$. So $g$ satisfies Cauchy's functional equation $g(s+t)=g(s)+g(t)$, and this implies that $g(r^2)=cr^2$, as desired (for example, take the $s$ derivative at $s=0$ in the functional equation to conclude that $g'$ is constant).

Answer 2

The cited authors assume that there are only three linearly independent summation invariants,$^\ast$ $Q_1(\mathbf{v})=1$, $Q_2(\mathbf{v})=\mathbf{v}$, and $Q_3(\mathbf{v})=|\mathbf{v}|^2$. Then if the function $\log f(\mathbf{v})$ is a summation invariant, it must be a linear combination of $Q_1$, $Q_2$, and $Q_3$ [otherwise, $\log f$ would be a fourth independent summation invariant, counter to the assumption].

This assumption that $Q_1$, $Q_2$, and $Q_3$ are the only linearly independent summation invariants is phrased in the cited text as "the only conserved quantities are energy and momentum and constants". If the physics of the problem allows for a fourth summation invariant, for example $Q_4=|\mathbf{v}|$, then $\log f$ could also contain terms linear in $|\mathbf{v}|$.

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$^\ast$ A summation invariant is a function $F(\mathbf{v})$ which is conserved upon collision, $F(\mathbf{v}_1)+F(\mathbf{v}_2)=F(\mathbf{v}'_1)+F(\mathbf{v}'_2)$.