MULTIPLICATIVE PROPERTY OF THE EXPONENTIAL FUNCTION

An article in the September issue of this journal¹ presents a derivation of the Boltzmann distribution function (BDF) for introductory students. However, a key proof in the derivation is left to a graduate text in mathematics. A simpler proof is presented here. It is based on the idea of separation of variables for partial differential equations, but is here applied to a basic example of a function of two variables. The BDF is presented in our curriculum in the third course in the introductory sequence for physics majors (after the first two courses in mechanics and in electricity and magnetism), while separation of variables is introduced in the fourth and final course (to solve Schrödinger's equation for the hydrogen atom).² It therefore does not seem too much of a stretch to make use of the technique in the derivation of the BDF. In fact, it could serve as a gentle preparation for follow-on courses where separation of variables is applied to more difficult problems.

The goal of this letter is to prove that the only nonzero real and differentiable function that satisfies

for the two independent variables x and y is the exponential function $f(u)=exp(cu)$ where c is an arbitrary constant. To do so, first take the partial derivative of both sides of Eq. (1) with respect to x to get

Next define $z≡x+y$ so that

and noting that $∂f(x)/∂x=df(x)/dx$ and $∂z/∂x=1$⁠, Eq. (2) can be rewritten as

Repeating these steps for the derivative of Eq. (1) with respect to y, one obtains

Since the right-hand sides of Eqs. (4) and (5) are equal, the left-hand sides must also be equal, which leads to

However, the left-hand side of this equation is only a function of x, while the right-hand side is only a function of the independent variable y. The only way this can be true is if each side is a constant, call it c, so that

which integrates to

(For the special case of $c=1$⁠, this result follows from the fact that the exponential is defined as that continuous function equal everywhere to its derivative.³) Substituting Eq. (8) back into Eq. (1) shows that $A2=A$ whose only nonzero solution is $A=1$⁠, thereby completing the proof.