The Maxwell distribution of velocities is:

$$p (v) = (\frac{m}{2\pi K_b T})^{\frac{3}{2}} e^{\frac{-mv^2}{2 k_b T}}$$

I want to understand how to obtain the average value of the velocity. The distribution function has already been normalised, then I just have to solve for the average of the velocity:

$$< v > = \beta m \int_0^{\infty} v^2 e^{-\frac{\beta m v^2}{2}} dv$$

What I do not get is the following equality:

$$< v > = \beta m \int_0^{\infty} v^2 e^{-\frac{\beta m v^2}{2}} dv = -2\beta \frac{\partial}{\partial \beta} \int_0^{\infty} e^{-\frac{\beta m v^2}{2}} dv$$

Why $v^2$ is not in the integral anymore and the operator $\frac{\partial}{\partial \beta}$ shows up?

I have been trying to find the reason based on the definition of kinetic energy but got nothing.

  • 1
    $\begingroup$ Just evaluate the derivative on the right hand side of the equation and see that the factor of $v^2$ comes out. It's just a rewriting of the equation to make things easier. $\endgroup$
    – enumaris
    Commented Oct 1, 2018 at 19:06

1 Answer 1


First, note that $$\frac{\partial}{\partial \beta} e^{- \beta mv^2 / 2} = - \frac{mv^2}{2} e^{-\beta mv^2/2}.$$ Thus, substituting this fact into $\beta m \int_0^\infty v^2 e^{-\beta mv^2/2} dv$ gives the right hand side of the equation you wrote in the original post.


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