# The average velocity of a particle

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.

• 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. – enumaris Oct 1 '18 at 19:06

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.