1
$\begingroup$

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.

$\endgroup$
  • 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 Oct 1 '18 at 19:06
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.