B.1 Rant on units management

(September 1, 2023)

Prof. T and most professional physicists care a lot about units. If you have a dimensionful integral you can’t do, that is bad. If you can turn the integral to something with overall units times a dimensionless integral (which is a number like $\sqrt{2}$ ) that isn’t so bad.

Suppose, for example, the integral integral you are trying to compute is an integral over position:

I=\int_{0}^{\infty}{\rm d}x\,x^{4}e^{-x^{2}/\ell^{2}}

(B.1)

where $\ell$ has units of length. Then $I\propto\ell^{5}$ times a dimensionless number, which turns out to be $0.66467$ . You should be able to show the $\ell^{5}$ without doing any integrals, by simply switching the integration variable from the dimensionful variable $x$ to a dimensionless variable $u=x/\ell$ (the position in units of $\ell$ ). Here are the steps

$\displaystyle I=$	$\displaystyle\int_{0}^{\infty}dxx^{4}\exp(-x^{2}/\ell^{2})$	(B.2)
$\displaystyle=$	$\displaystyle\ell^{5}\int_{0}^{\infty}\frac{dx}{\ell}\frac{x^{4}}{\ell^{4}}%\exp(-x^{2}/\ell^{2})$	(B.3)
$\displaystyle=$	$\displaystyle\ell^{5}\times\int_{0}^{\infty}duu^{4}\exp(-u^{2})$	(B.4)
$\displaystyle=$	$\displaystyle\ell^{5}c$	(B.5)

where $c$ is an order one constant. I think that we can agree that

I=c\ell^{5}

(B.6)

shows a great deal more insight than Eq. (B.1).

The fact that the proportionality constant is $c=\Gamma(5/2)/2=3\sqrt{\pi}/8\simeq 0.66467$ doesn’t seem so important¹¹ 1 This value of $c$ follows by a change of variables, defining $y=u^{2}$ in Eq. (B.4)., and I would be happy with $I=c\ell^{5}$ as a result. Finding $c$ requires doing a dimensionless integral, which is the only kind of integral you should ever try to do!