Is Gaussian function dimensional or non-dimensional?

Question

The Gaussian function is $$g(x,y,\sigma) = \frac{1}{2\pi \sigma^2}\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)$$, from a perspective of dimensional analysis, can I say that $x,y,\sigma$ have the same dimension $L$ for length? If so, $g$ should have dimension $L^{-2}$ because of term $\dfrac{1}{2\pi\sigma^2}$, right?

But I kinda think that $g$ should be dimensionless, because $g(x,y,\sigma)$ is just the probability which is a ratio without any dimension, right?

Could you help me this problem? Any idea is appreciated.

UPDATE

The Laplacian of Gaussian operator $$LoG(x,y,\sigma)=\frac{\partial^2g}{\partial x^2} +\frac{\partial^2g}{\partial y^2}$$, since $x,y$ have dimension $L$, $g$ has dimension $L^{-2}$, then $LoG$ should have dimension $L^{-4}$, right? In order to make $LoG$ dimensionless, should I multiply it by $\sigma^4$?

score 1 · Answer 1 · answered Sep 15 '13 at 09:30

1

$g$ is the probability density - in this case, probability per unit area. Thus the dimensions $L^{-2}$ (i.e. inverse area) makes perfect sense!

answered Sep 15 '13 at 09:30

Anthony Carapetis

35,361

upvoted for probability per unit area, I updated my post, see my update please. – avocado Sep 15 '13 at 09:42
@loganecolss: the Laplacian would indeed have dimension $L^{-4}$. – Anthony Carapetis Sep 15 '13 at 15:49

Is Gaussian function dimensional or non-dimensional?

1 Answers1