Laplacian of Gaussian formula for 2d case is $$\operatorname{LoG}(x,y) = \frac{1}{\pi\sigma^4}\left(\frac{x^2+y^2}{2\sigma^2} - 1\right)e^{-\frac{x^2+y^2}{2\sigma^2}},$$ in scale-space related processing of digital images, to make the Laplacian of Gaussian operator invariant to scales, it is always said to normalize $LoG$ by multiplying $\sigma^2$, that is $$\operatorname{LoG}_\text{normalized}(x,y) = \sigma^2\cdot \operatorname{LoG}(x,y) = \frac{1}{\pi\sigma^2}\left(\frac{x^2+y^2}{2\sigma^2} - 1\right)e^{-\frac{x^2+y^2}{2\sigma^2}}.$$ I wonder why multiply by $\sigma^2$ not $\sigma^4$ or anything else?


Thanks to comments from @achille. From the perspective of dimensional analysis, in the Laplacian of Gaussian operator $$LoG(x,y,\sigma)=\frac{\partial^2g}{\partial x^2} +\frac{\partial^2g}{\partial y^2}$$, I think $x,y$ are variables with dimension $L$, $\sigma$ is a parameter with dimension $L$. But what about $g$? Since $g$ is a function of $x,y,\sigma$, $$g(x,y,\sigma)=\frac{1}{2\pi \sigma^2}exp(-\frac{x^2+y^2}{2\sigma^2})$$, and $x,y,\sigma$ are of the same dimension $L$, so I guess in $g$, the term $exp(-\frac{x^2+y^2}{2\sigma^2})$ is dimensionless, isn't it? And the term $\frac{1}{2\pi \sigma^2}$ is of dimension $L^{-2}$, right? So $g$ is actually of dimension $L^{-2}$, isn't it?

Now come back to $LoG$, it should have dimension $L^{-4}$?


Laplacian operator is $$\nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$, and a Gaussian function's scale is $\sigma$, right? If I apply $\nabla^2$ on a Gaussian function $g(x,y,\sigma)$, what is difference of applying the dimensionless $\sigma^2*\nabla^2$?

  • $\begingroup$ Laplacian = $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ has dimension $Length^{-2}$ and $\sigma^2$ has the dimension $Length^2$. This is what one need to bring the "Laplacian operator" dimensionless. $\endgroup$ – achille hui Sep 9 '13 at 13:25
  • $\begingroup$ @achillehui, sorry I don't understand what do you mean by Laplacian has dimension $Length^{-2}$? $\endgroup$ – avocado Sep 10 '13 at 5:05
  • 1
    $\begingroup$ Notice Laplacian are linear combination of second derivatives. If you apply it to a function of typical length scale $L$ (e.g. a spherical symmetric bump peaked at origin and falls off to 1/2 its peak value at radius $L$ ), then the Laplacian of that function will be proportional to $\frac{1}{L^2}$. In general, every time you take derivative of a spatial coordinate, you changed the dimension by a factor $Length^{-1}$. This is part of the so called Dimensional analysis which help you to cross check the validity of any equation. $\endgroup$ – achille hui Sep 10 '13 at 5:20

First, let me try to give you some intuition of why you have to normalize by scale at all. As you go from finer to coarser scales you blur the image. That makes the intensity surface more and more smooth. That, in turn, means that the amplitude of image derivatives gets smaller as you go up the scale volume. This is a problem for finding interest points, because you are looking for local extrema over scale. Without normalization you will always get the maximum at the finest scale and the minimum at the coarsest scale, and that's not what you want.

So, image derivatives are attenuated as $\sigma$ increases. In fact, the derivatives decrease exponentially as a function of $\sigma$. To compensate for that you have to normalize them by multiplying the $n$-th derivative by $\sigma^n$. Since the LoG is a combination of second derivatives, you have to multiply it by $\sigma^2$.

You can find the derivation and a better explanation of this in this paper by Toni Lindeberg.

  • $\begingroup$ Thanks, and I understand the intuition of why it's necessary to normalize by scale. I'll read the paper later, one more thing, the scale here is just the $\sigma$ of a Gaussian function used to convolve the image, but my intuitive understanding of scale is that, scaling a image is resizing the image, so how could the convolution by a Gaussian make up scale? $\endgroup$ – avocado Sep 16 '13 at 14:19
  • $\begingroup$ @loganecolss, The Gaussian and resizing the image are very much related. Before reducing the size of the image you need to blur it with the Gaussian of the appropriate $\sigma$ to remove the high frequencies to avoid strange aliasing artifacts. $\endgroup$ – Dima Sep 16 '13 at 14:23
  • $\begingroup$ @loganecolss, Another way to look at this: blurring removes fine-grained features from the image. This is equivalent to zooming out or moving the camera away from the scene. $\endgroup$ – Dima Sep 16 '13 at 14:27
  • $\begingroup$ If the image1 is first blurred by the Gaussian and then downsampled to another image2 half-size of the original, then can I say image1 is rescaled to image2? The $\sigma$ should be 2, and it's the scale of image2? $\endgroup$ – avocado Sep 16 '13 at 14:27
  • $\begingroup$ @loganecolss. That's right. But even if you don't downsample, you still have the notion of scale. If you blur the image with a Gaussian of $\sigma = \sqrt{2}$, you remove any features that are smaller than $\sqrt{2}$. So your scale is $\sqrt{2}$. $\endgroup$ – Dima Sep 16 '13 at 14:29

EDIT As far as I can (now) tell, your problem is unrelated to dimensional analysis, so yesterday's input (below) is obsolete. Now, your LoG function is the Laplacian of $$ G(x,y;\sigma) = \frac{1}{2\pi\sigma^2} {\rm e}^{-\frac{x^2+y^2}{2\sigma^2}} , $$ whence $$ {\rm LoG}(x,y;\sigma) = \Delta_{(x,y)}G(x,y;\sigma) = \frac{\partial^2 G(x,y;\sigma)}{\partial x^2} + \frac{\partial^2 G(x,y;\sigma)}{\partial y^2} = \frac{1}{\pi\sigma^4} \left(\frac{x^2+y^2}{2\sigma^2}-1\right){\rm e}^{-\frac{x^2+y^2}{2\sigma^2}} . $$ Multiplying the latter by $\sigma^2$ is equivalent to multiplying the former by the same number, so in fact you're working with $$ G_{\rm norm}(x,y;\sigma) = \frac{1}{2\pi} {\rm e}^{-\frac{x^2+y^2}{2\sigma^2}} $$ and $$ {\rm LoG}_{\rm norm}(x,y;\sigma) = \Delta_{(x,y)}G_{\rm norm}(x,y;\sigma) = \frac{1}{\pi\sigma^2} \left(\frac{x^2+y^2}{2\sigma^2}-1\right){\rm e}^{-\frac{x^2+y^2}{2\sigma^2}} . $$ Note that both $G$ and $G_{\rm norm}$ are rescalings of the standard (i.e., unit $L^1-$norm) Gaussian. Knowing nothing about image processing, the one interesting thing that strikes me about ${\rm LoG}_{\rm norm}$ is the following identity: $$ \Delta_{(x,y)}G_{\rm norm}(\sigma x,\sigma y;\sigma) = \Delta_{(x,y)}G_{\rm norm}(x,y;1) . $$ The non-normalized version unsurprisingly has a factor of $\sigma^2$ in front of the LHS. Make what you want of this i.t.o. image processing, I'm afraid I can't help without a clear mathematical objective.

OBSOLETE There's a fundamental misunderstanding here related to what you write: 'I guess in g, the term ${\rm exp}\left(−\frac{x^2+y^2}{2σ^2}\right)$ is dimensionless [...]' No, this does not need be the case: the unit in front of the exponential - the one ($1$) multiplying it - is the one carrying the units of $g$.

To close with a simple example, recall that a body traveling with 'constant velocity equal to one' travels a distance $s(t)=t$ after time $t$. Does that mean that distance $s$ has the units of time $t$? Of course not; it means that $s(t)=vt$, with the velocity $v$ here being equal to one in the chosen system of units. Change units and $v$ will become something else: for example, if $v=1{\rm m/sec}$ and $t$ is measured in seconds, then $s(t)=t$ as long as $s$ is measured in meters: after $t$ seconds, the body has traveled $t$ meters. Change from meters to millimeters, so that $v=1000{\rm mm/min}$ and $s(t)=1000t$: after $t$ seconds, the body has traveled $1000t$ millimeters. To students learning the concept of dimensionality, it tends to be clearer that $1000$ carries the units (${\rm mm/sec}$) in the latter problem than that the unit does the same in the former one - that's because they forget that the unit is even there.

  • $\begingroup$ I don't understand why $g_{xx} + g_{yy} = 0$? And how to get the characteristic quantity $[g]$? $\endgroup$ – avocado Sep 16 '13 at 10:21
  • $\begingroup$ Oh man, did I screw up - I thought this was a 2-D kernel for Laplace's eq., ouch... I edited the post above by removing the paragraph on Laplace's eq.; the rest still holds. $\endgroup$ – automaton 3 Sep 16 '13 at 11:02
  • $\begingroup$ I understand your simple example, but I still don't understand how the Laplacian operator is normalized? $\endgroup$ – avocado Sep 16 '13 at 11:29
  • $\begingroup$ I'll get back to you later today/tomorrow - the key is the LoG's behavior under scaling of the independent variables, I'm working it out. $\endgroup$ – automaton 3 Sep 16 '13 at 14:10
  • $\begingroup$ Thank you so much, I'll be right here waiting ;-) $\endgroup$ – avocado Sep 16 '13 at 14:11

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