# Laplacian operator and relation to the Laplace Transform

I'm trying to understand why the Laplacian operator is used in blob detection in image analysis. I must admit that in trying to figure out why the Laplacian is useful in this application, I've really confused myself with the different uses of the word 'Laplace.' For instance, Wikipedia has many articles on this, and the ones I'm having trouble unifying conceptually are the Laplace Transform and the Laplace Operator.

From co-workers and some reading on the internet, I have come to very shallowly think of my Laplacian convolutions on images as performing something similar to the second derivative, where the most quickly changing areas on the image are what become highlighted in the new, convolved image. From the page on the Laplace Operator this makes a lot of sense. This doesn't make sense to me from the page on the Laplace Transform. My question then, I think, is how are the Laplace Operator and the Laplace Transform related? If I can see, from the definition, that the Laplace Operator is basically doing the second derivative, I would think I should be able to see something similar from the Laplace Transform. But I don't. Am I mistaken in thinking that the Laplace Transform and the Laplace operator are the same thing? How are they related?

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They are certainly not the same thing.

You might sometimes see them appear in the same context because transforms of Laplace-Fourier type are immensely useful for analyzing linear differential operators like the Laplacian. But the Fourier transform has better analytic properties, so that's the one you are more likely to see used.

Here's some intuition you might find helpful.

The discrete Laplacian computes the difference between a node's averaged neighbors and the node itself. It's often used in image processing and that gives an easy way to visualize it. The 1D case where the kernel is [1 -2 1] is especially simple:

In an area of constant color the Laplacian is zero. Indeed, even if you have linear variation it remains zero, e.g. in the neighborhood [1 2 3] the Laplacian's value at the center point is

$$1 \cdot 1 + (-2) \cdot 2 + 3 \cdot 1 = 0.$$

But quadratic and higher-order variation excites the Laplacian and results in non-zero values. Thus it's especially useful for detecting 'jumps'. That's why it's the weapon of choice in edge detection. It's often combined with a Gaussian to pre-filter out any small-scale features or noise that might cause spurious edges to be detected.

I should mention that the Laplacian in two dimensions and higher is significantly richer than the one-dimensional case might suggest. For one, not all two-dimensional images with a uniformly zero Laplacian are linear. But qualitatively a lot of the same intuition holds true as to how the Laplacian reacts to variation.

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so when you do a Laplacion convolution, you're not actually doing a Laplacian Transform of the image (similar to how you do a Fourier transform) but instead are convolution with the Laplacian operator? So convolution with the laplacian operator is different then applying a Laplacian Transformation to the image? – Nick Aug 2 '10 at 12:06
Strictly speaking, the Laplacian is only a convolution operator in the discrete case. But yes, it is absolutely not the same thing as the Laplace transform (which is never called the Laplacian transform, by the way). – Per Vognsen Aug 2 '10 at 12:10
ah, thank you very much. – Nick Aug 2 '10 at 12:15

There is no relation on the basic level between the Laplace operator and Laplace transform. From the point of view of learning about them, put the coincidence of names out of your mind.

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so the operator is not actually derived from the equation, and the thing they share in common is that they were discovered/used by Laplace? – Nick Aug 2 '10 at 12:03
What they certainly share in common is that both are named after Laplace. Whether he discovered them is a tougher question. – Michael Hardy Aug 2 '10 at 21:20