Suppose, a 2D image is represented by g(x,y).

Then, what does it mean by "Gradient of g(x,y)", "Divergence of g(x,y)" and "Curl of g(x,y)"?

Please, avoid formal math for explanations.

Please, use layman's terms and informal discussions.

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    $\begingroup$ Google is your friend: gradient, etc. $\endgroup$ – Matt L. Mar 19 '16 at 11:44
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    $\begingroup$ The deleted answer was a perfect answer to your question. If you want to know something else, you should formulate your question in a way that people here can actually understand what it is that you need to know. Nobody wants to waste time on answering a badly formulated question. $\endgroup$ – Matt L. Mar 19 '16 at 12:43
  • $\begingroup$ @MattL., I updated my question. $\endgroup$ – user18425 Mar 19 '16 at 12:49

Original Image

enter image description here

Image Gradient

The gradient of a scalar function $f(x_1, x_2, x_3, \dots, x_n)$ is denoted by $\vec{\nabla}$ f where $\nabla$ denotes the vector differential operator, del.

$$ \nabla f = \frac{\partial f}{\partial x_1 }\mathbf{n}_1 + \cdots + \frac{\partial f}{\partial x_n }\mathbf{n}_n$$

  • $\mathbf{n}_i$ are the orthogonal unit vectors pointing in the coordinate directions.

An image gradient is a directional change in the intensity or color in an image. Image gradients may be used to extract information from images. This aproach have a lot of application such as edge detection, image enhancement. As image not have a defined function, then the image processing field developed "operators" that aproximate the image gradient performing convolution with a kernel as: sobel, prewitt, kirsch and more.

In color image, the image gradient is a three directional vector of the derivatives.

In layman's terms:

The gradient shows how quickly the color (or greyscale) changes from pixel to pixel. In very "busy" images, the gradient will change much. In areas of less color change, the gradient will be close to zero.

In the image above, the petals don't change color very much so the gradient is close to zero (or small). At the edges of the petals where the color changes from purple to green, the gradient is large.

enter image description here

Image Divergent

Being $ \mathbf{F} = M\hat{\mathbf{i}} + N\hat{\mathbf{j}} + O\hat{\mathbf{k}} $ is an vector field in $\mathbb{R}^3$ , then the divergent of $\mathbf{F}$, denoted by $\operatorname{div}\mathbf{F}$, is the scalar field given by the dot product of the $\nabla$ and $\mathbf{F}$.

$$ \operatorname{div} \mathbf{F} = \nabla \cdot \mathbf{F} $$

In the sense of image processing, it is the inverse of gradient, this is a indicate of minor gray scale variation.

In this aproach i used the channel Red and Green as vector field

enter image description here

Image Curl

Being $ \mathbf{F} = M\hat{\mathbf{i}} + N\hat{\mathbf{j}} + O\hat{\mathbf{k}} $ is an vector field in $\mathbb{R}^3$ , then the divergent of $\mathbf{F}$, denoted by $\operatorname{curl}\mathbf{F}$, is the scalar field given by the vector product of the $\nabla$ and $\mathbf{F}$.

$$ \operatorname{curl} \mathbf{F} = \nabla \times \mathbf{F} $$

Its a vector operator that describes the infinitesimal rotation of a 3-dimensional vector field. At every point in the field, the curl of that point is represented by a vector.

In the sense of image, it is like a circular motion of variation of pixels

In layman's terms:

The curl operation measures the tendency of rotation around the point itself.

enter image description here

  • $\begingroup$ Bro, no offence, but, your formal texts are already available in DIP books. I am here to have some informal knowledge. $\endgroup$ – user18425 Mar 19 '16 at 12:26
  • $\begingroup$ Wait im editing still, this concept need much abstraction. $\endgroup$ – Darleison Rodrigues Mar 19 '16 at 13:06
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    $\begingroup$ Great answer! If you could elaborate and add intuition, it will be perfect. Especially the practical meaning of each symbol. $\endgroup$ – Royi Apr 15 '16 at 8:27

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