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My main doubt is :
"How do you visualize finding out the divergence of a vector field?"

I will first tell you what I think can be a nice way of visualizing it.
(There is obviously a mistake in my method as we get a contradiction which I will discuss below)

Step 1 : Sketch the vector field (length of the vector is proportional to its magnitude)
Step 2 : Place a particle at the tail of all the vectors visible in the sketch
Step 3 : Assume that the vectors indicate the velocity of the particles
Step 4 : Let the particles move for one second. At the end of unit time(one second), the particles will reach the arrow head of the vector
Step 5 : See whether the density has increased in the surrounding of the point, or has decreased

If the density has increased, it has negative divergence and vice versa.

1-D case

I hope you can understand my one dimensional (1/r^2) vector field sketch and my method of calculating the divergence at a point p
More particles enter the rectangular surrounding of p than leave.

2-D case

Here I have made the diagram for 2 dimensions.

As we can see from the diagrams, more particles come into the neighbourhood of p.
This suggests negative divergence at p and that is indeed the case. The math checks out.

Now, doing the same for 3 dimensional vector field (1/r^2) gives us trouble.

We know by calculating that the divergence of this field is 0, but it does not look like that according to my method of visualisation.

3-D case

The dotted circular line shows the spherical neighbourhood of the point p.
The solid ellipse depicts a circular intersection when sliced by a plane passing through origin.

We can break the entire spherical surrounding as infinite such circular discs.
Now we know the 2-D case. More points enter the disc than leave. This means more points enter the sphere than leave.
This should make the divergence at p negative, but it turns out to be 0.

SO MY QUESTION IS, "WHAT IS THE MISTAKE IN MY METHOD OF VISUALISATION? IS THERE A BETTER WAY TO VISUALISE THE DIVERGENCE AT A POINT?"

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1 Answer 1

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The flaw here is the way you extend your argument to 3D from 2D.

The thing is, a 3D object cannot be constructed out of 2D planes. They have exactly zero thickness. You need to take elements with some thickness, and then you can let that thickness tend to zero. The difference might not be apparent, but its a subtle point.

When you slice the sphere with planes pasing through the origin, any particular plane has no information of the way particles not on the plane (even those which are infinitesimally away from it) are behaving. But to calculate divergence in 3D, you need to know the behavior of particles in volume around that point. So you will have to give some thickness to the planes, however small.

A better method could be to visualize divergence using fluid flow. Make a small volume element at the point you are interested in and calculate the net amount of water coming out of that box. For more details please see the derivation of divergence on Wikipedia: Check the derivation part

P.S. - I hope you understand my point. A circular disc cannot be constructed from rings of zero thickness. You have to taken annular rings of elemental width. Also as you might have guessed, the way we extend the argument from 1D to 2D is similarly flawed. But as the sign still remains negative, we got the correct answer. :)

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