Divergence of an inverse square field

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The discussion centers on the divergence of an inverse square vector field, specifically in the context of Griffith's electrodynamics. It highlights that when calculating the divergence in spherical coordinates, the result is (-2)/r^3, while using Cartesian coordinates yields a divergence of 0. This discrepancy arises from the need to consider the three-dimensional nature of the field, as the increasing surface area of a sphere compensates for the diminishing magnitude of the vector field. Participants emphasize the importance of visualizing the problem in three dimensions to avoid misconceptions that can arise from simplified diagrams. Understanding the correct application of divergence formulas in different coordinate systems is crucial for accurate calculations.
Maharshi Roy
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Reference to Griffith electrodynamics question:- 1.16
Compute the divergence of an inverse square vector field.

Now gradient is (∂/∂r)(r cap)
Hence upon taking divergence of inverse square field (r cap)/r^2...We don't get 0.
In fact we get (-2)/r^3.

But if we write the vector field and the gradient both in terms of x, y & z components, and then compute gradient then it comes out to be 0.
Where is the glitch?
 
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Now gradient is (∂/∂r)(r cap)
Hence upon taking divergence of inverse square field (r cap)/r^2...We don't get 0.
In fact we get (-2)/r^3.

The divergence is calculated differently in spherical coordinates than in Cartesian coordinates:

http://hyperphysics.phy-astr.gsu.edu/hbase/diverg.html#c3

See also section 1.4.1 in Griffiths (3rd edition), specifically equation (1.71).
 
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Maharshi Roy said:
Reference to Griffith electrodynamics question:- 1.16
Compute the divergence of an inverse square vector field.

Now gradient is (∂/∂r)(r cap)
Hence upon taking divergence of inverse square field (r cap)/r^2...We don't get 0.
In fact we get (-2)/r^3.

But if we write the vector field and the gradient both in terms of x, y & z components, and then compute gradient then it comes out to be 0.
Where is the glitch?

One thing to keep in mind is it's in 3 dimensions. It's easy to accidentally think of it as just 1D and just f=1/r^2 or to just look at the 2D drawing he uses in the book in which cases it would not turn out to be 0 away from the origin. The quick drawing makes it look like you can just look at a single line with arrows and see the arrows fading off so of course if they are changing like that as you go out then you have a divergence. But, it's actually really in 3D and as you have a 1/r^2 dropping off you also have the surrounding surface area of an enveloping sphere increasing it's surface area as r^2 so it balances out and you can see that there is actually no divergence if you just remember to picture it in your mind as in 3 dimensions and to see think about how it's really working out.

It's interesting to note that changing the dimensions of the world you are thinking about can radically change what happens with the physics of that world if you accidentally forget to change the laws as you bring them over.

And it's good to keep in mind how a quick diagram can actually end up misleading one at times. I mean you quickly glance at the drawing, since the arrows shrinking in size (and thus changing in magnitude in a radial direction so of course there is divergence boom. Only not so fast. You need to really think about and visualize in your mind what is really going on in full.

(or as said above, if you just want to plug in and do math, you have to realize that it wasn't written as like f=1/x^2 but it is f=r(spherical radial basis vector)/r^2 so when you take derivatives you need to use the proper formula for spherical divergence which is different and you have to add an r^2 factor times the r component of the function before taking partial/partial r)
 
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