Divergence in spherical coordinate system

Click For Summary
Divergence in spherical and cylindrical coordinate systems is complex due to the changing basis vectors that depend on position, necessitating differentiation. The del operator's form can be derived from the Jacobian matrix and its Cartesian counterpart. Resources like Boas' "Mathematical Methods of the Physical Sciences" provide a concise overview of curvilinear coordinates and related infinitesimals. Understanding the relationships between differentials and partial derivatives is crucial for mastering these concepts. Careful exploration of these relations can lead to a deeper comprehension of divergence in non-Cartesian systems.
PrakashPhy
Messages
35
Reaction score
0
I have been studying gradience; divergence and curl. I think i understand them in cartesian coordinate system; But i don't understand how do they get such complex stuffs out of nowhere in calculating divergence in spherical and cylindrical coordinate system. Any helps; links or suggestion referring del operator in these coordinate system would be appreciable.
 
Physics news on Phys.org
The simplest answer is that in spherical coordinates, the basis vectors change based on position, so they get differentiated too. The exact form of del can be computed using the Jacobian matrix and the expression for del in Cartesian.
 
Ooh playing with rates (differentials and partial derivatives, aka covectors and vectors) is really cool in changes of coordinates.

In Boas' "Mathematical Methods of the Physical Sciences", there is a pretty quick treatment (two or three sections) of curvilinear coordinates and these infinitesimals (although I've never gotten around to figuring out how to marry Boas' notation with Jack Lee's in Introduction to Smooth Manifolds, which is harder to dig through quickly, but seems to have a more consistent development).

I conjecture that the main idea is finding the relation between dx and dθ, and ∂/∂x and ∂/∂θ. I'm not sure without trying it out this moment, but you might be able to find these relations yourself.

And the rest follows -maybe- if you're careful. ∇=(∂x,∂y). Hmm, I've got to work on some other stuff, so I have to leave it there.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K