Divergence of curl in spherical coordinates

Click For Summary
SUMMARY

The divergence of the curl of any vector field \( f \) is zero, expressed mathematically as \( \nabla \cdot (\nabla \times F) = 0 \). This holds true in spherical coordinates, provided that the vector field \( F \) is well-behaved. A common error arises from neglecting that basis vectors in spherical coordinates are functions of position, unlike in Cartesian coordinates. Proper understanding of these conditions ensures the validity of the divergence-curl relationship in spherical systems.

PREREQUISITES
  • Understanding of vector calculus concepts, specifically divergence and curl.
  • Familiarity with spherical coordinate systems and their basis vectors.
  • Knowledge of well-behaved vector fields in mathematical physics.
  • Proficiency in mathematical notation and operations involving gradients, divergences, and curls.
NEXT STEPS
  • Study the properties of vector fields in spherical coordinates.
  • Learn about the implications of basis vector dependency in non-Cartesian systems.
  • Explore examples of well-behaved vector fields and their applications in physics.
  • Review advanced vector calculus topics, including theorems related to divergence and curl.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with vector fields, particularly in spherical coordinates. This discussion is beneficial for those seeking to deepen their understanding of vector calculus and its applications in various scientific fields.

member 428835
Hey pf!

I was thinking about how div(curl(f)) = 0 for any vector field f. However, is this true for div and curl in spherical coordinates? It doesn't seem to be.

If not, what needs to happen for this to be true in spherical coordinates??

Thanks all!
 
Physics news on Phys.org
div and curl do not depend on coordinates so the result holds for all coordinates including spherical

Keep in mind you omitted some conditions, f must be well behaved for that to be true
 
joshmccraney said:
Hey pf!

I was thinking about how div(curl(f)) = 0 for any vector field f. However, is this true for div and curl in spherical coordinates? It doesn't seem to be.

If not, what needs to happen for this to be true in spherical coordinates??

Thanks all!

If you don't get \nabla \cdot (\nabla \times F) = 0 for well-behaved F in spherical coordinates then you are making an error in your calculations, such as forgetting that the basis vectors are functions of position and not constant as in the Cartesian case.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
6
Views
2K