Trouble/Question with/about Maxwell Equations

Click For Summary
The discussion revolves around solving Maxwell's equations for electric fields in spherically symmetric cases, particularly for a solid sphere with uniform charge distribution. The user derived the electric field using Gauss's Law but encountered confusion when applying the divergence operator in spherical coordinates with Mathematica. Key points of confusion included the treatment of the electric field vector components and the proper formulation of Gauss's Law. Participants clarified that the electric field should only have a radial component in this case, while the angular components should be zero, emphasizing the importance of correctly specifying the coordinate system. The conversation highlights the nuances of vector fields in spherical coordinates and the correct application of mathematical principles in physics.
  • #31
If E is not a function of theta or phi, we can write: E(r) = <E(r)_r, 0, 0>
This is not always true for a general vector function A. If you have a vector function A of r then its possible to have a A(r)phi not equal to zero.

Pete
 
Physics news on Phys.org
  • #32
Andrew Mason said:
Ques: What is the field at a point \vec r = (r, \theta_r, \phi_r)?

Ans: The unit vector at \vec r is \hat r = (1, \theta_r, \phi_r).

AM

This is incorrect, the radial unit vector is always (1, 0, 0) (By definition). The radial unit vector is not constant (like the Cartesian unit vectors) but depends on position.

Claude.
 
  • #33
pmb_phy said:
No. The notation E = <E_r, E_theta, E_phi> quite literally means that E = E_r e_r + E_theta e_theta + E_phi e_phi (I'm being sloppy at the moment). Since E = E_r e_r it follows that E_phi = 0 and E_theta e_theta = 0. I'm afraid that you also got e_r (what you call "r hat") incorrect. If e_r is a unit vector pointing in the direction of increasing r then the way you expressed it is incorrect. You've already written it on the left side so there is no other expression, i.e. the right side is incorrect. Notice that on the left you have e_r and on the right you have e_r + (other stuff) which means (other stuff) = 0. (other stuff) does not have the value you think it does. You seem to be confusing the expansion of e_r in terms of spherical unit vectors with e_r expressed in terms of cartesian unit vectors. Also e_r does not equal (1, theta_r, phi_r).

Pete


What Pete (and Claude) is saying makes sense to me. That's why I was questioning AM about what he said.

P.S. Hurkyl...I'm afraid I am too dense to understand your point, despite the fact you've said it three times. Sorry...
 
Last edited:
  • #34
cepheid said:
What Pete (and Claude) is saying makes sense to me. That's why I was questioning AM about what he said.

P.S. Hurkyl...I'm afraid I am too dense to understand your point, despite the fact you've said it three times. Sorry...
I am afraid that I have been adding confusion rather than clarification or help.

In reviewing vectors in spherical co-ordinates, I see that the convention is to define a unique orthogonal coordinate system at each point, consisting of the basis vectors: \hat r (perpendicular to surface of sphere of radius r), \hat \theta (longitudinal unit vector) and \hat \phi (perpendicular to the other two - colatitudinal). I was defining the vector in terms of the original spherical co-ordinates.

So the vector for the field would be \vec E = (|E|, 0 , 0) in terms of these basis vectors. My mistake was in thinking of E in spherical coordinates rather than in terms of these orthogonal basis vectors. I can see why it is done this way because one cannot do vector addition or other operations on a vector in spherical coordinates.

Again, sorry for adding confusion. I have learned something and perhaps others have as well.

AM
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
506
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K