Sph Harmonics Homework: Find Potential for r>a, Contribution Using Superposition

Click For Summary
SUMMARY

The discussion focuses on determining the potential for a sphere of radius a, where the potential V is zero except in the region defined by 0 < θ < π/2 and 0 < φ < π. The participants analyze the contributions of spherical harmonics for values of m and discuss the implications of symmetry in the problem. Key insights include the importance of recognizing that the potential is independent of φ and the utility of superposition to simplify the analysis of contributions, particularly for l = 2.

PREREQUISITES
  • Spherical harmonics and their properties
  • Understanding of potential theory in physics
  • Knowledge of superposition principle in electrostatics
  • Basic integration techniques in spherical coordinates
NEXT STEPS
  • Study the properties of spherical harmonics, particularly Y_lm functions
  • Learn about the superposition principle in electrostatics and its applications
  • Explore the implications of symmetry in potential problems
  • Investigate the derivation of potential expressions in spherical coordinates
USEFUL FOR

Students and professionals in physics, particularly those studying electrostatics, potential theory, and mathematical methods in physics. This discussion is beneficial for anyone looking to deepen their understanding of spherical harmonics and their applications in potential problems.

Faizan Samad
Messages
12
Reaction score
1

Homework Statement


A sphere of radius a has V = 0 everywhere except between 0 < θ < π/2 and 0 < φ < π. Write an expression in spherical harmonics for the potential for r > a. For which values of m are there contributions? Determine the contributions through l= 2. How could you determine the leading contribution using just superposition?

Homework Equations


Again I will be uploading images. I am only looking for answer in terms of a general formula. Not a fully drawn out calculation.
Screen Shot 2018-09-17 at 12.24.04 AM.png


The Attempt at a Solution


I will be uploading a solution. Again the solution they are looking for is very general. What I am confused about is that this problem implies that for certain values of m there are no contributions but that does not seem to be the case with these bounds of integration. Also I do not understand the superposition part of this question.
IMG_2450.jpg
 

Attachments

  • IMG_2450.jpg
    IMG_2450.jpg
    22.4 KB · Views: 605
  • Screen Shot 2018-09-17 at 12.24.04 AM.png
    Screen Shot 2018-09-17 at 12.24.04 AM.png
    21.4 KB · Views: 1,059
Physics news on Phys.org
Look at the symmetry of the problem and compare it with the symmetry of the spherical harmonics.
 
bobob said:
Look at the symmetry of the problem and compare it with the symmetry of the spherical harmonics.
I’m integrating all the phi terms they don’t seem to go to 0.
 
Faizan Samad said:
I’m integrating all the phi terms they don’t seem to go to 0.

That's why you don't just throw math a problem that involves physics. It appears your potential independent of \phi, correct? If so,which m values should you not get zero answers for? Second, you were asked how you could guess a term or terms by superposition. Look at the symmetry of the Y_lm you can have after figuring out which ones those are (that you got by symmetry). You have the potential on a sphere which is V on the upper hemisphere and 0 on the lower hemisphere. It might help to shift the potential by -V/2 so that the symmetry is more obvious. Then you have a constant V/2 on the entire sphere and term and and a tem with +V/2 on the top hemisphere and -V/2 on the bottom hemisphere. When you superpose them you get V on the top half and zero on the bottom half. First thing you should always do in any problem is shift the origin or shift something to make as much symmetry explicit as possible. Then when you are done, you can shift back.
 
bobob said:
It appears your potential independent of \phi, correct?
That's not correct. The potential is non-zero for only ##0 < \phi < \pi##.
 
vela said:
That's not correct. The potential is non-zero for only ##0 < \phi < \pi##.
My mistake. Sorry about misreading the phi boundary condition. In that case, you should still look for symmetry, but adding -V/2 to everything will not be useful.
 
  • Like
Likes   Reactions: Delta2

Similar threads

Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
8K
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K