Strange Laplacian of 1/|x-x’| ?

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Discussion Overview

The discussion revolves around the application of Gauss' theorem in evaluating the Laplacian of the function \( \frac{1}{|x-x'|} \). Participants explore the reasoning behind using a spherical surface for integration, particularly in the context of physics, and question whether other surfaces could be used instead.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of using a spherical surface around the origin for integration, seeking a general explanation.
  • Another participant asserts that while Gauss' theorem applies to any closed surface, a spherical surface is chosen for its symmetry, which simplifies the integration process.
  • A third participant reiterates the importance of choosing surfaces that align with the symmetry of the problem at hand, suggesting that other surfaces may be appropriate in different contexts.

Areas of Agreement / Disagreement

Participants generally agree on the validity of Gauss' theorem for any closed surface, but there is a lack of consensus on the necessity of using a spherical surface specifically, with differing views on the implications of surface choice.

Contextual Notes

The discussion does not resolve the question of whether other surfaces could be used effectively in this context, leaving the implications of surface choice open to interpretation.

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Strange Laplacian of 1/|x-x’| !?

Please read the file first (http://www.pa.msu.edu/courses/2007fall/PHY481/lectures/lecture08.pdf" ) ..


and look into page 8

there is a sentence like this "Evaluate right side with sphere, radius R around origin"

Now there comes up against a question :

Why do we have to use "sphere around origin" (← That is often used in physic. But I need a general elucidate...)

how about any surface around the origin?
 
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Of course Gauss' theorem is valid for any closed surface, however, one chooses the surface to simplify the process of integration. Since the Laplacian is spherically symmetric, it makes sense to use a spherical surface. Obviously other problems will have other symmetries and you should choose your surface appropriately.
 
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Hootenanny said:
Of course Gauss' theorem is valid for any closed surface, however, one chooses the surface of integration to simplify the process of integration. Since the Laplacian is spherically symmetric, it makes sense to use a spherical surface. Obviously other problems will have other symmetries and you should choose your surface appropriately.

Umm,thank you!

I need to go to bed now (because the zone here is diffirent from yours >< (Taiwan))

although I want to learn more things from you...


THX again!
 


abcdefg10645 said:
Please read the file first (http://www.pa.msu.edu/courses/2007fall/PHY481/lectures/lecture08.pdf" ) ..


and look into page 8

there is a sentence like this "Evaluate right side with sphere, radius R around origin"

Now there comes up against a question :

Why do we have to use "sphere around origin" (← That is often used in physic. But I need a general elucidate...)

how about any surface around the origin?



We use sphere so that the resulting surface integral can be done.

See post #10 in

www.physicsforums.com/showthread.php?t=200580


regards

sam
 
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