Why does the following function equate to a delta in classical feild theory

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SUMMARY

The discussion centers on the mathematical representation of the delta function in classical field theory, specifically the equation $\nabla^2 \frac{1}{|r|} = \delta(r)$. Participants seek a derivation or argument supporting this representation, referencing Jackson's work for guidance. The equation indicates that the divergence of the gradient of the function $\frac{1}{|r|}$ is proportional to the delta function, highlighting its significance in potential theory.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and gradient operations.
  • Familiarity with the delta function and its properties in mathematical physics.
  • Knowledge of classical field theory concepts.
  • Access to "Classical Electrodynamics" by John David Jackson for reference.
NEXT STEPS
  • Review the derivation of the delta function in "Classical Electrodynamics" by John David Jackson.
  • Study the properties and applications of the delta function in physics.
  • Explore the implications of the Laplacian operator in three-dimensional space.
  • Investigate the role of singularities in potential theory and their mathematical representations.
USEFUL FOR

Students and professionals in physics, particularly those studying classical field theory, mathematical physics, or vector calculus, will benefit from this discussion.

Fwahnak
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Homework Statement



Can anyone remember a decent argument/derivation for the following representation of the delta function.


Homework Equations



[tex]$ \nabla^2 \frac{1}{|r|} =\delta(r)$[/tex]

(probally up to some multipicative constant [tex]$\frac{1}{2\pi}$[/tex] or something

The Attempt at a Solution



I know I've seen an agrument in jackson but I don't have a copy.
 
Last edited:
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I've no confidence that I've latexed that correctly and my work computer won't show me the symbols so here's the equation in equation again

div(grad(1/|r|)) is proportional to deltafunction(r)
 

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