Delta function representation from EM theory

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SUMMARY

The discussion centers on the proof of the identity \nabla \cdot \frac{\hat{e}_r}{r^2}=4\pi\delta^3(\vec{x}), which is crucial in electromagnetic theory. The relationship is derived from the electric field of a point charge, expressed as \vec{E}=\frac{q}{r^2}\hat{e}_r, leading to \nabla \cdot \vec{E}=4\pi q\delta^3(\vec{x}). The challenge lies in demonstrating that the delta function relation holds specifically at the origin. A referenced resource from MIT provides foundational explanations and examples relevant to this topic.

PREREQUISITES
  • Vector calculus, specifically divergence operations
  • Electromagnetic theory, particularly concepts of electric fields
  • Understanding of delta functions in mathematical physics
  • Familiarity with integral calculus, especially in spherical coordinates
NEXT STEPS
  • Study the derivation of the electric field from point charges in electromagnetic theory
  • Learn about the properties and applications of the Dirac delta function
  • Examine vector calculus identities and their proofs in physics contexts
  • Review the MIT resource on electromagnetic theory for detailed examples and calculations
USEFUL FOR

This discussion is beneficial for physics students, researchers in electromagnetism, and anyone interested in advanced mathematical proofs related to electric fields and delta functions.

pellman
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Claim:

[tex]\nabla \cdot \frac{\hat{e}_r}{r^2}=4\pi\delta^3(\vec{x})[/tex]

Anyone know of a proof of this? (or a reference which covers it?) We need to show that

[tex]\frac{1}{4\pi}\int_0^R{(\nabla \cdot \frac{\hat{e}_r}{r^2})f(r)dr=f(0)[/tex].



The claimed identity can be seen in the solution for the electric field of a point charge in EM theory, where

[tex]\vec{E}=\frac{q}{r^2}\hat{e}_r[/tex]

is the solution to

[tex]\nabla \cdot \vec{E}=4\pi q\delta^3(\vec{x})[/tex]

It is easy to show in this case that [tex]\nabla \cdot \vec{E}=0[/tex] everywhere but the origin, but I don't know how to show that the delta function relation holds at the origin.
 
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