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Three dimensional ##\delta## function

  • Thread starter Apashanka
  • Start date
387
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Problem Statement
Three dimensional ##\delta## function
Relevant Equations
Three dimensional Delta function
##r,\theta,\phi## are the usual spherical polar coordinate system.
##\int_v\nabla•(\frac{\hat r}{r})dv## over a spherical volume of radius ##R## reduces to ##\int_s(\frac{\hat r}{r})•\vec ds=4\pi R##
Now ##r## runs from 0 to ##R,\theta## from 0 to ##\pi## and ##\phi## from 0 to ##2\pi##.
In terms of ##\delta## function ##\int_{0}^{R} \delta (r-r')dr=1## where ##r'## lies within 0 to R ,and ##\ne 0,R##,similarly for ##\theta## and ##\phi## also where ##\int_{0}^{\pi} \delta(\theta-\theta')d\theta=1## where ##\theta'## lies ##{0,\pi}## and ##\ne0,\pi## similarly for ##\phi## also.
Putting them
##\int_v(\nabla•\frac{\hat r}{r})dv=4\pi R\int_0^R\int_0^\pi \int_0^{2\pi} \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')drd\theta d\phi##
can anyone please tell me how can the RHS be reduced to volume integral form??
 
2,783
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Actually, the Dirac delta in spherical coordinates is: ## \delta(\vec r)= \frac 1 {r^2} \delta(r-r')\delta(\cos\theta-\cos\theta')\delta(\phi-\phi')=\frac 1 {r^2\sin \theta} \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi') ##

So you just need to multiply the RHS by ## \frac {r^2\sin\theta}{r^2\sin\theta} ##.

P.S.
You may find this useful.
 
387
13
Actually, the Dirac delta in spherical coordinates is: ## \delta(\vec r)= \frac 1 {r^2} \delta(r-r')\delta(\cos\theta-\cos\theta')\delta(\phi-\phi')=\frac 1 {r^2\sin \theta} \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi') ##

So you just need to multiply the RHS by ## \frac {r^2\sin\theta}{r^2\sin\theta} ##.

P.S.
You may find this useful.
Thanks
 

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