Three dimensional ##\delta## function

  • Thread starter Thread starter Apashanka
  • Start date Start date
  • Tags Tags
    Function
Click For Summary
The discussion focuses on the evaluation of integrals involving the divergence of a vector field in spherical coordinates. It highlights that the integral over a spherical volume simplifies to a surface integral, yielding a result of 4πR. The use of the Dirac delta function in spherical coordinates is emphasized, showing how it can be expressed in terms of the radial, polar, and azimuthal components. The conversation seeks clarification on how to reduce the right-hand side of an integral involving delta functions to a volume integral form. The key takeaway is the necessity of multiplying by a specific factor to achieve the desired form in spherical coordinates.
Apashanka
Messages
427
Reaction score
15
Homework 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??
 
Physics news on Phys.org
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.
 
ShayanJ said:
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
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
14
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K