Three dimensional ##\delta## function

[Apashanka]

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??

 

[ShayanJ]

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.

 

[Apashanka]

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