Module 1 : A Crash Course in Vectors
Lecture 6 : Laplacian
Consider an infinitisimally small sphere of radius $r_0$ with the centre at the origin. Using divergence theorem, we have,
 
\begin{displaymath}\int_V\nabla^2\left(\frac{1}{r}\right)d^3r= \int_V\nabla\cd... ...}{r}\right)d^3r= \int_S\nabla\left(\frac{1}{r}i\right)\cdot dS\end{displaymath}
  where the last integral is over the surface $S$ of the sphere defined above. As the gradient is taken at points on the surface for which $r\ne 0$, we may replace $\nabla(1/r)$ with $-1/r_0^2$ at all points on the surface. Thus the value of the integral is
 
\begin{displaymath}-\frac{1}{r_0^2}\int_SdS = -\frac{1}{r_0^2}\cdot 4\pi r_0^2 = -4\pi\end{displaymath}
  Hence
 
\begin{displaymath}\nabla^2 \left(\frac{1}{r}\right) = -4\pi\delta(r)\end{displaymath}
   
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