Saturday, 7 July 2018

electrostatics - Divergence of $frac{ hat {bf r}}{r^2}$ , what is the 'paradox'?


I just started in Griffith's Introduction to electrodynamics and I stumbled upon the divergence of $\frac{ \hat r}{r^2}$ , now from the book, Griffiths says:


enter image description here


Now what is the paradox, exactly? Ignoring any physical intuition behind this (point charge at the origin) how are we supposed to believe that the source of $\vec v$ is concentrated at the origin mathematically? Or are we forced to believe that because there was a contradiction with the divergence theorem?



Also how would the situation differ if $\vec v$ was the same vector function but not for a point charge? Or is it impossible?



Answer




Now what is the paradox, exactly?



The paradox is that the vector field $\vec{v}$ considered obviously points away from the origin and hence seems to have a non-zero divergence, however, when you actually calculate the divergence, it turns out to be zero.


enter image description here



How are we supposed to believe that the source of $\vec v$ is concentrated at the origin mathematically?




Most important point to observe is that $\nabla.\vec v = 0$ everywhere except at the origin. The diverging lines appearing are from the origin. Our calculations cannot account for that since $\vec v$ blows up at $r = 0$. Moreover, eq. (1.84) is not even valid for $r = 0$. In other words, $\nabla.\vec v \rightarrow \infty$ at that point.


However, if you apply the divergence theorem, you will find $$\int \nabla.\vec v \ \text{d}V = \oint \vec v.\text{d}\vec a = 4 \pi$$ Irrespective of the radius of a sphere centred at the origin, we must obtain the surface integral as $4 \pi$. The only conclusion is that this must be contributed from the point $r = 0$.


This serves as the motivation to define the Dirac delta function: a function which vanishes everywhere except blowing up at a point and has a finite area under the curve.


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