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A vector function

[tex] V(\vec{r}) = \frac{ \hat r}{r^2} [/tex]

If we

[tex] \nabla \cdot \vec{V} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r^2} \right) = 0 [/tex]

However, by divergence theorem, the surface integral is [tex]4\pi [/tex]. This paradox can be solved by Dirac Delta function.

My problem is, it seems like we can't

But it seem to me that a vector function [tex] V'(\vec{r}) =\frac{ \hat r}{r} [/tex] also has the same problem, which also blows up at r=0, and we will also divide the zero when

[tex] \nabla \cdot \vec{V'} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r} \right) [/tex]

But we will get the accurate answer !? Why both of the function V and V' has the same problem but we only have to use Dirac Delta function for V ??

[tex] V(\vec{r}) = \frac{ \hat r}{r^2} [/tex]

If we

*calculate*it's divergence directly:[tex] \nabla \cdot \vec{V} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r^2} \right) = 0 [/tex]

However, by divergence theorem, the surface integral is [tex]4\pi [/tex]. This paradox can be solved by Dirac Delta function.

My problem is, it seems like we can't

*calculate*V's divergence*directly*because V blow up at r=0, and we unwittingly divide the zero when [tex] \left( r^2 \frac{1}{r^2} \right) [/tex]But it seem to me that a vector function [tex] V'(\vec{r}) =\frac{ \hat r}{r} [/tex] also has the same problem, which also blows up at r=0, and we will also divide the zero when

[tex] \nabla \cdot \vec{V'} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r} \right) [/tex]

But we will get the accurate answer !? Why both of the function V and V' has the same problem but we only have to use Dirac Delta function for V ??

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