Is there a wayIn his book introduction to representelectrodynamics, Griffiths uses derives the vector field r−r′|r−r′|32 where r′ is a constant vector and r isidentity ∇⋅ˆrr2=4πδ3(r) Using the position vectorformula for divergence in polar coordinates in terms of the basis vectors \hat{\mathbf{r}},\hat{\mathbf{\theta}}\boldsymbol{\hat{\phi}}?
The reason I'm asking this is for deriving the identity \nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}\right)=4\pi\delta^3(\mathbf{r-r'}).
I know how to show. He then states that "more generally" \nabla \cdot \frac{\mathbf{\hat{r_s}}}{r_s^2} = 4\pi\delta^3(\mathbf{r_s}) Where \nabla\cdot\frac{\hat{\mathbf{r}}}{r^2}\mathbf{r_s} is \delta^3(\mathbf{r}) using polar coordinates, and I loosely understand how changing the argument to \mathbf{r-r'} should "shift" the divergence becauseseperation vector, \frac{\partial }{\partial x}\left(f(x-a)\right)=f_x(x-a)\mathbf{r}-\mathbf{r'} and I can show this\mathbf{r'} is the casea constant. I'm trying to find a way to derive this identity using cartesianpolar coordinates, but I was wondering how to use polar coordinatesam struggling to solve this. My main issue is findingfind a way to relate the valuesrepresent \frac{\mathbf{r-r'}}{{|r-r'|^3}} in terms of the vector field to the basis vectors, since in the case wherepolar coordinates \mathbf{r'}=\mathbf{0} they always point(\mathbf{\hat{r}}, \hat{\boldsymbol{\phi}},\hat{\boldsymbol{\theta}}) as v_r\mathbf{\hat{r}}+v_\theta\boldsymbol{\hat{\theta}}+v_\phi\boldsymbol{\hat{\phi}} for some scalar functions v_r, v_{\theta}, v_{\phi} to use in the radial direction, butformula for divergence in general I don't see how to derive a relationship between the vector field and the polar basis vectorscoordinates.