- #1
henpen
- 50
- 0
[tex]\nabla \cdot \frac{\mathbf{r}}{|r|^3}=4 \pi \delta ^3(\mathbf{r})[/tex]
What's the proof for this, and what's wrong with the following analysis?
The vector field
[itex]\frac{\mathbf{r}}{|r|^3}=\frac{1}{r^2}\hat{r}[/itex] can also be written [tex]\mathbf{F}=\frac{x}{\sqrt{x^2+y^2+z^2}^3}\hat{x}+ \frac{y}{\sqrt{x^2+y^2+z^2}^3}\hat{y}+ \frac{z}{\sqrt{x^2+y^2+z^2}^3}\hat{z}[/tex]
Take the z-part of the divergence of F:
[tex]\frac{\partial}{\partial z}\mathbf{F_z}=\frac{\partial}{\partial z}\frac{z}{\sqrt{x^2+y^2+z^2}^3}=\frac{1}{\sqrt{x^2+y^2+z^2}^3}+\frac{-3}{2}\frac{z}{\sqrt{x^2+y^2+z^2}^5}2z[/tex]
[tex]\frac{1}{\sqrt{x^2+y^2+z^2}^5}(x^2+y^2+z^2-3z^2)=\frac{x^2+y^2-2z^2}{\sqrt{x^2+y^2+z^2}^5}[/tex]
Evoking symmetry to find the other parts of the divergence is easy:
[tex]\frac{(x^2+y^2-2z^2)+(x^2+z^2-2y^2)+(z^2+y^2-2x^2)}{\sqrt{x^2+y^2+z^2}^5}=0[/tex]
Why am I incorrect at [itex]x,y,z=0[/itex]?
What's the proof for this, and what's wrong with the following analysis?
The vector field
[itex]\frac{\mathbf{r}}{|r|^3}=\frac{1}{r^2}\hat{r}[/itex] can also be written [tex]\mathbf{F}=\frac{x}{\sqrt{x^2+y^2+z^2}^3}\hat{x}+ \frac{y}{\sqrt{x^2+y^2+z^2}^3}\hat{y}+ \frac{z}{\sqrt{x^2+y^2+z^2}^3}\hat{z}[/tex]
Take the z-part of the divergence of F:
[tex]\frac{\partial}{\partial z}\mathbf{F_z}=\frac{\partial}{\partial z}\frac{z}{\sqrt{x^2+y^2+z^2}^3}=\frac{1}{\sqrt{x^2+y^2+z^2}^3}+\frac{-3}{2}\frac{z}{\sqrt{x^2+y^2+z^2}^5}2z[/tex]
[tex]\frac{1}{\sqrt{x^2+y^2+z^2}^5}(x^2+y^2+z^2-3z^2)=\frac{x^2+y^2-2z^2}{\sqrt{x^2+y^2+z^2}^5}[/tex]
Evoking symmetry to find the other parts of the divergence is easy:
[tex]\frac{(x^2+y^2-2z^2)+(x^2+z^2-2y^2)+(z^2+y^2-2x^2)}{\sqrt{x^2+y^2+z^2}^5}=0[/tex]
Why am I incorrect at [itex]x,y,z=0[/itex]?