- #1
Mike400
- 59
- 6
Let:
##\nabla## denote dell operator with respect to field coordinate (origin)
##\nabla'## denote dell operator with respect to source coordinates
The electric field at origin due to an electric dipole distribution in volume ##V## having boundary ##S## is:
\begin{align}
\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3} dV
&=-\nabla \left( \int_V \dfrac{\mathbf{M} \cdot \hat{r}}{r^2} dV \right)\\
&=-\nabla \left[ \int_V \mathbf{M} \cdot \nabla' \left( \dfrac{1}{r} \right) \right]\\
&=-\nabla \left[ - \int_V \dfrac{\nabla' \cdot \mathbf{M}}{r} dV
+ \int_V \nabla' \cdot \left(\dfrac{\mathbf{M}}{r} \right) dV \right] \\
&=-\nabla \left( \int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r} dV + \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r} dS \right)\\
&=\int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r^2}(\hat{r}) dV + \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS\\
\end{align}
In the above equation, we applied the vector identity ##\mathbf{A} \cdot \nabla\psi=-(\nabla \cdot \mathbf{A})\psi + \nabla \cdot (\psi \mathbf{A})## and the divergence theorem.
If the origin point is not on boundary ##S## but inside ##V##:
\begin{align}
LHS&=\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3} dV\\
&=\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3}\ r^2 \sin\theta\ d\theta\ d\phi\ dr\\
&=\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r}\ \sin\theta\ d\theta\ d\phi\ dr\\
\end{align}
##(1)## Here the integrand contains a singular point and hence the integral diverges.
\begin{align}
RHS&=\int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r^2}(\hat{r}) dV + \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS\\
&=\int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r^2}(\hat{r})\ r^2 \sin\theta\ d\theta\ d\phi\ dr
+ \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS\\
&=-\int_V \nabla' \cdot \mathbf{M}\ (\hat{r})\ \sin\theta\ d\theta\ d\phi\ dr
+ \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS
\end{align}
##(2)## Here both the integrands contain no singular point and hence the integrals converge.
##(1)## and ##(2)## contradicts. Why is there such a contradiction?
##\nabla## denote dell operator with respect to field coordinate (origin)
##\nabla'## denote dell operator with respect to source coordinates
The electric field at origin due to an electric dipole distribution in volume ##V## having boundary ##S## is:
\begin{align}
\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3} dV
&=-\nabla \left( \int_V \dfrac{\mathbf{M} \cdot \hat{r}}{r^2} dV \right)\\
&=-\nabla \left[ \int_V \mathbf{M} \cdot \nabla' \left( \dfrac{1}{r} \right) \right]\\
&=-\nabla \left[ - \int_V \dfrac{\nabla' \cdot \mathbf{M}}{r} dV
+ \int_V \nabla' \cdot \left(\dfrac{\mathbf{M}}{r} \right) dV \right] \\
&=-\nabla \left( \int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r} dV + \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r} dS \right)\\
&=\int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r^2}(\hat{r}) dV + \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS\\
\end{align}
In the above equation, we applied the vector identity ##\mathbf{A} \cdot \nabla\psi=-(\nabla \cdot \mathbf{A})\psi + \nabla \cdot (\psi \mathbf{A})## and the divergence theorem.
If the origin point is not on boundary ##S## but inside ##V##:
\begin{align}
LHS&=\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3} dV\\
&=\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3}\ r^2 \sin\theta\ d\theta\ d\phi\ dr\\
&=\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r}\ \sin\theta\ d\theta\ d\phi\ dr\\
\end{align}
##(1)## Here the integrand contains a singular point and hence the integral diverges.
\begin{align}
RHS&=\int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r^2}(\hat{r}) dV + \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS\\
&=\int_V \dfrac{-\nabla' \cdot \mathbf{M}}{r^2}(\hat{r})\ r^2 \sin\theta\ d\theta\ d\phi\ dr
+ \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS\\
&=-\int_V \nabla' \cdot \mathbf{M}\ (\hat{r})\ \sin\theta\ d\theta\ d\phi\ dr
+ \oint_S \dfrac{\mathbf{M} \cdot \hat{n}}{r^2} (\hat{r}) dS
\end{align}
##(2)## Here both the integrands contain no singular point and hence the integrals converge.
##(1)## and ##(2)## contradicts. Why is there such a contradiction?