Why Does the Electric Field Calculation Diverge Inside the Volume?

[Mike400]

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?

 

[kuruman]

I am a little confused about your use of coordinates. Source coordinates are usually primed and field coordinates unprimed. When you say

The electric field at origin due to an electric dipole distribution in volume $V$ having boundary $S$ is:
$$\int_V \dfrac{(\mathbf{M} \cdot \hat{r}) (\hat{r})}{r^3} dV,$$

how did you derive that expression and what are you integrating over? If the integral is over unprimed coordinates, as implied by the way it is written, then the result of the integration should be a function of $\mathbf{r'}$ because $\mathbf{M}$, being the source, is a function of $\mathbf{r'}$.

 

[Mike400]

I am a little confused about your use of coordinates. Source coordinates are usually primed and field coordinates unprimed. When you say
how did you derive that expression and what are you integrating over? If the integral is over unprimed coordinates, as implied by the way it is written, then the result of the integration should be a function of $\mathbf{r'}$ because $\mathbf{M}$, being the source, is a function of $\mathbf{r'}$.

Sorry for the confusion. Actually all the integration are with respect to source (primed) coordinates.

The equation I mention is reached by taking the negative gradient of potential due to continuous electric dipole distribution.

 

[kuruman]

Sorry for the confusion. Actually all the integration are with respect to source (primed) coordinates.

But then the result will have no dependence on the field (unprimed) coordinates. You need both $\mathbf{r}$ and $\mathbf{r'}$ in your starting integral. What you should have is
$$\varphi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int_V \frac{\mathbf{M}(\mathbf{r'})\cdot (\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3}dV'$$which, after a series of transformations found in most decent intermediate E&M textbooks, becomes
$$\varphi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\oint_S\frac{\mathbf{M}\cdot \hat n}{|\mathbf{r}-\mathbf{r'}|} dA' + \frac{1}{4\pi\epsilon_0}\int_V\frac{(-\mathbf{\nabla '}\cdot \mathbf{M})}{|\mathbf{r}-\mathbf{r'}|} dV'.$$This is sort of what you have and is applicable to points outside the distribution where the dipolar approximation is valid. I think your problem is trying to apply it at points inside the distribution where, unavoidably, there will be dipoles too close for the dipolar approximation to be valid.

 

[Mike400]

But then the result will have no dependence on the field (unprimed) coordinates. You need both $\mathbf{r}$ and $\mathbf{r'}$ in your starting integral. What you should have is
$$\varphi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int_V \frac{\mathbf{M}(\mathbf{r'})\cdot (\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3}dV'$$which, after a series of transformations found in most decent intermediate E&M textbooks, becomes
$$\varphi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\oint_S\frac{\mathbf{M}\cdot \hat n}{|\mathbf{r}-\mathbf{r'}|} dA' + \frac{1}{4\pi\epsilon_0}\int_V\frac{(-\mathbf{\nabla '}\cdot \mathbf{M})}{|\mathbf{r}-\mathbf{r'}|} dV'.$$

Both your equations are correct and that was what I was referring to.

This is sort of what you have and is applicable to points outside the distribution

Mathematically speaking, it is applicable inside the distribution also. See the below two references:

(1) Reflections in Maxwell's treatise. Section 4.2
(2) Electromagnetic theory. (by Alfred) Chapter II Section 2, The para below equation (2.4)