Why ignoring the contribution from point r=0 in eq (1) and (2)?

[Mike400]

The potential of a dipole distribution at a point $P$ is:

$\psi=-k \int_{V'} \dfrac{\vec{\nabla'}.\vec{M'}}{r}dV' +k \oint_{S'}\dfrac{\vec{M'}.\hat{n}}{r}dS'$

If $P\in V'$, the integrand is discontinuous (infinite) at the point $r=0$. So we need to use improper integrals by removing a small cavity $\delta$ from $V'$:

$\psi=k \left[ -\lim \limits_{\delta \to 0} \int_{V'-\delta} \dfrac{\vec{\nabla'}.\vec{M'}}{r} dV'+ \lim \limits_{\delta \to 0} \oint_{S'+\Delta} \dfrac{\vec{M'}.\hat{n}}{r}dS' \right] \tag 1$

It must be noted that we ignored the contribution to the volume integral from the point $r=0$ (if there is any).

$\nabla\psi=k \left[ -\lim \limits_{\delta \to 0} \int_{V'-\delta} (\vec{\nabla'}.\vec{M'})\ \nabla \left( \dfrac{1}{r} \right) dV'+ \lim \limits_{\delta \to 0} \oint_{S'} (\vec{M'}.\hat{n})\ \nabla \left( \dfrac{1}{r} \right)dS' \right] \tag 2$

Here also, it must be noted that we ignored the contribution to the volume integral from the point $r=0$ (if there is any).

\begin{align}
\nabla^2 \psi&=k \left[ -\lim \limits_{\delta \to 0}
\int_{V'-\delta} (\vec{\nabla'}.\vec{M'})\ \nabla^2 \left( \dfrac{1}{r} \right) dV'+
\lim \limits_{\delta \to 0}
\oint_{S'} (\vec{M'}.\hat{n})\ \nabla^2 \left( \dfrac{1}{r} \right)dS'
\right]\\
&\bbox[yellow]{-4\pi\ k\ (\vec{\nabla}.\vec{M'})}\\
&=-4\pi\ k\ (\vec{\nabla}.\vec{M'})\\ \tag 3
\end{align}

Here, it must be noted that we *did not* ignore the contribution to the volume integral from the point $r=0$.

So why is it that we are ignoring the contribution to the volume integral from the point $r=0$ in equation $(1)$ and $(2)$?

 

[Charles Link]

What you are doing here is computing a potential $\psi$ so that $-\nabla \psi=\vec{H}$ if there are no contributions to $\vec{H}$ from currents in conductors. This is the "pole" method of magnetism. Both the pole method and the surface current method get the same answer for $\vec{B}$, and the pole method uses the equation $\vec{B}=\mu_o \vec{H}+\vec{M}$. (In some units they use $\vec{B}=\mu_o \vec{H}+\mu_o \vec{M}$).
The surface current method is presently taught more so than the pole method. For a simple introduction to both methods, see: https://www.physicsforums.com/threa...a-ferromagnetic-cylinder.863066/#post-5433500
In spherical coordinates $dV=r^2 \sin{\theta} \, d \theta \, d \phi$. IMO, it is unnecessary to exclude the point at $r=0$. The $r^2$ of $dV$ in the numerator will remove any divergence from $r=0$ in the denominator.

 

[Mike400]

In spherical coordinates $dV=r^2 \sin{\theta} \, d \theta \, d \phi$. IMO, it is unnecessary to exclude the point at $r=0$. The $r^2$ of $dV$ in the numerator will remove any divergence from $r=0$ in the denominator.

In equations $(1)$ and $(2)$, your statement works fine in volume integrals.

For equation $(1)$:

\begin{equation}
\dfrac{\vec{\nabla'}.\vec{M'}}{r} dV'
=\dfrac{\vec{\nabla'}.\vec{M'}}{r} r^2 \sinθ\ dθ\ dϕ\ dr\
\end{equation}

Now putting $r=0$, our expression becomes:

\begin{equation}
(\vec{\nabla'}.\vec{M'}) \dfrac{0}{0}\ 0\ \sinθ\ dθ\ dϕ\ dr=\text{indeterminate times zero}=0
\end{equation}

That is, the contribution from $r=0$ is zero.

For equation $(2)$:

\begin{equation}
(\vec{\nabla'}.\vec{M'})\ \nabla \left( \dfrac{1}{r} \right) dV'
=(\vec{\nabla'}.\vec{M'})\ \left( \dfrac{\hat{r}}{r^2} \right) r^2 \sinθ\ dθ\ dϕ\ dr\
\end{equation}

Now putting $r=0$, our expression becomes:

\begin{equation}
(\vec{\nabla'}.\vec{M'}) (\hat{r}) \dfrac{0}{0}\ \dfrac{0}{0}\ \sinθ\ dθ\ dϕ\ dr=\text{indeterminate times infinitesimal}=0
\end{equation}

That is, the contribution from $r=0$ is zero.

Now what about the contributions from $r=0$ due to surface integrals? How can we show that they both are zero?

 

[Charles Link]

The surface integrals cover the outer boundary of the solid, and in general don't contain $r=0$. If it does, that's why the prescription for ignoring $r=0$ may be useful.

 

[Mike400]

The surface integrals cover the outer boundary of the solid, and in general don't contain $r=0$. If it does, that's why the prescription for ignoring $r=0$ may be useful.

I get your point. By the way, is my interpretation correct i.e. are equations (5) and (7) mathematically valid. I have never seen such equations neither in mathematics nor physics texts. I should say that equations (5) and (7) are expressions of my intuition.

 

[Charles Link]

I get your point. By the way, is my interpretation correct i.e. are equations (5) and (7) mathematically valid. I have never seen such equations neither in mathematics nor physics texts. I should say that equations (5) and (7) are expressions of my intuition.

When you have $\frac{r}{r}$, it equals $1$, regardless of whether $r=0$ or not. It is not indeterminant.
Meanwhile, so long as $\nabla \cdot M$ remains finite, the contribution at $r=0$ to $\int \nabla \cdot M \sin{\theta} \, d \theta d\phi \, dr$ remains infinitesimal at $r=0$ and you don't need to exclude $r=0$ in the integral.
Your prescription says take $\int\limits_{0^+}^{+\infty}$, but I don't think it is necessary to have $r=0^+$ with a $+$ on the zero of the lower limit of $r$.