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

• I

## Main Question or Discussion Point

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)##?

Homework Helper
Gold Member
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.

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)##:

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

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

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

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

For equation ##(2)##:

(\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\

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

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

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?

Homework Helper
Gold Member
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.

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.