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

  • #1
58
6

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)##?
 

Answers and Replies

  • #2
Charles Link
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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.
 
  • #3
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6
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?
 
  • #4
Charles Link
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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.
 
  • #5
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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.
 
  • #6
Charles Link
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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 ##.
 

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