Why do we ignore the contribution to a surface integral from the point r=0?

[Mike400]

I am a little puzzled by the diagram of post 1. The region where $\vec{M}$ is non-zero is presumably the region below the paraboloid, and you want to find the contribution to the integral at a point $\vec{r}$ that lies on the paraboloid. The integral over $R$ is ok, but the $\vec{r}$ in the diagram needs to be $\vec{r}'$, and the $r$ in the denominator is $r=|\vec{r}-\vec{r}'|$. This $r$ is really a poor choice of letters, and perhaps should be called letter $s$ . It's very clumsy to work with these complicated surface integral formulas and prove anything meaningful. It's much easier to simplify what's going on, and to prove things for the simplified case.

I get it

 

[Charles Link]

@Mike400 See also post 27. I don't think you saw it yet.

 

[Mike400]

You might consider the general case to be composed of super-positions of a couple of basic building blocks. Perhaps that would complete the proof.
Even near the surface of a small disc of surface charge density $\sigma$, we find $E=\frac{\sigma}{2 \epsilon_o}$. The electric field does not experience a singularity here.

I know that electric field of an infinite sheet of charge in $x$-$y$ plane is $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{k})$. From this fact, how can we conclude that electric field very close to the surface point of a $\sigma$ distribution is approximately $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{n})$ where $\hat{n}$ is normal to surface point.

 

[vanhees71]

I've no clue, what the volume and surface is you want to integrate over in #1. You have to specify your problem first before one can help to solve the integrals. Also I don't understand your notation. I can only guess you mean the solution for a continuous magnetic-dipole distribution. One correct form of the formula reads
$$\phi(\vec{x})=-\vec{\nabla}_x \cdot \int_{V} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
The surface term can usually be neglected (except if there are singularities in $\vec{M}$ on the surface).

If $\vec{x} \notin V$, there's no singularity to think about. If $\vec{x} \in V$ and $\vec{M}$ has no singularities anywhere, there's also no singularity. To see this take a little sphere around $\vec{x}$ and introduce spherical coordinates,
$$\vec{x}'=\vec{x}+r' \begin{pmatrix} \cos \varphi' \sin \vartheta' \\ \sin \varphi' \sin \vartheta ' \\ \cos \vartheta' \end{pmatrix}.$$
Then $\mathrm{d}^3 x' = \mathrm{d} r' \mathrm{d} \vartheta' \mathrm{\varphi'} r^{\prime 2} \sin \vartheta'$, and the singularity due to the Green's-function factor in the integrand $1/|\vec{x}-\vec{x}'|=1/r'$ is safely compensated by the volume element, i.e., under these conditions the singularity is integrable.

As a simple example it's a nice exercise to calculate the magnetic field of a homogeneously magnetized sphere.

 

[vanhees71]

I know that electric field of an infinite sheet of charge in $x$-$y$ plane is $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{k})$. From this fact, how can we conclude that electric field very close to the surface point of a $\sigma$ distribution is approximately $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{n})$ where $\hat{n}$ is normal to surface point.

Of course, there is a singularity. The jump of the normal component of $\vec{D}=\epsilon_0 \vec{E}$ is the surface charge density $\sigma$. You have to be careful with the directions of the normal vector. It originates from the integration over a little volume with two sides (you can simply use a "pill box") parallel to the surface and using Gauss's integral theorem. The normal vectors are always pointing outside the integration volume, i.e., the normal has to be taken in opposite directions on the both sides of the surface.

It's also intuitive: Due to the surface charge, the electric field always points away (towards) the surface if $\sigma>0$ ($\sigma<0$), i.e., into opposite directions on both sides of the surface.

 

[Mike400]

I know that electric field of an infinite sheet of charge in $x$-$y$ plane is $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{k})$. From this fact, how can we conclude that electric field very close to the surface point of a $\sigma$ distribution is approximately $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{n})$ where $\hat{n}$ is normal to surface point.

I think I have got the answer. Please check its validity.

Consider a $\sigma$ distribution and a point $P$ infinitely close to it. Then:

$\displaystyle\mathbf{H}(P)_1=\int_{S'} \dfrac{\sigma}{r^2}(\hat{\mathbf{r}})\ dS'$

Now increase the linear dimensions of the system infinitely. Then:

$\displaystyle\mathbf{H}(P)_{\infty}=\lim\limits_{n \to \infty} \int_{S'} \dfrac{\sigma}{(nr)^2} \left( \dfrac{n\mathbf{r}}{nr} \right) n^2 dS' =\int_{S'} \dfrac{\sigma}{r^2}(\hat{\mathbf{r}})\ dS'=\mathbf{H}(P)_1$

Now after the increment, $\sigma$ distribution is an infinite sheet near point $P$. Point $P$ is also at a finite distance from $\sigma$ distribution (infinite sheet). Therefore:

$\mathbf{H}(P)_1=\mathbf{H}(P)_{\infty}=\dfrac{\sigma}{2 \epsilon_0} (\hat{\mathbf{n}})$

 

[Charles Link]

I think I have got the answer. Please check its validity.

Consider a σσ distribution and a point PP infinitely close to it. Then:

H(P)1=∫S′σr2(^r) dS′H(P)1=∫S′σr2(r^) dS′

Now increase the linear dimensions of the system infinitely. Then:

$\displaystyle\mathbf{H}(P)_{\infty}=\lim\limits_{n \to \infty} \int_{S'} \dfrac{\sigma}{(nr)^2} \left( \dfrac{n\mathbf{r}}{nr} \right) n^2 dS' =\int_{S'} \dfrac{\sigma}{r^2}(\hat{\mathbf{r}})\ dS'=\mathbf{H}(P)_1$

Now after the increment, σσ distribution is an infinite sheet near point PP. Point PP is also at a finite distance from σσ distribution (infinite sheet). Therefore:

H(P)1=H(P)∞=σ2ϵ0(^n)H(P)1=H(P)∞=σ2ϵ0(n^)

The equation actually reads $\vec{H}(\vec{r})=\int \frac{\sigma_m (\vec{r}')}{4 \pi \mu_o r^3} (\vec{r}-\vec{r}') dS'$ where $r=|\vec{r}-\vec{r}'|$.
I'm defining $\vec{M}$ so that $\vec{B}=\mu_o \vec{H}+\vec{M}$. $\\$ $\vec{H}=\frac{\sigma_m}{2 \mu_o}$.

 

[Mike400]

Now is there a similar way to show that $\displaystyle\psi=\int_{S'} \dfrac{\sigma}{r} dS'$ does not blow up near the surface?

 

[Charles Link]

The easiest way I know of is that $\psi=\pm \int H \cdot dr$. At $r=0$, the $H$ field remains finite. Like $E$ it is conservative, (when there are no currents in conductors) and $\nabla \times H=0$. Thereby, the equation $\psi=\int H \cdot dr$ can be used.

 

[Mike400]

The easiest way I know of is that $\psi=\pm \int H \cdot dr$. At $r=0$, the $H$ field remains finite. Like $E$ it is conservative, (when there are no currents in conductors) and $\nabla \times H=0$. Thereby, the equation $\psi=\int H \cdot dr$ can be used.

That is what I thought too.

 

[Mike400]

Is there a way to show:

$(1)\displaystyle\int_{S'} \dfrac{\sigma}{r} dS'$ is continuous all over space

$(2)\displaystyle\int_{S'} \dfrac{\sigma}{r^2}\ \hat{\mathbf{r}}\ dS'$ is continuous everywhere except at $S'$

 

[Charles Link]

Is there a way to show:

$(1)\displaystyle\int_{S'} \dfrac{\sigma}{r} dS'$ is continuous all over space

$(2)\displaystyle\int_{S'} \dfrac{\sigma}{r^2}\ \hat{\mathbf{r}}\ dS'$ is continuous everywhere except at $S'$

By Gauss' law, $E_{1 \, ||}-E_{2 \, ||}=\frac{\sigma}{\epsilon_o}$. and similarly for $H$. (parallel to $\hat{n}$). Meanwhile, the perpendicular $E_1$ and $E_2$ are equal. (parallel to the surface).

 

[Mike400]

By Gauss' law, $E_{1 \, ||}-E_{2 \, ||}=\frac{\sigma}{\epsilon_o}$. and similarly for $H$. (parallel to $\hat{n}$). Meanwhile, the perpendicular $E_1$ and $E_2$ are equal. (parallel to the surface).

How shall we show field is continuous at other points?