Stratton-Chu solution, special case

[Unconscious]

I will try to ask the question, saving as much calculations as possible, so as not to weigh down those who want to try to help me.
Starting from the general electromagnetic problem in empty space, taken as a domain a volume V delimited by a closed surface S, Elliot (1) shows how the field (i.e. the solution of the problem) at each internal point in V depends only on the sources inside V and from the field values on S:

$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S\left (\mathbf{1}_n\cdot\mathbf{E} \right )\nabla'\psi+\left(\mathbf{1}_n\times\mathbf{E}\right )\times\nabla'\psi-j\omega\psi\left(\mathbf{1}_n\times\mathbf{B}\right )\mathrm{d}S$$

$$\mathbf{B}(x,y,z)=\frac{1}{4\pi}\int_{V}\frac{\mathbf{J}}{\mu_0^{-1}}\times\nabla'\psi \; \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S\left (\mathbf{1}_n\cdot\mathbf{B} \right )\nabla'\psi+\left(\mathbf{1}_n\times\mathbf{B}\right )\times\nabla'\psi+\frac{j\omega\psi}{c^2}\left(\mathbf{1}_n\times\mathbf{E}\right )\mathrm{d}S$$where:

1. $\rho$ e $\mathbf{J}$ are the sources, so they are known,

2. $\psi=\frac{e^{-jkR}}{R}$ and $R=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$,

3. $\nabla'$ operates on variables $x',y',z'$,

4. $\mathbf{1}_n$ is the unit vector perpendicular to the surface S in all its points, inward in V.That said, let's detail this solution when S is a sphere of radius R (large enough to contain all the sources existing in the universe):

$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}+\frac{\mathbf{E}}{c} \right )+ \frac{\mathbf{E}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S$$$$\mathbf{B}(x,y,z)=\frac{1}{4\pi}\int_{V}\frac{\mathbf{J}}{\mu_0^{-1}}\times\nabla'\psi \; \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S \left[ \frac{j\omega}{c^2} \left ( \mathbf{1}_n\times\mathbf{E}+c\mathbf{B} \right )+ \frac{\mathbf{B}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S$$In this particular case $-\mathbf{1}_n$ is the unit radial vector as in spherical coordinates.
Now, if the field respects the Sommerfeld conditions, so (as a consequence) if by hypothesis:

$$\lim_{R\to\infty} \int_S (...)\mathrm{d}S=\mathbf{0}\quad (*)$$

then, since the volume integrals do not depend on S (I assumed that in the sphere from which we started there were already all the sources inside) and since (*) holds, we can pass to the limit to both members (both for the field E that for H) obtaining:

$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V$$$$\mathbf{B}(x,y,z)=\frac{1}{4\pi}\int_{V}\frac{\mathbf{J}}{\mu_0^{-1}}\times\nabla'\psi \; \mathrm{d}V$$that is, obtaining that the two surface integrals (the one for E and the one for H) are both null, both for the starting sphere considered, and for any other sphere with a larger radius. Since it applies to an infinity of spheres, then the integrands must be null, that is:

$$\left\{\begin{matrix}
j\omega \left ( -\mathbf{1}_n\times\mathbf{B}+\frac{\mathbf{E}}{c} \right )+ \frac{\mathbf{E}}{R}=\mathbf{0}\\
\frac{j\omega}{c^2} \left ( \mathbf{1}_n\times\mathbf{E}+c\mathbf{B} \right )+ \frac{\mathbf{B}}{R}=\mathbf{0}
\end{matrix}\right.$$

that written in a better way:

$$\left\{\begin{matrix}
\mathbf{1}_n\times\mathbf{B}= \left (\frac{1}{j\omega R}+\frac{1}{c} \right )\mathbf{E}\\
-\mathbf{1}_n\times\mathbf{E}= \left (\frac{c^2}{j\omega R}+c \right )\mathbf{B}\\
\end{matrix}\right.$$

from which I conclude, multiplying the first from the left by
$\mathbf{1}_n\times$:

$$-\mathbf{B}_{\perp}= -\left (\frac{1}{j\omega R}+\frac{1}{c} \right )\left (\frac{c^2}{j\omega R}+c \right )\mathbf{B}\implies \mathbf{B}=\mathbf{0}$$

which cannot be true. Did I make a mistake?

References:

(1) Robert S. Elliot, Antenna theory and Design, Wiley, IEEE Press, 2003, sect. 1.7, p. 17, ..., 21.

 

[Paul Colby]

Did I make a mistake?

Clearly.

In the universe you describe the fields far from all sources are radiating AKA obeying the Sommerfeld condition. At large distance the fields are falling off as $1/R$, the distance to the sources. At each point in this distant region the fields look like plane waves with $\mathbf{E}$ and $\mathbf{B}$ orthogonal and orthogonal to the direction of propagation, $\hat{\mathbf{R}}$, the outward directed unit vector. So,

$\mathbf{E}\times\mathbf{B} \sim \hat{\mathbf{R}}$

Now, this only holds in the far field, not for every sphere.

 

[Unconscious]

Did you read my mathematical steps?

I obviously know that in far field the EM wave is a spherical wave that appears locally plane, and so on...
My question is about mathematics, in particular I cannot see any error in the steps in the first message.Thanks for the reply, anyway.

 

[Paul Colby]

Your last equation is wrong. It should read,

$-\mathbf{1}_n \times (\mathbf{1}_n \times \mathbf{B}) = \mathbf{B} \ne 0$

you can drop the $\frac{1}{R}$ terms.

in slow motion.

$-\mathbf{1}_n\times(\mathbf{1}_n\times\mathbf{B}) = -\mathbf{1}_n\times(\frac{1}{c}\mathbf{E}) = c \frac{1}{c} \mathbf{B} = \mathbf{B} \ne 0$

 

[Unconscious]

you can drop the 1R\frac{1}{R} terms.

Mathematically speaking, why?

 

[Paul Colby]

$\mathbf{B} \sim \frac{1}{R}$

$\frac{1}{R}\mathbf{B} \sim \frac{1}{R^2} << \frac{1}{R}$

even better argument $\frac{1}{R} << c$

 

[Unconscious]

Apparently I can't explain myself, I apologize for this.
I retry in this way: in all my math steps, there is no approximation, no far field hypothesis, no other non-rigorous things. All the steps are exact in the strict sense. We can't introduce 'apriori' the fact that $\mathbf{B} \sim \frac{1}{R}$, hopefully it should pop up as a consequence.
So, staying in mathematical accuracy without introducing approximations, it seems that one eventually gets:

$$\left (\frac{1}{j\omega R}+\frac{1}{c} \right )\left (\frac{c^2}{j\omega R}+c \right )=1,\quad \forall R>R_0$$

that is an absurd.
Once here, it makes no sense to introduce ad-hoc hypotheses to get the accounts back (since I repeat, these steps are exact, they must be coherent without any 'help').
I don't know if this time I was able to explain myself, I hope so.

 

[Paul Colby]

When do you apply the limit in equation (*) of #1? The Sommerfeld condition is an asymptotic one based on assumptions about the geometry. Math can't make physical arguments for you.

 

[Paul Colby]

$\lim_{R\rightarrow \infty} \left(\frac{1}{j\omega R}+\frac{1}{c}\right)\left(\frac{c^2}{j\omega R}+c\right) = 1$

How is this a'priori or ad-hoc? Seems to be just true, right?

 

[Unconscious]

When do you apply the limit in equation (*) of #1?

When I pass from these:
$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}+\frac{\mathbf{E}}{c} \right )+ \frac{\mathbf{E}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S$$$$\mathbf{B}(x,y,z)=\frac{1}{4\pi}\int_{V}\frac{\mathbf{J}}{\mu_0^{-1}}\times\nabla'\psi \; \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S \left[ \frac{j\omega}{c^2} \left ( \mathbf{1}_n\times\mathbf{E}+c\mathbf{B} \right )+ \frac{\mathbf{B}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S$$to these:

$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V$$$$\mathbf{B}(x,y,z)=\frac{1}{4\pi}\int_{V}\frac{\mathbf{J}}{\mu_0^{-1}}\times\nabla'\psi \; \mathrm{d}V$$

The Sommerfeld condition is an asymptotic one based on assumptions about the geometry. Math can't make physical arguments for you.

In fact I use the Sommerfeld condition only when I do the limit $R\to\infty$ to both members of each equality.
After the limit, the new two equalities obtained are exact, not asymptotics.

 

[Unconscious]

$\lim_{R\rightarrow \infty} \left(\frac{1}{j\omega R}+\frac{1}{c}\right)\left(\frac{c^2}{j\omega R}+c\right) = 1$

How is this a'priori or ad-hoc? Seems to be just true, right?

Yes, but there is no need to do another time the limit $R\to\infty$. I did the first time the limit $R\to\infty$ only to prove that $\forall R>R_0$ ($R_0$ is a radius of a sphere containing all sources) the surface integrals are zero.

 

[Paul Colby]

I think you need to study limits and asymptotic arguments. I don't think I can teach you this.

 

[Unconscious]

Let's take one of the two, to make a more detailed explanation. We are in a sphere S of radius $R_0$ in which all sources are present, and we know that:

$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V+$$
$$+\frac{1}{4\pi}\int _S \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}+\frac{\mathbf{E}}{c} \right )+ \frac{\mathbf{E}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S$$

Now:

$$\underbrace{\lim_{R\to\infty}\mathbf{E}(x,y,z)}_{(1)}=\underbrace{\lim_{R\to\infty}\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V}_{(2)}+$$
$$+\underbrace{\lim_{R\to\infty}\frac{1}{4\pi}\int _S \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}+\frac{\mathbf{E}}{c} \right )+ \frac{\mathbf{E}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S}_{(3)}$$

$(1) = \mathbf{E}(x,y,z)$ because $\mathbf{E}(x,y,z)$ is independent of the radius R of the sphere.
$(2)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V$ because the sources are all inside the sphere of radius $R_0$, so the integral in not sensitive to the limit.
$(3) = 0$ for Sommerfeld.Therefore, in summary:

$$\mathbf{E}(x,y,z)=\frac{1}{4\pi}\int_{V}\left(\frac{\rho}{\epsilon_0}\nabla'\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right) \mathrm{d}V$$

that is not an asymptotic equality for the field E, it is valid for every (x,y,z).

Because, at the same time, also the first equation is valid (in particular it is valid for all (x,y,z) inside the sphere of radius $R_0$), we must have (as you can see, R is inside the integral, it is not the same R in the limit, it is not the radius of any sphere but it is $R=\sqrt{(x−x′)^2+(y−y′)^2+(z−z′)^2}$; I should have used a more detailed notation):

$$\frac{1}{4\pi}\int _S \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}+\frac{\mathbf{E}}{c} \right )+ \frac{\mathbf{E}}{R}\right ] \frac{e^{-jkR}}{R}\mathrm{d}S=0$$

for any spherical surface S of radius bigger than $R_0$.

 

[Unconscious]

I think you need to study limits and asymptotic arguments. I don't think I can teach you this.

Trust me, I'm not that ignorant.
I tried to explain myself better in my last message.

 

[Paul Colby]

One parting comment.

The Summerfeld condition only hold asymptotically and you've assumed (or think you've proven) it for all $R > R_o$ for which it clearly doesn't hold.

What you may have proven is that assuming the Summerfeld condition holds for all $R > R_o$ really means $\mathbf{B}=0$.

 

[Paul Colby]

Your last equation is false. It only holds as an equality in the limit $R\rightarrow \infty$.

Nope, I take it back. I believe your argument. What's your question?

 

[Paul Colby]

I think I've seen the identity though not exactly in the form given. There is an integral identity that holds between two source free solutions of ME on a region. If the region is multi connected, say that between two spheres, then one may show the inner integral vanishes if both solutions obey the radiation condition at infinity. I think in what's shown the Green function acts as the second solution which obeys the radiation condition by construction while the first obeys the radiation condition by assumption.

 

[Paul Colby]

The identity is as follows assuming time harmonic fields.

let $F = E_1 \times H_2 - E_2\times H_1$

It follows from Maxwell's equations that $\nabla \cdot F = 0$.

Therefore, for any region, $V$

$\iiint_V \nabla\cdot F dV = \iint_{\partial V} F\cdot dS = 0$

by the divergence theorem. Let $V$ be the sphere (any surface will do) I describe and let the outer sphere expand to infinity and beyond. On this distant sphere, $F=0$, point wise asymptotically.

 

[Unconscious]

Nope, I take it back. I believe your argument.

I am not sure if I have understood this: now do you see my question more reasonable?

Anyway, I think we can approach this problem in a more visual form, maybe... now I try.

The initial problem is this:

where, as we know, we can find the field (let's talk only about the electric field for greater readability) as:

$$\mathbf{E}_0(\mathbf{r})=\mathbf{E}_0(x,y,z)=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\frac{1}{4\pi}\int _{S_0} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}(\mathbf{r}')+\frac{\mathbf{E}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'\quad (0)$$

that is valid for every P=(x,y,z) in the interior (strict sense) of $S_0$.
So, as you can see, it is necessary to know the value of the field in every point of the surface $S_0$ (not necessarly a spherical surface), otherwise the surface integral cannot be solved. Let us assume that the field at any point on the surface is known. This information, in addition to solving the problem by allowing integrals to be carried out, allows the uniqueness theorem to be applied in the volume enclosed by $S_0$.
So, the field that will be calculated with the formula (0) will be unique in $V_0$.
I think you have no problem agreeing with me so far.

Next step. Let's consider a new surface, bigger than $S_0$, call it $S_1$:

with the same reasoning, we can conclude that in every point (x,y,z) in the interior of $S_1$, that is, in $V_1\supset V_0$, the following formula holds:

$$\mathbf{E}_1(\mathbf{r})=\frac{1}{4\pi}\int_{V_1}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\frac{1}{4\pi}\int _{S_1} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}(\mathbf{r}')+\frac{\mathbf{E}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'\quad (1)$$

and the field, knowing its value on the surface $S_1$ is unique in $V_1$.
In particular, if in the first problem ($S_0,V_0$) we use as boundary conditions of the field $\mathbf{E}_0(x,y,z)$ on $S_0$ those obtained by using (1) at points (x, y, z) of $S_0$, we obtain, for the uniqueness that also existed in the first problem, that for all the points of $V_0$ both (0) and (1) must give the same values. So, in other words, $\mathbf{E}_1(x,y,z)=\mathbf{E}_0(x,y,z)$ for all (x,y,z) in $V_0$.
Note also that in (1), $\int_{V_1}=\int_{V_0}$, because sources are all in $V_0$.
After these considerations, (0) can be more precisely written as:

$$\mathbf{E}_0(\mathbf{r})=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\frac{1}{4\pi}\int _{S_0} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_1(\mathbf{r}')+\frac{\mathbf{E}_1(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_1(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'\quad (0)$$

Take a new surface, do the same thing:

$$\mathbf{E}_2(\mathbf{r})=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\frac{1}{4\pi}\int _{S_2} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}(\mathbf{r}')+\frac{\mathbf{E}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'\quad (2)$$

and rewrite (1) as:

$$\mathbf{E}_1(\mathbf{r})=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\frac{1}{4\pi}\int _{S_1} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_2(\mathbf{r}')+\frac{\mathbf{E}_2(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_2(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'\quad (1)$$

Then $\mathbf{E}_2(\mathbf{r})=\mathbf{E}_1(\mathbf{r})$ for every (x,y,z) in $V_1$ and also $\mathbf{E}_2(\mathbf{r})=\mathbf{E}_1(\mathbf{r})=\mathbf{E}_0(\mathbf{r})$ for every (x,y,z) in $V_0$.Doing this thing n times, we have:

$$\mathbf{E}_n(\mathbf{r})=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\frac{1}{4\pi}\int _{S_n} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_{n+1}(\mathbf{r}')+\frac{\mathbf{E}_{n+1}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_{n+1}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'\quad (n)$$

where:

$\mathbf{E}_n(\mathbf{r})=\mathbf{E}_{n-1}(\mathbf{r}), \quad \forall \mathbf{r}\in V_{n-1}$,
$\mathbf{E}_n(\mathbf{r})=\mathbf{E}_{n-1}(\mathbf{r})=\mathbf{E}_{n-2}(\mathbf{r}), \quad \forall \mathbf{r}\in V_{n-2}$,
...
$\mathbf{E}_n(\mathbf{r})=\mathbf{E}_{n-1}(\mathbf{r})=...=\mathbf{E}_{0}(\mathbf{r}), \quad \forall \mathbf{r}\in V_{0}$.

Now, let's take (x,y,z) inside an arbitrary choosen $S_k$, with $k<n$. Formula (n) is valid to evaluate the field in every point of $V_k$. So, the field in every point of $V_k$ is indifferent to which formula (n) (with n>k) I will use to calculate it.
In particular, bringing $n$ to infinity we have:

$$\mathbf{E}(\mathbf{r})|_{\mathbf{r}\in V_k}=\lim_{n\to\infty}\mathbf{E}_n(\mathbf{r})|_{\mathbf{r}\in V_k}=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'+$$
$$+\lim_{n\to\infty}\frac{1}{4\pi}\int _{S_n} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_{n+1}(\mathbf{r}')+\frac{\mathbf{E}_{n+1}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_{n+1}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'$$

The last quantity is, by hypothesis that the field satisfies Sommerfeld conditions, null. So we have

$$\mathbf{E}(\mathbf{r})|_{\mathbf{r}\in V_k}=\frac{1}{4\pi}\int_{V_0}\left(\frac{\rho(\mathbf{r}')}{\epsilon_0}\nabla'\psi(\mathbf{r},\mathbf{r}')-j\omega\psi(\mathbf{r},\mathbf{r}')\frac{\mathbf{J}(\mathbf{r}')}{\mu_0^{-1}}\right) \mathrm{d}V'$$

Comparing this formula with the formula (k), we must have:

$$\int _{S_k} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_{k+1}(\mathbf{r}')+\frac{\mathbf{E}_{k+1}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_{k+1}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}\mathrm{d}S'=\mathbf{0}$$

Do you agree so far?
It seems to me that the reasoning is now crystal clear.

 

[Paul Colby]

Do you agree so far?

Yes, as I have said. Mystery solved. I make the following 3 observations.

1) The last identity which you derived in #19 is valid, but only as an integral relation.

2) Your original conclusion in #1 that the integrand vanishes point wise is clearly wrong in general.

3) In the extreme far field the integrand does vanish point wise but only in an asymptotic sense.

3.01) Oh, after skimming your post #19 more carefully your "limiting" argument seems kind of silly. Just use 2 points, one in the near field at $S_o$ and the other in the extreme far field, $S_\infty$. Then subtract the resulting expressions. Observe that $S_\infty$ vanishes asymptotically. Conclude the last expression in #19 holds at $S_o$.

 

[Paul Colby]

Do you agree so far?

Okay, some additional comments are needed. $E_0(x,y,z)$ is just $E(x,y,z)$ the field at a point. Equation (0) is an integral identity. The field at this point doesn't care where your integrations are performed provided the evaluation point is interior to the region. For the volume integrals it's important to realize the integrands are only non-zero where the sources are non-zero. Making a bigger surrounding sphere doesn't change the volume integrals at all.

 

[Unconscious]

For the volume integrals it's important to realize the integrands are only non-zero where the sources are non-zero. Making a bigger surrounding sphere doesn't change the volume integrals at all.

In fact, this is exactly what I used in #19.

Yes, as I have said.

Ok, wonderful.

2) Your original conclusion in #1 that the integrand vanishes point wise is clearly wrong in general.

Ok, right. Then, I think that rigorous mathematical way to view what happens in far fields is the following. Let's take as start point the last expression in #19, for $S_0$ for example, in the particular case in which $S_0$ is a sphere. Let's then write the surface integral in terms of solid angles:

$$\int _{4\pi} \left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_{1}(\mathbf{r}')+\frac{\mathbf{E}_{1}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_{1}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}R_0^2\mathrm{d}\Omega'=\mathbf{0}$$

We can take the limit $R_0\to\infty$ on both sides, and we can also note that the limit can pass the integral sign because it acts on a different variable:

$$\int _{4\pi} \lim_{R_0\to\infty}\left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_{1}(\mathbf{r}')+\frac{\mathbf{E}_{1}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_{1}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}R_0^2\mathrm{d}\Omega'=\mathbf{0}$$

Then we must have:

$$\lim_{R_0\to\infty}\left[ j\omega \left ( -\mathbf{1}_n\times\mathbf{B}_{1}(\mathbf{r}')+\frac{\mathbf{E}_{1}(\mathbf{r}')}{c} \right )+ \frac{\mathbf{E}_{1}(\mathbf{r}')}{R(\mathbf{r},\mathbf{r}')}\right ] \frac{e^{-jkR(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')}R_0^2=\mathbf{0}$$

otherwise the integral would not be zero.
Now we note that:
$$R(\mathbf{r},\mathbf{r}')=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=\sqrt{||\mathbf{r}-\mathbf{r}'||^2}=\sqrt{(\mathbf{r}-\mathbf{r}')\cdot (\mathbf{r}-\mathbf{r}')}=\sqrt{||\mathbf{r}'||^2+||\mathbf{r}||^2-2\mathbf{r}'\cdot \mathbf{r}}=||\mathbf{r}'||\sqrt{1+||\mathbf{r}||^2/||\mathbf{r}'||^2-2\mathbf{r}'\cdot \mathbf{r}/||\mathbf{r}'||^2}\underset{R_0\to\infty}{\sim} ||\mathbf{r}'||$$

because $||\mathbf{r}'||=R_0$ in the surface integral.
Turning back to the limit, and using this last fact, we obtain:

$$\left\{\begin{matrix}\lim_{R_0\to\infty}\left[ j\omega R_0\left ( -\mathbf{1}_n\times\mathbf{B}_{1}(\mathbf{r}')+\frac{\mathbf{E}_{1}(\mathbf{r}')}{c} \right )\right ] =\mathbf{0} \\ \lim_{R_0\to\infty}\mathbf{E}_{1}(\mathbf{r}')=\mathbf{0}\end{matrix}\right.$$

The first in particular can be equivalently written as:

$$j\omega \left( -\mathbf{1}_n\times\mathbf{B}_{1}(\mathbf{r}')+\frac{\mathbf{E}_{1}(\mathbf{r}')}{c}\right ) \underset{ R_0\to\infty }{=} o(1/R_0) $$

that is:

$$\mathbf{E}_{1}(\mathbf{r}')\underset{ R_0\to\infty }{\sim}c \left(\mathbf{1}_n\times\mathbf{B}_{1}(\mathbf{r}')\right)$$which is wonderfully beautiful and rigurous.

Thanks for letting me know where I was wrong.