Is the Sommerfeld radiation condition equivalent to the requirement that the Poyting vector points outward?

In the context of classical electromagnetism with spatially bounded time-varying sources $$J(x)$$, the usual boundary condition that we impose at spatial infinity is the Sommerfeld radiation condition $$\lim_{r \to \infty} \left [ r \left( \frac{\partial}{\partial r} - i k \right) A({\bf x}) \right] = 0,$$ where $$A(x)$$ is the electromagnetic four-potential. Physically, this means that electromagnetic waves must be propagating outward rather than inward at spatial infinity.

It seems to me that a much more intuitive way to formalize this physical requirement is in terms of the direction of the Poynting vector. Physically, the Poynting vector should point outward rather than inward at spatial infinity. We can encode this requirement by requiring that the pseudoscalar triple product $$({\bf E} \times {\bf B}) \cdot \left (r^2 {\bf r} \right)$$ be positive-semidefinite at spatial infinity: $$\liminf_{r \to \infty} \left[ r^2 ({\bf E} \times {\bf B}) \cdot {\hat r} \right] = 0.$$

(It doesn't quite work to just require that $$({\bf E} \times {\bf B}) \cdot {\bf r}$$ be nonnegative outside of a sufficiently large radius, because the Poynting vector for an EM plane wave vanishes on certain hypersurfaces that extend arbitrarily far from the origin. The subleading $$O(1/r^3)$$ terms dominate near those hypersurfaces, and (I think) they can have either sign. But these (usually) subleading corrections will decay as $$1/r$$ in the limit above and so will not affect the lim inf.)

Is this boundary condition equivalent to the Sommerfeld radiation condition? If so, is there any reason why the SRC is preferable?

This is not a full answer to your question but it is too long, especially because of the equations, for a comment.

The Sommerfeld radiation conditions, see [1], are used to ensure that a pair of surface integrals $$\int_{\partial \mathcal V(r)}\left(\mathfrak j \omega \mu_0 \left[\hat {\mathbf r} \times \mathbf H+\eta_0 \mathbf E\right]+\frac{\mathbf E}{r}\right) \frac{e^{-\mathfrak j k r}}{r}dS \tag{1}\label{1}$$ and $$\int_{\partial \mathcal V(r)}\left(\mathfrak j \omega \epsilon_0 \left[\hat {\mathbf r} \times \mathbf E-\zeta_0 \mathbf H\right]+\frac{\mathbf H}{r}\right) \frac{e^{-\mathfrak j k r}}{r}dS \tag{2}\label{2}$$ be convergent as $$r \to \infty$$ where $$\mathcal V(r)$$ is a sphere of radius $$r$$ centered at the origin and $$\zeta_0 =\eta_0^{-1}= \sqrt{ \frac{\mu_0}{\epsilon_0} } =120\pi\Omega$$ is the impedance of space.

The Sommerfeld radiation condition is then a sufficient condition for these integrals be convergent. For $$\eqref{1}$$ the condition is the pair $$\lim_{r\to\infty} r\mathbf E = \text{finite} \tag{3a}\label{3a}$$ $$\lim_{r\to\infty}r\left[\hat {\mathbf r} \times \mathbf H+\eta_0 \mathbf E\right]=0 \tag{3b}\label{3b}$$ and for $$\eqref{2}$$ the condition is the pair $$\lim_{r\to\infty} r\mathbf H = \text{finite} \tag{4a}\label{4a}$$ $$\lim_{r\to\infty}r\left[\hat {\mathbf r} \times \mathbf E-\zeta_0 \mathbf H\right]=0 \tag{4b}\label{4b}$$ and neither the $$\eqref{3a},\eqref{3b}$$ nor $$\eqref{4a}, \eqref{4b}$$ is necessary mathematically, although I do not know a single physical case where they would not hold.

Take the scalar product of $$\eqref{3b}$$ and $$\eqref{4b}$$ with $$\hat {\mathbf r}$$, and get $$\lim_{r\to\infty}\mathbf r \cdot \mathbf E=0 \tag{5b}\label{5b}$$ $$\lim_{r\to\infty}\mathbf r \cdot \mathbf H=0 \tag{6b}\label{6b}$$

Evidently, these $$\eqref{5b}, \eqref{6b}$$ are stronger than $$\eqref{3a}, \eqref{4a}$$, resp., and they express the asymptotically plane wave nature of the outgoing radiation. Hence, conventionally, the $$\eqref{3b}\eqref{4b}$$ are taken as the Sommerfeld radiation conditions that are sufficient to assure that the surface integrals $$\eqref{1},\eqref{2}$$ be convergent.

It seems to me that your condition $$\lim \inf_{r \to \infty} r^2(\mathbf E \times \mathbf H) \cdot \hat {\mathbf r}=0 \tag{7}$$ is also another sufficient condition but I admit I do not know how to show that it be so for either of the integrals $$\eqref{1}, \eqref{2}$$.

[1] Silver: Microwave Antenna Theory and Design, sections 3.8 and 3.9 https://archive.org/details/dli.ernet.15390/page/83/mode/2up?view=theater

• What's the motivation for equations (1) and (2), and where do those integrands come from? Do you have a reference? Commented Dec 15, 2023 at 14:59
• my bad, see sections 3.8 and 3.9 , eqs 108 - 113 in archive.org/details/dli.ernet.15390/page/83/mode/… Commented Dec 15, 2023 at 16:28

No, they are not just inequivalent, but your choice would not even capture the correct physics (but it should be fix-able).

The SRC is written with the $$\dfrac{\partial\ }{\partial r}-ik$$; this means that, at infinity, the behaviour is dominated by $$\dfrac{e^{+ikr}}r$$, which can be picked up in the integration over the surface at infinity. You needed the $$-ik$$ to make the SRC zero, and also pick out one specific wavelength of the outgoing wave.

Instead, what $$\liminf_{r\to\infty}[r^2(\vec E\times\vec B)\cdot\hat{\vec r}]=0$$ implies, is that there is no net radiation at all. It says that the Poynting vector integrated over the spherical surface at infinity is zero. You need to fix this in some way to assert that the energy is going out.

Actually, it should be rather easy for you to simply compute $$\vec E$$ and $$\vec B$$ from a $$\vec A$$ satisfying SRC, and thereby figure out the correct replacement condition that will be equivalent. Whether you would continue to consider it as SRC or not, you can choose. I do agree, however, that Poynting vector version of SRC would be interesting to have at hand.

• I don't think this is correct - I didn't say "$\lim$", I said "$\liminf$". The "inf" is critical. I believe that the triple product inside the limit varies like $\cos^2(kr - \omega t)$ at large distances. So it doesn't have a limit at all, but it has a lim sup and a lim inf - and you want the whole limiting range to be nonnegative, which requires that the lim inf must be zero since it can't be positive. Commented Dec 15, 2023 at 5:41
• Note that even for simple EM plane wave, which certainly transmits net radiation, there are infinitely many planes perpendicular to the wave vector at which the Poynting vector vanishes. Commented Dec 15, 2023 at 5:44
• Ok, I see what you are trying to do, but then you would lose the simplicity of having monochromatic waves of defined $k$. As for your simple EM plane wave, it is a superposition of ingoing and outgoing waves, so it does not satisfy the SRC. It is not clear to me what it is you think you are getting as extra goodness out of this. Commented Dec 15, 2023 at 5:48
• Why was this downvoted? His answer maybe right or maybe wrong but it deserves at least a cogent comment to explain the downvote. Commented Dec 15, 2023 at 16:36