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Silver-Müller Radiation Condition

Silver-Müller Radiation Condition

Definition1

Let’s define EE as the electric field. The following condition is called the Silver-Mueller radiation condition. When xR3,r=xx\in \mathbb{R}^{3}, r = \left| x \right|,

limrr[(×E)×xrE]=0 \lim \limits_{r \to \infty} r \left[ (\nabla \times E) \times x - r E \right] = 0

Explanation

The Silver-Mueller radiation condition is a necessary condition for the scattering problem of electromagnetic waves for electromagnetic waves to satisfy.

Explanation

The Sommerfeld radiation condition, proposed in the paper Die greensche Funktion der Schwingungsgleichung (The Green’s function of the wave equation) by the German physicist Sommerfeld in 1912, is a condition that a physically possible solution to the Helmholtz equation must satisfy. In other words, if a solution to the Helmholtz equation does not meet the Sommerfeld radiation condition, it can be considered physically meaningless.

This condition implies that the wave must move away from the source, meaning, waves should only propagate forward and never reverse. Consider throwing a pebble into a lake. Then, ripples will spread out from the point where the pebble falls. It’s evident that ripples do not suddenly make a U-turn and start moving towards the center in reality.

The Sommerfeld radiation condition imposes restrictions to ensure that solutions adhere to the described scenario. Even if some u(x)u(x) mathematically solves the Helmholtz equation, if it describes a wave moving in reverse, it lacks physical meaning. Let’s apply this condition to a simple situation.

Example

If u(r)u(r) is called a [spherical wave], then the following two solutions solve the Helmholtz equation Δu+k2u=0\Delta u + k^{2}u=0.

u(r)=e±ikrr \begin{equation} u(r) = \dfrac{e^{\pm ikr}}{r} \end{equation}

This can be easily verified by substitution. In the spherical coordinate system, the Laplacian is as follows.

2u=Δu=1r2r(r2ur)+1r2sinθθ(sinθuθ)+1r2sin2θ2u2ϕ \nabla ^{2} u = \Delta u = \frac{1}{r^{2}}\frac{\partial}{\partial r} \left( r^{2}\frac{\partial u}{\partial r} \right) + \frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial u}{\partial \theta} \right) + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2} u}{\partial^{2} \phi}

Since uu is called a spherical wave, it’s irrelevant to angles, leaving only the term related to radius in the Laplacian.

2u=Δu=1r2ddr(r2dudr)=2rdudr+d2udr2 \nabla ^{2} u = \Delta u = \frac{1}{r^{2}}\frac{d}{d r} \left( r^{2}\frac{d u}{d r} \right) = \dfrac{2}{r}\dfrac{du}{dr} + \dfrac{d^{2}u}{dr^{2}}

Now, substituting (1)(1) into the Helmholtz equation shows that it is a solution.

Δu+k2u= 2rdudr+d2udr2+k2u= 2rddr(e±ikrr)+d2dr2(e±ikrr)+k2(e±ikrr)= (±2ike±ikrr22e±ikrr3)+ddr(±ike±ikrre±ikrr2)+k2e±ikrr= (±2ike±ikrr22e±ikrr3)+(k2e±ikrrike±ikrr2ike±ikrr2+2e±ikrr3)+k2e±ikrr= (±2ike±ikrr22e±ikrr3)+(k2e±ikrrike±ikrr2ike±ikrr2+2e±ikrr3)+k2e±ikrr= 0 \begin{align*} & \Delta u + k^{2}u \\ =&\ \dfrac{2}{r}\dfrac{du}{dr} + \dfrac{d^{2}u}{dr^{2}} + k^{2}u \\ =&\ \dfrac{2}{r}\dfrac{d}{dr}\left( \dfrac{e^{\pm ikr}}{r} \right) + \dfrac{d^{2}}{dr^{2}}\left( \dfrac{e^{\pm ikr}}{r} \right) + k^{2}\left( \dfrac{e^{\pm ikr}}{r} \right) \\ =&\ \left( \pm 2ik\dfrac{e^{\pm ikr}}{r^{2}} -2\dfrac{e^{\pm ikr}}{r^{3}} \right) + \dfrac{d}{dr}\left( \pm ik \dfrac{e^{\pm ikr}}{r} - \dfrac{e^{\pm ikr}}{r^{2}} \right) + k^{2}\dfrac{e^{\pm ikr}}{r} \\ =&\ \left( \pm 2ik\dfrac{e^{\pm ikr}}{r^{2}} -2\dfrac{e^{\pm ikr}}{r^{3}} \right) + \left( - k^{2} \dfrac{e^{\pm ikr}}{r} \mp ik \dfrac{e^{\pm ikr}}{r^{2}} \mp ik \dfrac{e^{\pm ikr}}{r^{2}} +2 \dfrac{e^{\pm ikr}}{r^{3}} \right) + k^{2}\dfrac{e^{\pm ikr}}{r} \\ =&\ \left( {\color{red}\cancel{\color{black}\pm 2ik\dfrac{e^{\pm ikr}}{r^{2}}}} {\color{blue}\bcancel{\color{black}-2\dfrac{e^{\pm ikr}}{r^{3}}}} \right) + \left( {\color{green}\cancel{\color{black}- k^{2} \dfrac{e^{\pm ikr}}{r}}} {\color{red}\cancel{\color{black}\mp ik \dfrac{e^{\pm ikr}}{r^{2}}}} {\color{red}\cancel{\color{black}\mp ik \dfrac{e^{\pm ikr}}{r^{2}}}} +{\color{blue}\bcancel{\color{black}2 \dfrac{e^{\pm ikr}}{r^{3}}}} \right) {\color{green}\cancel{\color{black}+ k^{2}\dfrac{e^{\pm ikr}}{r} }} \\ =&\ 0 \end{align*}

In this context, u+(r)=eikrru_{+}(r)=\dfrac{e^{ikr}}{r} in the complex wave function is a wave that progresses in the direction where rr increases, and u(r)=eikrru_{-}(r)=\dfrac{e^{-ikr}}{r} progresses in the direction where rr decreases. Thus, u+u_{+} represents a wave moving straight, and uu_{-} represents a wave moving backwards. If only u+u_{+} satisfies the radiation condition, it’s reasonable to say the radiation condition well describes the nature of a physical solution. It can be verified as follows.

limrr(du+driku+)= limrr(ddreikrrikeikrr)= limrr(ikeikrreikrr2ikeikrr)= limrr(eikrr2)= limreikrr= limrcoskr+isinkrr= 0 \begin{align*} \lim \limits_{r \to \infty} r \left( \dfrac{du_{+}}{dr} - ik u_{+} \right) =&\ \lim \limits_{r \to \infty} r \left( \dfrac{d}{dr}\dfrac{e^{ikr}}{r} - ik \dfrac{e^{ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} r \left( ik\dfrac{e^{ikr}}{r} - \dfrac{e^{ikr}}{r^{2}} - ik \dfrac{e^{ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} r \left( - \dfrac{e^{ikr}}{r^{2}} \right) \\ =&\ \lim \limits_{r \to \infty} - \dfrac{e^{ikr}}{r} \\ =&\ - \lim \limits_{r \to \infty} \dfrac{\cos kr + i \sin kr}{r} \\ =&\ 0 \end{align*}

Therefore, u+u_{+} satisfies the radiation condition.

limrr(dudriku)= limrr(ddreikrrikeikrr)= limrr(ikeikrreikrr2ikeikrr)= limr(2ikeikreikrr)= limr(2ikeikr)limr(eikrr)= limr(2ikeikr)= 2iklimr(coskrisinkr) \begin{align*} \lim \limits_{r \to \infty} r \left( \dfrac{du_{-}}{dr} - ik u_{-} \right) =&\ \lim \limits_{r \to \infty} r \left( \dfrac{d}{dr}\dfrac{e^{-ikr}}{r} - ik \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} r \left( -ik\dfrac{e^{-ikr}}{r} - \dfrac{e^{-ikr}}{r^{2}} - ik \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} \left( -2ike^{-ikr} - \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} \left( -2ike^{-ikr} \right) - \lim \limits_{r \to \infty} \left( \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} \left( -2ike^{-ikr} \right) \\ =&\ -2ik\lim \limits_{r \to \infty} \left( \cos kr -i \sin kr \right) \end{align*}

Since the above equation diverges, uu_{-} does not satisfy the radiation condition.


  1. David Colton and Rainer Kress, Inverse Acoustic and Electromagnetic Scattering Theory (4th Edition, 2019), p3-5 ↩︎