Sommerfeld Radiation Condition

Definition¹

Let’s refer to $u$ as a time-harmonic wave. The following condition is known as the Sommerfeld radiation condition.

$$ \lim \limits_{r \to \infty} r \left( \dfrac{\partial u}{\partial r} - ik u \right) = 0 $$

Explanation

The Sommerfeld radiation condition is a criterion that physically feasible solutions of the Helmholtz equation must satisfy. It was proposed by the German physicist Sommerfeld in his 1912 paper Die greensche Funktion der Schwingungsgleichung (The Green’s Function of the Vibrational Equation). In other words, if a solution of the Helmholtz equation does not meet the Sommerfeld radiation condition, it can be considered as physically meaningless.

This condition means that waves must move away from the source; they should not turn back. Consider throwing a stone into a lake. Waves will spread out from the point where the stone fell into the water. We know from reality that waves suddenly making a U-turn and moving backward toward the center never happens.

The Sommerfeld radiation condition imposes constraints to ensure solutions accurately reflect the situation described above. Even if a certain $u(x)$ theoretically solves the Helmholtz equation, it would not hold physical meaning if it describes waves moving in the reverse direction. Let’s apply this condition to a simple scenario.

Example

If $u(r)$ is referred to as a [spherical wave], the following two solutions satisfy the Helmholtz equation $\Delta u + k^{2}u=0$.

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

It is easy to verify by substitution. In spherical coordinates, the Laplacian is as follows.

$$ \nabla ^{2} u = \Delta u = \frac{1}{r^{2}}\frac{\partial}{\partial r} \left( r^{2}\frac{\partial u}{\partial r} \right) + \frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial u}{\partial \theta} \right) + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2} u}{\partial^{2} \phi} $$

Since $u$ is referred to as a spherical wave, it is independent of the angle, leaving only the term related to the radius.

$$ \nabla ^{2} u = \Delta u = \frac{1}{r^{2}}\frac{d}{d r} \left( r^{2}\frac{d u}{d r} \right) = \dfrac{2}{r}\dfrac{du}{dr} + \dfrac{d^{2}u}{dr^{2}} $$

Substituting $(1)$ into the Helmholtz equation shows it is a solution.

$$ \begin{align*} & \Delta u + k^{2}u \\ =&\ \dfrac{2}{r}\dfrac{du}{dr} + \dfrac{d^{2}u}{dr^{2}} + k^{2}u \\ =&\ \dfrac{2}{r}\dfrac{d}{dr}\left( \dfrac{e^{\pm ikr}}{r} \right) + \dfrac{d^{2}}{dr^{2}}\left( \dfrac{e^{\pm ikr}}{r} \right) + k^{2}\left( \dfrac{e^{\pm ikr}}{r} \right) \\ =&\ \left( \pm 2ik\dfrac{e^{\pm ikr}}{r^{2}} -2\dfrac{e^{\pm ikr}}{r^{3}} \right) + \dfrac{d}{dr}\left( \pm ik \dfrac{e^{\pm ikr}}{r} - \dfrac{e^{\pm ikr}}{r^{2}} \right) + k^{2}\dfrac{e^{\pm ikr}}{r} \\ =&\ \left( \pm 2ik\dfrac{e^{\pm ikr}}{r^{2}} -2\dfrac{e^{\pm ikr}}{r^{3}} \right) + \left( - k^{2} \dfrac{e^{\pm ikr}}{r} \mp ik \dfrac{e^{\pm ikr}}{r^{2}} \mp ik \dfrac{e^{\pm ikr}}{r^{2}} +2 \dfrac{e^{\pm ikr}}{r^{3}} \right) + k^{2}\dfrac{e^{\pm ikr}}{r} \\ =&\ \left( {\color{red}\cancel{\color{black}\pm 2ik\dfrac{e^{\pm ikr}}{r^{2}}}} {\color{blue}\bcancel{\color{black}-2\dfrac{e^{\pm ikr}}{r^{3}}}} \right) + \left( {\color{green}\cancel{\color{black}- k^{2} \dfrac{e^{\pm ikr}}{r}}} {\color{red}\cancel{\color{black}\mp ik \dfrac{e^{\pm ikr}}{r^{2}}}} {\color{red}\cancel{\color{black}\mp ik \dfrac{e^{\pm ikr}}{r^{2}}}} +{\color{blue}\bcancel{\color{black}2 \dfrac{e^{\pm ikr}}{r^{3}}}} \right) {\color{green}\cancel{\color{black}+ k^{2}\dfrac{e^{\pm ikr}}{r} }} \\ =&\ 0 \end{align*} $$

In complex wave functions, $u_{+}(r)=\dfrac{e^{ikr}}{r}$ represents a wave moving in the direction where $r$ increases, while $u_{-}(r)=\dfrac{e^{-ikr}}{r}$ represents a wave moving in the direction where $r$ decreases. Therefore, $u_{+}$ means a wave moving straightforward, and $u_{-}$ represents a reverse-moving wave. If only $u_{+}$ satisfies the radiation condition, it can be said that the radiation condition properly describes the nature of physical solutions. Verification shows as follows.

$$ \begin{align*} \lim \limits_{r \to \infty} r \left( \dfrac{du_{+}}{dr} - ik u_{+} \right) =&\ \lim \limits_{r \to \infty} r \left( \dfrac{d}{dr}\dfrac{e^{ikr}}{r} - ik \dfrac{e^{ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} r \left( ik\dfrac{e^{ikr}}{r} - \dfrac{e^{ikr}}{r^{2}} - ik \dfrac{e^{ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} r \left( - \dfrac{e^{ikr}}{r^{2}} \right) \\ =&\ \lim \limits_{r \to \infty} - \dfrac{e^{ikr}}{r} \\ =&\ - \lim \limits_{r \to \infty} \dfrac{\cos kr + i \sin kr}{r} \\ =&\ 0 \end{align*} $$

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

$$ \begin{align*} \lim \limits_{r \to \infty} r \left( \dfrac{du_{-}}{dr} - ik u_{-} \right) =&\ \lim \limits_{r \to \infty} r \left( \dfrac{d}{dr}\dfrac{e^{-ikr}}{r} - ik \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} r \left( -ik\dfrac{e^{-ikr}}{r} - \dfrac{e^{-ikr}}{r^{2}} - ik \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} \left( -2ike^{-ikr} - \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} \left( -2ike^{-ikr} \right) - \lim \limits_{r \to \infty} \left( \dfrac{e^{-ikr}}{r} \right) \\ =&\ \lim \limits_{r \to \infty} \left( -2ike^{-ikr} \right) \\ =&\ -2ik\lim \limits_{r \to \infty} \left( \cos kr -i \sin kr \right) \end{align*} $$

Since the equation diverges, $u_{-}$ does not satisfy the radiation condition.