User:Zhennan/3DsommerfeldRadiationCondition

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Sommorfeld Radiation condition (3D) is

as . is the upper surface of of closed surface , and is its radius.

Prove

The gradient of 3D causal free space Green function in spherical coordinate has the following form,

where is the radial distance from surface element to the origin, is the depression angle of and is its azimuthal angle. As , both and infinitely approaches the origin from the view of far field and thus at far field the outgoing reference wave,

\begin{equation} G_0^+(\textbf{r}',\textbf{r},\omega)=-\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'-\textbf{r}|}}{|\textbf{r}'-\textbf{r}|} \rightarrow -\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'-\textbf{0}|}}{|\textbf{r}'-\textbf{0}|}=-\frac{1}{4\pi}\frac{e^{ikr'}}{r'}=o(\frac{1}{r'}) \end{equation}

So at infinity, the inner product of and normal unit vector of becomes

\begin{equation} \nabla' G_0^+ \cdot \hat{n}'=\nabla ' G_0^+ \cdot \hat{r'}=\frac{\partial G_0^+}{\partial r'}=-\frac{1}{4\pi}(ik\frac{e^{ikr'}}{r'}-\frac{e^{ikr'}}{r'^2})=ikG_0^++o(\frac{1}{r'^2}) \end{equation}

Meanwhile, according to Lippmann-Schwinger equation, the field anywhere is

\begin{equation} P(\textbf{r}',\omega)=\int_\infty \rho(\textbf{r}_s')G_0^+(\textbf{r}',\textbf{r}_s',\omega)d\textbf{r}_s'=\int_\infty \rho(\textbf{r}_s')(-\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'-\textbf{r}_s'|}}{|\textbf{r}'-\textbf{r}_s'|})d\textbf{r}_s' \end{equation}

where is distribution of all sources including airguns and passive sources due to medium perturbation of air and earth. Since at infinitely far field, all sources' location and all sources are in a finite region, is finite. So

\begin{equation} P(\textbf{r}',\omega)\rightarrow \int_\infty \rho(\textbf{r}_s')(-\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'|}}{|\textbf{r}'|})d\textbf{r}_s'=(-\frac{1}{4\pi}\frac{e^{ikr'}}{r'}) \int_\infty \rho(\textbf{r}_s')d\textbf{r}_s'=o(\frac{1}{r'}) \end{equation}

So

\begin{equation} \nabla ' P(\textbf{r}',\omega)\cdot \hat{n'}\\ \nonumber =\nabla ' P(\textbf{r}',\omega)\cdot \hat{r'}\\ \nonumber =\int_\infty \rho(\textbf{r}_s')(-\frac{1}{4\pi}\frac{\partial}{\partial r'}\frac{e^{ik|\textbf{r}'-\textbf{r}_s'|}}{|\textbf{r}'-\textbf{r}_s'|})d\textbf{r}_s'\\ \nonumber \rightarrow \int_\infty \rho(\textbf{r}_s')[ikG_0^+(\textbf{r}',\textbf{r}_s',\omega)+o(\frac{1}{r'^2})]d\textbf{r}_s'\\ \nonumber =ikP(\textbf{r}',\omega)+o(\frac{1}{r'^2})\int_\infty \rho(\textbf{r}_s')d\textbf{r}_s'\\ \nonumber =ikP(\textbf{r}',\omega)+o(\frac{1}{r'^2}) \end{equation}

Therefore, at infinitely far field,

\begin{align} &\int_{h.s.} [P(\textbf{r}',\textbf{r}_s,\omega)\nabla ' G_0^{+}(\textbf{r}',\textbf{r},\omega)-G_0^{+}(\textbf{r}',\textbf{r},\omega)\nabla ' P(\textbf{r}',\textbf{r}_s,\omega)] \cdot \hat{n}' dS'\\ \nonumber &=\int_{h.s.} [P(ikG_0^++o(\frac{1}{r'^2}))-G_0^+(ikP+o(\frac{1}{r'^2}))]dS' \\ \nonumber &=\int_{h.s.} o(\frac{1}{r'^3}) dS' \end{align}

This means the integrand has the order of . Because the semiphere surface has the order of , ultimately the integration has the order of and vanishes as .