Sommerfeld Radiation Condition: Difference between revisions

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This is a condition for the [[Frequency Domain Problem]] that the scattered wave is only
This is a condition for the [[Frequency Domain Problem]] that the scattered wave is only
outgoing at infinity. It depends on the convention regarding whether the time dependence
outgoing at infinity. It depends on the convention regarding whether the time dependence
is <math>\exp (i\omega t)\,</math> or <math>\exp (-i\omega t)\,</math>
is <math>\exp (i\omega t)\,</math> or <math>\exp (-i\omega t)\,</math>.
Assuming the former (which is the standard convention on this wiki)  
Assuming the former (which is the standard convention on this wiki).
In two-dimensions the condition is  
In two dimensions the condition is  
<center>
<center>
<math>
<math>
\left(  \frac{\partial}{\partial|x|}+{i}k\right)
\left(  \frac{\partial}{\partial|x|}+\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
</math>
</math>
</center>
</center>
Line 15: Line 15:
is the wave number.
is the wave number.


In three-dimensions the condition is  
In three dimensions the condition is  
<center>
<center>
<math>
<math>
\sqrt{|\mathbf{r}|}\left(  \frac{\partial}{\partial|\mathbf{r}|}+{i}k\right)
r^{1/2}\left(  \frac{\partial}{\partial r}+\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.}
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}r\rightarrow\infty\mathrm{.}
</math>
</math>
</center>
</center>


If the time dependence is assumed to be <math>\exp (-i\omega t)\,</math> then we  
If the time dependence is assumed to be <math>\exp (-i\omega t)\,</math>, then we  
have in two-dimensions  
have in two dimensions  
<center>
<center>
<math>
<math>
\left(  \frac{\partial}{\partial|x|}-{i}k\right)
\left(  \frac{\partial}{\partial|x|}-\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
</math>
</math>
</center>
</center>
and in three-dimensions  
and in three dimensions  
<center>
<center>
<math>
<math>
\sqrt{|\mathbf{r}|}\left(  \frac{\partial}{\partial|\mathbf{r}|}-{i}k\right)
r^{1/2}\left(  \frac{\partial}{\partial r}-\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.}
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}r\rightarrow\infty\mathrm{.}
</math>
</math>
</center>
</center>


[[Category:Linear Water-Wave Theory]]
[[Category:Linear Water-Wave Theory]]

Latest revision as of 04:55, 4 September 2012


This is a condition for the Frequency Domain Problem that the scattered wave is only outgoing at infinity. It depends on the convention regarding whether the time dependence is [math]\displaystyle{ \exp (i\omega t)\, }[/math] or [math]\displaystyle{ \exp (-i\omega t)\, }[/math]. Assuming the former (which is the standard convention on this wiki). In two dimensions the condition is

[math]\displaystyle{ \left( \frac{\partial}{\partial|x|}+\mathrm{i}k\right) (\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.} }[/math]

where [math]\displaystyle{ \phi^{\mathrm{{In}}} }[/math] is the incident potential and [math]\displaystyle{ k }[/math] is the wave number.

In three dimensions the condition is

[math]\displaystyle{ r^{1/2}\left( \frac{\partial}{\partial r}+\mathrm{i}k\right) (\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}r\rightarrow\infty\mathrm{.} }[/math]

If the time dependence is assumed to be [math]\displaystyle{ \exp (-i\omega t)\, }[/math], then we have in two dimensions

[math]\displaystyle{ \left( \frac{\partial}{\partial|x|}-\mathrm{i}k\right) (\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.} }[/math]

and in three dimensions

[math]\displaystyle{ r^{1/2}\left( \frac{\partial}{\partial r}-\mathrm{i}k\right) (\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}r\rightarrow\infty\mathrm{.} }[/math]