Template:Derivation of reflection and transmission in two dimensions - Revision history

Meylan at 09:33, 27 November 2009

2009-11-27T09:33:15Z

← Older revision		Revision as of 09:33, 27 November 2009
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	<center><math>		<center><math>
	\iint_{\Omega}(\phi\Delta\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\Delta\phi)\mathrm{d}x\mathrm{d}z		\iint_{\Omega}(\phi\Delta\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\Delta\phi)\mathrm{d}x\mathrm{d}z
	= \int_{\partial\Omega}(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}l = 0,		= \int_{\partial\Omega}(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}s = 0,
	</math></center>		</math></center>
	This means that (using the far field behaviour of the potential <math>\phi</math>)		This means that (using the far field behaviour of the potential <math>\phi</math>)
	<center><math>		<center><math>
	\int_{\partial\Omega_{B}}		\int_{\partial\Omega_{B}}
	(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}l		(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}s
	+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0,		+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0,
	</math></center>		</math></center>
	For the present case the body is present only on the surface and we therefore have		For the present case the body is present only on the surface and we therefore have
	<center><math>		<center><math>
	~~\phi_0(0)~~ \int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x		\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x
	+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0		+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0
	</math></center>		</math></center>

Meylan at 09:25, 27 November 2009

2009-11-27T09:25:13Z

← Older revision		Revision as of 09:25, 27 November 2009
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	We assume that <math>A=1</math>. This gives us		We assume that <math>A=1</math>. This gives us
	<center><math>		<center><math>
	\iint_{\Omega}(\phi\~~nabla^2~~\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\~~nabla^2~~\phi)\mathrm{d}x\mathrm{d}z		\iint_{\Omega}(\phi\Delta\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\Delta\phi)\mathrm{d}x\mathrm{d}z
	= \int_{\partial\Omega}(\phi\~~frac{\partial~~\phi^{\rm I~~}}{\partial n~~} - \phi^{\rm I}\~~frac{\partial~~\phi~~}{\partial n}~~)\mathrm{d}l = 0,		= \int_{\partial\Omega}(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}l = 0,
	</math></center>		</math></center>
	This means that (using the far field behaviour of the potential <math>\phi</math>		This means that (using the far field behaviour of the potential <math>\phi</math>)
			<center><math>
			\int_{\partial\Omega_{B}}
			(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}l
			+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0,
			</math></center>
			For the present case the body is present only on the surface and we therefore have
	<center><math>		<center><math>
	\phi_0(0) \int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x		\phi_0(0) \int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x

Meylan at 09:09, 27 November 2009

2009-11-27T09:09:29Z

← Older revision		Revision as of 09:09, 27 November 2009
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	This means that (using the far field behaviour of the potential <math>\phi</math>		This means that (using the far field behaviour of the potential <math>\phi</math>
	<center><math>		<center><math>
	\phi_0(0) \int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x		\phi_0(0) \int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x
	+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0		+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0
	</math></center>		</math></center>
	Therefore		Therefore
	<center><math>		<center><math>
	R = -\frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }		R = -\frac{\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }
	{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.		{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.
	</math></center>		</math></center>
	and using a wave incident from the right we obtain		and using a wave incident from the right we obtain
	<center><math>		<center><math>
	T = 1 - \frac{\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }		T = 1 - \frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }
	{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.		{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.
	</math></center>		</math></center>
	Note that an expression for the integral in the denominator can be found in		Note that an expression for the integral in the denominator can be found in
	[[Eigenfunction Matching for a Semi-Infinite Dock]]		[[Eigenfunction Matching for a Semi-Infinite Dock]]

Meylan at 09:08, 27 November 2009

2009-11-27T09:08:28Z

← Older revision		Revision as of 09:08, 27 November 2009
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	<center><math>		<center><math>
	\phi_0(0) \int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x		\phi_0(0) \int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x
	- 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0		+ 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0
	</math></center>		</math></center>
	Therefore		Therefore
	<center><math>		<center><math>
	R = \frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }		R = -\frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }
	{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.		{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.
	</math></center>		</math></center>
	and using a wave incident from the right we obtain		and using a wave incident from the right we obtain
	<center><math>		<center><math>
	~~1 +~~ T = \frac{\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }		T = 1 - \frac{\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }
	{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.		{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.
	</math></center>		</math></center>
			Note that an expression for the integral in the denominator can be found in
			[[Eigenfunction Matching for a Semi-Infinite Dock]]

Meylan at 08:47, 27 November 2009

2009-11-27T08:47:29Z

← Older revision		Revision as of 08:47, 27 November 2009
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	= \int_{\partial\Omega}(\phi\frac{\partial\phi^{\rm I}}{\partial n} - \phi^{\rm I}\frac{\partial\phi}{\partial n})\mathrm{d}l = 0,		= \int_{\partial\Omega}(\phi\frac{\partial\phi^{\rm I}}{\partial n} - \phi^{\rm I}\frac{\partial\phi}{\partial n})\mathrm{d}l = 0,
	</math></center>		</math></center>
			This means that (using the far field behaviour of the potential <math>\phi</math>
	<center><math>		<center><math>
	= \phi_0(0) \int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x		\phi_0(0) \int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x
	- 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z.		- 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0
	</math></center~~><br\>~~		</math></center>
	~~where <math> k_0 \,</math> is the first imaginery root of the dispersion equation and the incident wave is of the form: <math> \phi^I=\phi_0(z)e^{-ikx} \,</math~~>		Therefore
	Therefore~~, in the case of a floating plate (where z=0):~~
	<center><math>		<center><math>
	R = \frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }		R = \frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }

Meylan at 08:41, 27 November 2009

2009-11-27T08:41:15Z

← Older revision		Revision as of 08:41, 27 November 2009
Line 1:		Line 1:
	The Reflection and Transmission Coefficients represent the ratio of the amplitude of the reflected or transmitted wave to the amplitude of the incident wave. ~~They hold the property~~ that <math> \|R\|^2+\|T\|^2=1\,</math> ~~(and may often contain an imaginery element)~~.		The Reflection and Transmission Coefficients represent the ratio of the amplitude of the reflected or transmitted wave to the amplitude of the incident wave. Conservation of energy means that <math> \|R\|^2+\|T\|^2=1\,</math>.

	[[Image:Square_volume.png\|600px\|right\|thumb\|frame\|A diagram depicting the area <math>\Omega\,</math> which is bounded by the rectangle <math>\partial \Omega \,</math>. The rectangle <math>\partial \Omega \,</math> is bounded by <math> -h \leq z \leq 0 \,</math> and <math>-\infty \leq x \leq \infty \,</math> or <math>-N \leq x \leq N\,</math>]]		[[Image:Square_volume.png\|600px\|right\|thumb\|frame\|A diagram depicting the area <math>\Omega\,</math> which is bounded by the rectangle <math>\partial \Omega \,</math>. The rectangle <math>\partial \Omega \,</math> is bounded by <math> -h \leq z \leq 0 \,</math> and <math>-\infty \leq x \leq \infty \,</math> or <math>-N \leq x \leq N\,</math>]]

	We can calculate the Reflection and Transmission coefficients ~~as follows:~~		We can calculate the Reflection and Transmission coefficients by
	~~Applying~~ Green's theorem to <math>\phi\,</math> and <math>\phi^{\mathrm{I}}\,</math> gives:		applying Green's theorem to <math>\phi\,</math> and <math>\phi^{\mathrm{I}}\,</math>
			{{incident plane wave 2d definition}}
			We assume that <math>A=1</math>. This gives us
	<center><math>		<center><math>
	~~0 =~~ \iint_{\Omega}(\phi\nabla^2\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\nabla^2\phi)\mathrm{d}x\mathrm{d}z		\iint_{\Omega}(\phi\nabla^2\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\nabla^2\phi)\mathrm{d}x\mathrm{d}z
	= \int_{\partial\Omega}(\phi\frac{\partial\phi^{\rm I}}{\partial n} - \phi^{\rm I}\frac{\partial\phi}{\partial n})\mathrm{d}l,		= \int_{\partial\Omega}(\phi\frac{\partial\phi^{\rm I}}{\partial n} - \phi^{\rm I}\frac{\partial\phi}{\partial n})\mathrm{d}l = 0,
	</math></center>		</math></center>
	<center><math>		<center><math>
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	- 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z.		- 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z.
	</math></center><br\>		</math></center><br\>
	where <math> k_0 \,</math> is the first imaginery root of the dispersion equation and the incident wave is of the form: <math> \phi^I=\phi_0(z)e^{-ikx} \,</math~~><br\><br\~~>		where <math> k_0 \,</math> is the first imaginery root of the dispersion equation and the incident wave is of the form: <math> \phi^I=\phi_0(z)e^{-ikx} \,</math>
	Therefore, in the case of a floating plate (where z=0):		Therefore, in the case of a floating plate (where z=0):
	<center><math>		<center><math>

Meylan at 00:14, 17 September 2009

2009-09-17T00:14:07Z

New page

The Reflection and Transmission Coefficients represent the ratio of the amplitude of the reflected or transmitted wave to the amplitude of the incident wave. They hold the property that <math> |R|^2+|T|^2=1\,</math> (and may often contain an imaginery element).

[[Image:Square_volume.png|600px|right|thumb|frame|A diagram depicting the area <math>\Omega\,</math> which is bounded by the rectangle <math>\partial \Omega \,</math>. The rectangle <math>\partial \Omega \,</math> is bounded by <math> -h \leq z \leq 0 \,</math> and <math>-\infty \leq x \leq \infty \,</math> or <math>-N \leq x \leq N\,</math>]]

We can calculate the Reflection and Transmission coefficients as follows:
Applying Green's theorem to <math>\phi\,</math> and <math>\phi^{\mathrm{I}}\,</math> gives:
<center><math>
0 = \iint_{\Omega}(\phi\nabla^2\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\nabla^2\phi)\mathrm{d}x\mathrm{d}z
= \int_{\partial\Omega}(\phi\frac{\partial\phi^{\rm I}}{\partial n} - \phi^{\rm I}\frac{\partial\phi}{\partial n})\mathrm{d}l,
</math></center>
<center><math>
= \phi_0(0) \int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x
- 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z.
</math></center><br\>
where <math> k_0 \,</math> is the first imaginery root of the dispersion equation and the incident wave is of the form: <math> \phi^I=\phi_0(z)e^{-ikx} \,</math><br\><br\>
Therefore, in the case of a floating plate (where z=0):
<center><math>
R = \frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }
{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.
</math></center>
and using a wave incident from the right we obtain
<center><math>
1 + T = \frac{\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x }
{2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}.
</math></center>