Template:Derivation of reflection and transmission in two dimensions

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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]\displaystyle{ |R|^2+|T|^2=1\, }[/math].

A diagram depicting the area [math]\displaystyle{ \Omega\, }[/math] which is bounded by the rectangle [math]\displaystyle{ \partial \Omega \, }[/math]. The rectangle [math]\displaystyle{ \partial \Omega \, }[/math] is bounded by [math]\displaystyle{ -h \leq z \leq 0 \, }[/math] and [math]\displaystyle{ -\infty \leq x \leq \infty \, }[/math] or [math]\displaystyle{ -N \leq x \leq N\, }[/math]

We can calculate the Reflection and Transmission coefficients by applying Green's theorem to [math]\displaystyle{ \phi\, }[/math] and [math]\displaystyle{ \phi^{\mathrm{I}}\, }[/math] [math]\displaystyle{ \phi^{\mathrm{I}}\, }[/math] is a plane wave travelling in the [math]\displaystyle{ x }[/math] direction,

[math]\displaystyle{ \phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \, }[/math]

where [math]\displaystyle{ A }[/math] is the wave amplitude (in potential) [math]\displaystyle{ \mathrm{i} k }[/math] is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form [math]\displaystyle{ \exp(-\mathrm{i}\omega t) }[/math]) and

[math]\displaystyle{ \phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h} }[/math]

We assume that [math]\displaystyle{ A=1 }[/math]. This gives us

[math]\displaystyle{ \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, }[/math]

This means that (using the far field behaviour of the potential [math]\displaystyle{ \phi }[/math])

[math]\displaystyle{ \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]

For the present case the body is present only on the surface and we therefore have

[math]\displaystyle{ \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 }[/math]

Therefore

[math]\displaystyle{ 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]

and using a wave incident from the right we obtain

[math]\displaystyle{ 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}. }[/math]

Note that an expression for the integral in the denominator can be found in Eigenfunction Matching for a Semi-Infinite Dock