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Region 0

According to (5), the solution at any point $x_0$ in Region 0 thus has the form
$\displaystyle s_0(x_0 \le 0,t)$ $\textstyle =$ $\displaystyle t_0(x_0-c_0 t) + R_{0 \rightarrow 1} t_0(-(x_0+c_0 t))$  
  $\textstyle =$ $\displaystyle {\mbox{Re\,}}\left(\underline{A} e^{ik_0(x_0-c_0 t)}\right) + R_{0
\rightarrow 1} {\mbox{Re\,}}\left( \underline{A} e^{ik_0(-(x_0+c_0 t))}\right)$  
  $\textstyle =$ $\displaystyle {\mbox{Re\,}}\left(\underline{A} e^{ik_0x_0} e^{-i k_0 c_0 t} + R_{0
\rightarrow 1} \underline{A} e^{-ik_0x_0} e^{-i k_0 c_0 t}\right)$  
  $\textstyle =$ $\displaystyle {\mbox{Re\,}}\left( e^{-i \omega t} \underline{A} \left( e^{ik_0x_0} + R_{0
\rightarrow 1} e^{-ik_0x_0} \right) \right)$  
  $\textstyle =$ $\displaystyle {\mbox{Re\,}}\left( e^{-i \omega t} \underline{Q}_0(x_0) \right),$ (8)

where the wave frequency is $\omega = k_0
c_0$, and we define a complex amplitude for the wave solution in Region 0 to be
\begin{displaymath}
\underline{Q}_0(x_0 \le 0) \equiv \underline{A} \left(
e^{ik_0x_0} + R_{0 \rightarrow 1} e^{-ik_0x_0} \right).
\end{displaymath} (9)

Conveniently, we find the total solution in Region 0 to be of the form (2) of simple harmonic motion at the incoming wave frequency. This is just the general consequence of the fact that the response of a system driven at frequency $\omega$ is motion at that same frequency. Given this knowledge, the complex amplitude $\underline{Q}_0(x_0)$ then determines entirely the motion for Region 0.

To write this result down directly using the complex representation, we consider the motion at each point $x_0$ relative to the motion associated with the incoming wave at point $a$, which from our previous discussion we know to be ${\mbox{Re\,}}\left(e^{-i\omega t} \underline{A} e^{ik_0a}\right)$ for some complex amplitude $\underline{A}$. We expect two waves contribute to the motion at each point $x_0<0$, the incoming wave and the reflected wave. (See Figure 1.) By the principle of superposition, the final motion will be the sum each of these.

The motion due to the incoming wave at point $x_0$, a distance $L=(x_0-a)$ from the point $a$, will include an additional propagation factor of $e^{i k_0 (x-0-a)}$. The first contribution to the motion at $x_0$ is thus

\begin{displaymath}
{\mbox{Re\,}}\left( e^{-i\omega t} \underline{A} e^{ik_0a} \cdot e^{ik_0(x_0-a)}
\right).
\end{displaymath}

Figure 1: Scattering (reflection and transmission) from a change in medium from Region 0 ($x<0$) to Region 1 ($x>0$).
\includegraphics[scale=0.75]{refltrans.eps}

The second contribution comes from the reflected wave. To compute this contribution at point $x_0$, we make a number of relative comparisons. First, we compare the motion of the incoming wave at $a$ to the motion of the same wave at the interface, $x=0$. The distance propagated is now $L=(0-a)=-a$, and so we must include a factor of $e^{ik_0(-a)}$ to find the motion of the incoming wave at $x=0$. Next, we know that the motion of the reflected wave at the interface matches the motion of the incoming wave at the interface, except for the reflection amplitude factor $R_{0\rightarrow
1}$. Thus, the motion of the reflected wave at $x=0$ is just the motion of the reference point times two correction factors, $e^{ik_0(-a)}$ and $R_{0\rightarrow
1}$. Finally, relative to the motion of the reflected wave at $x=0$, the motion induced by the reflected wave at $x_0$, must contain one final propagation factor $e^{ik_0(-x_0)}$. Thus, the reflected wave contribution to the motion at $x_0$ is

\begin{displaymath}
{\mbox{Re\,}}\left( e^{-i\omega t} \underline{A} e^{ik_0a} \...
..._0(-a)}
\cdot R_{0\rightarrow 1} \cdot e^{ik_0(-x_0)}
\right).
\end{displaymath}

Adding our two contributions together, we find that we are able to directly write a result equivalent to (8),

\begin{displaymath}
s_0(x_0,t)=
{\mbox{Re\,}}\left( e^{-i\omega t} \underline{A}...
...\cdot R_{0\rightarrow 1} \cdot e^{ik_0(-x_0)} \right] \right).
\end{displaymath}


next up previous contents
Next: Region 1 Up: Solutions for waves at Previous: Solutions for waves at   Contents
Tomas Arias 2003-10-27