Fourier transforms in retarded coordinates

Given a pressure field P(t,x,z), we may Fourier transform it with respect to any or all of its independent variables (t,x,z). Likewise, if the pressure field is specified in retarded coordinates, we may Fourier transform with respect to (t' ,x' ,z' ). Since the Fourier dual of (t,x,z) is $( \omega , k_x , k_z )$, it seems appropriate for the dual of (t' ,x' ,z' ) to be $( \omega' , {{ k_x }' } , {{ k_z }' } )$.Now the question is, how are $( \omega' , {{ k_x }' } , {{ k_z }' } )$ related to the familiar $( \omega , k_x , k_z )$? The answer is contained in the chain rule for partial differentiation. Any expression like (33)

$${\partial\ \over \partial z}\ \ =\ \ -\ {1 \over \bar v }\ ... ...ial\ \ \over \partial t' }\ +\ {\partial\ \ \over \partial z' }$$

on Fourier transformation says  

$$i\,k_z\ \ =\ \ -\ {-\,i\,\omega' \over \bar v } \ +\ i\, {{ k _z }'}$$ (36)

Computing all the other derivatives, we have the transformation

$$\begin{eqnarray} \omega\ \ &=&\ \ \omega' \\ k_x\ \ &=&\ \ {{ k_x }' } \\ k_z\ \ &=&\ \ {{ k_z }' } \ +\ {\omega' \over \bar v \,}\end{eqnarray}$$ (37) (38) (39)

Recall the dispersion relation for the scalar wave equation:  

$${\omega^2 \over v^2 }\ \ =\ \ k_x^2\ +\ k_z^2$$ (40)

Performing the substitutions from (37), (38) and (39) into (40) we have the expression of the scalar wave equation in retarded time, namely,  

$${\left( {\omega' \over v \,} \, \right)}^2 \eq {{ k_x }' }^... ...ft( {{ k_z }' } \ +\ { \omega' \over \bar v \, } \, \right)}^2$$ (41)

These two dispersion relations are plotted in Figure 4 for the retardation velocity chosen equal to the medium velocity.

 

rdisper Figure 4 Dispersion relation of the wave equation in usual coordinates (left) and retarded time coordinates (right).

Figure 4 graphically illustrates that retardation can reduce the cost of finite-difference calculations. Waves going straight down are near the top of the dispersion curve (circle). The effect of retardation is to shift the circle's top down to the origin. Discretizing the x- and z-axes will cause spatial frequency aliasing on them. The larger the frequency $\omega$, the larger the circle. Clearly the top of the shifted circle is further from folding. Alternately, $\Delta z$ may be increased (for the sake of economy) before k' _z exceeds the Nyquist frequency $\pi / \Delta z$.