Retarded Muir recurrence

The k_z square root may be computed with the square root function in your computer or by Muir's expansion. For Fourier domain calculations incorporating causality, you must use a complex square root function. This will also take care of the evanescent region automatically--you no longer have the discontinuity between evanescent and nonevanescent regions. The square root of a complex number is multivalued, so you better first check that your computer chooses the phase as described in Figure . Mine did. But I found that limited numerical accuracy prevented me from achieving strict positivity of the real part of the impedance until I replaced the expression $\sqrt{s^2 \,+\, v^2 k^2}\,-\,s$ by its algebraic equivalent $v^2 k^2 / ( \sqrt { s^2 \,+\, v^2 k^2} \,+\,s)$.

For finite differencing we will need the Muir recurrence. Let r₀ define the cosine of the angle that starts the Muir recurrence, often 0$^\circ$ or 45$^\circ$.This is another free parameter for optimization. This angle is also an angle of exact fit for all orders of the recurrence.  

$$s \eq - \, i\, \hat \omega$$ (52)

Starting from $R_0 = r_0 \,s$ the Muir recursion is equation (53):  

$$R_{n+1} \eq s\ +\ { v^2 \, k_x^2 \over s \ +\ R_n }$$ (53)

For a diffraction program we will be evaluating $\exp ( -Rz)$.Since R can be proven to have a positive real part, the exponential should never grow. Finite difference calculations are normally done with retarded time. To retard time, $\exp ( -Rz)$ is expressed as  

$$e^{ -\,R\,z/v } \eq e^{ -\,( R\,-\,s)\,z/v} \ \ e^{ - s\,z/v }$$ (54)

As discussed earlier in this chapter, you probably don't want the time shift of retardation to be associated with viscous effects. So you will probably want to downward continue instead with  

$$e^{ -\,[ R( -i \hat \omega ) \,+\, i \hat \omega ]\,z/v } \ \ e^{ + i \omega\,z/v}$$ (55)

Notice the signs and distinction of $\omega$ from $\hat \omega$.

From equation (53) we see that R-s should have a positive real part. I found that numerical roundoff sometimes prevented it. So the Muir recurrence was reorganized to incorporate the retardation. Let  

$$R ' \eq R \ -\ s$$ (56)

Equation (53) becomes  

$$R ' _{n+1} \eq { v^2 \, k_x^2 \over 2\,s \ +\ R ' _n }$$ (57)

From Muir's rules, you can see that R ' will always have a positive real part if we start it that way, so we start it from  

$${R ' }_1 \eq { v^2 \,k_x^2 \over s\, ( 1 \ +\ r_0 ) }$$ (58)

(Combining (58) and (56) gives the same 15$^\circ$ equation as does (53).) Mathematically (55) is identical to  

$$e^{ - \, R ' \, z/v} \ \ e^{ + i \omega \,z/v }$$ (59)

but numerically the exponential in (59) is assured to decay in z.