Page 278 - Phase Space Optics Fundamentals and Applications

Rays and Waves 259

Let us also define the vector
∂

= (HP,H) = H (P,z) (8.71)
˙
∂z
With this, the transport equation in Eq. (8.57b) can be written as

2
˜ ∇· B
= 0 (8.72)
0
This equation can be solved by integrating it over the volume in the
(p,z) spaceoccupiedbyaninfinitesimalbundleofraysbetween z 0 and
z, and using Gauss’ theorem. Equation (8.71) states that
is locally
parallel to the rays in this space, so the contributions to the surface
integral from the sides of the bundle vanish, as in the position repre-
sentation case. By following steps analogous to those in Sec. 8.3, we
find
9
−1
H(z 0 , ) [P(z 0 , )] [P(z, )]
B 0 [P(z, ),z] = B 0 [P(z 0 , ),z 0 ]
H(z, ) ( ) ( )
(8.73)
Notice that this solution has problems when the Jacobian between
braces vanishes. This happens when contiguous rays in the family
have the same transverse momentum. (For a homogeneous medium,
this means that the rays are locally parallel.) That is, the field estimate
that results from this derivation also has problems, but these are dif-
ferent from those for the estimate found in the position representation,
associated with caustics. The location of these new problems, i.e., the
places where contiguous rays have the same momentum, are called
momentum caustics. As in the case of the amplitude of the position rep-
resentation estimate, one must be careful when choosing the sign of
the square root in Eq. (8.73).

8.6.4 Field Estimate
The field estimate is obtained by approximating B ≈ B 0 , that is,
˜ U(P,z) ≈ B 0 exp(ikT). To obtain the field in the position represen-
tation, the inverse Fourier transform of this estimate must be taken.
Because p is parameterized, the Fourier transform integral must be
done parameterically by inserting a Jacobian factor:

k (P)
r
U( ) ≈ B 0 (P,z) exp(ikT) exp(ikx · P) d 1 d 2
2 ( )

k H(z 0 , ) [P(z 0 , )] [P(z, )]
= B 0 [P(z 0 , ),z 0 ]
2 H(z, ) ( ) ( )
× exp(ik{L(z, ) + [x − X(z, )] · P(z, )})d 1 d 2 (8.74)