�
�
�
�
where we will now proceed to solve for the coefficients a j . The
kinetic energy of a particle on axis is
a 0 (z) = φ(0, z) ≡ Φ(z).
(2.121)
From (2.119, 2.120, 2.121) it follows that
Φ
�� 2
1 Φ
IV 4
φ(r, z) = Φ −
1
4
r + 64
r + . . . ,
(2.122)
where primes indicate differentiation with respect to z. Expanding
the scalar kinetic momentum p we obtain (2.116, 2.122):
p(r, z) =
2φ + φ 2

= p −
1
4
p
−1 Φ
�� (1 + Φ) r
2

1 −1 Φ
IV
1 −3 Φ
�� 2
4
+
p
(1 + Φ) − p
r + . . . ,
64
32
(2.123)
where we have defined a quantity p(z) as the scalar kinetic momentum on axis as follows (2.121, 2.122):
√
p(z) ≡ p(0, z) = 2Φ + Φ 2 .
(2.124)
Separately, the magnetic field B is given in terms of the magnetic
vector potential A as
∂A θ
∂A θ A θ
B = v × A = −ˆ r
+ ˆ
z
+
.
(2.125)
∂z
∂r
r
In the absence of space current, Maxwell’s equation is
v × B = 0,
(2.126)
from which it follows that (2.125, 2.126)
∂
2
1 ∂
1
∂
2
−θ ˆ
+
− 2 +
A θ = 0.
(2.127)
∂r 2 r ∂r r
∂z 2
We assume a series representation for A θ as follows:
A θ (r, z) = b 1 (z) r + b 3 (z) r
3 + b 5 (z) r
5 + . . . ,
(2.128)
52
Chapter 2. Geometrical optics