Page 535 - Electromagnetics

Wronskians and cross products

2
J ν (z)N ν+1 (z) − J ν+1 (z)N ν (z) =− (E.88)
πz
4
(2)
(1)
(2)
(1)
H ν (z)H ν+1 (z) − H ν (z)H ν+1 (z) = (E.89)
jπz
1
I ν (z)K ν+1 (z) + I ν+1 (z)K ν (z) = (E.90)
z
1

I ν (z)K (z) − I (z)K ν (z) =− (E.91)
ν ν
z
2 j

(1)
(1)

J ν (z)H (z) − J (z)H (z) = (E.92)
ν ν ν
πz
2 j

(2)
(2)

J ν (z)H ν (z) − J (z)H ν (z) =− (E.93)
ν
πz
4 j

(1)
(1)
(2)
(2)
H ν (z)H ν (z) − H ν (z)H ν (z) =− (E.94)
πz
1
j n (z)n n−1 (z) − j n−1 (z)n n (z) = (E.95)
z 2
2n + 1
j n+1 (z)n n−1 (z) − j n−1 (z)n n+1 (z) = (E.96)
z 3
1

j n (z)n (z) − j (z)n n (z) = (E.97)
n n 2
z
2 j

(1)
(2)
(2)
(1)
h (z)h n (z) − h n (z)h (z) =− (E.98)
n
n
z 2
✦ ✦ ✦
✁
r ✦ ✦ ψ ✁ R, r, ρ, φ, ψ as shown.
✦
✦ ✦ ✁ R
2
2
Summation formulas ✦ ✦ φ ✁ R = r + ρ − 2rρ cos φ.
ρ
∞ π

e jνψ Z ν (zR) = J k (zρ)Z ν+k (zr)e jkφ , ρ < r, 0 <ψ < (E.99)
2
k=−∞
∞
jkφ
jnψ
e J n (zR) = J k (zρ)J n+k (zr)e (E.100)
k=−∞
∞

k
e jzρ cos φ = j (2k + 1) j k (zρ)P k (cos φ) (E.101)
k=0
For ρ< r and 0 <ψ <π/2,
∞
e jzR jπ (1)
= √ (2k + 1)J 1 (zρ)H 1 (zr)P k (cos φ) (E.102)
R 2 rρ k+ 2 k+ 2
k=0
∞
e − jzR jπ (2)
=− √ (2k + 1)J 1 (zρ)H 1 (zr)P k (cos φ) (E.103)
R 2 rρ k+ 2 k+ 2
k=0
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