Classical Electrodynamics Pro ject. Waveguides
Xabier García Andrade
10th of October 2018
Contents
1
Introduction
2
2
Exercise a
2
3
Exercise b
5
4
Exercise c
6
5
Bibliography
8
1
Waveguides
1
Xabier G. Andrade
Introduction
Wave Guides A resonant cavity consists of the empty space between two perfectly conducting,
concentric spherical shells, the smaller having an outer radius a and the larger an inner radius
b. As shown in Section 8.9, the azimuthal magnetic eld has a radial dependence given by
spherical Bessel functions, jl (kr) and nl (kr), where k = w/c.
2
Exercise a
Write down the transcendental equation for the characteristic frequencies of the
cavity for arbitrary l.
If we are only interested in the lowest frequencies, we can focus our attention on the TM modes,
with only tangential magnetic elds. Considering our spherical symmetry, TM modes notation
refers to the absence of radial magnetic eld components. Also, we will consider that the elds
are independent of the azimuthal angle φ. In spherical harmonics, the dominant term is l
~ does not depend on φ , from
instead of m, so it is a suitable consideration. Since Br = 0 and B
the divergence equation we obtain:
0
r
1 r2 · ∂B
∇ · B = 2 r ∂r
>
0
7
1 ∂Bφ
1 ∂Bθ sinθ
+
+
=0
rsinθ ∂φ
rsinθ ∂θ
(1)
From this equation we obtain that if the elds are nite at θ = 0, then the only term that
would survive is Bφ . From Faraday law it follows that:
∇×E =−
∂Bφ
∂B
=
φ̂
∂t
∂t
(2)
With this we can conclude that Eφ also has to vanish. In summary, our problem would only
consist of nding Er , Eθ and Bφ . We still have to use the other curl equations, which will
bring (assuming a time dependence of e−iωt and taking permeabilities as unity):
ω2
B−∇×∇×B =0
c2
(3)
We can write down the equation in terms of the φ component:
ω2
∂2
1 ∂
1 ∂
(rBφ ) + 2 (rBφ ) + 2
(sinθrBφ ) = 0
c2
∂r
r ∂θ sinθ ∂θ
(4)
Rewriting the angular part:
∂
1 ∂
1 ∂
∂(rBφ )
rBθ
(sinθrBφ ) =
sinθ
−
∂θ sin θ ∂
sinθ ∂θ
∂θ
sin2 θ
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We can compare the previous equation with the following one:
1 d
sinθ dθ
dP
m2
sinθ
+ l(l + 1) −
P =0
dθ
sin2 θ
(6)
Thus we can easily see that the solution to equation (5) is given by the associated Legendre
polynomials with m = ±1. We then try a solution of the form:
Bφ (r, θ) =
ul (r) 1
Pl (cosθ)
r
(7)
We then substitute directly into (4):
2
d2 ul (r)
ω
l(l + 1)
ul (r) = 0
+ 2 −
r2
c
r2
(8)
The solutions of this equation can be expressed in terms of spherical Bessel functions. First we
need to nd the boundary conditions for ul (r) :
From (7) we can obtain radial and tangential electric elds:
ul (r)
ic2
∂
ic2
Pl (cosθ)
Er =
= − l(l + 1)
ωrsinθ ∂Bφ
ωr
r
ic2 ∂
ic2 ∂ul (r)
Eθ = −
(rBφ ) = −
Pl (cosθ)
ωr ∂r
ωr ∂r
Since Eθ must vanish at the surface of both shells, we can obtain the boundary condition for
ul r :
dul (r)
=0
dr
(9)
f or r = a and r = b
Now, we can write the solutions as follows:
ul (r) = rAl jl (kr) + rBl nl (kr)
with
dul (r)
=0
dr
at r = a and r = b
(10)
Now we need to dierentiate with respect to r to obtain the trascendental equations that
characteristic frequencies satisfy:
dul (r)
= Al jl (kr) + rAl jl 0(kr) + Bl nl (kr) + rBl nl 0(kr)
dr
(11)
By using (9) we end up with the following system of equations:
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(
Al [jl (ka) + kajl 0(ka)] + Bl [nl (ka) + kanl 0(ka)] = 0
Al [jl (kb) + kbjl 0(kb)] + Bl [nl (kb) + kbnl 0(kb)] = 0.
(12)
We can transform this into a matrix equation:
jl (ka) + kajl 0(ka) nl (ka) + kanl 0(ka) Al
0
=
jl (kb) + kbjl 0(kb) nl (kb) + kbnl 0(kb) Bl
0
(13)
Which from Kramer´s theorem we can deduce that non-trivial solutions will happen only when
the range of the matrix is less than 2. Then we need the determinant to vanish:
[jl (ka) + kajl 0(ka)] · [nl (kb) + kbnl 0(kb)] − [nl (ka) + kanl 0(ka)] · [jl (kb) + kbjl 0(kb)] = 0 (14)
Which can be better written as the following quotient, known as the trascendental equation:
nl (kb) + kbnl 0(kb)
jl (ka) + kajl 0(ka)
=
nl (ka) + kanl 0(ka)
jl (kb) + kbjl 0(kb)
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3
Xabier G. Andrade
Exercise b
For l = 1 use the explicit forms of the spherical Bessel functions to show that the
characteristic frequencies are given by
1
(k 2 + ab
)
tankh
= 2
1
2
kh
k + ab(k − a2 )(k 2 − b12 )
where
h = b − a.
Now we have to consider l = 1 to obtain the exact form of the spherical Bessel functions. First
we consider spherical bessel functions of rst kind:
j1 (z) =
sinz − zcosz
sinz cosz
=
−
2
z
z
z2
(16)
n1 (z) = −
−cosz − zsinz
cosz sinz
−
=
2
z
z
z2
(17)
z 2 cosz − 2zsinz −zsinz − cosz
z 2 sinz + 2zcosz − 2sinz
−
=
z4
z2
z3
(18)
−z 2 cosz + 2zsinz + 2cosz
z 2 sin(z) + 2zcos(z) −zcosz − sinz
+
=
z4
z2
z3
(19)
Second kind:
Their respective derivatives:
j1 0(z) =
n1 0(z) =
Replacing each one in the transcendental equation we obtain:
kasin(ka) − k2 a2 cos(za) + a3 k3 sin(ka) + 2k2 a2 cos(ka) − 2kasin(ka)
−kacos(ka) − k2 a2 sin(ka) − k3 a3 cos(ka) + 2k2 a2 sin(ka) + 2kacos(ka)
Simplifying the equation:
=
kbsin(kb) − k2 b2 cos(zb) + b3 k3 sin(kb) + 2k2 b2 cos(kb) − 2kbsin(kb)
−kbcos(kb) − k2 b2 sin(kb) − k3 b3 cos(kb) + 2k2 b2 sin(kb) + 2kbcos(kb)
(20)
(a2 k 2 − 1)sin(ka) + kacos(ka)
(b2 k 2 − 1)sin(kb) + kbcos(kb)
=
(−k 2 a2 + 1)cos(ka) + kasin(ka)
(−k 2 b2 + 1)cos(kb) + kbsin(kb)
(21)
Now we can continue operating:
2 2
2 2
2 2
2 2
2
−(a k − 1)(b k − 1)sin(ka)cos(kb) − ka(b k − 1)cos(kb)cos(ka) + kb(a k − 1)sin(kb)sin(ka) + k bacos(ka)sin(kb) =
2 2
2 2
2 2
2 2
2
(22)
−(b k − 1)(a k − 1)cos(ka)sin(kb) + ka(b k − 1)sin(ka)sin(kb) − kb(a k − 1)cos(ka)cos(kb) + k abcos(kb)sin(ka)
Grouping similar terms together:
−(a2 k 2 − 1)(b2 k 2 − 1)[sin(ka)cos(kb) − cos(ka)sin(kb)] + k 2 ab[sin(kb)cos(ka) − sin(ka)cos(kb)] =
ka(b2 k 2 − 1)[cos(kb)cos(ka) + sin(ka)sin(kb)] − kb(a2 k 2 − 1)[cos(ka)cos(kb) + sin(ka)sin(kb)]
(23)
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Now using the following trigonometric relations:
(
sin(α − β) = sinαcosβ − cosαsinβ
cos(α − β) = cosαcosβ + sinβsinα
(24)
We get to the following expression:
−(a2 k 2 − 1)(b2 k 2 − 1)sin(ka − kb) + k 2 absin(kb − ka) =
ka(b2 k 2 − 1)cos(kb − ka) − kb(a2 k 2 − 1)cos(ka − kb)
(25)
Now using the fact that sin(−α) = −sin(α) and cos(−α)) = cos(α) and grouping terms:
(a2 k 2 − 1)(b2 k 2 − 1) + k 2 ab sin(kb − ka) = ka(b2 k 2 − 1) − kb(a2 k 2 − 1) cos(kb − ka) (26)
Now dening h = b − a we get:
tan(hk) =
k 3 ba(b − a) + k(b − a)
ka(b2 k 2 − 1) − kb(a2 k 2 − 1)
=
(a2 k 2 − 1)(b2 k 2 − 1) + k 2 ab
a2 b2 k 4 − a2 k 2 − b2 k 2 + 1 + k 2 ab
(27)
k 2 ba + 1
abk 2 + a2 b2 k 4 − a2 k 2 − b2 k 2 + 1
(28)
tan(hk) = kh
Then we get to the nal solution:
1
k 2 + ab
tan(hk)
= 2
hk
k + ab k 2 − a12 k 2 − b12
4
(29)
Exercise c
Write down the transcendental equation for the characteristic frequencies of the
cavity for arbitrary l.
Now we must consider the following approximation : h/a << 1
Rewriting (29) in terms of h/a:
1
k 2 + a2 (1+
h
)
a
1
k 2 + a2 (1 + h/a)(k 2 − a12 ) k 2 − a2 (1+
h 2
)
(30)
a
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Now we must perform a Taylor expansion both in the numerator and the denominator. We
will only consider rst order terms because the following ones are neglible.
1
h
≈1−
h
a
1+ a
(31)
1
h
≈1−2
h 2
a
(1 + a )
(32)
k 2 + a12 − a12 ha
tankh
= 2
kh
k + (1 + ha )(k 2 a2 − 1)(k 2 − a12 + a22 ha )
(33)
Plugging this result in (30):
Multiplicating factors in the denominator:
k 2 + a12 − a12 ha
tankh
= 2
kh
k + k 4 a2 − k 2 + k 2 ha + k 4 ah − k 2 + a12 − a12 ha − k 2 ha
(34)
k 2 + a12 − a12 ha
tankh
kh k 2 (k 2 a2 − 1) + ha (k 4 a2 − a12 )
(35)
Now we can also expand the left-hand term considering again h/a << 1:
tankh
≈ 1 + O((h/a)2 )
kh
(36)
Where again we can disregard second order terms. Now writing it all together:
1
h
k (k a − 1) 2 +
a
a
2
2 2
1
k a − 2
a
4 2
= k2 +
1
1 h
− 2
2
a
a a
(37)
Doing further manipulations we arrive to:
k2 =
2
a(a + h)
(38)
We can relate k to the frequency by means of the speed of light c (ω = ck):
ω 2 = c2 k 2 = 2c2
1
1
2c2
= 2
a(a + h)
a (1 + ha )
(39)
Using (31) to expand the denominator as we did before:
ω2 =
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2c2
h
(1 − )
2
a
a
(40)
Classical Electrodynamics
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Xabier G. Andrade
√ r
2c
h
ω=
1−
a
a
(41)
We need to perform another Taylor expansion to retrieve the result from Jackson:
r
h
h
=1−
a
2a
(42)
√
√ c
2h
ω= 2 −
a
2 a
(43)
1−
Then , the value of the frequency would be:
We can compare this to the result given in Jackson´s equation (8.105):
ωl ≈
p
c
l(l + 1)
a
(44)
√
For l = 1 , we obtain ω1 = 2 ac . Thus, we can conclude that the secon term in (43) is the rst
order correction.
5
Bibliography
ˆ Jackson, John David. Classical Electrodynamics, 3rd edition, 1998.
ˆ
WolframMathWorld http://mathworld.wolfram.com/SphericalBesselFunctionoftheFirstKind.html
ˆ
WolframMathWorld http://mathworld.wolfram.com/SphericalBesselFunctionoftheSecondKind.html
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