Physics 506 Winter 2008

Homework Assignment #2 — Solutions

Textbook problems: Ch. 8: 8.5, 8.6, 8.7

8.5 A waveguide is constructed so that the cross section of the guide forms a right trianglewith sides of length a, a,

√2a, as shown. The medium inside has µr = εr = 1.

a) Assuming infinite conductivity for the walls, determine the possible modes ofpropagation and their cutoff frequencies.

In general, to solve a problem like this, we need to consider the Dirichlet or Neu-mann problem for a boundary without any ‘standard’ (ie rectangular or circular)symmetry. In particular, this means there is no natural coordinate system to usefor the two-dimensional Helmholtz equation [∇2

t + γ2]ψ = 0 that both allows forseparation of variables and respects the symmetry of the boundary surface (whichwould allow a simple specification of the boundary data). A general problem ofthis form (with no simple boundary symmetry) is quite unpleasant to solve.

In this case, however, we can think of the triangle as ‘half’ of a square.

a

ax

y

In particular, the key step to this problem is to note that the triangle may be ob-tained from the square by imposing reflection symmetry along the x = y diagonal.This symmetry is a Z2 reflection on the coordinates of the form

Z2 : x→ y, y → x

Eigenfunctions ψ(x, y) can then be classified as either Z2-even or Z2-odd

Z2 : ψ(x, y)→ ±ψ(y, x)

The odd functions vanish along the diagonal, so they automatically satisfy Dirich-let conditions ψ(x = y) = 0 on the diagonal. Similarly, the even functions havevanishing normal derivative on the diagonal and hence automatically satisfy Neu-mann conditions. We will use this fact to construct TM and TE modes for thetriangle.

We begin with the TM modes. Using rectangular coordinates, it is natural towrite solutions of the Helmholtz equation [∂2

x + ∂2y + γ2]ψ = 0 as ψ ∼ ei(kxx+kyy)

where k2x + k2

y = γ2. This means we may expand the eigenfuctions in terms ofsines and cosines. For TM modes satisfying the Dirichlet condition ψS = 0, westart with eigenfunctions on the square

ψ ∼ sinmπx

asin

nπy

a

which automatically satisfy the boundary conditions on the four walls of thesquare. This gives

γmn =π

a

√m2 + n2

so the cutoff frequencies are

ωmn =π

√µ0ε0a

√m2 + n2 =

πc

a

√m2 + n2 (1)

In order to satisfy the Dirichlet condition on the diagonal, we take the Z2-oddcombination

(TM) ψmn = sinmπx

asin

nπy

a− sin

nπx

asin

mπy

a

It is simple to verify that ψ(x, 0) = ψ(a, y) = ψ(x, x) = 0, so that all boundaryconditions on the triangle are indeed satisfied. The cutoff frequencies are givenby (1). Note here that the Z2 projection removes the m = n modes and alsoantisymmetrizes m with n. As a result, the integer labels m and n may be takento satisfy the condition m > n > 0.

The analysis for TE modes is similar. However, for Neumann conditions, we takecosine combinations as well as a Z2-even eigenfunction. This gives

(TE) ψmn = cosmπx

acos

nπy

a+ cos

nπx

acos

mπy

a

with identical cutoff frequencies as in (1). This time, however, the labels m andn may be taken to satisfy m ≥ n ≥ 0 (except m = n = 0 is not allowed).

b) For the lowest modes of each type calculate the attenuation constant, assumingthat the walls have large, but finite, conductivity. Compare the result with thatfor a square guide of side a made from the same material.

The attenuation coefficients are determined by power and power loss. We beginwith TM modes. For the power, we need to compute∫

A

|ψ|2 da =∫A

[sin kmx sin kny − sin knx sin kmy

]2da (2)

It is perhaps easiest to compute this by integrating over the square and thendividing by two for the triangle. This is because the integration separates into xand y integrals, and we may use orthogonality∫ a

0

sin kix sin kjx dx =a

2δi,j

(where kj =

jπ

a

)This gives ∫

A

|ψ|2 da =12× 2

(a2

)2

=a2

4

The factor of 1/2 is for the triangle, while the factor of 2 is because two non-vanishing terms arise when squaring the integrand in (2). (Recall that m 6= n forTM modes.) This gives an expression for the power

P =12

√ε

µ

(ω

ωmn

)2(1− ω2

mn

ω2

)1/2 ∫A

|ψ|2 da

=12

√ε

µ

(ω

ωmn

)2(1− ω2

mn

ω2

)1/2A

2

where A = a2/2 is the area of the triangle. Calculating the power loss involvesintegrating a normal derivative ∮

C

∣∣∣∣∂ψ∂n∣∣∣∣2 dl

We break this into three parts: along y = 0, along x = a and along the diagonalx = y. Along the y = 0 wall, we have n = y and

∂ψ

∂y

∣∣∣∣y=0

=π

a

[n sin kmx−m sin knx

]As a result ∫ a

0

∣∣∣∣∂ψ∂y∣∣∣∣2 dx =

(πa

)2 a

2(m2 + n2) =

π2

2a(m2 + n2) (3)

A similar calculation, or use of symmetry, will result in an identical expressionfor the integral along the x = a wall. For the diagonal, we use n = 1√

2(x− y) to

compute

∂ψ

∂n=

1√2

[∂ψ

∂x− ∂ψ

∂y

]y=x

=√

2π

a

[m cos kmx sin knx− n cos knx sin kmx

]=√

22π

a

[(m− n) sin km+nx− (m+ n) sin km−nx

]

This gives∫ √a0

∣∣∣∣∂ψ∂n∣∣∣∣2 dl =

√2∫ a

0

∣∣∣∣∂ψ∂n∣∣∣∣2 dx =

√2

12

(πa

)2 a

2[(m− n)2 + (m+ n)2

]=√

2π2

2a(m2 + n2)

Combining this diagonal with (3) for the sides, we obtain∮C

∣∣∣∣∂ψ∂n∣∣∣∣2 dl = C

π2

2a2(m2 + n2) =

C

2γ2mn

where C = a + a +√

2a is the circumference of the triangle. This gives a TMmode power loss of

−dPdz

=1

2σδ

(ω

ωmn

)2 1µ2ω2

mn

∮C

∣∣∣∣∂ψ∂n∣∣∣∣2 dl

=1

2σδ

(ω

ωmn

)2 1µ2ω2

mn

C

2γ2mn =

12σδ

(ω

ωmn

)2ε

µ

C

2

The attenuation coefficient is thus

βmn = − 12P

dP

dz=

1σδ

√ε

µ

(1− ω2

mn

ω2

)−1/2C

2A

so that the geometrical factor ξmn = 1 is trivial. Note that the energy losscalculation along the diagonal of the triangle gives the same result as along thesquare edges. As a result, the geometrical factor ξmn = 1 is independent ofwhether the waveguide is square or right triangular. This is why the triangularTM result is identical to the square TM result, at least up to the ratios C/A =2(2 +

√2)/a ≈ 6.83/a for the triangle and C/A = 4/a for the square.

The power loss for the TE modes is somewhat harder to deal with because of thepossibility of special cases. Consider

ψ = cos kmx cos kny + cos knx cos kmy (4)

where m ≥ n ≥ 0. If n = 0, we end up with

ψ = cos kmx+ cos kmy (m > 0)

In this case∫A

|ψ|2da =12

∫ a

0

dx

∫ a

0

dy[cos kmx+ cos kmy

]2 =12× 2( 1

2a2) =

a2

2= A

while the perimeter integrals are∫ a

0

dx∣∣ψ(y = 0)

∣∣2 =∫ a

0

dx[1 + cos kmx

]2 = a(1 + 12 ) =

3a2

√2∫ a

0

dx∣∣ψ(y = x)

∣∣2 =√

2∫ a

0

dx[2 cos kmx

]2 = 4√

2( 12a) = 2

√2a

which gives ∮C

∣∣ψ∣∣2dl = (3 + 2√

2)a

and ∫ a

0

dx∣∣n× ~∇tψ

∣∣2 =∫ a

0

dx∣∣y × ~∇tψ

∣∣2 =∫ a

0

dx∣∣−z∂xψ∣∣2y=0

=∫ a

0

dxπ2

a2m2∣∣sin kmx∣∣2 =

π2

2am2

√2∫ a

0

dx∣∣n× ~∇tψ

∣∣2y=x

=√

2∫ a

0

dx∣∣ 1√

2z(∂y + ∂x)ψ

∣∣2y=x

=√

22

∫ a

0

dxπ2

a2m2∣∣2 sin kmx

∣∣2 =√

2π2

am2

which gives ∮C

∣∣n× ~∇tψ∣∣2dl = (1 +

√2)π2

am2 = (1 +

√2)aγ2

m0

Using

P =12

õ

ε

(ω

ωmn

)2(1− ω2

mn

ω2

)1/2 ∫A

|ψ|2da

and

−dPdz

=1

2σδ

(ω

ωmn

)2 ∮C

[1γ2mn

(1− ω2

mn

ω2

)|n× ~∇tψ|2 +

ω2mn

ω2|ψ|2

]dl

with the above integrals gives an attenuation coefficient

βm0 = − 12P

dP

dz

=1

2σδ

√ε

µ

(1− ω2

m0

ω2

)−1/2 [(1 +

√2)(

1− ω2m0

ω2

)+ω2m0

ω2(3 + 2

√2)]a

A

=1σδ

√ε

µ

(1− ω2

m0

ω2

)−1/2[

1 +√

22 +√

2+ω2m0

ω2

]C

2A

where C = (2 +√

2)a and A = a2/2. Here the geometrical factors are

ξm0 =1 +√

22 +√

2, ηm0 = 1 (m > n = 0)

For the rectangular waveguide, one has instead

ξm0 =a

a+ b→ 1

2, ηm0 =

2ba+ b

→ 1 when b→ a

This is different because the power loss calculation is no longer universal, givingdifferent coefficients along the diagonal as along the square edges. The remainingTE cases to consider are modes (4) where m = n > 0 and m > n > 0. Here wesimply state the results. For m = n > 0 we have

ψ = cos kmx cos kmy

(we have removed an unimportant factor of two) so that∫A

|ψ|2da =a2

8=A

4∮C

|ψ|2dl =

(1 +

3√

28

)a

∮C

|n× ~∇tψ|2dl =

(1 +√

24

)π2

am2 =

(12

+√

28

)aγ2mm

This gives

ξmm =4 +√

24 + 2

√2, ηmm = 1 (m = n > 0)

On the other hand, for the general case m > n > 0 we find∫A

|ψ|2da =a2

4=A

2∮C

|ψ|2dl = (2 +√

2)a = C∮C

|n× ~∇tψ|2dl = (2 +√

2)π2

2a(m2 + n2) =

C

2γ2mn

which yieldsξmn = 1, ηmn = 1 (m > n > 0)

In all cases, ηmn = 1, which is the same for the triangle or the square waveg-uide. For ξmn, the factor is essentially a geometric combination of contributions

along the perimeter of either 1 or 1/2 depending on the particular mode and itsdegeneracies.

8.6 A resonant cavity of copper consists of a hollow, right circular cylinder of inner radiusR and length L, with flat end faces.

a) Determine the resonant frequencies of the cavity for all types of waves. With(1/√µεR) as a unit of frequency, plot the lowest four resonant frequencies of each

type as a function of R/L for 0 < R/L < 2. Does the same mode have the lowestfrequency for all R/L?

This cavity is essentially covered in Section 8.7 of the textbook. It is also similarto the waveguide problem 8.4, but with endcaps to form a resonant cavity. Thenormal modes are either TM or TE modes. The TM modes are given by

ψ(ρ, φ) = E0Jm(γmnρ)e±imφ, γmn =xmnR

(5)

where xmn are the zeros of the Bessel functions Jm. The resonant frequencies arethus

(TM) ωmnp =1

√µεR

√x2mn +

(pπR

L

)2

(p ≥ 0)

with

x01 = 2.405, x11 = 3.832, x21 = 5.136, x02 = 5.520

The lowest four resonant frequncies are plotted as follows

0.5 1 1.5 2

1

2

3

4

5

6ω

R/LNote that the p = 0 modes are independent of R/L. Clearly the same modedoes not always have the lowest frequency. The cross-over points are accidentaldegeneracies, and are not related to any particular symmetry of the cylinder.

The TE modes are given by

ψ(ρ, φ) = H0Jm(γmnρ)e±imφ, γmn =x′mnR

where x′mn are the zeros of J ′m. The TE resonant frequencies are

(TE) ωmnp =1

√µεR

√x′ 2mn +

(pπR

L

)2

(p > 0)

with

x′11 = 1.841, x′21 = 3.054, x′01 = 3.832, x′31 = 4.201

In this case, the lowest four resonant frequencies are

0.5 1 1.5 2

1

2

3

4

5

6

7ω

R/L

b) If R = 2 cm, L = 3 cm, and the cavity is made of pure copper, what is thenumerical value of Q for the lowest resonant mode?

For this geometry, it turns out the lowest mode is the TM010 mode. We thuscalculate the Q factor for TMmn0 modes. We start with the stored energy

U =Lε

2

∫A

|ψ|2da (6)

where ψ is given by (5). Note that this is double the p 6= 0 result. The powerloss expression for TMmn0 modes is

Ploss =ε

σδµ

(1 + ξmn

CL

2A

)∫A

|ψ|2da (7)

where we have taken p = 0 into account. Here C = 2πR and A = πR2 are thecircumference and cross-sectional area of the cylinder. The geometrical factorξmn is the same as the waveguide result, which was obtained in Problem 8.4 asξmn = 1. More directly, we may start with the definition∮

C

∣∣∣∣dψdn∣∣∣∣2 dl = ξmnγ

2mn

C

A

∫A

|ψ|2da

Using (5), this statement is equivalent to

C(γ2mnJ

′m(xmn)2

)= ξmnγ

2mn

C

A

(2π∫ R

0

Jm(xmnρ/R)2ρ dρ

)

= ξmnγ2mn

C

A

(πR2Jm+1(xmn)2

)

where the second line is obtained by the Bessel function normalization condition.This results in

ξmn =(

J ′m(xmn)Jm+1(xmn)

)2

However, using the Bessel recursion relation Jm+1(ξ) = (m/ξ)Jm(ξ)−J ′m(ξ) andletting ξ = xmn be a zero of Jm, we obtain simply Jm+1(xmn) = −J ′m(xmn).This proves that the geometrical factor is simply ξmn = 1. Finally, using (6) and(7) with ξmn = 1 gives

Qmn0 = ωmnU

Ploss=

µ

µc

L

δ

(1 +

CL

2A

)−1

=µ

µc

L

δ

(1 +

L

R

)−1

Since copper is non-ferromagnetic, we may take µc = µ0. Furthermore, we assumethe interior of the cavity has µ = µ0 and ε = ε0. Substituting in R = 2 cm,L = 3 cm then yields

Qmn0 =1.2× 10−2 m

δ

We calculate the lowest resonant frequency to be

ω010 =x01c

R=

2.405cR

= 3.61× 1010 s−1

or ν010 = 5.94 GHz, where we have used R = 3 cm. At this frequency, the skindepth for copper is

δ =6.52× 10−2 m√

νmnp(Hz)= 8.6× 10−7 m

This gives a cavity Q ofQ010 = 1.4× 104

8.7 A resonant cavity consists of the empty space between two perfectly conducting, con-centric spherical shells, the smaller having an outer radius a and the larger an innerradius b. As shown in Section 8.9, the azimuthal magnetic field has a radial depen-dence given by spherical Bessel functions, jl(kr) and nl(kr), where k = ω/c.

a) Write down the transcendental equation for the characteristic frequencies of thecavity for arbitrary l.

According to the analysis of Section 8,9, the azimuthal magnetic field is given by

Bφ(r, θ) =ul(r)r

P 1l (cos θ)

where ul(r) satisfies [d2

dr2+ω2

c2− l(l + 1)

r2

]ul(r) = 0

The solution for ul(r) is of the form

ul(r) = r[Aljl(kr) +Blnl(kr)]

where jl and nl are spherical Bessel functions and Al and Bl are constants to bedetermined from the boundary conditions

dul(r)dr

∣∣∣∣r=a

=dul(r)dr

∣∣∣∣r=b

= 0

Sincedul(r)dr

= Al[jl(kr) + kr j′l(kr)] +Bl[nl(kr) + kr n′l(kr)]

the homogeneous boundary conditions may be given in terms of a matrix equation(jl(ka) + ka j′l(ka) nl(ka) + kan′l(ka)jl(kb) + kb j′l(kb) nl(kb) + kb n′l(kb)

)(AlBl

)= 0

This only admits non-trivial solutions when the determinant vanishes. This cor-responds to the transcendental equation

jl(ka) + kaj′l(ka)nl(ka) + kan′l(ka)

=jl(kb) + kb j′l(kb)nl(kb) + kb n′l(kb)

(8)

where k = ω/c.

b) For l = 1 use the explicit forms of the spherical Bessel functions to show that thecharacteristic frequencies are given by

tan khkh

=

(k2 +

1ab

)k2 + ab

(k2 − 1

a2

)(k2 − 1

b2

)where h = b− a.

For l = 1, we have

j1(x) =sinxx2− cosx

x, n1(x) = −cosx

x2− sinx

x

So thatj1(x) + xj′1(x)n1(x) + xn′1(x)

=x cosx− (1− x2) sinxx sinx+ (1− x2) cosx

As a result, for l = 1, (8) may be rewritten as

ka cos ka− (1− (ka)2) sin kaka sin ka+ (1− (ka)2) cos ka

=kb cos kb− (1− (kb)2) sin kbkb sin kb+ (1− (kb)2) cos kb

which can be rearranged to give[(ka)(kb) + (1− (ka)2)(1− (kb)2)

]sin(kb− ka)

+[ka(1− (kb)2)− kb(1− (ka)2)

]cos(kb− ka) = 0

Defining h = b− a, we end up with

tan kh =kb(1− (ka)2)− ka(1− (kb)2)

(ka)(kb) + (1− (ka)2)(1− (kb)2)

=kh(1 + (ka)(kb))

(ka)(kb) + (1− (ka)2)(1− (kb)2)

= khk2 +

1ab

k2 + ab

(k2 − 1

a2

)(k2 − 1

b2

)(9)

c) For h/a� 1, verify that the result of part b yields the frequency found in Section8.9, and find the first order correction in h/a.

For h/a� 1, we writeb = a+ h = a(1 + h/a)

and expand for small h/a. In this case, (9) becomes

tan khkh

=k2 +

1a2− 1a2

h

a+O((h/a)2)

k4a2 − k2 +1a2

+(k4a2 − 1

a2

)h

a+O((h/a)2)

=(ka)2 + 1

(ka)4 − (ka)2 + 1− (ka)2((ka)4 + 2(ka)2 − 2)

((ka)4 − (ka)2 + 1)2h

a+O((h/a)2)

Expanding the left-hand-side for h/a� 1 gives

tan khkh

= 1 +O((h/a)2)

As a result, the characteristic frequency equation becomes

0 = − (ka)2((ka)2 − 2)(ka)4 − (ka)2 + 1

− (ka)2((ka)4 + 2(ka)2 − 2)((ka)4 − (ka)2 + 1)2

h

a+O((h/a)2) (10)

This demonstrates that the zeroth order solution is given by

ka =√

2 ⇒ ω1 =√

2c

a

which agrees with the approximate result found in Section 8.9.

To get the first order correction, we write

ka =√

2 + δ(ka)

where δ(ka) is the desired correction term. Substituting this into (10) and ex-panding to first order in δ(ka) gives

0 = −2√

2δ(ka)(ka)2

(ka)4 − (ka)2 + 1

∣∣∣∣ka=√

2

− (ka)2((ka)4 + 2(ka)2 − 2)((ka)4 − (ka)2 + 1)2

∣∣∣∣ka=√

2

h

a

or0 = −2

√2δ(ka)

23− 4

3h

a

which gives

δ(ka) = − 1√2h

a

As a result, we have

ka =√

2− 1√2h

a+ · · · =

√2(1− h/2a+ · · ·)

or

ω1 =√

2c

a(1− h/2a+ · · ·) =

√2

c

a(1 + h/2a+ · · ·)=√

2c

a+ h/2 + · · ·

Note thata+ h/2 =

a+ b

2is the average of the radii of the inside and outside spheres.