emii1sol_13.pdf

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Physics 214 Solution Set 1 Winter 2013

1. The energy and the linear momentum of a distribution of electromagnetic fields in vacuumis given (in SI units) by

U =02

d3r (E2 + c2 B 2) , (1)

P = 0

d3r E B , (2)

where the integration is over all space. Consider an expansion of the electric field in terms ofplane waves:

E(r, t) =

d3k(2)3

E0(k, ) (k) e

i(krt)

+ c.c.

, (3)

where E0(k, ) is a complex amplitude and c.c. stands for complex conjugate of the precedingterm. The polarization vector satisfies:

(k) =

(k) . (4)

(a) Show that P can be written as

P =20

c

d3k

(2)3|E0(k, )|2 k . (5)

Note that all time dependence has canceled out. Explain.

Consider the Coulomb gauge, where A = 0 [cf. eq. (6.21) of Jackson]. In the absence of

external sources ( = J= 0), we also have = 0 [cf. eq. (6.23) of Jackson]. Using eq. (6.9) ofJackson, it therefore follows that the electric and magnetic fields are given by,

E= A

t, B = A . (6)

If we write

A(r, t) =

d3

k(2)3

a(k) ei(krt) + a(k) ei(krt)

,

where = kc anda(k) =

a(k)(k) ,

then the vector amplitudes in the plane wave expansion of the electric and magnetic fields,obtained from eq. (6), are given by:

E0(k) = ikca(k) , B0(k) = ika(k) =1

ck E0(k) , (7)

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whereE0(k) =

E0(k, )(k) .

That is,

B(r, t) =1

c

d3k

(2)3

E0(k, ) k (k) e

i(krt) + c.c.

. (8)

Inserting eqs. (3) and (8) into eq. (2) [taking care to employ different dummy variables inthe sums and integrals], and expanding out the resulting expression,

P =0

(2)6 c

d3k d3k d3r

E0(k, )E0(k

, ) (k) [k

(k

)]ei(

k+k)r ei(+

)t

+E0 (k, )E

0 (k, ) (k) [k

(k

)]ei(

k+k)r ei(+

)t

+E0(k, )E

0 (k

,

) (k) [k

(k

)]ei(kk

)r

ei()t

+E0 (k, )E0(k, ) (k) [k

(k

)]ei(

kk

)r ei()t

, (9)

where kc and kc. In our notation, k |k| and k |k|.We may now perform the integral over r, using

1

(2)3

d3r ei(

kk

)r = 3(k k) , (10)

and then use the delta function to integration overk

. Then eq. (9) reduces to

P =0c

d3k

(2)3

E0(k, )E0(k, ) (k) [k (k)]e2it

E0 (k, )E0 (k, ) (k) [k (k)]e2it

+E0(k, )E

0 (k, ) (k) [k

(k)]

+E0 (k, )E0(k, ) (k) [k (k)]

, (11)

where we have used eq. (4) to write:1

(k)3(k + k

) = (k)3(k + k) = (k)3(k +k) . (12)

We can now use the vector identity,

(k) [k

(k)] = k[(k)

(k)] (k)[k (k)] = k , (13)1Recall that for any well-behaved function f(k,k

) we have f(k,k

)3(k k) = f(k,k)3(k k), due to

the presence of the delta function. For example, 3(k k) = kc 3(k k) = kc 3(k k) = 3(k k),since | k| = k.

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using the properties of the polarization vector,

k (k) = 0 , and (k)

(k) = . (14)

Using eq. (13) allows us to perform the sum over

in eq. (11), which yields

P =0c

d3k

(2)3k

E0(k, )E0(k, )e2it E0 (k, )E0 (k, )e2it + 2|E0(k, )|2

.

(15)

Noting that = kc and k k/k, it follows thatd3k kE0(k, )E0(k, )e2it = 0 ,

since the integrand is an odd function under k k. Hence, eq. (15) yields

P =20

c

d3k

(2)3k |E0(k, )|2 , (16)

which confirms the result of eq. (5).

Note that P given in eq. (16) is explicitly time-independent. This is simply an expression

of the conservation of momentum, dP/dt = 0. This is a consequence of eq. (6.122) of Jackson.

Since = J= 0 for a free electromagnetic field, we have Pmech = 0, in which case

dP

dt=

Pfield

dt= Sda n

T= 0 ,

where

T is the Maxwell stress tensor. The unit vector n is the outward normal to the surface S,where S is the surface of infinity. For any finite energy field configuration, the stress tensorvanishes at the surface of infinity and we recover dP/dt = 0 as expected.

(b) Obtain the corresponding expression for the total energy U. Employing the photon

interpretation for each mode (k, ) of the electromagnetic field, justify the statement thatphotons are massless.

The total energy is given (in SI units) by

U =02

d3r (E

2+ c2 B

2) . (17)

We first computeE

2d3r =

1

(2)6

d3kd3kd3r

E0(k, )E0(k

, ) (k) (k

) ei(

k+k

)r ei(+)t + c.c.

+

E0(k, )E

0 (k

, ) (k)

(k) ei(

kk)r ei(

)t + c.c.

,

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where the computation is similar to that of part (a). Integrating over r and using eq. (12) aswe did in part (a), it follows that

E

2

d

3

r =

d3k

(2)3

E0(k, )E0(

k,

) (k)

(k) e

2it

+E0(k, )E

0 (k,

) (k)

(k) +

Summing over using eq. (14), we obtainE

2d3r =

d3k

(2)3

E0(k, )E0(k, ) e

2it + c.c.

+ 2

d3k

(2)3|E0(k, )|2 . (18)

Next, we compute

c2 B2d3r. The only difference in the computation compared to the one

above is that (k) is replaced by k (k) and (k) is replaced by k

(k

). Thus,

instead of obtaining the factor (k) (k) 3(k + k

) after the integration over r, we now

have [cf. footnote 1]:

[k (k)] [k (k

)] 3(k +k

) = [k k][(k) (k

)] [k (k)][k (k)] 3(k +k

)

=(k) (k) + [k (k)][k (k)]3(k +k)

= (k) (k) 3(k + k

) ,

after using eqs. (12) and (14). Similarly, instead of obtaining the factor (k) (k) 3(kk)

after the integration over r, we now have:

[k (k)] [k (k

)] 3(k

k

) = (k)

(k) 3(k

k

) .

Hence, it follows that:c2 B

2d3r =

d3k

(2)3

E0(k, )E0(k, ) e

2it + c.c.

+ 2

d3k

(2)3|E0(k, )|2 . (19)

Adding eqs. (18) and (19) yields

U = 20

d3k

(2)3|E0(k, )|2 . (20)

Note that U given in eq. (20) is explicitly time-independent. This is simply an expressionof the conservation of momentum, dU/dt = 0. This is a consequence of eq. (6.111) of Jackson.

Since = J= 0 for a free electromagnetic field, we have Pmech = 0, in which case

dU

dt=

Ufielddt

= S

da n S= 0 ,

where S is the Poynting vector. For any finite energy field configuration, the Poynting vectorvanishes at the surface of infinity and we recover dU/dt = 0 as expected.

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Finally, consider a fixed wave number vector k0, for which E0(k, ) E0() 3(k k0).Then, eqs. (16) and (20) yield

U = 20 |E0()|

2

,P =

k0

20

c |E0()|

2

=

k0

c U .

That is, U = P c. Comparing this result to the relativistic relation between the energy andmomentum of a particle, E =

p2c2 + m2c4, we conclude that photons are massless.

2. [Jackson, problem 7.27] The angular momentum of a distribution of electromagnetic fieldsin vacuum (in SI units) is given by

L =1

0c

2 d3rr (E B) , (21)where the integration is over all space.

(a) For fields produced a finite time in the past (and so localized to a finite region of space)

show that, provided the magnetic field is eliminated in favor of the vector potential A, theangular momentum can be written in the form

L =1

0c2

d3r

E A +

3=1

E(r )A

. (22)

The first term above is sometimes identified with the spin of the photon and the second with

the orbital angular momentum because of the presence of the angular momentum operatorLop = i(r ).

The magnetic field can be written in terms of the vector potential, B = A. Hence, weneed to evaluate r [E ( A)]. Using the Einstein summation convention, where thereis an implicit summation over a pair of identical indices, we can write2

(ab)i = ijkajbk ,

where the indices take on the values i,j,k = 1, 2, 3 and there is an implicit sum over j and k.The Levi-Civita tensor is defined as

ijk =

+1 , if (i,j,k) is an even permutation of (1, 2, 3),

1 , if (i,j,k) is an odd permutation of (1, 2, 3),0 , otherwise.

Thus, it follows thatr[E( A)]

i

= ijkxj[E( A)]k = ijkxjkmE( A)m = ijkxjkmEmpqpAq ,2In this calculation, all vectors of Euclidean three-vectors. Consequently, we do not distinguish between

upper and lower indices. All indices will be written as subscripts in what follows.

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where r (x1, x2, x3) and p /xp. We now employ the following -identity,

kmmpq = kpq kqp .

Hence, it follows thatr [E ( A)]

i

= ijkxjE(kpq kqp)p = ijkxjEkA ijkxjEAk . (23)

We recognize ijkxjEkA = E(r )iA which corresponds to the second term in eq. (22).To obtain the first term in eq. (22) will require an integration by parts. That is, we first write:

ijkxjEAk = ijk [(xjEAk) Ak(xjEk)] ,

which is an identity that follows from the rule for differentiating products. Next, we note that

ijkAk(xjE) = ijkAk [xj(E) + E(xj)] = ijkAkEj = ijkAkEj = (E A)i ,

where we used xj xj/x = j and E = E= 0 (in vacuum). Thus, eq. (23) yieldsthe vector identity,

r [E ( A)]i

= E(r )iA + ( E A)i ijk(xjEAk) , (24)

where there is an implicit sum over the repeated index .When we integrate over all of space, we can use the divergence theorem [given in the inside

cover of Jacksons textbook]:

V d

3

r ijk(xjEAk) = Sda ijknxjEAk = Sda n E(r A)i = 0 , (25)where n is the outward normal at the surface S and S is the surface of infinity. Since thefields are assumed to be localized to a finite region of space, the integral above vanishes. Hence,inserting the results of eqs. (24) and (25) into eq. (21) [after putting B = A] immediatelyyields

d3rr (E B) =

d3r

E A +

3=1

E(r )A

.

Therefore, eq. (22) is proven.

REMARK: The identification of

Lspin =1

0c2

d3r E A , (26)

as the spin angular momentum is problematical, as eq. (26) is not invariant under gauge trans-formations. In fact, a gauge-invariant expression for the spin angular momentum can be con-structed that reduces to eq. (26) in the radiation (Coulomb) gauge.3

3See e.g., Iwo Bialynicki-Birula and Zofia Bialynicki-Birula, Journal of Optics 13, 064014 (2011) and refer-ences therein.

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(b) Consider an expansion of the vector potential in the radiation (Coulomb) gauge in termsof plane waves,

A(r, t) =

d3k

(2)3 (k)a(k)e

i(krit) + c.c. . (27)The vectors (k) are conveniently chosen as the positive and negative helicity polarizationvectors4

= 12

(1 i2) , (28)where 1 and 2 are the real orthogonal vectors in the plane whose positive normal is in thedirection ofk.

Show that the time average of the first (spin) term of L can be written as

Lspin =2

0c

d3k

(2)3k|a+(k)|2 |a(k)|2

.

Can the term spin angular momentum be justified from this expression? Calculate the energyof the field in terms of the plane wave expansion of A and compare.

In the Coulomb gauge, the electric field is (in SI units):

E(r, t) = A

t= i

d3k

(2)3(k)a(k)e

i(krit) c.c.

, (29)

where = ck and k |k|. Note that due to the overall factor of i, we must subtract thecomplex conjugate inside the square brackets in order to ensure that E(r, t) is a real field.Inserting eqs. (27) and (29) into eq. (26) and expanding out the integrand, we obtain:

Lspin =1

0c2i

(2)6

d3k d3k d3r

[(k)(k

)]a(k)a(k

)ei(

k+k)r ei(+

)t

+[(k)

(k)]a(k)a

(k)ei(

kk )r ei(

)t

[(k)(k)]a(k)a(k)ei(kk

)r ei()t

[(k)(k)]a(k)a(k)ei(k+k )r ei(+

)t

,

where = kc and = kc.We may now perform the integral over r, using eq. (10), and then use the delta function to

integrate over k. The end result is

Lspin =i

0c2

d3k

(2)3

[(k)

(k)]a(k)a

(k) [(k)(k)]a(k)a(k)

+[(k)(k)]a(k)a(k) e2it [(k)(k)]a(k)a(k) e2it

.

(30)

4Jackson omits the overall factor of in the definition of . I prefer to maintain this phase convention,but you are free to choose any convention that suits you.

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However, the last two terms above vanish when integrated over k, since the correspondingintegrands are odd functions ofk. For example, under k k,

[(k)(k)]a(k)a(k) e2it

[(k)(k)]a(k)a(k) e2it ,

=

[(k)(k)]a(k)a(k) e2it ,

=

[(k)(k)]a(k)a(k) e2it ,

where we interchanged and in the penultimate step (which is justified since these aredummy labels that are being summed over), and used the antisymmetry of the cross product

in the final step. Note that = |k|c does not change sign when k k. Hence, eq. (30)simplifies to

Lspin =

i

0c2

d3k(2)3

[(

k)

(k)]a(

k)a

(k) [

(k)

(k)]a

(k)a

(k)

. (31)

Using the definition of the polarization vectors given in eq. (28), it is straightforward to verifythat5

(k)

(k) = i k , for , = . (32)This result allows us to sum over in eq. (31). Both terms in eq. (31) contribute equally andthe end result is:

Lspin =2

0c2

d3k

(2)3k|a+(k)|2 |a(k)|2 , (33)

after using = kc and k = kk. Note that Lspin is time-independent and thus conserved. This

is a stronger condition than the conservation of angular momentum, which only requires thatthe sum L = Lorbital + Lspin is conserved. Eq. (33) implies that the spin angular momentumof the electromagnetic field is separately a constant of the motion.6 If we interpret each mode(k, ) as a photon, then the two possible photon spin states (in a spherical basis) correspond

to positive and negative helicity, i.e. states of definite spin angular momentum in which Lspinpoints in a direction parallel or antiparallel to the direction of propagation k, respectively.

It is instructive to consider the energy of the electromagnetic fields, which was obtained inproblem 1. In particular, eq. (20) yields

U = 20

d3k

(2)32|a(k)|2 , (34)

5To prove eq. (32), use the fact that 1 2 = 2 1 = k and 1 1 = 2 2 = 0.6Indeed, Jackson only asks that we show that the time-average of Lspin is given by eq. (33). In such a

calculation, the last two terms in eq. (30) are immediately set to zero when taking the time-average since thetime-averaged values

e2it = 1T

T

0

e2it dt = 0 , when = 0 ,

where T = 2/ is the time for one oscillation cycle. The case of = 0 corresponds to k = 0, in which case thelast two terms in eq. (30), when summed over and , are each manifestly equal to zero, since eq. (28) implies

that (k)(k) = 0 for = (and the cross-terms vanish). However, our result above is more general sinceno time-averaging is required to obtain eq. (33).

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where we have used eq. (7) to write E0(k, ) = ia(k). Consider a fixed mode of positive

helicity (k0, = +1). Then, a(k) = a+(k0)(k k0),+1, in which case eq. (34) yields

U =

2020

(2)3 |a(k0)|

2

,

andLspin =

2

0c 1

(2)3k0|a(k0)|2 = 200

(2)3k0|a(k0)|2 ,

after using 00 = 1/c2 and k0 = (0/c)k0. That is,

Lspin = U

0k0 , for = +1 . (35)

For a fixed mode of negative helicity (k0, =

1), we again obtain eq. (35) with =

1. For

a single photon of frequency 0, quantum mechanics states that U = 0, and eq. (35) yields

Lspin = k0 ,corresponding to a spin-one particle of helicity 1, with its spin parallel or antiparallel to thedirection of propagation k0.

3. [Jackson, problem 8.4] Transverse electric and magnetic waves are propagated along a hollow,right circular cylinder with inner radius R and conductivity .

(a) Find the cutoff frequencies of the various TE and TM modes. Determine numericallythe lowest cutoff frequency (the dominant mode) in terms of the tube radius and the ratio ofcutoff frequencies of the next four higher modes to that of the dominant mode. For this partassume that the conductivity of the cylinder is infinite.

For the TM modes, we must solve [cf. eqs. (8.34)(8.36) of Jackson]:

(2 +

2)Ez = 0 , where EzS

= 0 , (36)

and 2 = 2 k2 > 0 (in SI units). In cylindrical coordinates, x = r cos and y = r sin ,with r =x2 + y2 and

2 = 1r

r

r

r

+ 1

r22

2.

Employing the separation of variables technique,

Ez(r, t) = R(r)()eikzit ,

eq. (36) is re-expressed as

1

R

d2R

dr2+

1

r

dR

dr

+

1

r2

d2

d2+ 2 = 0 .

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Separating variables,

1

d2

d2= 1

R r2 d

2R

dr2+ r

dR

dr 2r2 = m2 ,

where m2 is the separation constant.The solution to the equation for is

() = Aeim , where m = 0, 1, 2, . . . ,

where n is an integer due to the requirement of single-valueness, ( + 2) = (), and A is aconstant.

The equation for R is then given by

d2R

dr2

+1

r

dR

dr

+2 m2

r2 R = 0 ,

which we recognize as Bessels equation. We reject Nm(r) as a solution since the Besselfunction of the second kind is singular at the origin. Thus, R(r) = CJm(r) where C is aconstant. We conclude that

Ez(r,,z,t) = EmJm(r)eimikzit , where m = 0, 1, 2, . . . , (37)

is the most general solution to eq. (36) before imposing the boundary condition, where Em isa constant. Note that since Jm(r) = (1)mJm(r), eq. (37) can be rewritten as

Ez(r,,z,t) = Jm(r) [Em1 sin m + Em2 cos m] eikzit , where m = 0, 1, 2, . . . , (38)

where Em1 and Em2 are constants.The boundary condition is given by

Ez(R,,z,t) = 0 .

Imposing this condition on eq. (37) yields

Jm(R) = 0 .

Thus, R = xmn, where xmn > 0 are the positive zeros of the Bessel function Jm. That is,

Jm(xmn) = 0, for n = 1, 2, 3, . . ., where n labels the zeros of the Bessel function for fixed m.Hence, the allowed values for the eigenvalues are

nm =xmn

R, where m = 0, 1, 2, . . . , and n = 1, 2, 3, . . . ,

which specify the allowed TM modes.For the TE modes, we must solve:

(2 +

2)Hz = 0 , whereHzn

S

= 0 , (39)

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We now insert eq. (44) into eq. (45) and evaluate the resulting integrals. The required integralover r is7

R

0

r dr J0 x01r

R 2

= 12

R2[J1(x01)]2 .

The end result is

P = 12 |E0|2 A

01

21

2

201

1/2[J1(x01)]

2 ,

where A = R2 is the cross-sectional area of the waveguide and 01 =1

x01R

[cf. eq. (41)].

Next, eq. (8.59) of Jackson yields

dPlossdz

=1

22201

01

2 C

Ezdn

2

d . (46)

In this problem, n = r, so

Ezdn

= n Ez =Ezr

=E0x01

rJ0

x01rR

ei(kzt) .

Hence, noting that d = Rd and x01 = 01R

, it follows that

C

Ezdn

2

d =|E0|2x201

R2

20

R d

J0

x01rR

2r=R

= 2R|E0|2201[J1(x01)]2 ,

after using the relation J0(x) = J1(x). Hence, eq. (46) yields:

dPlossdz

= 12 |E0|2C

01

2

[J1(x01)]2 ,

where C = 2R is the circumference of the cross-sectional area of the waveguide.The attenuation constant defined in eq. (43) is therefore given by:

01 =1

2

C

A

1

201

2

1/2. (47)

The skin depth, which is given by eq. (8.8) of Jackson, can be rewritten as:

=

2

c

1/2

=01

1/201 , (48)

where

2

c

1/2(49)

7See e.g. formula 6.5615 on p. 676 of I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, andProducts (7th edition), edited by Alan Jeffrey and Daniel Zwillinger (Academic Press, Elsevier, Inc., Burlington,MA, 2007).

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is the skin depth corresponding to the mode frequency . Then, we can rewrite eq. (47) as:

01 =1

201

C

A

01

1/21

2

201

1/2.

Comparing this result with eq. (8.63) of Jackson,

=

1

C

2A

1/21

2

2

1/2 +

2, (50)

we see that for the TM01 mode of this problem, we have 01 = 1 and 01 = 0.

Case 2: TE11

In part (a), we obtained

Hz(r,,z,t) = H0J1 y11rR ei ei(kzt) , (51)

for the longitudinal magnetic field of the TE mode with ( m, n) = (1, 1). Assuming no losses,the power is given by eq. (8.51) of Jackson,

P =1

2

11

2 1

2

211

1/2 R0

r dr

20

d |Hz(r,,z,t)|2 . (52)

We now insert eq. (51) into eq. (52) and evaluate the resulting integrals. The required integralover r is8 R

0

r dr

J1y11r

R

2= 12 R

2

1 1

y211

[J1(y11)]

2 .

The end result is

P = 12|H0|2 A

11

21

2

211

1/21 1

y211

[J1(y11)]

2 ,

where A = R2 is the cross-sectional area of the waveguide and 11 =1

y11R

[cf. eq. (42)].

Next, eq. (8.59) of Jackson yields

dPlossdz

=1

2

11

2 C

1

211

1

211

2

|n Hz|2 +

211

2|Hz|2

d .

In this problem, n = r, so

n Hz = rx

x

+ y

y

Hz = (x cos + y sin )

x

x

+ y

y

Hz

= z

cos

y sin

x

Hz = z

1

r

Hz

= iH0r

J1

y11rR

ei ei(kzt) . (53)

8See e.g. eq. (5.14.9) on p. 130 of N.N. Lebedev, Special Functions and Their Applications (Dover Publica-tions, Inc., Mineola, NY, 1972).

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Hence, noting that d = Rd and

d = 2R = C, it follows that

dPlossdz

= 12|H0|2 C

112

[J1(y11)]2

1

R2211 1 2112 +

2112 .

We can simplify this expression using y11 = R11

. The expression in brackets above is:

1

R2 211

1

211

2

+

2112

=1

y211

1

211

2

+

2112

=1

y211+

2112

1 1

y211

.

Hence,dPloss

dz= 1

2|H0|2 C

11

2[J1(y11)]

2

1

y211+

2112

1 1

y211

.

The attenuation constant defined in eq. (43) is therefore given by:

11 =1

2

CA

1 211

2

1/2

1y211 1

+ 211

2

. (54)

The skin depth can be rewritten as [cf. eq. (48)]

=

2

c

1/2=11

1/211 ,

where is defined in eq. (49). Then, we can rewrite eq. (54) as:

11 =

1

211

C

A

111/2

1 211

21/2 1

y211 1 +211

2

.

Comparing this result with eq. (8.63) of Jackson [cf. eq. (50)], we see that for the TE11 modeof this problem, we have

11 =1

y211 1 0.419 , 11 = 1 .

The graphs of the attenuation constants, 01 for the TM01 mode and 11 for the TE11 mode,are very similar to Figure 8.6 on p. 366 of Jackson, so we will not elaborate further here.

4. [Jackson, problem 8.5] A waveguide is constructed so that the cross section of the guideforms a right triangle with sides of length a, a and

2a, as shown in Figure 1. The medium

inside has r = r = 1.

(a) Assuming infinite conductivity for the walls, determine the possible modes of propagationand their cutoff frequencies.

Since the waveguide consists of a single hollow conductor of infinite conductivity, there areno TEM waves. Hence, we consider separately the cases of TM waves and TE waves.

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y

xa

aa2

Figure 1: The cross-section of a triangular waveguide, projected onto the xy plane.

TM waves

Following the lecture notes, the TM waves are solutions to

2

+2

c2 k2

Ez = 0 , where EzS

= 0 , (55)

after putting r = r =1 . Thus, we must solve2

x2+

2

y 2+ 2

Ez = 0 , where EzS

= 0 , (56)

where

2 2

c2 k2 .

The general solution to eq. (56) is of the form,

Ez(x, t) = Ez(x, y) eikzit ,

subject to the boundary conditions,

Ez(x, 0) = Ez(a, y) = Ez(x, x) = 0 , for 0 x , y a .In class, we showed that the solutions for TM waves in the case of a square cross-section (wherea is the length of a side of the square), which differs only in the boundary conditions, is

Ez(x, y) = E0 sinmx

a

sinny

a

,

where

2mn =2

a2

m2 + n2

, for n, m = 1, 2, 3, . . . .

These solutions satisfy two of the three boundary conditions for the triangular waveguide. Butthe condition Ez(x, x) = 0 is not satisfied. However, a simple linear combination of two solutionsdoes satisfy this third boundary condition,

Ez(x, y) = E0

sinmx

a

sinny

a

sin

nxa

sinmy

a

. (57)

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Having found the solutions that satisfy the boundary conditions, we can invoke the uniquenesstheorem for solutions to Laplaces equation in two-dimensions to argue that the most generalsolution for TM waves consists of arbitrary linear combinations of solutions of the form givenin eq. (57) for any two positive integer m and n, assuming that m

= n. (The case ofm = n is

rejected since in this case Ez(x, y) = 0.)The cutoff frequencies (after putting r = r =1) are given by:

mn = cmn =c

a(n2 + m2)1/2 , for n = m and n, m = 1, 2, 3, . . . .

TE waves

Following the lecture notes, the TE waves are solutions to

2 +

2

c2 k2

Bz = 0 , whereBzn

S

= 0 , (58)

after putting r = r =1 . Thus, we must solve2

x2+

2

y2+ 2

Bz = 0 , whereBzn

S

= 0 , (59)

where once again,

2 2

c2 k2 .

The general solution to eq. (59) is of the form,

Bz(x, t) = Bz(x, y) eikzit ,

subject to the boundary conditions,

Bzy

(x, 0) =Bzx

(a, y) = 0 ,

and along the diagonal y = x, where

n =1

2(x + y) and

n= n =

12

x+

y

, (60)

we have

Bzx

(x, x) + Bzy

(x, x) = 0 . (61)

In class, we showed that the solutions for TM waves in the case of a square cross-section (wherea is the length of a side of the square), which differs only in the boundary conditions, is

Bz(x, y) = B0 cosmx

a

cosny

a

, (62)

where

2mn =2

a2

m2 + n2

, for n, m = 0, 1, 2, 3, . . . ,

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where the case ofn = m = 0 is not allowed.9 These solutions satisfy two of the three boundaryconditions for the triangular waveguide. But the boundary condition given by eq. (61) is notsatisfied. However, a simple linear combination of two solutions does satisfy this third boundarycondition,

Bz(x, y) = E0

cosmx

a

cosny

a

+ cos

nxa

cosmy

a

. (63)

Having found the solutions that satisfy the boundary conditions, we can invoke the uniquenesstheorem for solutions to Laplaces equation in two-dimensions to argue that the most generalsolution for TE waves consists of arbitrary linear combinations of solutions of the form given ineq. (57) for any two non-negative integer m and n, unless m = n = 0 which must be rejected.9

Note that if n and m are positive integers, the case of n = m is a valid solution for the TEcase, in contrast to the TM case treated above.

The cutoff frequencies (after putting r = r =1) are given by:

mn = cmn =c

a (n2+m2)1/2 , for n, m = 0, 1, 2, 3, . . . (the case n = m = 0 is not allowed) .

(b) For the lowest modes of each type, calculate the attenuation constant, assuming thatthe walls have large, but finite, conductivity. Compare the result with that for a square guideof side a made from the same material.

The attenuation constant is defined in eq. (8.57) of Jackson,

= 12P

dP

dz,

where labels the mode. To evaluate this, we employ eq. (8.51) of Jackson for P and eq. (8.59)of Jackson for dP/dz. It is convenient to define

2

c

1/2=

1/2, (64)

using the frequency dependence of the skin depth given in eq. (8.8) of Jackson.

9It appears that Bz is a constant (independent of position) in the case of m = n = 0 [cf. eq. (62)]. But, thisconstant must be zero due to Faradays law,

C

E d = i S

B n da ,

for harmonic fields (where d/dt i and the factors eit have been stripped off). Choose a surface S thatlies in the x-y plane (in which case n = z), and whose boundary C lies inside the metallic walls of the conductor.

Since E= 0 inside the conductor and Bz is a constant inside the waveguide, it follows that

0 =

S

B n da = Bz

S

da = ABz .

where A is the cross-sectional area of the waveguide. Hence, Bz = 0 as claimed. Thus, any mode thatcorresponds to m = n = 0 must be purely transverse, i.e. a TEM mode. However, TEM waves cannot besupported by a single hollow conductor of infinite conductivity, so we conclude that there are no non-trivialsolutions in the case ofn = m = 0.

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Since the medium inside has r = r = 1, we may put = 0, = 0 and 00 = 1/c2.

Then, using eqs. (8.51), (8.57) and (8.59) of Jackson [along with eq. (64) above], one obtainsthe attenuation constant for TM and TE modes respectively.

For TM modes,

=

00

1

2

(/)1/2

(1 2/2)1/2

C

c2

2

Ezn

2

dA

|Ez|2 da. (65)

For TE modes,

=0

0

1

2

(/)1/2

(1 2

/2

)

1/2

C

c2

2(1 2/2)|n tBz|2 + (2/2)|Bz|2

d

A

|Bz|2

da

. (66)

Attenuation of the lowest TM mode

The lowest TM modes correspond to (m, n) = (1, 2) and (2, 1). Due to the symmetry of theproblem under the interchange of the x and y coordinates, the attenuation constant is the samefor both modes. For definiteness, we focus on the case of (m, n) = (1, 2). The correspondingfrequency is

=

5c

a

. (67)

Using eq. (57), the z-component of the electric field is given by:

Ez(x, y) = E0

sinx

a

sin

2y

a

sin

2x

a

siny

a

. (68)

Hence, it follows thatA

|Ez|2 da = E20a

0

dx

x0

dy

sin2x

a

sin2

2y

a

+ sin2

2x

a

sin2y

a

2sinxa sin2ya sin2xa sinya . (69)If we interchange the order of integration, it follows that

A

|Ez|2 da = E20a

0

dy

ay

dx

sin2x

a

sin2

2y

a

+ sin2

2x

a

sin2y

a

2sinx

a

sin

2y

a

sin

2x

a

siny

a

. (70)

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Hence,

C

Ezn

2

d = 2E20

42

a2 a

2+

2

a2 a

2

+ 2

2 4

2

a2 a

0

sin6

x

a dx

=52E20

a

1 +

12

.

Using eqs. (65) and (67), we end up with

=

00

2 +

2

a

(/)1/2

(1 2/2)1/2, (72)

where =

5c/a. Comparing with eq. (8.63) of Jackson, we see that C = a(2 +

2) andA = 12 a

2 so that eq. (72) corresponds to = 1 and = 0.We can repeat the above analysis for the square guide. Note that the mode of the triangular

guide given in eq. (68) is also a mode of the square guide. 10 Thus, we only need to make minormodifications of the above computations. First, we note that

A

|Ez|2 da = a2E202

,

since the area of integration is twice as large for the square guide. Next, we obtainC

Ezn

2

d = 4E20

42

a2 a

2+

2

a2 a

2

=

102E20a

,

since we replace the integration over the diagonal with the integration over the other two sides

of the square. Thus, =

00

2

a

(/)1/2

(1 2/2)1/2.

Since C = 4a and A = a2 for the square, we again recover eq. (8.63) of Jackson with = 1and = 0. Consequently, the relative attenuation of the triangular guide and the squareguide is simply governed by the quantity C/A. Thus, the attenuation of the TM mode with(m, n) = (1, 2) of the triangular guide is 1 + 12

2 1.71 times larger than that of the square

guide.

Attenuation of the lowest TE mode

The lowest TE modes correspond to (m, n) = (1, 0) and (0, 1). Due to the symmetry of theproblem under the interchange of the x and y coordinates, the attenuation constant is the samefor both modes. For definiteness, we focus on the case of (m, n) = (1, 0). The correspondingfrequency is

=c

a. (73)

10Note that this mode is not the lowest mode of the square guide, since (m,n) = (1, 1) yields a lower frequencyfor the square guide. This latter mode is not permitted for the triangular guide as noted in part (a) of thisproblem.

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Using eq. (63), the z-component of the magnetic field is given by:

Bz(x, y) = E0

cosx

a

+ cos

ya

. (74)

Following the analysis of the lowest TM mode above, we first computeA

|Bz|2 da = E20

2

a0

dx

a0

dy

cos2x

a

+ cos2

ya

+ 2 cos

xa

cosy

a

= 12 E

20 a

2 .

Next, we evaluate the relevant integrals over the closed triangular contour C. Our analysisabove showed that the integration of an arbitrary function f(x, y) is given by

C

f(x, y) d =

a0

f(x, 0) dx +

a0

f(0, y) dy +

2

a0

f(x, x) dx .

Applying this result to integrate |Bz|2 using eq. (74), we haveC

|Bz|2 d = E20

a

0

cos2

xa

+ 2 cos

xa

+ 1

dx

+

a0

cos2y

a

+ 2 cos

ya

+ 1

dy +

2

a0

4cos2x

a

dx

= (3 + 2

2)E20 a .

Likewise, using

t = x

x

+ y

y

,

it follows thatC

|n tBz|2 dx =a

0

dx

Bzx

2

y=0

+

a0

dy

Bzy

2

x=a

+1

2

a0

dx

Bzdx + Bzdy

2

x=y

,

since n = y in the first integral, n = x in the second integral and n = (x + y)/2 in thethird integral. Plugging in eq. (74) then yieldsC

|n tBz|2 dx = 2E20a2

a0

sin2x

a

dx +

a0

sin2y

a

dy

+

2

2E20 2

a2

a0

sin2x

a

dx

=2E20

a

1 +

2

.

Hence eq. (66) yields

=

00

1

a

(/)

1/2

(1 2/2)1/2

1 2

2

(1 +

2) +

22

(3 + 2

2)

=

00

2 +

2

a

(/)

1/2

(1 2/2)1/2

12

+22

. (75)

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Comparing with eq. (8.63) of Jackson, we see that C = a(2 +

2) and A = 12 a2 so that eq. (75)

corresponds to = 1/

2 and = 1.We can again repeat the above analysis for the square guide. Note that the mode of the

triangular guide given in eq. (74) is also a mode of the square guide. Thus, we only need tomake minor modifications of the above computations. First, we note that

A

|Bz|2 da = a2E20 ,

since the area of integration is twice as large for the square guide. Next, we obtain

C

|Bz|2 d = 2E20a

0

cos2x

a

+2cos

xa

+1

dx+

a0

cos2y

a

+2 cos

ya

+1

dy

= 6E20 a

andC

|ntBz|2 dx =a

0

dx

Bzx2y=0

+

a0

dy

Bzy2x=a

+

a0

dx

Bzx2y=a

+

a0

dy

Bzy2x=0

=22E20

a

Hence eq. (66) yields

=

00

1

2a

(/)

1/2

(1 2/2)1/2

2

1

2

2

+

622

=

00

1

a

(/)

1/2

(1 2/2)1/2

1 +222

. (76)

Comparing with eq. (8.63) of Jackson, we see that C = 4a and A = a2 so that eq. (76)corresponds to =

12

and = 1. Thus, the attenuation of the TE mode with (m, n) = (1, 0)of the triangular guide is larger than that of the square guide for all values of the frequency.

5. [Jackson, problem 8.6] A resonant cavity of copper consists of a hollow, right circular cylinderof inner radius R and length L, with flat end faces.

(a) Determine the resonant frequencies of the cavity for all types of waves. With (1/

R)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 lowest frequency for all R/L?

The resonant frequencies for TM modes are given by eq. (8.81) of Jackson. Defining theunit of frequency by

0 1 R

, (77)

the resonant frequencies are given by:

mnp = 0

x2mn +

p22R2

L2, where m, p = 0, 1, 2, 3, . . . and n = 1, 2, 3, . . . ,

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and xmn is the nth zero of Jm.For p = 0, we have (independent of the value of R/L),

mn0

0

= xmn ; x01 = 2.405 , x11 = 3.832 , x21 = 5.136 , x02 = 5.520 , . . . .

For p = 1, we have

0110

=

(2.405)2 + 2R2/L2 ; 2.405

emii1sol_13.pdf

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