k ŷ E0 cos kz sin ωt =
ω ŷ B0 cos kz sin ωt. Dividing out the
common factor ŷ cos kz sin ωt gives
kE0 = ωB0 or (since c = ω/k)
√
E0 = cB0 .
QED
The University of British Columbia
Physics 401 Assignment # 6:
Electromagnetic Waves
SOLUTIONS:
(c)
Wed. 8 Feb. 2006 — finish by Wed. 22 Feb.
1.
CMBR: Most of the electromagnetic energy
in the universe is in the cosmic microwave
background radiation (CMBR), sometimes
referred to as the 3◦ Kelvin background.
Penzias and Wilson discovered the CMBR in
1965 using a radio telescope, and subsequently
received the Nobel Prize for this discovery. This
background radiation has wavelength
λ ∼ 1.1 mm. The energy density of the CMBR
is about 4.0 × 10−14 J/m3 . What is the rms
electric field strength of the CMBR?
ANSWER: If
huEM i =p
ǫ0 hE 2 i =p
4.0 × 10−14 J/m3 , then
2
E
qrms ≡ hE i = huEM i/ǫ0 =
(d)
4.0×10−14
0.88541878×10−11
= 0.0672 V/m . (The
wavelength, while interesting, is irrelevant to the
question.)
2.
3.
STANDING WAVES: Consider standing
electromagnetic waves:
~
E
~
B
(a)
(b)
=
=
E0 (sin kz sin ωt) x̂
B0 (cos kz cos ωt) ŷ .
with
Show that these satisfy the wave equation
(9.2). ANSWER: When we’re taking the
spatial derivatives, the t-dependent factor is
just part of the amplitude, and vice versa.
Thus ∇2 sin kz = −k 2 sin kz and
∇2 cos kz = −k 2 cos kz;
∂/∂t sin ωt = −ω 2 sin ωt and
∂/∂t cos ωt = −ω 2 cos ωt; so
~ − (1/c2 )∂ E/∂t
~
~
∇2 E
= (−k 2 + ω 2 /c2 )E
~
and similarly for B. But
(−k 2 + ω 2 /c2 ) = −k 2 [1 − (ω 2 /k 2 )/c2 ] = 0,
~ = 0 and
since c = ω/k. Thus 2 E
√
~
similarly for B.
QED
Show that we must also have c = ω/k and
E0 = cB0 . ANSWER: Since c = ω/k is
a universal property of all solutions of The
Wave Equation (TWE), that’s a given.
Applying Faraday’s law,
~ ×E
~ = −∂ B/∂t,
~
∇
gives
~ × (sin kz x̂) =
E0 sin ωt ∇
−B0 cos kz ∂(cos ωt)/∂t ŷ or
Show that the time-averaged power flow
across any area will be zero.
~ =E
~ ×H
~ = (x̂ × ŷ)
ANSWER: S
(E0 B0 /µ) (sin kz sin ωt) (cos kz cos ωt) =
ẑ (E0 B0 /µ) (sin kz cos kz) (sin ωt cos ωt).
Looking only at the t-dependence to get the
time average, we note that
sin ωt√cos ωt = 21 sin(2ωt) which averages to
zero.
QED
Show that the Poynting vector will also be
zero, i.e. there is no net energy flow.
ANSWER: I must apologize for a
defective question. [The hazards of using
someone else’s problem!] As explained
~ = (E0 B0 /4µ) sin 2kz · sin 2ωt ẑ.
above, S
This is only zero where sin 2kz = 0, i.e. at
z = 0 and 2kz = nπ (where n is any
integer). That is, for z = nλ/4. At any
~ oscillates in the ±ẑ
other position, S
direction, averaging to zero.
(p. 386, Problem 9.14) — REFLECTED &
TRANSMITTED POLARIZATION: In
Eqs. (9.76) and (9.77) it was tacitly assumed
that the reflected and transmitted waves have
the same polarization as the incident wave,
namely along the x̂ direction. Prove that this
must be so. [Hint: Let the polarization vectors
of the reflected and transmitted waves be
n̂T
n̂R
= cos θT x̂ + sin θT ŷ
= cos θR x̂ + sin θR ŷ
and
and prove from the boundary conditions that
~k
θT = θR = 0.] ANSWER: We must have E
continuous across the boundary. Since the normal
~ k is constituted of x and y
direction is k̂ = ẑ, E
~I +E
~R=E
~ T or
components. Thus E
EI + ER cos θR = ET cos θT [1] and
~ k must be
ER sin θR = ET sin θT [2]. Similarly, H
continuous across the boundary, and, as always,
~ = k̂ × E,
~ giving EI −ER cos θR = ET cos θT [3]
vB
µ1 v1
µ2 v2
ER sin θR
θT
and µ1 v1 = − ETµsin
[4]. If β ≡ µµ21 vv21 ,
2 v2
Eq. [4] reads ER sin θR = −βET sin θT , which we
can combine with Eq. [2] to conclude that
ET sin θT = −βET sin θT , which can be true only
if ET = 0 (trivial case) or θT = 0 (mod 2π).
Equation [2] then also requires θR = 0
√
2π).
QED
4.
(mod
(p. 392, Problem 9.15) — COMPLEX
ALGEBRA EXERCISE: Suppose that we
2
=
have six nonzero constants A, B, C, a, b, c such
than Aeiax + Beibx = Ceicx for all x. Prove that
a = b = c and A + B = C. ANSWER: The
first part is easy: if it were not true that
a = b = c then even if the equation were satisfied
at some position in x, it would √not be satisfied at
some nearby x. So a = b = c.
The second √part
is even easier: at x = 0, A + B = C. Done.
5.
(p. 392, Problem 9.17) — DIAMOND: The
index of refraction of diamond is 2.42.
Construct the graph analogous to Figure 9.16
for the air/diamond interface. (Assume
µ1 = µ2 = µ0 .) ANSWER: Fresnel’s
equations read
Ẽ0R
Ẽ0T
α−β
2
,
=
=
α+β
α+β
Ẽ0I
Ẽ0I
where
r
h
i2
p
1 − nn12 sin θI
2
cos θT
1 − sin θT
α≡
=
=
cos θI
cos θI
cos θI
µ1 n2
µ1 v1
=
. In this case β = 2.42 (we
and β ≡
µ2 v2
µ2 n1
assume the light is entering
q the diamond rather
2
1 − (sin θI /2.42)
.
than emerging) and α =
cos θI
You can use your favourite spreadsheet or other
plotting software to produce the graph below. (I
used http://musr.org/muview/, a free Java
spreadsheet applet we built at TRIUMF.)
1−5.8564
(1/5.8564)−5.8564
sin θB = 0.9242 ⇒
(c)
(a)
the amplitudes at normal incidence;
ANSWER: For θI = 0, α = 1, giving
1 − 2.42 I
Ẽ0R =
Ẽ or Ẽ0R = −0.4152Ẽ0I
1 + 2.42 0
2
and Ẽ0T =
Ẽ I or
1 + 2.42 0
Ẽ0T = 0.5848Ẽ0I .
(b)
Brewster’s angle;
ANSWER: sin2 θB =
1 − β2
(n1 /n2 )2 − β 2
θB = 67.55◦
.
and the “crossover” angle at which the
reflected and transmitted amplitudes are
equal. ANSWER: Rather than try to
read this off the graph, let’s calculate it
exactly:
q The condition is α − β = 2 or
2
1 − (sin θI /2.42)
= 4.42 or
cos θI
2
1 − (sin θI /2.42) = 19.5364 cos2 θI or
5.8564 − 1 + cos2 θI = 114.413 cos2 θI or
4.8564 = 113.413 cos2 θI or
cos2 θI = 4.8564/113.413 = 0.04282 or
cos θI = 0.20693 ⇒ θC = 78.06◦ .
α=
6.
PLANE WAVE STRESS TENSOR: Find
all the elements of the Maxwell stress tensor of
a monochromatic plane wave traveling in the
z-direction, polarized in the x-direction:
~
E(z,
t) =
E0 cos(kz − ωt + δ)x̂
~
B(z,
t) =
E0
cos(kz − ωt + δ)ŷ
c
ANSWER: Recall Eq. (8.19) on p. 352:
`
´ `
´
Tij = ǫ0 Ei Ej − δ( ij)E 2 /2 + Bi Bj − δ( ij)B 2 /2 /µ0 .
Here Ei = δi1 E where E ≡ E0 cos(kz − ωt + δ)
and Bi = δi2 B where B ≡ Ec0 cos(kz − ωt + δ)
= E/c, so all off-diagonal elements
are zero. We
have T11 = ǫ0 E 2 − E 2/2 − B 2 /2µ0 =
2
/2
or
ǫ0 E 2 /2 − E 2 /2ǫ0 µ0 c2 = ǫ0 E 2 /2 − E
2
2
T11 = 0, T22 = −ǫ0 E 2 /2 +
B
−
B
/2
µ
=
0
ǫ0 −E 2 /2 + E 2 /2ǫ0 µ0 c2 = ǫ0 −E 2 /2 + E 2 /2
or T22 = 0 and T33 = −ǫ0 E 2 /2 − B 2 /2µ0 or (only
nonzero element!)
In particular, calculate
= 0.85415 or
T33 = −ǫ0 E 2 = −uEM
.
In what direction does this EM wave transport
momentum? Does this agree with the form of
the Maxwell stress tensor you just deduced?
ANSWER: If Tij represents the force per unit
area acting in the x̂i direction on a surface whose
normal is in the x̂j direction, then the diagonal
elements are pressures and T33 is the radiation
pressure on a surface normal to ẑ. In the same
way −T33 represents the the momentum current
density transported by the fields, and is (as
expected) in the same direction as k̂ and is, in
~
fact, equal to S/c.
3
7.
(p. 412, Problem 9.33) — SPHERICAL WAVES: Suppose that
A sin θ
1
~
E(r, θ, φ, t) =
cos (kr − ωt) −
sin (kr − ωt) φ̂
r
kr
with c = ω/k, as usual. [This is, incidentally, the simplest possible spherical wave.
let (kr − ωt) ≡ u in your calculations.]
(a)
For notational convenience,
~ obeys all four of Maxwell’s equations, in vacuum, and find the associated magnetic field.
Show that E
~ = E φ̂ and E does not depend on φ, Gauss’ law reads (in spherical coordinates)
ANSWER: Since E
√
1 ∂E
=0.
r sin θ ∂φ
1 ∂
∂E
∂E
1
1∂
1
~
~
∇×E =
E cos θ + sin θ
E+r
r̂ −
θ̂ .
(E sin θ) r̂ −
(rE) θ̂ =
r sin θ ∂θ
r ∂r
r sin θ
∂θ
r
∂r
1
∂E
A cos θ
cos θ
cos u −
Now,
=
sin u = E
∂θ
r
kr
sin θ
~ ·E
~ =
∇
and
∂E
∂r
=
=
~ ×E
~ = A
so ∇
r2
A sin θ
sin u k cos u
sin u
A sin θ
+
−k sin u +
cos u −
−
−
r2 kr
r
kr2
kr
A sin θ
sin u
−2 cos u + 2
− kr sin u
r2
kr
1
1
sin u θ̂ .
2 cos θ cos u −
sin u r̂ + sin θ cos u + kr −
kr
kr
In order to satisfy Faraday’s law we must therefore have (within a constant of integration)
Z ~ ×E
~ dt = − A 2 cos θ C − 1 S r̂ + sin θ C + kr − 1 S θ̂
~ =−
∇
B
r2
kr
kr
Z
Z
sin u
cos u
where C ≡ cos u dt = −
and S ≡ sin u dt =
.
(Note: ω = ck .)
ω
ω
1
1
A
~
cos u θ̂
2 cos θ sin u +
cos u r̂ + sin θ sin u − kr −
Thus B =
ckr2
kr
kr
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
~ = Br r̂ + Bθ θ̂ where
or B
2A cos θ
1
Br =
cos u
sin u +
ckr2
kr
A sin θ
1
and Bθ =
cos u .
sin u − kr −
ckr2
kr
~ ·B
~ = 1 ∂ r2 Br + 1 ∂ (sin θBθ )
∇
2
r ∂r
r sin θ ∂θ
1
1
2A cos θ ∂
A
∂
sin
u
+
sin
u
+
sin2 θ
=
cos
u
+
+
kr
cos
u
2
3
ckr ∂r
kr
ckr sin
kr
∂θ
θ
2A cos θ
1
1
1
1
=
sin u − kr −
cos u
k cos u − 2 cos u − sin u +
ckr2 kr
r
r
kr
√
1
1
1
1
2A cos θ
k cos u − 2 cos u − sin u + sin u − k cos u + 2 cos u = 0 .
=
2
ckr
kr
r
r
kr
~ ×B
~ = 1 ∂ (rBθ ) − ∂ Br φ̂ or
It remains only to check Ampère’s law: ∇
r ∂r
∂θ
1
1 A sin θ
~
~
cos u
sin u − kr −
∇×B =
r
ckr2 kr A sin θ
1
1
+
k cos u − k + 2 cos u + k kr −
sin u
ckr
kr
kr
(9)
This should satisfy Gauss’ law too:
(10)
4
(b)
2A sin θ h
cos u i
+
sin u +
φ̂
2
ckr
kr
1
A sin θ
cos
u
−
sin
u
−
kr
−
=
3
ckr
kr
1
+ kr cos u − kr +
cos u + k 2 r2 − 1 sin u
kr
h
cos u io
φ̂
+ 2 sin u +
kr
~ ×B
~ = A sin θ (cos u + kr sin u) φ̂ .
(11)
giving ∇
cr2
~
~
~
1 ∂E
k2 ∂ E
k ∂E
Now, if we’re to get any joy from this, it had better be equal to 2
= 2
=
c ∂t
ω ∂t
cω ∂t
h
k A sin θ ∂
sin u
cos u i
A sin θ k
=
cos u −
ω sin u + ω
φ̂ =
φ̂
cω r ∂t
kr
r cω
kr
√
A sin θ
(kr sin u + cos u) φ̂ .
QED
(12)
=
2
cr
Thus the proposed function does satisfy all of Maxwell’s equations as advertised and is therefore also a valid
solution of TWE (The Wave Equation). And this is the simplest possible spherical wave! (Don’t you just love
curvilinear coordinates?)
~ over a full cycle to get the intensity vector ~I. Does ~I point in
Calculate the Poynting vector. Average S
the expected direction? Does it fall off like r−2 , as it should? ANSWER:
~
~
~ = E × B = 1 E φ̂ × Br r̂ + Bθ θ̂ = − 1 EBr θ̂ + EBθ r̂
S
µ0
µ0
µ0
sin u cos u A2
2 sin θ cos θ cos u −
θ̂
sin u +
= −
3
µ0 ckr kr
kr sin u
cos u + sin2 θ cos u −
sin u − kr cos u +
r̂
kr
kr
2
A2
cos2 u sin u sin u cos u
= −
2
sin
θ
cos
θ
cos
u
sin
u
+
θ̂
−
−
µ0 ckr3
kr
kr
kr
cos2 u sin2 u
sin u cos u
2
2
+ sin θ cos u sin u − kr cos u +
r̂
−
+ sin u cos u −
kr
kr
k 2 r2
2
cos2 u − sin u
1
A2
+
2 sin θ cos θ cos u sin u 1 −
= −
θ̂
3
µ0 ckr
kr
kr
cos2 u − sin2 u
1
2
2
r̂ .
(13)
+ sin θ cos u sin u 2 − 2 2 − kr cos u +
k r
kr
~ has a non-radial component may seem alarming, but let’s check the time average: all of
The fact that S
sin u cos u, sin2 u and cos2 u oscillate in time, but only the first averages to zero; the other two average to 12 ,
but their difference does average to zero. Thus
2
2
2
2
~I ≡ hSi
~ = A sin θ kr r̂ = A sin θ r̂ ,
µ0 ckr3 2
2µ0 c r2
(c)
(14)
which points radially outward and falls off like 1/r2 , as expected.
Integrate ~I · d~a over a spherical surface to determine the total power radiated.
[You should get P = 4πA2 /3µ0 c.]
ANSWER:
ZZ
Z
Z
Z
π
2
πA2 π 3
πA2 −1
sin2 θ
2
~I · d~a = A
P =
2πr
sin
θdθ
=
1 − cos2 θ d(cos θ)
sin
θdθ
=
−
2
2µ0 c 0
r
µ0 c 0
µ0 c 1
Z
√
4πA2
πA2
2
πA2 +1
P =
.
(1 − x2 )dx =
2−
or
or P =
µ0 c −1
µ0 c
3
3µ0 c
This was a tedious problem; it took me all day to get it right. I will be duly impressed if you managed to grind through
it successfully. Now you know why we like our plane waves so much, Huygens’ principle notwithstanding!