W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 359

whereωis the frequency of the oscillating dipole source that is producing thewave^104.

We are fortunate in this is actually a function of the formf(z±vt)! To see this, let’s

factor the argument:

Ex(z, t) =E 0 xsin(k(z±

ω

k

t) =E 0 xsin (k(z±ct)) (862)

which has the desired form ifc=ω/k. Indeed, if you substitute this harmonic wave into

the wave equation, you get:

d^2

dz^2

E 0 xsin(kz±ωt) =−k^2 E 0 xsin(kz±ωt) =^1

c^2

d^2

dt^2

E 0 xsin(kz±ωt)

= −

1

c^2

ω^2 E 0 xsin(kz±ωt) (863)

or (dividing out)

c^2 =ω

2

k^2

(864)

andc=ω/kas promised.

Again recalling our work with harmonic waves, we expect that in theseequations:

k=^2 π

λ

(865)

is thewave numberof the wave, the “spatial angular frequency” in terms of thewave-

lengthof the waveλ, just as:

ω=

2 π

T

(866)

is thetemporalangular frequency of the wave in terms of its periodT. Thus:

c=

ω

k=

2 π

T

λ

2 π=

λ

T=f λ (867)

areall useful ways of relating the frequency, wavelength, angular frequency, wave number,

period, and speedof the wave. Yes, you can remember just one of these and figure out

the rest, but on an examspeedcounts and I recommend learningallof these forms so

that they are second nature and you don’t have to think about them.

We expect that:

By(z, t) =B 0 ysin(kz±ωt+φ) (868)

where we cannot yet assume thatExandByhave the same phase, although we do insist

(since they are parts of the same wave) that they have the same frequency. Now let’s

work some magic. We’ll restrict our interest for the moment to a wave propagating to

theright:

Ex(z, t) = E 0 xsin(kz−ωt) (869)

By(z, t) = B 0 ysin(kz−ωt+φ) (870)

We substitute these two forms into (your choice of) Ampere’s or Faraday’s Law in dif-

ferential form. Let’s choose Faraday as being marginally simpler:

d

dz

E 0 xsin(kz−ωt) = −

d

dt

B 0 ysin(kz−ωt+φ) (871)

kE 0 xsin(kz−ωt) = ωB 0 ysin(kz−ωt+φ) (872)

E 0 xsin(kz−ωt) =

ω

k

B 0 ysin(kz−ωt+φ) (873)

E 0 xsin(kz−ωt) = cB 0 ysin(kz−ωt+φ) (874)

(^104) Note well that we could have equally well usedE 0 xcos(kz±ωt+φ) for some arbitrary phase angleφ, or better
yetE 0 xeikze±iωtwhereE 0 x=|E 0 x|eiφis an arbitrary complex amplitude. We choose to use sin(kz±ωt) for no
other reason than to have something specific to work with, butthese all satisfy the wave equation and are equally
valid possibilities. The phase angleφin particular corresponds to determining simply the shape of the wave when we
start the “clock” of our harmonic wave in our particular reference frame.