9. Electromagnetic Waves

Waves along a rope

Stadium waves (medium is people)

• It appears like a ripple is moving:

– but particles in the medium only do a local cyclic movement.

– only the disturbance (wave) is moving.

Wave motion

A transverse wave

A longitudinal wave

Suppose a displacement of a wave propagating in z is given by 𝑓 𝑧, 𝑡

the initial value of the wave (t=0) is 𝑓 𝑧, 0 = 𝑔 𝑧

If the velocity of the wave is 𝑣 , at time t it has traveled a distance 𝑣𝑡

So 𝑓 𝑧, 𝑡 = 𝑓 𝑧 − 𝑣𝑡, 0 = 𝑔(𝑧 − 𝑣𝑡)

represents a wave of fixed shape travelling in the z direction at speed v

e.g. 𝑓1 𝑧, 𝑡 = 𝐴𝑠𝑖𝑛 𝑎 𝑧 − 𝑣𝑡 ; . 𝑓2 𝑧, 𝑡 = 1 𝑏 𝑧−𝑣𝑡 2 are waves

Waves on a stretched string

Net transverse force on the string segment of Δz =ΔF

2

2

2 2 2

2 2 2

sin sin tan tan

If mass per unit length is , ( )

z z z

F F fF T T T T T T z

z z z

f f fF z

t z T t

:Wave equation it admits a solution of the foam

𝑓 𝑧, 𝑡 = 𝑔(𝑧 − 𝑣𝑡)

2 2

2 2

f f

z T t

2 2 2

2 2 2

2 2 2 2 22

2 2 2 2 2 2

-

1 and =

f dg u dg f dg u dgv u z vt

z du z du t du t du

f dg d g u d g

z z du du z du

f dg d g u d g f f Tv v v v

t t du du t du z v t

also has a solution of the form 𝑓 𝑧, 𝑡 = ℎ 𝑧 + 𝑣𝑡which represents a wave travelling in –z direction

2 2

2 2 2

1

f f

z v t

So the general solution to wave equation

𝑓 𝑧, 𝑡 = 𝑔 𝑧 − 𝑣𝑡 + ℎ 𝑧 + 𝑣𝑡

22

2 2

1

ff

v t

In 3 dimensions wave is

𝑓 𝑟, 𝑡 = 𝑔 𝑟 − 𝑣𝑡 + ℎ 𝑟 + 𝑣𝑡

9.1.2 Sinusoidal Waves

pattern repeats itself after a distance equal to the wavelength λ

k𝜆 = 2𝜋 ⇒ 𝜆 =2𝜋

𝑘

• Period T =2𝜋

𝜔=2𝜋

𝑘𝑣frequency = ν =

1

𝑇=

𝑘𝑣

2𝜋=

𝑣

𝜆

• angular frequency ω = 𝑘𝑣 =2𝜋

𝜆𝑣 = 2𝜋𝜈

• For the wave in –z direction : 𝑓 𝑧, 𝑡 = 𝐴𝑐𝑜𝑠 𝑘𝑧 + 𝜔𝑡 + 𝛿

=𝐴𝑐𝑜𝑠 −𝑘𝑧 − 𝜔𝑡 + 𝛿wave with –k propagates in -z

𝑓 𝑧, 𝑡 = 𝐴𝑐𝑜𝑠 𝑘 𝑧 − 𝑣𝑡 + 𝛿= 𝐴𝑐𝑜𝑠 𝑘𝑧 − 𝜔𝑡 + 𝛿

amplitude

phase

wave number angular frequency

k =2p

l

v =w

k

w = 2p f

These are purely

kinematic, they

work for any wave.

v = flHow far the wave

moves per cycle

How many cycles per second

θ

z = x + iy

z* = x - iy

z2

= zz* = (x + iy)(x - iy) = x2 + y2real

Imaginary

z = reiq

r

2 3 4 5

2 3 4 5

2 4 3 5

( ) ( ) ( ) ( )1 ( ) ...

2! 3! 4! 5!

1 ...2! 3! 4! 5!

1 ...2! 4! 3! 5!

cos sin

i i i i ie i

i i i

i

i

Complex number notation

Since 𝑒𝑖𝜃= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃

( )

( )

( , ) cos( ) Re[ ]

introduce complex wave function ( , )

with complex amplitude

then ( , ) Re( ( , ))

i kz t

i kz t

i

f z t A kz t Ae

f z t Ae

Ae

f z t f z t

3 1 2 1 2 3

( ) ( ) ( )3 1 2 3 3 1 2

Example:

combining two waves f f +f Re(f ) Re(f ) Re(f )

( , ) + = where

i kz t i kz t i kz tf z t Ae A e A e A A A

i.e. just add the complex amplitudes. Combined wave still have the same frequency and wavelength

Boundary conditions:

• Consider an incident wave 𝑓𝐼 𝑧, 𝑡 = 𝐴𝐼𝑒𝑖(𝑘1𝑧−𝜔𝑡) 𝑧 < 0

• Gives rise to a reflected wave 𝑓𝑅 𝑧, 𝑡 = 𝐴𝑅𝑒𝑖(−𝑘1𝑧−𝜔𝑡) 𝑧 < 0

• And a transmitted wave 𝑓𝑇 𝑧, 𝑡 = 𝐴𝑇𝑒𝑖(𝑘2𝑧−𝜔𝑡)

𝑓𝐼 𝑧, 𝑡

𝑓𝑇 𝑧, 𝑡

𝑓𝑅 𝑧, 𝑡

1 2 1

2 1 2

k v

k v

All waves have to have the same frequency, but on two sides wave lengths and velocities are different

1( ) ( )

( )

net disturbance is:

+ for 0

( , )

for 0

R

T

i k z t i k z tI R

i k z tT

A e A e z

f z t

A e z

• At the boundary (0 , ) (0 , )f t f t

0 0

and f f

z z

+ = I R TA A A1 1 2 = I R Tk A k A k A

1 2 1 2 1 2

1 2 1 2 2 1 1 2

2 2 = , = or = , = R I T I R I T I

k k k v v vA A A A A A A A

k k k k v v v v

2 1 2

2 1 1 2

2= , = R I T Ii i i i

R I T I

v v vA e A e A e A e

v v v v

In terms of real amplitudes and phases

If the 2nd string is lighter than the 1st 𝜇2 <𝜇1 ⇒ 𝑣2 > 𝑣1 all

have the same phase angle 𝛿𝐼 = 𝛿𝑇 = 𝛿𝑅2 1 2

2 1 1 2

2= , = R I T I

v v vA A A A

v v v v

If the 2nd string is heavier than the 1st 𝑣2< 𝑣1 so the reflected wave is out

of phase by 𝜋 to incident 𝛿𝐼 = 𝛿𝑇 = 𝛿𝑅 + 𝜋

)))

since cos( ) cos( )I Ikz t kz t

The reflected wave is upside down12