Physics 306: Waves Lecture 5 2/7/2008

Wave Superposition Principle

It is quite a common situation for two or more waves to arrive at the same point in space or to exist together along the same direction. We will consider today several important cases of the combined effects of two or more waves.

Let us consider where two waves with displacements given by Ψ1 and Ψ2 are present. What is the net displacement?

Superposition Principle: Ψ = Ψ1 + Ψ2

The principle is valid for linear PDE’s.

The wave equation is:

∂ 2Ψ 1 ∂ 2Ψ = ∂x 2 v 2 ∂t 2

If Ψ1 and Ψ2 are solutions of the above PDE, then

Ψ = aΨ1 + bΨ2,

(where a and b are constants) is also a solution.

This makes “interference” possible

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Constructive (waves combine “in step”)

=

Destructive (waves combine “out of step”)

=

Ψ1 and Ψ2 have the same frequency

Something in between

Get some Ψ with the same frequency =

Let us develop the mathematical formalism for combining two waves:

Beginning with the wave equation

∂ 2Ψ 1 ∂ 2Ψ = ∂x 2 v 2 ∂t 2

Let Ψ1 and Ψ2 be solutions.

Then try the sum of solutions as a possible solution: Ψ = Ψ1 + Ψ2. We get

∂ 2Ψ ∂ 2Ψ 1 ∂ 2Ψ 1 ∂ 2Ψ 1 + 2 = 1 + 2 ∂x 2 ∂x 2 v 2 ∂t 2 v 2 ∂t 2

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Therefore, Ψ is a solution. In general, we may form a solution as a linear combination of solutions:

N Ψ()x,t = ∑Ci Ψi ()x,t i=1

But suppose

∂ 2Ψ 1 ∂ 2Ψ + cΨ 2 = ∂x 2 v 2 ∂t 2

2 2 2 For solutions Ψ1, Ψ2, and Ψ = Ψ1 + Ψ2, the cΨ is c(Ψ1 + Ψ2 + 2Ψ1Ψ2 ). So,

LHS ≠ RHS

So the principle of superposition is only valid for linear PDEs (where the function and its derivatives appear only in first order.)

The Addition of Waves of the Same Frequency

As a starting point, let us say that we have two 1-D waves that are harmonic. We can describe these waves as:

Ψ(x,t) = Ψ0 sin(ωt − (kx + ε ))

If α = kx + ε, then we can write

Ψ1 (x,t) = Ψ01 sin(ωt −α1 ) Note, we are and assuming they have the same frequency Ψ2 (x,t) = Ψ02 sin(ωt −α 2 )

Let us suppose these two waves coexist in space. The resultant disturbance is the linear superposition of these waves

Ψ = Ψ1 + Ψ2

= Ψ01 sin()ωt −α1 + Ψ02 sin ()ωt −α 2

= Ψ01 (sin()ωt cos ()α1 + cos ()ωt sin ()α1 )+ Ψ02 ()sin ()ωt cos ()α 2 + cos ()ωt sin ()α 2

Therefore,

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Ψ = ()Ψ01 cos()α1 + Ψ02 cos(α 2 ) sin(ωt)+ (Ψ01 sin(α1 )+ Ψ02 sin(α 2 ))()cos ωt

independent of time

This can be simplified if we define

Ψ0 cos(α ) = (Ψ01 cos(α1 )+ Ψ02 cos(α 2 )) (1) and

Ψ0 sin(α ) = (Ψ01 sin(α1 )+ Ψ02 sin(α 2 )) (2)

If we now square the equations above and add 1 2 2 2 2 Ψ0 = ()cos α + sin α Ψ0 2 2 2 2 = Ψ01 cos α1 + Ψ02 cos α 2 + Ψ01Ψ02 cosα1 cosα 2 2 2 2 2 + Ψ01 sin α1 + Ψ02 sin α 2 + Ψ01Ψ02 sinα1 sinα 2

Ψ 2 = Ψ 2 + Ψ 2 + 2Ψ Ψ cos(α −α ) ⇒ 0 01 02 01 02 2 1 (3)

If we divide equation (2) by (1) we get

Ψ sinα + Ψ sinα tan()α = 01 1 02 2 Ψ01 cosα1 + Ψ02 cosα 2

Mathematical Aside:

It is often convenient to make use of the complex representation of waves when dealing with the superposition of waves.

Recall that

Ψ = Ψ sin(ωt − (kx + ε )) 1 01 1 = Ψ01 sin()ωt +α1

can be written as:

i(ωt+α1 ) Ψ1 = Ψ01e

where the wave is the imaginary part of this equation.

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Similarly, we can write:

i(ωt+α2 ) Ψ2 = Ψ02e

Let

i(ωt+α ) Ψ = Ψ0e = Ψ1 + Ψ2

iωt What are Ψ0 and α? Factor out e , which cancels.

Re(Ψ) = Re(Ψ1 )+ Re(Ψ2 )

Im(Ψ) = Im(Ψ1 )+ Im(Ψ2 )

Im(Ψ) tanφ = Re()Ψ

2 * * * Ψ0 = Ψ0 Ψ0 = (Ψ1 + Ψ2 )(Ψ1 + Ψ2 ) * * * * = Ψ1 Ψ1 + Ψ2 Ψ2 + Ψ2 Ψ1 + Ψ1 Ψ2 2 2 * * = Ψ1 + Ψ2 + Ψ2 Ψ1 + Ψ1 Ψ2

2 2 −iα2 iα1 −iα1 iα2 = Ψ01 + Ψ02 + Ψ02e Ψ01e + Ψ01e Ψ02e

2 2 i()α1 −α2 −i ()α1−α2 = Ψ01 + Ψ02 + Ψ01Ψ02 ()e + e

2 2 2 Ψ0 = Ψ01 + Ψ02 + 2Ψ01Ψ02 cos(α1 −α 2 )

which is identical to Eq. (3), and

Ψ sinα + Ψ sinα tan()α = 01 1 02 2 Ψ01 cosα1 + Ψ02 cosα 2

We can generalize this to N waves:

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N Ψ = ∑Ψ j j=1 N i()ωt+α j = ∑Ψ0 j e j=1 ⎡ N ⎤ iα j iωt = ⎢∑Ψ0 j e ⎥e ⎣ j=1 ⎦ iα iωt = Ψ0e ⋅e

where

N iα iα j Ψ0e = ∑Ψ0 j e j=1

is defined as the source complex amplitude of the superpositional wave. The intensity of the resultant wave is given by:

2 iα iα * Ψ0 = ()Ψ0e (Ψ0e ) N N N 2 = ∑Ψ0i + 2∑∑Ψ0i Ψ0 j cos()α j −α1 i=1 j>=i i 1

and

N ∑Ψ0i sinα i i=1 tanα = N ∑Ψ0i cosα i i=1

We can also use equations (1) and (2) to write Ψ as

Ψ = Ψ cosα sinωt + Ψ sinα cosωt 0 0 = Ψ0 sin()ωt +α

What does this last equation say? It shows us that a single wave results from the superposition of the original two. This new wave is harmonic and of the same frequency as the original wave, although its amplitude and phase are different.

An important consequence of this is that we can superposition any number of harmonic waves having a given frequency, and get a resultant wave which is harmonic as well.

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Interference

The intensity of a wave is proportional to the square of its amplitude. Thus from Eq. (3), we see that the intensity of resulting wave is not just the sum of flux densities of individual original waves, but that there is an additional term:

2Ψ01Ψ02 cos(α 2 −α1 )

This is an interference term, which is a function of the difference in phase between the two original waves.

The crucial factor is δ, where

δ = α2 – α1

When

δ = 0, ± 2π, ± 4π, … → resultant amplitude is max

δ = ± π, ± 3π, … → resultant amplitude is min

Recall that

α = -kx + ε

2π where k = . λ

So we can write:

2π δ = α −α = ()()kx + ε − kx + ε = ()()x − x + ε + ε 2 1 1 1 2 2 λ 1 2 1 2 where x1 and x2 are the distances from the sources of the two waves to the point of observation and λ is the wavelength.

Let us suppose the two waves are initially in phase: ε1 = ε2, then

2π δ = ()x − x λ 1 2

If we define the index of refraction

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wavelength in vacuum λ n = 0 , λ

Then we can write:

2π δ = n()x2 − x1 λ0

this quantity is known as the optical path difference (OPD) ≡ Λ

δ = k0Λ

Definitions: waves for which ε1 - ε2 = constant are coherent waves.

Random and Coherent Sources

From a previous lecture, we define the source intensity as ∝ time-average of the amplitude of wave squared:

2 I ∝ Ψ0

For two waves, using Eq. (3) for the resultant wave, we have time-average

I = I1 + I 2 + 2 Ψ1Ψ2 cos(α 2 −α1 )

If the square of the waves 1 and 2 have random phase (≡ incoherent), then:

cos(α 2 −α1 ) = 0 i.e. the time-average will be zero.

Then,

I = I1 + I2,

Or, more generally:

N I = ∑ I i = NI1 i=1

Page 8 of 15 Physics 306: Waves Lecture 5 2/7/2008 if you have N randomly phased sources of equal amplitude and frequency. (resultant intensity arising from N coherent sources is determined by the sum of the individual intensities)

[What does this mean? If you have two light bulbs that emit light with random phase, the result will have an intensity equal to the sum of the intensities of each bulb → no interference effect is observed.]

If

α1 - α2 = constant → coherent waves then

cos(α 2 −α1 ) ≠ 0

But varies from 1 and -1, so I varies between

I = I1 + I 2 + 2 I1I 2 (cos = 1) and

I = I1 + I 2 − 2 I1I 2 (cos = -1)

If

I1 = I2, (5) then I varies between

4I1 ⇒ constructive interference and

0 ⇒ destructive interference

If you have N sources that are in phase with αi = αj and equal amplitude, then

2 iα iα * I = Ψ0 = (Ψ0e )(Ψ0e ) N N N 1 2 = ∑Ψ0i + 2∑∑Ψ0i Ψ0 j cos()α j −α1 i=1 j>=i i 1

N 2 ⎛ ⎞ 2 2 2 I = ⎜∑Ψ0i ⎟ = ()NΨ0i = N Ψ0i ⎝ i=1 ⎠

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if each amplitude is the same = Ψ01.

So here, the amplitudes are added first and then squared to determine the resulting intensity.

Standing Waves

We already covered this briefly. Here we assume that we have two harmonic waves traveling with some frequency and traveling in opposite directions.

Let us assume the two waves have equal amplitude.

E1 = E0 sin(ωt + kx) (left)

E2 = E0 sin(ωt − kx) (right)

Using complex representation

i(kx+ωt ) i(−kx+ωt ) E = E0e + E0e iωt ikx −ikx = E0e ()e + e iωt = 2E0e cos()kx where the resultant wave is the real part:

E = 2E0 cos(kx)cos(ωt)

This is the equation for a standing or stationary wave as opposed to a traveling wave. Its profile does not move through space since it is not of the form f(x±vt).

A snapshot would look like

at various times, standing waves will look like sinusoidal waves of various amplitudes

Where is coskx = 0? (called “nodes”)

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cosθ = 0 for θ = π , 3π , 5π , … 2 2 2

kx = (n + ½)π

n is an integer

Note: In the treatment of lasers (later in class), we will find that that laser light is generated in laser cavities, which take the form of two highly reflecting mirrors surrounding something called a gain medium. The light in such a cavity then consists of counter-propagating EM waves that form standing waves. It is typically the case that the EM boundary conditions at mirror surfaces require z = 0 ⇒ nodes. This means that the wavelength is restricted to those supported by cavity dimensions to discrete values. d ⎛ mλ ⎞ d = ⎜ m ⎟ , where m = integer ≠ 0, ⎝ 2 ⎠ i.e. there is anteger of half- wavelengths that “fit” in the cavity length.

mirror mirror

Superposition of Waves of Different Frequency

Except in rare circumstances, we never encounter situations where we have superposition of waves of the same frequency.

Consider the superposition of two waves

Ψ1 = Ψ0 cos(k1x −ω1t) and

Ψ2 = Ψ0 cos(k2 x −ω2t)

For simplicity, we assume the waves have the same amplitude and initial phase = 0.

The resultant wave is:

Ψ = Ψ1 + Ψ2 = Ψ0 [cos(k1x −ω1t)+ cos(k2 x + ω2t)]

Using the identity

1 1 cosα + cosβ = 2cos 2 (α + β )cos 2 (α − β )

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This yields: 2k 2ω 2km 2ωm 1 1 Ψ = 2Ψ0 cos 2 [(k1 + k2 )x − (ω1 + ω2 )t]cos 2 [(k1 − k2 )(x − ω1 −ω2 )t] (4)

This wave can be simplified by defining:

1 ω = 2 ()ω1 + ω2 as the average angular frequency

1 k = 2 ()k1 + k2 as the average propagation number

1 ωm ≡ 2 ()ω1 −ω2 as the modulation angular frequency

1 km ≡ 2 ()k1 − k2 as the modulation propagation number

Using Eq. (4), we get:

Ψ()x,t = 2Ψ0 cos[kx −ωt]cos[km x −ωmt] or

Ψ(x,t) = ξ0 (x,t)cos[kx −ωt]

time varying amplitude where traveling wave with frequency

ξ0 (x,t) = 2Ψ0 cos[km x −ωmt] which can be thought of as a time varying amplitude.

Let us consider the case where ω1 and ω2 are large and ω1 ≈ ω2. Then, ω >> ωm and

ξ0 ()x,t will change slowly. However, Ψ(x,t) will vary rapidly. The intensity is

2 I ∝ ξ0 (x,t) 2 2 = 4Ψ0 cos ()km x −ωmt 2 = 2Ψ0 []1+ cos()2km x − 2ωmt where we used

cos 2θ =− cos22θθ sin

=−2cos2 θ 1

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2 Notice that I oscillates about 2Ψ0 with an angular frequency of 2ωm = (ω1 - ω2) = beat frequency

This is a low frequency envelope modulating a high frequency wave. That is, the resultant wave consists of a higher frequency carrier wave modulated by a cosine function.

How fast does the envelope move? This velocity is referred to as the group velocity.

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We know that if a waveform is of the form

ei(kx−ωt ) ,

ω then the velocity is v = . So, k

ω −ω ∆ω dω v = 1 2 = ≈ k1 − k2 ∆k dk

ω From v = , ω = kvp, where vp = phase velocity. k

Differentiating, ω = kvp with respect to k, we have:

dω dv v = = v + k p g dk p dk

dv p [Note, if the medium is dispersive, then vp depends on k, so vg ≠ vp, otherwise = 0 dk and vg = vp holds.]

If we write ω = kvp, then

dω dv v = = v + k p g dk p dk

We can show that:

⎡ λ dn ⎤ vg = v p 1+ ⎣⎢ n dλ ⎦⎥

c where n = v p

Proof:

dω dv v = = v + k p g dk p dk

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1 dv p dv p v p + = v p + ()− λ λ d 1 dλ ()λ

c Since v = , p n dv c dn p = − dλ n2 dλ

λ dn ⎛ λ dn ⎞ vg = v p + c = v p ⎜1+ ⎟ n2 dλ ⎝ n dλ ⎠

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