wave superposition principle -...

Physics 306: Waves Lecture 5 2/7/2008

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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:

2

2

22

2 1tvx ∂Ψ∂

=∂Ψ∂

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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Let us develop the mathematical formalism for combining two waves: Beginning with the wave equation

2

2

22

2 1tvx ∂Ψ∂

=∂Ψ∂

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

22

2

221

2

222

2

21

2 11tvtvxx ∂Ψ∂

+∂Ψ∂

=∂Ψ∂

+∂Ψ∂

=

Constructive (waves combine “in step”)

=

Destructive (waves combine “out of step”)

=

Something in between

Ψ1 and Ψ2 have the same frequency

Get some Ψ with the same frequency


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

( ) ( )∑=

Ψ=ΨN

iii txCtx

1,,

But suppose

2

2

22

2

2 1tv

cx ∂

Ψ∂=Ψ+

∂Ψ∂

For solutions Ψ1, Ψ2, and Ψ = Ψ1 + Ψ2, the cΨ2 is ( )21

22

21 2 ΨΨ+Ψ+Ψc . 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:

( ) ( )( )εω +−Ψ=Ψ kxttx sin, 0 If α = kx + ε, then we can write

( ) ( )1011 sin, αω −Ψ=Ψ ttx and

( ) ( )2022 sin, αω −Ψ=Ψ ttx Let us suppose these two waves coexist in space. The resultant disturbance is the linear superposition of these waves

( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )22021101

202101

21

sincoscossinsincoscossinsinsin

αωαωαωαωαωαω

tttttt

+Ψ++Ψ=−Ψ+−Ψ=

Ψ+Ψ=Ψ

Therefore,

Note, we are assuming they have the same frequency


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( ) ( )( ) ( ) ( ) ( )( ) ( )tt ωααωαα cossinsinsincoscos 202101202101 Ψ+Ψ+Ψ+Ψ=Ψ

This can be simplified if we define

( ) ( ) ( )( )2021010 coscoscos ααα Ψ+Ψ=Ψ (1) and

( ) ( ) ( )( )2021010 sinsinsin ααα Ψ+Ψ=Ψ (2) If we now square the equations above and add

( )

210201222

02122

01

210201222

02122

01

20

2220

sinsinsinsin

coscoscoscos

sincos

αααα

αααα

αα

ΨΨ+Ψ+Ψ+

ΨΨ+Ψ+Ψ=

Ψ+=Ψ

⇒ (3) If we divide equation (2) by (1) we get

( )202101

202101

coscossinsintan

αααα

αΨ+ΨΨ+Ψ

=

Mathematical Aside: It is often convenient to make use of the complex representation of waves when dealing with the superposition of waves. Recall that

( )( )( )101

1011

sinsin

αωεω

+Ψ=+−Ψ=Ψ

tkxt

can be written as:

( )1011

αω +Ψ=Ψ tie where the wave is the imaginary part of this equation.

( )1202012

022

012

0 cos2 αα −ΨΨ+Ψ+Ψ=Ψ

independent of time

1


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Similarly, we can write:

( )2022

αω +Ψ=Ψ tie Let

( )210 Ψ+Ψ=Ψ=Ψ +αωtie

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

( ) ( ) ( )21 ReReRe Ψ+Ψ=Ψ

( ) ( ) ( )21 ImImIm Ψ+Ψ=Ψ

( )( )ΨΨ

=ReImtanφ

( )( )

( ) ( )( )2121

2112

02012

022

01

020101022

022

01

2*

11*2

22

21

2*

11*22

*21

*1

21*2

*10

*0

20

αααα

αααα

−−−

−−

+ΨΨ+Ψ+Ψ=

ΨΨ+ΨΨ+Ψ+Ψ=

ΨΨ+ΨΨ+Ψ+Ψ=

ΨΨ+ΨΨ+ΨΨ+ΨΨ=

Ψ+ΨΨ+Ψ=ΨΨ=Ψ

ii

iiii

ee

eeee

( )210201

202

201

20 cos2 αα −ΨΨ+Ψ+Ψ=Ψ

which is identical to Eq. (3), and

( )202101

202101

coscossinsintan

αααα

αΨ+ΨΨ+Ψ

=

We can generalize this to N waves:


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( )

tii

tiN

j

ij

N

j

tij

N

jj

ee

ee

e

j

j

ωα

ωα

αω

⋅Ψ=

⎥⎦

⎤⎢⎣

⎡Ψ=

Ψ=

Ψ=Ψ

∑

∑

∑

=

=

+

=

0

10

10

1

where

∑=

Ψ=ΨN

j

ij

i jee1

00αα

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

( )( )( )∑∑∑

> ==

−ΨΨ+Ψ=

ΨΨ=ΨN

ij

N

ijji

N

ii

ii ee

1100

1

20

*00

20

cos2 αα

αα

and

∑

∑

=

=

Ψ

Ψ= N

iii

N

iii

10

10

cos

sintan

α

αα

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

( )αωωαωα

+Ψ=Ψ+Ψ=Ψ

ttt

sincossinsincos

0

00

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:

( )120201 cos2 αα −ΨΨ 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

where λπ2

=k .

So we can write:

( ) ( ) ( ) ( )21212211122 εελπεεααδ ++−=+−+=−= xxkxkx

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

( )212 xx −=λπδ

If we define the index of refraction


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λλ0=n ,

Then we can write:

( )120

2 xxn −=λπδ

Λ= 0kδ 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:

20Ψ∝I

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

( )122121 cos2 αα −ΨΨ++= III If the square of the waves 1 and 2 have random phase (≡ incoherent), then:

( ) 0cos 12 =−αα i.e. the time-average will be zero. Then,

I = I1 + I2, Or, more generally:

11

NIIIN

ii ==∑

=

wavelength in vacuum

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

time-average


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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

( ) 0cos 12 ≠−αα But varies from 1 and -1, so I varies between

2121 2 IIIII ++= (cos = 1) and

2121 2 IIIII −+= (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

( )( )( )∑∑∑

> ==

−ΨΨ+Ψ=

ΨΨ=Ψ=N

ij

N

ijji

N

ii

ii eeI

1100

1

20

*00

20

cos2 αα

αα

( ) 20

220

2

10 ii

N

ii NNI Ψ=Ψ=⎟⎠

⎞⎜⎝

⎛Ψ= ∑

=

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.

( )kxtEE += ωsin01 (left)

( )kxtEE −= ωsin02 (right) Using complex representation

( ) ( )

( )( )kxeE

eeeE

eEeEE

ti

ikxikxti

tkxitkxi

cos2 0

0

00

ω

ω

ωω

=

+=

+=−

+−+

where the resultant wave is the real part:

( ) ( )tkxEE ωcoscos2 0= 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

Where is coskx = 0? (called “nodes”)

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


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cosθ = 0 for θ = 2π , 3 2

π , 5 2π , …

kx = (n + ½)π

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.

⎟⎠⎞

⎜⎝⎛=

2mmd λ , where m = integer ≠ 0,

i.e. there is anteger of half-wavelengths that “fit” in the cavity length.

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

( )txk 1101 cos ω−Ψ=Ψ and

( )txk 2202 cos ω−Ψ=Ψ For simplicity, we assume the waves have the same amplitude and initial phase = 0. The resultant wave is:

( ) ( )[ ]txktxk 2211021 coscos ωω ++−Ψ=Ψ+Ψ=Ψ Using the identity

( ) ( )βαβαβα −+=+ 21

21 coscos2coscos

n is an integer

d

mirror mirror


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This yields:

( ) ( )[ ] ( ) ( )[ ]txkktxkk 212121

212121

0 coscos2 ωωωω −−−+−+Ψ=Ψ (4) This wave can be simplified by defining:

( )2121 ωωω += as the average angular frequency

( )212

1 kkk += as the average propagation number

( )2121 ωωω −≡m as the modulation angular frequency

( )212

1 kkkm −≡ as the modulation propagation number Using Eq. (4), we get:

( ) [ ] [ ]txktxktx mm ωω −−Ψ=Ψ coscos2, 0 or

( ) ( ) [ ]txktxtx ωξ −=Ψ cos,, 0 where

( ) [ ]txktx mm ωξ −Ψ= cos2, 00 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 ( )tx,0ξ will change slowly. However, ( )tx,Ψ will vary rapidly. The intensity is

( )

( )( )[ ]txk

txk

txI

mm

mm

ω

ω

ξ

22cos12

cos4

,

20

220

20

−+Ψ=

−Ψ=

∝

where we used

2 2

2

cos 2 cos sin2cos 1

θ θ θ

θ

= −

= −

k2 ω2 mk2 mω2

traveling wave with frequency time varying amplitude


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Notice that I oscillates about 202Ψ 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

( )tkxie ω− ,

then the velocity is k

v ω= . So,

dkd

kkkv ωωωω

≈∆∆

=−−

=21

21

From k

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

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

dkdv

kvdkdv p

pg +==ω

[Note, if the medium is dispersive, then vp depends on k, so vg ≠ vp, otherwise dkdvp = 0

and vg = vp holds.] If we write ω = kvp, then

dkdv

kvdkdv p

pg +==ω

We can show that:

⎥⎦⎤

⎢⎣⎡ +=

λλ

ddn

nvv pg 1

where pv

cn =

Proof:

dkdv

kvdkdv p

pg +==ω


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( ) ( )λ

λλ

λ ddv

vd

dvv p

pp

p −+=+1

1

Since ncvp = ,

λλ ddn

nc

ddvp

2−=

⎟⎠⎞

⎜⎝⎛ +=+=

λλ

λλ

ddn

nv

ddn

ncvv ppg 12

wave superposition principle -...

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