25. Standing Waves, Beats, and Group Velocity

Group velocity: the speed of information

Going faster than light... (but not really)

Superposition again

Standing waves: the sum of two oppositely

traveling waves

Beats: the sum of two different frequencies

Sometimes, the refractive index is less than one.

Example: a bunch of free electrons, above the plasma frequency

( )2

21 Pn

ωωω

≈ −

Thus, in the x-ray range (where ω >> ωP), the refractive index of most materials is slightly less than unity:

( ) ( )1n ω δ ω≈ − δ is small and positive

So, how fast do waves propagate in these situations? What is the speed of light?

To answer this question, we need to think more carefully about what we mean by “speed”.

(see lecture 10)

Superposition allows waves to pass through each other.

Otherwise they'd get screwed up while overlapping, and wouldn’t come out the same as they went in.

Recall:

If E1(x,t) and E2(x,t) are both solutions to the wave equation, then so is their sum.

Adding waves of the same frequency, but different initial phase, yields a wave of the same frequency.

This isn't so obvious using trigonometric functions, but it's easywith complex exponentials:

1 2 3

1 2 3

( , ) exp ( ) exp ( ) exp ( )

( ) exp ( )

= − + − + −= + + −

ɶ ɶ ɶ ɶ

ɶ ɶ ɶ

totE x t E j kx t E j kx t E j kx t

E E E j kx t

ω ω ωω

where all the phases (other than the kx−ωt) are lumped into E1, E2, and E3.

Adding waves of the same frequency, but opposite direction, yields a "standing wave."

0 0

0

( , ) exp ( ) exp ( )

exp( )[exp( ) exp( )]

= += − +

ɶ ɶ ɶ

ɶ

totE x t E j kx t E j kx t

E jkx j t j t

ω ωω ω

− +

0( , ) 2 cos( ) cos( )totE x t E kx tω=

Since we must take the real part of the field, this becomes:

(taking E0 to be real)

Standing waves are important inside lasers, where beams areconstantly bouncing back and forth.

Waves propagating in opposite directions:

0 2 exp( ) cos( )=ɶE jkx tω

A Standing Wave

0( , ) 2 cos( )cos( )totE x t E kx tω=

A Standing WaveYou’ve seen the previews. Now, the movie!

Question: what is the speed of this wave?

What is the speed of energy propagation?

A Standing Wave: Experiment

3.9 GHz microwaves

Note the node at the reflector at left.

Mirror

Input beam

The same effect occurs in lasers.

Interfering spherical waves also yield a standing wave

Antinodes

Two Point Sources

Different separations. Note the different node patterns.

When two waves of different frequency interfere, they produce beats.

0 1 0 2( , ) exp( ) exp( )= +ɶ ɶ ɶ

totE x t E j t E j tω ω

0 0

0

0

( , ) exp ( ) exp ( )

exp( )[exp( ) exp( )]

2 exp( )cos( )

So : = + ∆ + − ∆= ∆ + − ∆= ∆

ɶ ɶ ɶ

ɶ

ɶ

tot ave ave

ave

ave

E x t E j t t E j t t

E j t j t j t

E j t t

ω ω ω ωω ω ω

ω ω

1 2 1 2 2 2

Let and+ −= ∆ =ave

ω ω ω ωω ω

Taking the real part yields the product of a rapidly varying

cosine (ωave) and a slowly varying cosine (∆ω).

When two waves of different frequency interfere, they produce "beats."

Individual Waves with ω1 and ω2:

Sum:

Envelope:

Irradiance:

When two light waves of different frequency interfere, they produce beats.

1 2 1 2

1 2 1 2

2 2

2 2

Let and

Similiarly, and

+ −= ∆ =

+ −= ∆ =

ave

ave

k k k kk k

ω ω ω ωω ω

0 1 1 0 2 2( , ) exp ( ) exp ( )= − + −ɶ ɶ ɶ

totE x t E j k x t E j k x tω ω

[ ]0 0

0

0

( , ) exp ( ) exp ( )

exp ( ) exp ( ) exp{ ( )}

2 exp ( )cos( )

So:

Real part :

= + ∆ − −∆ + −∆ − + ∆= − ∆ −∆ + − ∆ −∆= − ∆ −∆

ɶ ɶ ɶ

ɶ

ɶ

tot ave ave ave ave

ave ave

ave ave

E x t E j k x kx t t E j k x kx t t

E j k x t j kx t j kx t

E j k x t kx t

ω ω ω ωω ω ωω ω

0 2 cos( )cos( )− ∆ −∆ave aveE k x t kx tω ω

Group velocity

The phase velocity comes from the rapidly varying part: v = ωave / kave

What about the other velocity—the velocity of the amplitude?

Light-wave beats (continued):

Etot(x,t) = 2E0 cos(kavex–ωavet) cos(∆kx–∆ωt)

This is a rapidly oscillating wave: [cos(kavex–ωavet)]

with a slowly varying amplitude: [2E0 cos(∆kx–∆ωt)]

Define the "group velocity:" vg ≡ ∆ω /∆k

In general, we define the group velocity as:=g

dv

dk

ω

Usually, group velocity is not equal to phase velocity, except in empty space.

0 1 0 2

1 1 2 2

∆≡∆

−=−

gvk

c k c k

n k n k

ωFor our example,

where the subscripts 1 and 2 refer to the values at ω1 and at ω2.k1 and k2 are the k-vectors in vacuum.

0 01 21 2

1 2

1 2

,

,

If phase velocity

If phase velocity

−= = = = =−

≠ ≠

g

g

c ck kn n n v

n k k n

n n v

vg ≡ dω /dk

Now, ω is the same in or out of the medium, but k = k0 n, where k0 is the k-vector in vacuum, and n is what depends on the medium. So it's easier to think of ω as the independent variable:

Calculating the Group velocity

[ ] 1/ −≡gv dk dω

So, the group velocity equals the phase velocity when dn/dω = 0, such as in vacuum. Otherwise, since n usually increases with ω(normal dispersion), dn/dω > 0 and so usually vg < vφ.

1g

vv

dn

n d

φ

ωω

=+

0g

cv

dnn

dω

ω

= +

So or

( ) ( )0 0

1 = = +

ndk d dnn

d d c c d

ω ωω ω

ω ω ωUsing , calculate: ( ) ( )

0

k nc

≡ ωω ω

Why is this important?

sum of 2 different frequencies

sum of 6 different frequencies

sum of many different frequencies

You cannot send information using a wave, unless you make it into some kind of pulse.

You cannot make a pulse without superposing different frequencies. Pulses travel at the group velocity.

Group velocity (vg) vs. phase velocity (vφ)

=gv vφ

>gv vφ

0

0

2

0 0 0 0 00 2 2

0 0 0

2 2 2

(2 / ) 2

Use the chain rule :

Now, , so :

=

− − −= = = =

ddn dn

d d d

c d c c

d c c

λω λ ω

π λ π π λλω ω ω π λ π

Calculating Group Velocity vs. Wavelength

We more often think of the refractive index in terms of wavelength,so let's write the group velocity in terms of the vacuum wavelength λ0.

0

2

0 0 0

0 0 0

/ 1

2/ 1

2

Recalling that :

we have:

= + − = +

g

g

c dnv

n n d

c c dnv

n n d c

ωω

π λλ λ π

0 0

0

/ 1 = −

g

c dnv

n n d

λλ

or:0

0

0

=−

c

dnn

dλ

λ

The group velocity is less than the phase velocity in regions of normal dispersion

In regions of normal dispersion, dn/dω is positive. So vg < c0/n < c0 for these frequencies.

0=+

g

cv

dnn

dω

ω

The group velocity often depends on frequency

We have seen that the phase velocity depends on ω, because n does.

( )0

cv

n=φ ω

( )0

g

cv

dnn

d

=+ω ω

ω

It should not be surprising that the group velocity also depends on ω.

When the group velocity depends on frequency, this is known as group velocity dispersion, or GVD.

Just as essentially all solids and liquids exhibit dispersion, they also all exhibit GVD. This property is crucially important in the design of, e.g., optical data transfer systems that use fiber optics.

GVD = 0

GVD = 0

GVD distorts the shape of a pulse as it propagates in a medium

vg(blue) < vg(red)

GVD means that the group velocity will be different for different wavelengths in a pulse.

Source:

http://web.bryanston.co.uk/physics/Applets/Wave%20animations/Sound%20waves/Dispersive%20waves.htm

The group velocity can exceed c0

whendispersion is anomalous

We note that absorption is strong in these regions. dn/dω is only steep when the resonance is narrow, so only a narrow range of frequencies has vg > c0. Frequencies outside this range have vg < c0.

0=+

g

cv

dnn

dω

ωdn/dω is negative in regions of anomalous dispersion, that is, near a resonance. So vg exceeds vφ, and can even exceed c0 in these regions!

For discussion, see: http://www.mathpages.com/home/kmath210/kmath210.htm

The group velocity can exceed c0

whendispersion is anomalous

There is a more fundamental reason why vg > c0 doesn’t

necessarily bother us.

The interpretation of the group velocity as the speed of

energy propagation is only valid in the case of normal

dispersion! In fact, mathematically we can superpose

waves to make any group velocity we desire - even zero!

In artificially designed materials, almost any behavior is possible

Here’s one recent example:

Science, vol. 312, p. 892 (2006)

A metal/dielectric

composite structure

experiment theory

Of course, relativity and causality are never violated.

In this material, a light pulse appears to exit the medium before entering it.