Lecture 16: Group and Phase Velocity
Up to now we have just talked about the speed (or velocity) v of a wave. Actually there are two
velocities: the Group Velocity vg and the Phase Velocity vp . They are given by the memorable formulæ:
vp =
ω
k
vg =
dω
dk
1. Phase velocity
A single (infinite) wave is described by the expression cos(ωt − kx) or sin[ 2π
λ (x − vt)] or equivalent.
λ
The pattern travels with a velocity (actually a speed) vp = T = f λ = ω/k
vp is what matters with interference. The refractive index n is defined as c/v and this means c/vp
2. Group velocity
An infinite wave is unrealistic.
A real wave has to have beginning and end.
The overall shape is called the envelope.
Various shapes are possible - abrupt or gentle.
vg = dω
dk is the velocity of the envelope.
3. Illustration
Consider two waves almost in step.
They have ω1 , k1 and ω2 , k2 (and λ1 , λ2 ...)
Write the means and differences
2
2
, k = k1 +k
ω = ω1 +ω
2
2
k1 −k2
ω1 −ω2
∆ω = 2 , ∆k = 2
the original quantities can be expressed
in terms of these
ω1 = ω + ∆ω, ω2 = ω − ∆ω etc
Adding the two waves gives a total wave
ei(ω1 t−k1 x) + ei(ω2 t−k2 x)
This can be written
ei(ωt+∆ωt−kx−∆kx) + ei(ωt−∆ωt−kx+∆kx)
Take out a common factor:
ei(ωt−kx) ei(∆ωt−∆kx) + ei(−∆ωt+∆kx)
iθ
Remembering cosθ = e +e
this is 2cos(∆ωt − ∆kx)ei(ωt−kx)
2
The first term is clearly the envelope. It has small wavenumber and frequency and so a long wavelength
and period. It travels with velocity vg = ∆ω/∆k.
This generalises: vg = dω
dk
−iθ
4. Finding vg
Often vp is known from measurements or from basic principles. Take the expression for vp and write
ω/k for vp in it. Turn all the λ and f etc terms into ω and k.
Then differentiate with respect to k. This gives an expression involving dω
dk from which vg can be
extracted.
As a trivial example, suppose vp is constant (i.e. independent of wavelength) with value c. Then ω = ck
and dω/dk = c. Group and phase velocity are the same in this case.
5. Example: Refractive Index
The velocity of light in a medium tends to depend on the wavelength. (Hence rainbows, prisms, etc.).
This is called dispersion. See the previous lecture for details.
People normally quote n as a function of λ rather than ω as a function of k. This contains the same
information, we need to manipulate it: what follows is mere algebra:
dn
dn dk
dλ = dk dλ
Take these two differentials separately.
dn
c
ck dω
First n = vcp = ck
ω so dk = ω − ω 2 dk
Lecture 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Group and Phase Velocity
dk
2π
k
and secondly k = 2π
λ so dλ = − λ2 = − λ
k c
Now put these together and get dn
dλ = − λ ( ω −
kc
ck2
c
c
dn
dλ = − ωλ + ω 2 λ vg = − vp λ + vp2 λ vg
Rearrange to get an expression for vg
vg =
ck
ω 2 vg )
vp2 λ
c
dn
c ( vp λ + dλ )
v λ
vp + pn dn
dλ = vp (1
vg =
+ nλ dn
dλ ) .
dvp
c
This can also be written: vg = vp (1 − nk dn
dk ) = vp − λ dλ = n+ω(dn/dω) . Of all these equally-valid
alternatives, only vg = dω
dk is memorable.
Note: if n is falling with λ, vg < vp . This is called normal dispersion. If n is rising, vg > vp and this is
called anomalous dispersion
6. Example
Suppose some glass has a slightly different refractive index for red light and blue light: n = 1.51 at 400
nm, n = 1.49 at 600 nm.
This is a fall of δn = 0.02 in δλ = 200 nm. The mean λ is 500 nm and the mean n is 1.5.
vp = 2 × 108 m/s
vg = 2 × 108 (1 − (500/1.5)(.02/200)) = 2 × 108 (1 − .0333) = 1.93 108 m/s
7. Refractive index of X rays
We sawpin the last lecture that at the highest frequencies(=shortest wavelengths, hence X rays) one can
write n = 1 − B/ω 2 where B is a positive number containing N and all the proportionality constants.
This n is less than 1 so vp > c (!)
q
Putting that worry on one side, let’s find the group velocity. Start from n =
c
vp
=
ck
ω
=
1−
B
ω2
2 2
Squaring: cωk2 = 1 − ωB2
Multiplying by ω 2 : c2 k 2 = ω 2 − B
Then differentiate wrt k: 2c2 k = 2ω dω
dk
This gives vg = c2 /vp
so vg < c and relativity is OK, as information (causal signals) travels with vg not vp .
In this case we have vg vp = c2 . The product of the group and phase velocities is equal to c2 . There are
many cases where this turns out to be true, but it is not universal and there are some where it isn’t.
8. A more general picture for deriving vg = dω
dk
The earlier (standard) example just considered two waves. If you’re happy with that, fine. For a more
general approach we need to bring in an extended concept of Fourier Series.
A periodic
Rπ
P function can be expressed as sum1 Rofπ sine and cosine terms
1
f (θ) = k ak sin(kθ) + bk cos(kθ)
ak = π −π f (θ)sin(kθ)dθ
bk = (2)π
−π f (θ)cos(kθ)dθ
This can
be
extended
to
non-periodic
functions
R∞
R∞
1
f (x)e−ikx dx
f (x) = −∞ F (k)eikx dk where F (k) = 2π
−∞
An infinite sine wave has a well-defined wavelength and thus a well-defined k. F (k) is a delta function.
If F (k) is broad, the wave is made up of lots of sine waves of different wavelengths and has a short
wavepacket (as the contributions all cancel away from the peak).
Suppose we have a long wave packet f (x). That means that there is a small spread in k. The function
F (k) will haveRa sharp peak about some central value k0 At some initial time t = 0 the wave can be written
∞
f (x, 0) = −∞ F (k)eikx dk
After timeR t the wave has evolved to
∞
f (x, t) = −∞ F (k)ei(kx−ω(k)t) dk
where ω(k) explicitly shows that different wavelength components have different frequencies.
Let ω(k) = ω0 + dω
dk (k − k0 ). This is the Taylor expansion to 1st order, and we’re using the fact that
the spread in k is small.
R
dω
Then f (x, t) = F (k)ei(k0 x+kx−k0 x−ω0 t−(k−ko ) dk t) dk
R
dω
f (x, t) = ei(k0 x−ω0 t) F (k)ei(k−k0 )(x− dk t) dk
The first part is the pure sine wave. The second part (the integral) describes the envelope. It looks
dω
messy, but all the x and t dependence is in the x − dω
dk t. So the envelope progresses with velocity dk ≡ vg .
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