GROUP VELOCITY
Group of waves with possibly different phase velocities:
1.
Pure sinusoidal wave: Y(x,t) = A cos(kx-t+o) = A cos(θ)
where θ = θ(x,t) = phase angle = kx-t+o
where k = 2π radians/λ (kx is then an angle)
where  = 2π radians/T (t is then an angle)
and o is the initial phase angle (when t=0)
[o = 0o for cosine wave; o= -90o for sine wave.
For crest of wave (phase angle of crest is constant at 90 = ½π rad):
θcrest = ½π = kxcrest- tcrest +o, or
xcrest = (½π + tcrest-o)/k .
For speed of crest of wave: vphase = vcrest = dxcrest/dtcrest , so
vphase = d([½π + t - o]/k)/dt = /k (here t = tcrest
for short);
vphase
vphase
2.
recall that k=2π/λ and =2π/T so that:
= (2π/T)/(2π/λ) = λ/T , and since f = (1/T)
= /k = λf .
Group of waves: A group of sine waves will add together to form
some pattern that also repeats (this is the Fourier Series in
reverse).
Ygroup(x,t) = A(x) cos(Kx - gt)
where
A(x) is the shape of the group,
K = 2π/λg where λg is the distance over which the pattern for the
group repeats, and
g = 2π/Tg where Tg is the time over which the pattern for the
group repeats.
At t=0 sec, Ygroup(x,0) = A(x) cos(Kx) where A(x) = nΣ bn sin(knx)
(here A(x) is expressed as a Fourier Series) , so
Ygroup(x,0) = nΣ bn sin(knx) cos(Kx) .
We can now use two trig identities
[sin(θ±φ) = sinθ cosφ ± cosθ sinφ]
to get sinθ cosφ = ½[sin(θ+φ) + sin(θ-φ)] ,
and with θ=knx and φ=Kx, we get
Ygroup(x,0) = nΣ ½ bn { sin[(kn+K)x] + sin[(kn-K)x] }
and since sin(-θ) = -sin(+θ) , we can write:
Ygroup(x,0) = nΣ ½ bn { sin[(K+kn)x] - sin[(K-kn)x] }
,
.
Now put in the time dependence such that wherever we had a Kx, we
put in an additional -t:
Ygroup(x,t) = nΣ ½ bn { sin[(K+kn)x-+t] - sin[(K-kn)x--t] }
where we use ± to indicate that  depends on k=(K±kn) .
[Recall that vphase =
/k, and vphase may not be constant but may depend on (vary with)
.]
Since  is a function of k [(k) = vphasek], we can expand (k) in a
Taylor Series about k=K:
(K±kn) = (K) ± (d/dk)K kn + higher order terms which we neglect ;
now let's let vg  (d/dk)K so that ± = (K±kn)  (K) ± vgkn ,
so
Ygroup(x,t) = nΣ ½ bn {sin[(K+kn)x - (+vgkn)t] - sin[(K-kn)x - (-vgkn)t] }
or re-grouping terms:
Ygroup(x,t) = nΣ ½ bn { sin[(Kx-t)+kn(x-vgt)] - sin[(Kx-t)-kn(x-vgt)] } .
We can again use our trig identity:
sin(θ+φ) + sin(θ-φ) = 2 sinθ sinφ ,
where θ = (Kx-t) and φ = kn(x-vgt) , to get:
Ygroup = nΣ bn sin(Kx-t) cos[kn(x-vgt)] ;
but here the sin(Kx-t) can come out of the summation, so
Ygroup = { nΣ bn cos[kn(x-vgt) } sin(Kx-t) = A(x-vgt) sin(Kx-t),
where we identify the function A(x-vgt) as the original
Fourier series with x replaced by (x-vgt); that is, the shape
moves through space with a speed of vg, hence the name
group velocity.
3.
Review:
vphase = /k = λf
(good for any pure sine wave [or cosine wave]
of wavelength λ and frequency f
[or wavevector k and angular speed ]) ;
vgroup = d/dk .
4.
Special case:
If vphase = constant, then  = vphasek , and so
vgroup = d/dk = d[vphasek]/dk = vphase .