Wave propagation
S.M.Lea
2017
1 A wave signal
An electromagnetic wave signal may be represented by a function of space and time
u(x,t) where u might be one component of the electric field, for example. Then we can
write this function in terms of its Fourier transform:
1
A(k,ω) exp ik · x − iωt dωdk
u(x,t) =
(2π)2 all ω and k space
The dispersion relation for the wave gives a relation (such as eqn 12 in the plasma wave
notes) between ω and k which allows us to write the integral in terms of k alone1 :
1
u(x,t) =
A(k) exp ik · x − iω k t dk
(2π)2 all k space
Now let’s simplify by choosing the x−axis along the direction of propagation and letting
the
√ whole problem be one-dimensional, so that u = u (x, t) . (We also lose two factors of
2π.)
+∞
1
u(x,t) =
A(k) exp [ikx − iω (k) t] dk
(1)
2π −∞
The Fourier amplitude A (k) is found from the value of u at some x and t, usually at t = 0.
Often we can make use of the fact that A (−k) = A (k)∗ if u is real, (see below) but this
requires that ω also be real, which, in general, it may not be. So we have to take
u(x, t) = Re
1
2π
+∞
−∞
+∞
A(k) exp [ikx − iω (k) t] dk
1
A (k) exp [ikx − iω (k) t] + A(k)∗ exp iω (k)∗ t − ikx dk(2)
4π −∞
Now we let κ = −k in the second term.
+∞
−∞
1
u (x, t) =
A(k) exp [ikx − iω (k) t] dk −
A(−κ)∗ exp iκx + iω (−κ)∗ t dκ
4π
−∞
+∞
=
=
1
1
4π
+∞
−∞
A(k) exp [ikx − iω (k) t] + A(−k)∗ exp ikx + iω (−k)∗ t
l
k
Effectively, A k, ω = A k δ ω − ω k , as in Lea Ch 7.
1
dk
Then, evaluating at t = 0, we get
1
1
√
u(x, 0) = √
2π 2 2π
+∞
A(k) + A(−k)∗ exp (ikx) dk
−∞
Inverting the transform:
1
1
∗
√
A(k) + A (−k) = √
2 2π
2π
If A(−k) = A (k)∗ , then
+∞
u(x,0) exp (−ikx) dx
(3)
−∞
+∞
A(k) =
u( x,0) exp (−ikx) dx
(4)
−∞
But we have an additional initial condition. Taking the time derivative of (2), we have
∂
u(x,t) =
∂t
1
4π
1
∂
u(x,0) =
∂t
4π
Inverting, we have:
−
+∞
−∞
+∞
−∞
−iωA(k) exp [ikx − iω (k) t] + A (k)∗ iω exp (−ikx − iωt) dk
−iω A(k) − A (−k)∗ exp (ikx) dk
iω
A(k) − A (−k)∗ =
2
+∞
−∞
∂
u(x, 0) exp (−ikx) dx
∂t
(5)
Combining the two relations (3) and (5) to eliminate A (−k)∗ , we have
+∞
1 ∂
u(x,0) exp (−ikx) dx
(6)
iω ∂t
−∞
√
which is Jackson equation 7.91 modulo a factor of 2π. (This arises because I started with
the transform over space and time, whereas J used the transform over space alone.) We
regain equation (4) if ∂u/∂t = 0 at t = 0.
A (k) =
u(x,0) −
2 Gaussian Pulse
∂
To simplify the algebra in this section, let’s assume ∂t
u(x,0) = 0. Generally the wave
is sent at a “carrier frequency” ω 0 with a corresponding k0 so that ω 0 = ω (k0 ) , and the
Fourier amplitude A (k) peaks at k0. For example, a Gaussian pulse with a carrier frequency
ω 0 is written:
x2
u (x, 0) = u0 cos (k0 x) exp − 2
(7)
a
2
Its transform is:
+∞
A (k) =
−∞
+∞
=
−∞
u0 cos (k0 x) exp −
x2
a2
exp (−ikx) dx
x2
u0 ik0 x
e
+ e−ik0 x exp − 2
2
a
exp (−ikx) dx
x2
u0 +∞ −i(k−k0 )x
e
+ e−i(k0 +k)x exp − 2 dx
(8)
2 −∞
a
So we have two integrals of identical form. To do each, we complete the square (see Lea
Example 7.2):
=
−
1
(k − k0 ) a2
x2
2
2
−
i
(k
−
k
)
x
=
−
+
i
(k
−
k
)
a
x
+
i
x
0
0
a2
a2
2
= −
1
a2
i
a2 (k − k0 )
+x
2
2
2
− i
(k − k0 ) a2
2
1
+ a4 (k − k0 )2
4
Thus the first term in the integral (8) is:
a2
u0
exp − (k − k0 )2
2
4
=
=
=
+∞
−∞
exp −
1
a2
i
a2 (k − k0 )
+x
2
2
dx
+∞+iγ
a2
u0
exp − (k − k0 )2 a
exp −v2 dv
2
4
−∞+iγ
√
a2
u0
exp − (k − k0 )2 a π
2
4
√
a2
π
au0 exp − (k − k0 )2
2
4
2
0)
+ x a1 = iγ + x/a,and used the fact that the integral of the
where we set v = i a (k−k
2
Gaussian is independent of the path between ±∞. (See Lea Ch 7 pg 328.) To obtain the
second term we replace k0 with −k0 . Thus the result for A (k) is two Gaussians, centered at
k = ±k0 , and each of width 2/a.
√ u0
a2
a2
2
2
A (k) = A1 (k) + A2 (k) = πa
exp − (k − k0 ) + exp − (k + k0 )
2
4
4
(9)
3 Group velocity
For a Gaussian pulse, the integral in (1) has two terms. To evauate the first term, we
write ω (k) in a Taylor series centered at k0 .
ω (k) = ω 0 + (k − k0 )
dω
dk
+
k0
3
1
d2 ω
(k − k0 )2
2
dk2
+ ...
k0
2
Substituting ω (k) into the expression for u (x, t) (1),




+∞
dω
A1 (k)

exp ikx − it ω0 + (k − k0 )
u1 =

2π
dk
−∞

drop
2
2
+
k0
(k − k0 ) d ω
2
dk2
k0
(10)
and dropping the term in ω , we have:
u1 (x,t)
1
dω
exp −i ω 0 − k0
2π
dk
u1 (x,t) = exp −i ω 0 − k0
= u x−
dω
dk
k0
dω
dk
+∞
A1 (k) exp ikx − ik
t
−∞
k0
1
2π
t
k0
dω
dk
+∞
−∞
t, 0 exp −i ω 0 − k0




+ .. dk


A1 (k) exp ik x −
dω
dk
t
k0
dω
dk
t dk+· · ·
t
dk
k0
(11)
k0
where we used eqn (1) again in the last step, with x → x − dω
dk k0 t and t → 0. . Thus to
first order in the expansion of ω (k) , u (x, t) equals a phase factor e−iφ times u (x − vg t, 0) ,
that is, at time t the signal looks like the pulse at t = 0 translated at speed vg , where
vg =
dω
dk
(12)
k0
is the group speed. The phase φ = (ω0 − vg k0 ) t indicates that the carrier wave shifts
within the Gaussian envelope.
To evaluate the second integral we expand ω (k) in a Taylor series centered at −k0 , and
obtain the same result.
4 Pulse spreading
For some waves with d2 ω/dk2 = 0, there is only a first order term in the Taylor series
for ω (k). But for other waves the higher order terms produce corrections to the first order
result. Generally these terms lead to both spreading and distortion of the pulse shape. For
example, consider the Whistler (Plasmawaves notes eqn 34):
k=
ωp
c
or
ω=Ω
The group velocity for this wave is:
vg =
ω
Ω
k2 c2
ω2p
kc2
dω
= 2Ω 2
dk
ωp
4
(13)
and the second derivative is
d2 ω
c2
=
2Ω
(14)
dk2
ω 2p
All further derivatives are zero. Thus from (10), the exact expression for u in this case is:
vg =
u (x, t) = exp −i ω 0 − k0
1
2π
t ×
k0
+∞
A(k) exp ik x −
−∞
u (x, t) = exp −i ω0 − k0
1
2π
dω
dk
dω
dk
k0
dω
dk
A(k) exp ik (x − vg t) −
= exp −i ω0 − k0 vg +
(k − k0 )2 d2 ω
2
dk2
t dk
k0
t ×
+∞
−∞
k0
t −i
k02 vg
2
t
i 2
k − 2kk0 + k02 vg t dk
2
1
2π
+∞
−∞
i
A(k) exp ik x + k0 vg t − vg t − k2 vg t dk
2
where vg and vg are given by equations (13) and (14) with k = k0 .
Now if we use the Gaussian pulse (9) as an example, we get:
u (x, t) = exp −i ω0 − k0 vg +
where
√
I= π
+∞
−∞

2
a2
au0  exp − 4 (k − k0 )
2
2
+ exp − a4 (k + k0 )2
k02 vg
2
t
1
I
2π
(15)

 exp ik x ± k0 vg t − vg t − i k2 vg t dk
2
Again there are two terms, differing only in the sign of k0 in the Gaussian, and we evaluate
these using the Taylor series about ±k0 .
√
π
au0 (I+ + I− )
I=
2
Looking at the first term:
+∞
I+
=
−∞
exp −
= exp −
a2 k02
4
a2 2
i
k − 2kk0 + k02 + ik x + k0 vg t − vg t − k2 vg t dk
4
2
+∞
−∞
exp −k2
i
k0 a2
a2
+ vg t + k i x + k0 vg t − vg t +
4
2
2
To simplify the notation, we let s = x − vg t and vg t = αa2 . Then the exponent in the
integrand is:
−k2
i
k0 a2
a2
+ αa2 + k i s + k0 αa2 +
4
2
2
2
is + k02a (1 + 2iα)
a2
k
= − (1 + 2iα) k2 +
a2
4
4 (1 + 2iα)
5
dk
Completing the square, the term in square brackets is
k2 + 2k
2is
2is
+ k0 + 2
+ k0
a2 (1 + 2iα)
a (1 + 2iα)
2
2is
2is
+ k0 − 2
+ k0
2
a (1 + 2iα)
a (1 + 2iα)
We can do the integral by making the change of variable:
=
2
−
2is
+ k0
a2 (1 + 2iα)
2
2
k+
ξ=
Then:
I+ = exp −
a2 k02
4
exp
2is
a√
+ k0
1 + 2iα k + 2
2
a (1 + 2iα)
2is
a2
(1 + 2iα) 2
+ k0
4
a (1 + 2iα)
2
2
√
a 1 + 2iα
+∞+iγ
−∞+iγ
√
where the path of integration is again moved off the real axis, but the result is still π. Thus:
√
−s2
a2 k02
2 π
k2 a2
√
+ 2
+ isk0 + 0 (1 + 2iα) (16)
exp −
I+ =
4
a (1 + 2iα)
4
a 1 + 2iα
√
2 π
s2 (1 − 2iα)
i 2 2
√
=
+ isk0 + αk0 a
exp − 2
(17)
a (1 + 4α2 )
2
a 1 + 2iα
And the exponent in u (15) is:
k02 a2 α
αk2 a2
s2 (1 − 2iα)
+ k0 s + 0
− 2
2
2
a (1 + 4α2 )
2
s (1 − 2iα)
= i (−ω0 t + k0 x) − 2
a (1 + 4α2 )
2iαs2
s2
= i (k0 x − ω0 t) − 2
+
a (1 + 4α2 ) a2 (1 + 4α2 )
i −ω0 t + k0 vg t −
Thus
u (x, t) =
=
√ au0 1
s2
exp [i (k0 x − ω0 t)] exp − 2
π
×
2 2π
a (1 + 4α2 )
√
2s2
2 π
√
+ (k0 → −k0 )
exp iα 2
a (1 + 4α2 ) a 1 + 2iα
u0 exp [i (k0 x − ω0 t)]
(x − vg t)2
exp
−
2
a2 (1 + 4α2 )
(1 + 4α2 )1/4
We may interpret this expression as folows:
• Initial wave form at frequency ω0 . This is the carrier.
e−iφ + (k0 → −k0 ) (18)
• Envelope peaks at x = vg t due to its travelling at the group speed.
√
• Envelope spread by factor 1 + 4α2 where α = vg t/a2
• Amplitude decreased by similar factor
6
2
e−ξ dξ
• Overall phase factor
φ = −α
tan−1 (2α)
2s2
+
2
+ 4α )
2
a2 (1
approximately proportional to α ∝ t.
At large times the pulse width grows roughly linearly with time. a (t)
2vg t/a (0) . Notice that (18) gives the correct result (7) at t = 0.
2a (0) α =
5 Arrival of a signal
The previous analysis shows the shape of the signal over all space as a function of time.
But this is not usually what we observe. Let us now consider a signal generated at x = 0
over a period of time. We want to see what kind of a signal arrives at a distant point x = X
as a function of time. So instead of using the dispersion relation in the form ω (k) we instead
think of it as k (ω) . Then:
+∞
1
A(ω) exp [ik (ω) x − iωt] dω
u(x,t) = √
2π −∞
where in general k(ω) is complex.
As we have seen, A (ω) is usually a smooth function, somewhat peaked around the
carrier frequency ω 0 . The exponential oscillates, so it tends to make the value of the integral
small. However, when the exponent stays almost constant over a range of ω, we will have a
subtantial contribution to the integral. This happens when the phase is stationary. (See Lea
Optional Topic D section 2)
d
[k (ω) x − ωt] = 0
dω
dk
−t = 0
x
dω
or
dω
x=
t = vg t
(19)
dk
Thus the major contribution to the integral is from the frequency that, at time t, has its group
speed equal to x/t. Put another way, the signal at that frequency, travelling at the group
speed, has just reached the observation point.
To find the received signal, we must evaluate the integral. Again we expand k (ω) in a
Taylor series, this time about the stationary frequency.as determined by equation (19).
k (ω) = ks + (ω − ω s )
dk
dω
+
ωs
1
d2 k
(ω − ω s )2
2
dω2
ωs
+ ···
where ks = k (ω s ) , giving
1
u(x,t) = √
2π
+∞
−∞
A(ω) exp i ks + (ω − ω s )
Now use the stationary phase condition (19):
7
dk
dω
+
ωs
(ω − ω s )2 d2 k
2
dω2
ωs
+ · · · x − iωt
dω
1
u(x,t) = √
2π
+∞
ωs 1
d2 k
+ (ω − ω s )2
vg
2
dω2
A(ω) exp i ks −
−∞
+ .. vg t
dω
ωs
Since the exponential guarantees that only frequencies near ωs contribute, we may pull out
the slowly varying amplitude, as well as the terms in the exponential that are independent of
ω:
A(ωs )
√
exp [i (ks vg − ω s ) t]
2π
u(x,t)
To do the integral, change variables to:
ξ = (ω − ωs )
−i d2 k
2 dω 2
ωs
+∞
i
d2 k
(ω − ωs )2
2
dω 2
exp
−∞
d2 k
dω2
vg t = e−iπ/4 (ω − ω s )
vg t dω
ωs
ωs
vg t
2
Then
+∞
I
exp
=
−∞
=
i
d2 k
(ω − ω s )2
2
dω 2
2π
d2 k
dω2 ωs
vg t
vg t dω =
ωs
+∞+iγ
eiπ/4
1
2
d2 k
dω2
ωs
2
e−ξ dξ
vg t
−∞+iγ
eiπ/4
and so, incorporating the result (19), we have
A(ω s )
π
u(x,t) =
exp i (ks x − ωs t) + i
2
4
d k
dω2 ωs x
(20)
Again we see the carrier wave, modulated by an envelope that changes in time, and with an
additional phase change. (Remember that ω s is a function of time.)
To see how this works out, again look at the whistler. The stationary phase condition is
(equations 13 and 19)
ωp
kc2
t=
ω 2p
c
√
ωΩ
= 2
ct
ωp
x = 2Ω
c2
ω
2Ω 2 t
Ω ωp
and thus
ω2p x2
ωs =
Ω 4c2 t2
which decreases from a large value toward zero as t increases. For this wave the signal (20)
is given by:
ω2
u(x,t) =
=
2
c2 A( Ωp 4cx2 t2 )
π
√
exp i (ks vg − ω s ) t + i
2Ω 2
ωp
4
x
c
ωp
π
2Ω ω 2p x2
A(
) exp i (ks vg − ωs ) t + i
x
Ω 4c2 t2
4
8
If A is a Gaussian as in (9), then

c
2Ω
A0 exp −η
u(x,t) =
ωp
x
ω 2p
2
x
− ω0
Ω 4c2 t2
2
where η is a constant.

 exp i (ks vg − ω s ) t + i π
4
At fixed x, we get something like this:
A
ct/x
The signal grows from zero to an asymptotic value, until the analysis breaks down at low
frequency (large t). (Remember: we neglected ion motion in getting the dispersion relation.)
Here’s another way to look at it. Consider the plasma waves (Plasmawave notes eqn 12):
c2 k2 = ω 2 − ω 2p
For this wave
dk
dω
dk
c
dω
2c2 k
= 2ω
= c
ω
=
kc2
which looks like this:
ω
=
ω 2 − ω2p
ω/ω p
(ω/ωp )2 − 1
5
4
3
k/dw
2
1
0
0.5
1
1.5
w/wp
9
2
2.5
3
The stationary phase condition is
dk
ct
=c
x
dω
There is no signal at a fixed x = x0 prior to t = t0 = x0 /c. The signal begins at infinite
frequency, and moves to lower frequency as time increases, asymptotically reaching ωp .
The horizontal line in the graph represents ct/x = 2 and the intersection of the two lines is
the observed frequency at this time.
For a more complicated dispersion relation, the signal can be more complicated.
Consider the RHC wave propagated along B. (Plasmawave notes eqn 28 with the minus
sign.) Then:
c2 k2
2c2 k
dk
dω
dk
dω
=
and thus:
c
ω 2p ω
ω−Ω
ω 2p
ω 2p ω
= 2ω −
+
ω − Ω (ω − Ω)2
ω 2p
ω
= 2ω −
1−
ω−Ω
ω−Ω
ω2p Ω
= 2ω +
(ω − Ω)2
= ω2 −
1
ck
ω+
1 ω 2p Ω
2 (ω − Ω)2
1
=
ω2 −
ω+
ω2p ω
ω−Ω
1
=
1−
ω2p
ω(ω−Ω)
1 ω2p Ω
2 (ω − Ω)2
1+
1 ω2p Ω/ω
2 (ω − Ω)2
Now let ωp /Ω = y and ω/Ω = z Then
c
dk
=
dω
1
1−
y2
z(z−1)
With y = 2, the diagram looks like:
10
1+
1
y2
2 z (z − 1)2
6
5
4
cdk/dw 3
2
1
0
1
2
z
3
4
5
As t increases we first obtain a signal at one frequency (dashed line) and later at 3
frequencies, two decreasing and one increasing (solid line) .
11