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
u(x,t) = A(k,) exp ik x it ddk
(2)2 all and k space
The dispersion relation for the wave gives a relation (such as eqn 12 in the plasma wave
notes) between and k which allows us to write the integral in terms of k alone1 :
1
u(x,t) = A(k) exp ik x i k t dk
(2)2 all k space
Now lets simplify by choosing the xaxis along the direction of propagation and letting
whole problem be one-dimensional, so that u = u (x, t) . (We also lose two factors of
the
2.)
+
1
u(x,t) = A(k) exp [ikx i (k) t] dk (1)
2
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 also be real, which, in general, it may not be. So we have to take
+
1
u(x, t) = Re A(k) exp [ikx i (k) t] dk
2
+
1
= A (k) exp [ikx i (k) t] + A(k) exp i (k) t ikx dk(2)
4
Now we let = k in the second term.
+
1
u (x, t) = A(k) exp [ikx i (k) t] dk A() exp ix + i () t d
4 +
+
1
= A(k) exp [ikx i (k) t] + A(k) exp ikx + i (k) t dk
4

    k  l
1
Effectively, A k, = A k k , as in Lea Ch 7.

1
Then, evaluating at t = 0, we get
+
1 1
u(x, 0) = A(k) + A(k) exp (ikx) dk
2 2 2

Inverting the transform:

+
1 1
A(k) + A (k) = u(x,0) exp (ikx) dx (3)
2 2 2
If A(k) = A (k) , then
+
A(k) = u( x,0) exp (ikx) dx (4)

But we have an additional initial condition. Taking the time derivative of (2), we have
+
1
u(x,t) = iA(k) exp [ikx i (k) t] + A (k) i exp (ikx it) dk
t 4
+
1
u(x,0) = i A(k) A (k) exp (ikx) dk
t 4
Inverting, we have:
+
i
A(k) A (k) = u(x, 0) exp (ikx) dx (5)
2 t
Combining the two relations (3) and (5) to eliminate A (k) , we have
+
1
A (k) = u(x,0) u(x,0) exp (ikx) dx (6)
i t

which is Jackson equation 7.91 modulo a factor of 2. (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.

2 Gaussian Pulse
To simplify the algebra in this section, lets assume t

u(x,0) = 0. Generally the wave
is sent at a carrier frequency 0 with a corresponding k0 so that 0 = (k0 ) , and the
Fourier amplitude A (k) peaks at k0. For example, a Gaussian pulse with a carrier frequency
0 is written:
x2
u (x, 0) = u0 cos (k0 x) exp 2 (7)
a

2
Its transform is:
+
x2
A (k) = u0 cos (k0 x) exp exp (ikx) dx
a2
+
u0 ik0 x x2
= e + eik0 x exp 2 exp (ikx) dx
2 a
u0 + i(kk0 )x x2
= e + ei(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):
2 2
x2 1 2 2 (k k0 ) a2 (k k0 ) a2
i (k k0 ) x = x + i (k k 0 ) a x + i i
a2 a2 2 2
2
1 a2 (k k0 ) 1
= i +x + a4 (k k0 )2
a2 2 4
Thus the first term in the integral (8) is:
+ 2
u0 a2 1 a2 (k k0 )
exp (k k0 )2 exp i +x dx
2 4 a2 2
++i
u0 a2
= exp (k k0 )2 a exp v2 dv
2 4 +i
u0 a2
= exp (k k0 )2 a
2 4

a2
= au0 exp (k k0 )2
2 4
2
where we set v = i a (kk
2
0)
+ x a1 = i + x/a,and used the fact that the integral of the
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 2 a2 2
A (k) = A1 (k) + A2 (k) = a 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 (k) in a Taylor series centered at k0 .
d 1 d2
(k) = 0 + (k k0 ) + (k k0 )2 + ...
dk k0 2 dk2 k0

3
Substituting (k) into the expression for u (x, t) (1),
drop

+
2 2


A1 (k) d (k k0 ) d
u1 = exp ikx it 0 + (k k0 ) + + .. dk
2
dk k0 2 dk2 k0

(10)
and dropping the term in , we have:
+
1 d d
u1 (x,t) exp i 0 k0 t A1 (k) exp ikx ik t dk+
2 dk k0 dk k0

+
d 1 d
u1 (x,t) = exp i 0 k0 t A1 (k) exp ik x t dk
dk k0 2 dk k0

d d
= u x t, 0 exp i 0 k0 t (11)
dk k0 dk k0

where we used eqn (1) again in the last step, with x x d dk k0 t and t 0. . Thus to
first order in the expansion of (k) , u (x, t) equals a phase factor ei 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
d
vg = (12)
dk k0
is the group speed. The phase = (0 vg k0 ) t indicates that the carrier wave shifts
within the Gaussian envelope.
To evaluate the second integral we expand (k) in a Taylor series centered at k0 , and
obtain the same result.

4 Pulse spreading
For some waves with d2 /dk2 = 0, there is only a first order term in the Taylor series
for (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):
p
k=
c
or
k2 c2
=
2p
The group velocity for this wave is:
d kc2
vg = = 2 2 (13)
dk p

4
and the second derivative is
d2 c2
vg = = 2 (14)
dk2 2p
All further derivatives are zero. Thus from (10), the exact expression for u in this case is:
d
u (x, t) = exp i 0 k0 t
dk k0

1 +
d (k k0 )2 d2
A(k) exp ik x t i t dk
2 dk k0 2 dk2 k0

d
u (x, t) = exp i 0 k0 t
dk k0
+
1 i 2
A(k) exp ik (x vg t) k 2kk0 + k02 vg t dk
2 2
k02 vg 1 +
i
= exp i 0 k0 vg + t A(k) exp ik x + k0 vg t vg t k2 vg t dk
2 2 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:
k02 vg 1
u (x, t) = exp i 0 k0 vg + t I (15)
2 2
where

+ a2 2
au0 exp 4 (k k0 ) exp ik x k0 vg t vg t i k2 vg t dk
I= 2
2 + exp a4 (k + k0 )2 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 .


I= au0 (I+ + I )
2
Looking at the first term:
+
a2 2 i
I+ = exp k 2kk0 + k02 + ik x + k0 vg t vg t k2 vg t dk
4 2
+
a2 k02 a2 i k0 a2
= exp exp k2 + vg t + k i x + k0 vg t vg t + dk
4 4 2 2
To simplify the notation, we let s = x vg t and vg t = a2 . Then the exponent in the
integrand is:
a2 i k0 a2
k2 + a2 + k i s + k0 a2 +
4 2 2
2
a2 is + k02a (1 + 2i)
= (1 + 2i) k2 + a2
k
4 4 (1 + 2i)

5
Completing the square, the term in square brackets is
2 2
2is 2is 2is
k2 + 2k + k0 + 2 + k0 + k0
a2 (1 + 2i) a (1 + 2i) a2 (1 + 2i)
2 2
2is 2is
= k+ 2
+ k0 2 + k0
a (1 + 2i) a (1 + 2i)
We can do the integral by making the change of variable:

a 2is
= 1 + 2i k + 2 + k0
2 a (1 + 2i)
Then:
2 ++i
a2 k02 a2 2is 2 2
I+ = exp exp (1 + 2i) 2 + k0 e d
4 4 a (1 + 2i) a 1 + 2i +i

where the path of integration is again moved off the real axis, but the result is still . Thus:

2 a2 k02 s2 k2 a2
I+ = exp + 2 + isk0 + 0 (1 + 2i) (16)
a 1 + 2i 4 a (1 + 2i) 4

2 s2 (1 2i) i 2 2
= exp 2 + isk0 + k0 a (17)
a 1 + 2i a (1 + 42 ) 2
And the exponent in u (15) is:
k02 a2 k2 a2 s2 (1 2i)
i 0 t + k0 vg t + k0 s + 0 2
2 2 a (1 + 42 )
2
s (1 2i)
= i (0 t + k0 x) 2
a (1 + 42 )
s2 2is2
= i (k0 x 0 t) 2 +
a (1 + 42 ) a2 (1 + 42 )
Thus
au0 1 s2
u (x, t) = exp [i (k0 x 0 t)] exp 2
2 2 a (1 + 42 )

2s2 2
exp i 2 + (k0 k0 )
a (1 + 42 ) a 1 + 2i
u0 exp [i (k0 x 0 t)] (x vg t)2
= exp ei + (k0 k0 ) (18)
2 (1 + 42 )1/4 a2 (1 + 42 )
We may interpret this expression as folows:
Initial wave form at frequency 0 . This is the carrier.
Envelope peaks at x = vg t due to its travelling at the group speed.

Envelope spread by factor 1 + 42 where = vg t/a2
Amplitude decreased by similar factor

6
 Overall phase factor
2s2 tan1 (2)
= +
a2 (1 2
+ 4 ) 2
approximately proportional to t.

At large times the pulse width grows roughly linearly with time. a (t) 2a (0) =
2vg t/a (0) . Notice that (18) gives the correct result (7) at t = 0.

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 (k) we instead
think of it as k () . Then:
+
1
u(x,t) = A() exp [ik () x it] d
2
where in general k() is complex.
As we have seen, A () is usually a smooth function, somewhat peaked around the
carrier frequency 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 , 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 () x t] = 0
d
dk
x t = 0
d
or
d
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 () in a
Taylor series, this time about the stationary frequency.as determined by equation (19).
dk 1 d2 k
k () = ks + ( s ) + ( s )2 +
d s 2 d2 s
where ks = k ( s ) , giving
1 +
dk ( s )2 d2 k
u(x,t) = A() exp i ks + ( s ) + + x it d
2 d s 2 d2 s

Now use the stationary phase condition (19):

7
 +
1 s 1 d2 k
u(x,t) = A() exp i ks + ( s )2 + .. vg t d
2 vg 2 d2 s

Since the exponential guarantees that only frequencies near s contribute, we may pull out
the slowly varying amplitude, as well as the terms in the exponential that are independent of
:

+
A(s ) i d2 k
u(x,t) exp [i (ks vg s ) t] exp ( s )2 vg t d
2 2 d 2 s

To do the integral, change variables to:

i d2 k d2 k vg t
= ( s ) vg t = ei/4 ( s )
2 d 2 s d2 s 2
Then
+ ++i
i d2 k ei/4 2
I = exp ( s )2 vg t d = e d
2 d 2 s 1 d2 k
vg t +i
2 d2 s

2
= d2 k
ei/4
d2 s vg t
and so, incorporating the result (19), we have
A( s )
u(x,t) = exp i (ks x s t) + i (20)
2
d k 4
d2 s x

Again we see the carrier wave, modulated by an envelope that changes in time, and with an
additional phase change. (Remember that 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)
kc2 p c2
x = 2 t= 2 2 t
2p c p


= 2 ct
p
and thus
2p x2
s =
4c2 t2
which decreases from a large value toward zero as t increases. For this wave the signal (20)
is given by:
2 2
c2 A( p 4cx2 t2 )
u(x,t) = 2 2 exp i (ks vg s ) t + i
p x 4
c 2 2p x2
= A( ) exp i (ks vg s ) t + i
p x 4c2 t2 4

8
If A is a Gaussian as in (9), then

2
c 2 2p
x 2
u(x,t) = A0 exp 0 exp i (ks vg s ) t + i
p x 4c2 t2 4

where is a constant.

At fixed x, we get something like this:

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.)
Heres another way to look at it. Consider the plasma waves (Plasmawave notes eqn 12):
c2 k2 = 2 2p
For this wave
dk
2c2 k = 2
d
dk / p
c = c = =
d kc2 2 2p (/p )2 1
which looks like this:
5

3
k/dw
2

0
0.5 1 1.5
w/wp 2 2.5 3

9
 The stationary phase condition is
ct dk
=c
x d
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 p .
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:
2p
c2 k2 = 2

dk 2p 2p
2c2 k = 2 +
d ( )2
2p
= 2 1

2p
= 2 +
( )2
and thus:
dk 1 1 2p
c = +
d ck 2 ( )2

1 1 2p
= +
2p 2 ( )2
2

1 1 2p /
= 1+
1
2p 2 ( )2
()

Now let p / = y and / = z Then

dk 1 1 y2
c = 1+
d 1 y2 2 z (z 1)2
z(z1)

With y = 2, the diagram looks like:

10
 6

cdk/dw 3

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