# 3.0 Wave Equation

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```					3.0WaveEquation.nb                                                                    Optics 505   −   James C. Wyant             1

3.0 Wave Equation
In deriving the wave equation we will first make use of the vector identity
” ∇ − L” ⋅ ∇H ∇ = L” ∇H ∇
E                                   E           2
E

However, from Maxwell's equations                                         L” ⋅ ∇H
E = 0 so
” ∂ H
”÷               2E
∇ µ − = ” ∇ − = L” ∇H ∇
E                       2
E               o               ε µ − = L”÷ ∇H ∂ µ − = ∂
o                 H    o    o
∂ t            t  ∂        ∂             t2
” ∂                          2E
2
ε µ−” ∇
E       o     o                      =      0
∂                           t2

Letting c     ####e####m###" =
1
Å ÅÅ ÅÅÅÅÅÅ
Å ÅÅ
o o
yields the wave equation

1
” ∂ 2E
−” ∇
2
E                               =       0
c2      ∂  t2

In one dimension we can write
2E                            2E
∂        y        1        ∂             y
−                                   =   0
∂    x2           c2       ∂     t2

3.1 Transverse waves
One solution to the one-dimensional wave equation is
ˆ
@ ”
E y x, t          =D    Eo j Cos k x              @           D ω−    t

which represents a plane wave travelling in the +x direction.

E Eoê
1

0.5

x
0.5                 1      1.5           2       2.5       3

-0.5

-1
3.0WaveEquation.nb                                                               Optics 505   −   James C. Wyant          2

One way of showing this is a solution is to substitute the solution into the wave equation to see if it checks.

ω +D ω−                2
−    Eo k2 Cos k x        @                        t              Eo Cos k x
@       t
=D ω−      0
c2

This is a solution if

k
ω =
c
2
We will let k             ÅÅpÅlÅÅÅ =
ÅÅ         .

3.2 Plane Wave
A solution to the three-dimensional wave equation

1
” ∂ 2E
−” ∇ 2
E                             =       0
c2         ∂ t2

is

@”
E x, y, z; t
”=D        Eo Cos k r     D ω − ” ⋅ ”@   t

where the position vector is
ˆ           ˆ                ˆ
r=”          xi   +   yj      +    zk

and the propagation vector which gives the direction of propagation is given by
ˆ            ˆ                    ˆ
=”
k            kx i    +   ky j      +       kz k

Then

= ω − ”⋅”
k r                   t       kx x          +      ky y   +   kz z       t
= ω−   constant

represents planes in space of constant phase. Hence we call this solution a plane wave.

3.3 Complex Representation
The use of the complex representation, rather than only the real representation, sometimes makes algebraic
computations simpler. Thus, instead of using

@”
E x, y, z; t
”=D        Eo Cos k r     D ω − ” ⋅ ”@   t

to represent a plane wave it is often easier to use

@”
E x, y, z; t
”=D        Eo          M ω − ”⋅”I
k r      t
3.0WaveEquation.nb                                                                    Optics 505   −   James C. Wyant           3

While the complex expression is itself a solution to the wave equation, it needs to be understood that for physical
reasons the real part is the actual quantity being represented.

3.4 Spherical Wave
Cos k r
”÷
A
w-”ÿ”÷I Â‰
t and k r t have constant have constant values on a sphere at any given radius r at a given time t,
w -”ÿ       E
M

however they are not solutions to the wave equation. It can be shown (see homework problems) that
1
Cos k r      t and 1
”÷
A   w - ”ÿ  kr
E
t               w-”ÿ”÷I Â‰
are solutions to the wave equation and they represent spherical waves
ÅÅÅÅ                 M
r                      r
ÅÅÅÅ

propagating outward from the origin.

3.5 Linear Superposition
Since the electric field is a vector quantity a vector summation must be performed to find the electric field resulting
from the summation of several electric fields.
”+ ”+ ”=”
E           E1       E2       E3   ∫+
This linear superposition is only approximately true in the presence of matter. Deviations from linearity are
observed at high intensities produced by lasers when the electric fields approach the electric fields comparable to
atomic fields (non-linear optics). In these notes we will consider only situations where linear superposition is
valid.

3.5.1 Sine waves having same frequency, but different amplitude and phase

ü   3.5.1.1 Two Sine Waves (same polarization)
φ       H   kx    t
E1 x, t @           =D    A1 Cos k x  @                    t
=D φ+ ω −         1        Re A1
@         1
D L ω−
φ       H   kx
ω−   t
E2 x, t @           =D    A2 Cos k x  @                    t
=D φ+ ω −         2        Re A2
@         2
DL

From now on we will forget about taking the Real part until the end of the calculation.

Eo x, t   @ =D   @        E1 x, t          +D     E2 x, t
@            D
kx       t
A1    + φ H=         1
A2           HL φ
2              H       ω−       LL
H    φ                     kx  L ω−t
Ao        =          o

Superposition of two harmonic waves of given frequency produces a harmonic wave of same frequency with a
given amplitude and phase.
3.0WaveEquation.nb                                                                   Optics 505                   −   James C. Wyant                                    4

Waves Having Same Frequency,
but Different Amplitude and Phase

α
a Cos(x+α)                                                                                                                          α
Cos(x) + a Cos(x+α)

Cos(x)

Now we will find the amplitude and phase.

A1          φ   1
+   A2              φ   2
=   Ao         φ
o

A1 Cos          + D φ@  1            A2 Cos            H + D φ@
2           A1 Sin        + D φ@
1           A2 Sin    = LD φ@
2           Ao Cos   + D φ@
o      Ao Sin
D φ@o

Since the real and imaginary parts must be equal it follows that the phase is given by

Ao Sin                                     A1 Sin                     A2 Sin
= D φ@
o                                  + D φ@
1
= D φ@  2
Tan     D φ@
o
Ao Cos               D φ@
o                   A1 Cos         + D φ@
1       A2 Cos      D φ@  2

Next we must calculate the amplitude.
2                                                                     2
Abs Ao  @                   φ   o
= D         Ao 2   =   Abs A1  @      φ  1
+   A2       φ   2
D

This yields the basic equation of two-beam interference.

Ao 2      =       A1 2    +       A2 2        +   2 A1 A2 Cos        D φ − φ@
2           1

If A1   =   A2
2
1
Ao 2    =   2 A1 2 1            + H             Cos    = LD φ − φ@
2       1         4 A1 2 Cos       A               EL φ − φH
2        1
2

Thus we can think of the linear combination giving a cosine intensity distribution, or equivalently a cosine squared
distribution.
3.0WaveEquation.nb                                                                                       Optics 505               −   James C. Wyant    5

ü   3.5.1.2 Combination of several sine waves
N
kx           t                                kx       t
E x, t
@              ‚=D        n 1 =
An               φ   n           H           L ω−     =   Ao        φ     o    H       L ω−

N
Ao           φ   o
‚=   n 1
=
An            φ   n

In the same manner as shown above for two sine waves
N
⁄                          n 1
=       An Sin               D φ@    n
Tan        = D φ@o                  N
⁄                          n 1
=       An Cos               D φ@    n

Now we need to calculate the amplitude.
N         N
Ao 2     ‚ ‚=                       An            φ   n
Am          φ −      m

n 1 m 1
=         =
N                      N                 N
‚=              An 2     ‚+                  ‚                   An Am                L φ − φH
n    m

n 1 =                      n 1
=           m 1,m n
=           ≠
N                       N                    N
‚ ‚+ ‚=         An 2                                                 An Am        H       φ− φH
n     m
+L   φ− φH −  n       m
LL
n 1 =                   n 1 =        m 1,m
=               <   n

N                       N                    N
‚=              An 2      ‚+                        ‚                   2 An Am Cos            D φ − φ@
n    m
n 1 =                   n 1 =        m 1,m
=               <   n

This last term is the interference term.

If N is large and                   f    is random the sine waves are incoherent and the add with no apparent interference to yield
N
Ao 2     ‚=              An 2
n 1 =

If all the An are equal

Ao 2     =   N An 2

For the in-phase coherent case

N        N                                   N                   2
i                          y
Ao   2
‚ ‚=                   An Am            ‚j =
j
j
j                  An      z
z
z
z
n 1 m 1
=        =                      = k          n 1             {
If all the An are equal

Ao 2     =   N 2 An 2
3.0WaveEquation.nb                               Optics 505    −   James C. Wyant                                              6

Note that for the coherent and incoherent cases the total energy does not change, but the distribution of the energy
does change.

Addition of 3 Waves of Given Frequency
Gives Another Wave of Same Frequency
α            β
Cos(x) + a Cos(x+α) + b Cos(x+β )

α
a Cos(x+α)                                                                         b Cos(x + β )

Cos(x)
3.0WaveEquation.nb                                                       Optics 505            −   James C. Wyant                   7

Always Larger Than Individual Waves
α
a Cos(x+α)                                                           α            β
Cos(x) + a Cos(x+α) + b Cos(x+β )
b Cos(x + β )

Cos(x)

3.5.2 Beats
We will now combine waves having different frequencies, but the same amplitude.

E1 x, t
@          =D   A Cos k1 x   @             ω −   1   t   D φ+
1

E2 x, t
@          =D   A Cos k2 x   @             ω −   2   t   D φ+
2

E    H = A Cos k1 x @          ω −      1   t   +D φ+1         Cos k2 x
@           ω −   2   t   LD φ +
2

But

1                                1
Cos      + Dα@      Cos   = Dβ@       2 Cos      A            ELβ + αH       Cos   A       ELβ − αH
2                                2

so

1
E    =   2 A Cos     A        HH   k1   +   k2 xL         ELL φ + φH + L ω + ωH −
1           2   t         1      2
2
1
Cos   A       HH   k1   −   k2 xL         ELL φ − φH + L ω − ωH −
1      2       t         1       2
2

Let
3.0WaveEquation.nb                                                                                Optics 505                −       James C. Wyant          8

1                                                        1
=¯
ω                L ω + ωH
1               2                    =¯
k                H   k1   +   k2      L
2                                                        2
1                                                        1
= ω  m                L ω − ωH
1               2                km   =           H   k1   −      k2   L
2                                                        2
1                                                        1
=α                L φ + φH
1               2                    =β             L φ − φH
1           2
2                                                        2

then

E    =       2 A Cos k x
¯@                       ¯
Dα + ω −     t              Cos km x  @                ω −       t   Dβ +
m

If   wº w1        2    then        w >> ê ê
w               m   and the second term changes much more slowly than the first term.
¯@               ¯                        2                                                     2
E2       =   4 A2 Cos k x                           ω −          t   @    Dα +   Cos km x               ω −         m   t   Dβ +
¯@               ¯                        2
=   2 A2 Cos k x                           ω −         t     + H Dα +       1        Cos 2 km x H @                ω − m      t   LDLβ +

Let

2
k1       =        = π          period of one signal and
λ   1
2
k2       =        = π          period of the second signal
λ   2

1            1
2 km         H=       k1   −   k2      Jπ =L        2                    −            N
λ   1       λ    2

Let the period of the modulation signal be                                                    l  eq .   Then

1                1
2    Jπ                −               π = λN      eq       2
λ   1            λ   2

1             1                   1                 λ λ = λ             1       2
=                −               ,             eq
λ
eq            λ   1               λ   2             λ− λ            2            1
3.0WaveEquation.nb                                      Optics 505       −   James C. Wyant                                       9

Waves Having Same Amplitude,
but Different Frequency (Beats)
ω φ
Cos(k1x-ω 1t+φ1)                      ω φ
Cos(k2x-ω 2t+φ2)                       ω φ                ω φ
Cos(k1x-ω 1t+φ1) + Cos(k2x-ω 2t+φ2)

E = 2 cos(kx − ω t + α )cos(km x − ω mt + β )
ω = 1 ( ω1 + ω 2 ), k = 1 ( k1 + k2 ), α = 1 (φ1 + φ 2 ), ω m = 1 ( ω 1 − ω 2 ), km = 1 ( k1 − k2 ), β = 2 ( φ1 − φ 2 )
2                   2                  2                    2                     2
1

Beats
cos (km x − ω mt + β )                                                                cos (k x − ω t + α )

π      λ 1λ 2
km = λ 1 − λ 2

Must resolve high frequency to see beats
3.0WaveEquation.nb                                              Optics 505         −   James C. Wyant   10

3.5.3 Standing Waves
If we have two waves of same amplitude and frequency travelling in the opposite directions

E x, t
@        =D   A Cos k x   @             t
+D φ+ ω −      1       A Cos k x
@             t
D φ+ ω +  2

We will now make use of the trig identity

Cos   = Dβ ± α @    Cos     Dα@     Cos
° Dβ@        Sin     Dα@    Sin   Dβ@

or

Cos   + Dβ − α @    Cos     = Dβ + α @    2 Cos        Dα@   Cos   Dβ@

We can let

=α    kx
φ+ φ +
1       2
and
φ− φ + ω = β
t      2        1
2                                             2

E x, t
@       =D    2 A Cos k x A     E φ+ φ + 1       2
Cos   E φ− φ + ωA
t     2       1
2                                 2

The first term is space dependent and the second term is time dependent.

Standing Waves

λ/2

E = A cos (kx − ωt + φ1 ) + A cos (kx + ωt + φ 2 )
     φ +φ               φ −φ 
= 2 A cos  kx + 1 2 2  cos  ωt + 1 2 2 
                               

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