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Maxwell's Equations and Light Waves
Longitudinal vs. transverse waves
Div, grad, curl, etc., and the 3D Wave equation
Derivation of the wave equation from Maxwell's Equations
Why light waves are transverse waves
Why we neglect the magnetic field
Irradiance, superposition and interference
Longitudinal vs. Transverse waves
Motion is along
the direction of
propagation
Motion is transverse
to the direction of
propagation
Space has 3 dimensions, of which 2 directions are transverse to
the propagation direction, so there are 2 transverse waves in ad-
dition to the potential longitudinal one.
Transverse:
Longitudinal:
Vector fields
Light is a 3D vector field.
A 3D vector field
assigns a 3D vector (i.e., an
arrow having both direction
and length) to each point in
3D space.
( )f r! !
A light wave has both electric and magnetic 3D vector fields:
x
y
A 2D vector field
2
2
2
2 2 2 2
2 2 2 2
0
0
EE
t
E E E E
x y z t
µ!
µ!
"# $ =
"
" " " "+ + $ =
" " " "
!! !
! ! ! !
The 3D vector wave equation for the
electric field
which has the vector field solution:
Note the vector symbol
over the E.
( ) ( ), expE r t A i k r t! "# $= % & &' (
!" !" " ""
#
( ) ( )0, expE r t E i k r t!" #= $ %& '
!" " """
# #
This is really just
three independent
wave equations,
one each for the
x-, y-, and z-
components of E.
Waves using complex vector amplitudes
We must now allow the complex field and its amplitude to be
vectors:
( ) ( )0, expE r t E i k r t!" #= $ %& '
! !! !!
" "
0 (Re{ } Im{ }, Re{ } Im{ }, Re{ } Im{ })x x y y z z
E E i E E i E E i E= + + +
!
"
The complex vector amplitude has six numbers that must be
specified to completely determine it!
0E!
E!
Note the arrows
over the E’s!
x-component y-component z-component
Div, Grad, Curl, and all that
Types of 3D vector derivatives:
The Del operator:
The Gradient of a scalar function f :
The gradient points in the direction of steepest ascent.
, ,x y z
! "# # #$ % & '# # #( )
!
, ,f f f
fx y z
! "# # #$ % & '# # #( )
!
If you want to know
more about vector
calculus, read this
book!
Div, Grad, Curl, and all that
The Divergence of a vector function:
yx zff f
fx y z
!! !" # $ + +
! ! !
!!
The Divergence is nonzero
if there are sources or sinks.
A 2D source with a
large divergence:
Note that the x-component of this function changes rapidly in the x
direction, etc., the essence of a large divergence.
x
y
Div, Grad, Curl, and more all that
The Laplacian of a scalar function :
The Laplacian of a vector function is the same,
but for each component of f:
2, ,
f f ff f
x y z
! "# # #$ % $&$ = $& ' (# # #) *
! ! !
2 2 2
2 2 2
f f f
x y z
! ! != + +
! ! !
2 2 22 2 2 2 2 2
2
2 2 2 2 2 2 2 2 2, ,
y y yx x x z z zf f ff f f f f f
fx y z x y z x y z
! "# # ## # # # # #$ = + + + + + +% &% &# # # # # # # # #' (
!
The Laplacian tells us the curvature of a vector function.
Div, Grad, Curl, and still more all that
, ,
x y z
y z
y yx xz z
x y z
f
f ff ff ff
y dz z dx x dy
fx
f f f
! !" #! !! !$% & ' ' '( )! ! !* +
, -. /! ! !. /$% = . /! ! !
. /
. /0 1
!
!!
" " "
!!
The of a vector function :
The curl can be treat
C
ed as a matrix determinant :
Functions that tend to
url
cu rl ar have large cound urls
A function with a large curl
( , , ) ( , ,0)
(1,0,0) (0,1,0)
(0,1,0) ( 1,0,0)
( 1,0,0) (0, 1,0)
(0, 1,0) (1,0,0)
, ,y yx xz z
f x y z y x
f
f
f
f
f ff ff ff
y z z x x y
= !
=
= !
! = !
! =
" "# $" "" "%& = ! ! !' " " " " " "( )
!
!
!
!
!
!!
( )
( )
0 0, 0 0, 1 ( 1)
0 , 0 , 2
2z
*
= ! ! ! !
=
"So this function has a curl of
x
y
The equations of optics are
Maxwell’s equations.
where is the electric field, is the magnetic field,! is the charge density, " is the permittivity, and µ is
the permeability of the medium.
/
0
BE E
t
EB B
t
! "
µ"
#$ % = $& = '
#
#$ % = $& =
#
!! ! ! !
!! ! ! !
!E
!B
Derivation of the Wave Equation
from Maxwell’s Equations
Take of:
Change the order of differentiation on the RHS:
!!"
BE
t
!"# = $
!
!! !
[ ] [ ]B
Et
!"# "# ="# $
!
!! ! ! !
[ ] [ ]E Bt
!"# "# = $ "#
!
! ! ! ! !
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
But:
Substituting for , we have:
Or:
EB
tµ!
"#$ =
"
!! !
!! "!B
[ ] [ ]E
Et tµ!
" "#$ #$ = %
" "
!! ! !
2
2[ ]
EE
tµ!
"#$ #$ = %
"
!! ! !
[ ] [ ]E Bt
!"# "# = $ "# %
!
! ! ! ! !
assuming that µ
and ! are constant
in time.
Lemma:
Proof: Look first at the LHS of the above formula:
Taking the 2nd yields:
x-component:
y-component:
z-component:
2[ ] ( )f f f!" !" = ! !# $ !! ! !! ! ! !
[ ] , ,y yx xz zf ff ff f
fy z z x x y
! !" #! !! !$% $% = $% & & &' (! ! ! ! ! !) *
!! ! !
!! "
2 2 2 2
2 2
22 2 2
2 2
2 2 22
2 2
y x x z
yx z z
y y xz
f f f f
x y y z x z
ff f f
x z x y z y
f f ff
z y z x x y
! "# ! "# # #= $ $ $% & % &% &# # # # # #' (' (
! "#! "# # #= $ $ $% &% & % &# # # # # #' ( ' (
! " ! "# # ##= $ $ $% & % &% & % &# # # # # #' ( ' (
Lemma (cont’d):
Proof (cont’d):
Now, look at the RHS:
2[ ] ( )f f f!" !" = ! !# $ !! ! !! ! ! !
2( )f f! !" # !! !! !
( ) ( )yx zff f
fx y z
!! !" "# =" + +
! ! !
!! ! !
2 22 22 2
2 2
22 2
2
( , ,
)
y yx xz z
yx z
f ff ff f
x x y x z x y y y z
ff f
x z z y z
! !! !! != + + + +
! ! ! ! ! ! ! ! ! !
!! !+ +
! ! ! ! !
2 2 22 2 22
2 2 2 2 2 2
2 2 2
2 2 2
( , ,
)
y y yx x x
z z z
f f ff f ff
x y z x y z
f f f
x y z
! ! !! ! !"# = " " " " " "
! ! ! ! ! !
! ! !" " "! ! !
!
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
Using the lemma,
becomes:
If we now assume zero charge density: ! = 0, then
and we’re left with the Wave Equation!
2
2[ ]
EE
tµ!
"#$ #$ = %
"
!! ! !
22
2( )
EE E
tµ!
"# #$ % # = %
"
!! ! ! !
0E!" =
! !
2
2
2
EE
tµ!
"# =
"
!!
Why light waves are transverse
Suppose a wave propagates in the x-direction. Then it’s a function
of x and t (and not y or z), so all y- and z-derivatives are zero:
Now, in a charge-free medium,
that is,
0y yz zE BE B
y z y z
! !! != = = =
! ! ! !
0 0E B!" = !" =
! ! ! !
0 0y yx xz zE BE BE B
x y z x y z
! !! !! !+ + = + + =
! ! ! ! ! !
0 0x xE B
x x
! != =
! !and
Substituting the zero
values, we have:
So the longitudinal fields are at most constant, and not waves.
and
The magnetic-field direction in a light wave
Suppose a wave propagates in the x-direction and has its electric field
along the y-direction [so Ex = Ez= 0, and Ey = Ey(x,t)].
What is the direction of the magnetic field?
Use:
So:
In other words:
And the magnetic field points in the z-direction.
0,0,yEB
t x
!" #!$ = % &! !' (
!
yzEB
t x
!!" =
! !
, ,y yx xz zE EE EB E E
Et y z z x x y
! !" #! !! ! !$ =%& = $ $ $' (! ! ! ! ! ! !) *
!! !
Suppose a wave propagates in the x-direction and has its electric
field in the y-direction. What is the strength of the magnetic field?
The magnetic-field strength in a light wave
0
( , ) ( ,0)
t
y
z z
EB x t B x dt
x
!= "
!#We can integrate:
Take Bz(x,0) = 0
Differentiating Ey with
respect to x yields an ik,
and integrating withrespect to t yields a 1/-i".
( ) ( )0, expyE r t E i kx t!= "# $% &!
"
yzEB
t x
!!" =
! !Start with: and
[ ]0( , ) exp ( )z
ikB x t E i kx t
i!
!= " "
" !
1( , ) ( , )z yB x t E x t
c=
So:
But " / k = c:
An Electromagnetic Wave
The electric field, the magnetic field, and the k-vector are all
perpendicular:
The electric and magnetic fields are in phase.
E B k! "
!! !
snapshot of the
wave at one time
The Energy Density of a Light Wave
The energy density of an electric field is:
The energy density of a magnetic field is:
Using B = E/c, and , which together imply that
we have:
Total energy density:
So the electrical and magnetic energy densities in light are equal.
UE=
1
2!E
2
UB=
1
2
1
µB
2
UB=
1
2
1
µE
2!µ( ) =
1
2!E
2 =UE
U =U
E+U
B= !E
2
c =1
!µ B = E !µ
Why we neglect the magnetic field
The force on a charge, q, is:
Taking the ratio of
the magnitudes
of the two forces:
Since B = E/c:
vF qE q B= + !
! ! !!
vmagnetic
electrical
F q B
F qE!
vmagnetic
electrical
F
F c!
So as long as a charge’s velocity is much less than the speed of light,
we can neglect the light’s magnetic force compared to its electric force.
where is the
charge velocityv!
electricalF!
magneticF!
v v sin
v
B B
B
!" =
#
!!
The Poynting Vector: S = c2 ! E x B
The power per unit area in a beam.
Justification (but not a proof):
Energy passing through area A in time #t:
= U V = U A c #t
So the energy per unit time per unit area:
= U V / ( A #t ) = U A c #t / ( A #t ) = U c = c " E2
= c2 " E B
And the direction is reasonable.E B k! "
!! !
A
c #t
U = Energy density
The Irradiance (often called the Intensity)
A light wave’s average power
per unit area is the irradiance.
Substituting a light wave into the expression for the Poynting vector,
, yields:
The average of cos2 is 1/2:
!S = c
2!!E "!B
2 2
0 0( , ) cos ( )S r t c E B k r t! " #= $ % & &!! ! !! !
/ 2
/ 2
1( , ) ( , ') '
t T
t T
S r t S r t dtT
+
!
= "! !! !
real amplitudes
2
0 0
( , ) ( , )
(1/ 2)
I r t S r t
c E B!
" = =
= #
!! !
! !
vector magnitude
The Irradiance (continued)
Since the electric and magnetic fields are perpendicular and B0 = E
0 / c,
21
0 02I c E B!= "
! !
0
2
12 ~
I c E!=!
2* * *
0 0 0 0 0 0 0x x y y z zE E E E E E E= + +
!
" " " " " " "
Remember: this formula only works when the wave is of the form:
where:
becomes:
( ) ( )0, Re expE r t E i k r t!" #= $ %
& '
!! !! !
"
that is, when all the fields involved have the same k r t!" #
! !
because the real amplitude
squared is the same as the
mag-squared complex one.
21
02I c E!=
!
or:
Computing the Irradiance (Intensity)
Sometimes, you’re given the electric field, in which case the
previous result works.
But sometimes, you’re given the energy (U), duration (#t), (or
power, P = U/ #t), and area (A). Then it’s easy:
I = U / (A #t)
or:
I = P / A
Sums of fields: Electromagnetism is linear,
so the principle of Superposition holds.
If E1(x,t) and E
2(x,t) are solutions to the wave equation,
then E1(x,t) + E
2(x,t) is also a solution.
Proof: and
This means that light beams can pass through each other.
It also means that waves can constructively or destructively interfere.
2 2 2
1 2 1 2
2 2 2
( )E E E E
x x x
! ! !
! ! !
+= +
2 2 2
1 2 1 2
2 2 2
( )E E E E
t t t
! ! !
! ! !
+= +
2 2 2 2 2 2
1 2 1 2 1 1 2 2
2 2 2 2 2 2 2 2 2
( ) ( )1 1 10
E E E E E E E E
x c t x c t x c t
! ! ! ! ! !
! ! ! ! ! !
" # " #+ +$ = $ + $ =% & % &
' ( ' (
Same polarizations (say ):
The irradiance of the sum of two waves
If they’re both proportional to , then the irradiance is:
* * * *1 1
0 0 0 0 0 0 0 02 2 x x y y z zI c E E c E E E E E E! ! " #= $ = + +% &
! !
" " " " " " " "
* *1
0 0 0 02 x x y y x yI c E E E E I I! " #= $ + $ = +% &
! ! ! !
{ }* * *1
1 1 1 2 2 222ReI c E E E E E E! " #= $ + $ + $% &! ! ! ! ! !
{ }1 2
*
1 2Rec E EI I I!= + " +! !
Different polarizations (say x and y):
The cross term is the origin of interference!
Interference only occurs for beams with the same polarization.
Therefore:Note the
cross term!
Intensities
add.
exp ( )i k r t!" #$ %& '
! !
0 1 2xE E E= +! ! !
The irradiance of the sum of two waves
of different color
Waves of different color (frequency) do not interfere!
21I II= +Different colors: Intensities add.
10 1 1 20 2 21 2cos( )cos( ) B k r tE k r t !! " "#$ % % $ % %!! !!! !
We can’t use the formula because the k’s and "’s are different.
So we need to go back to the Poynting vector, 2( , )S r t c E B!= "! ! !!
2
21 12( , ) ES r Bc Et B! " # " #= + $ +% & % &! !!! !!
{ }1 1 1 2
2
2 2 21BE Ec E BB E B!= " + " + " + "! ! !! ! !! !
This product averages to zero, as does2 1BE !
! !
Irradiance of a sum of two waves
{ }2
*
2
1
1Rec E E
I I I
!
= + +
"! !
Different
colors
Different polarizations
Same
colors
Same polarizations
1 2I I I= +
1 2I I I= +
1 2I I I= +
Interference only occurs when the waves have the same color and
polarization.