Maxwell's Equations and Light Waves
Longitudinal vs. Transverse waves
Motion is along
the direction of
propagation
Longitudinal:
Longitudinal vs. transverse waves
Div, grad, curl, etc., and the 3D Wave equation
Motion is transverse
to the direction of
propagation
Transverse:
Derivation of the wave equation from Maxwell's Equations
Why light waves are transverse waves
Space has 3 dimensions, of which 2 directions are transverse to
the propagation direction, so there are 2 transverse waves in addition to the potential longitudinal one.
Why we neglect the magnetic field
Irradiance, superposition and interference
Vector fields
The 3D vector wave equation for the
electric field
y
Light is a 3D vector field.
! !
A 3D vector field f (r )
assigns a 3D vector (i.e., an
arrow having both direction
and length) to each point in
3D space.
x
A 2D vector field
A light wave has both electric and magnetic 3D vector fields:
!
Note the vector symbol
!2 !
"2 E
over the E.
# E $ µ! 2 = 0
"t
!
!
!
!
This is really just
three independent
"2 E "2 E "2 E
"2 E
+ 2 + 2 $ µ! 2 = 0
wave equations,
2
one each for the
"x
"y
"z
"t
x-, y-, and z-
components of E.
which has the vector field solution:
!" "
!"
" "
E (r , t ) = A exp #i k % r & ! t & " $
'
(
#
!" "
" "
"
E (r , t ) = E0 exp "i k $ r % ! t #
&
'
#
#
(
)
(
)
Waves using complex vector amplitudes
We must now allow the complex field E and its amplitude E0 to be
!
!
vectors:
! !
! !
!
E (r , t ) = E0 exp "i k $ r % ! t #
&
'
"
"
(
)
Note the arrows
over the E’s!
The complex vector amplitude has six numbers that must be
specified to completely determine it!
x-component
y-component
Div, Grad, Curl, and all that
Types of 3D vector derivatives:
The Del operator:
The Gradient of a scalar function f :
!
! #f
#f
#f "
$f % &
,
,
'
( #x #y #z )
z-component
!
E0 = (Re{Ex } + i Im{Ex }, Re{E y } + i Im{E y }, Re{Ez } + i Im{Ez })
"
Div, Grad, Curl, and all that
!
! #
#
#"
$ % &
,
,
'
( #x #y #z )
If you want to know
more about vector
calculus, read this
book!
The gradient points in the direction of steepest ascent.
Div, Grad, Curl, and more all that
The Laplacian of a scalar function :
The Divergence of a vector function:
$2 f
! !
!f !f
!f
"# f $ x + y + z
!x !y !z
! ! #f
#f
#f "
= $&'
,
,
(
#
x
#
y
#z *
)
=
The Divergence is nonzero
if there are sources or sinks.
A 2D source with a
large divergence:
! !
% $ &$f
!2 f
!2 f
!2 f
+
+
2
2
!x
!y
!z 2
The Laplacian of a vector function is the same,
but for each component of f:
y
x
Note that the x-component of this function changes rapidly in the x
direction, etc., the essence of a large divergence.
2
2
2
! ! #2 f
#2 fx #2 fx # f y # f y # f y #2 fz #2 fz #2 fz
$ 2 f = % 2x +
+ 2 ,
+
+
,
+
+ 2
2
2
2
% #x
#y
#z
#x
#y
#z 2 #x 2
#y 2
#z
'
The Laplacian tells us the curvature of a vector function.
"
&&
(
Div, Grad, Curl, and still more all that
A function with a large curl
!
f ( x, y, z ) = (! y, x, 0)
!
The Curl of a vector function f :
!
f (1, 0, 0) = (0,1, 0)
!
f (0,1, 0) = (!1, 0, 0)
!
f (!1, 0, 0) = (0, !1, 0)
!
f (0, !1, 0) = (1, 0, 0)
! !
" !f !f !f !f !f
!f #
$% f & ( z ' y , x ' z , y ' x )
* !y dz !z dx !x dy +
The curl can be treated as a matrix determinant :
"
,x
.
! !
!
$% f = .
. !x
.
.0 f x
"
y
!
!y
fy
"
z /
!/
!z /
/
f z /1
! ! # "f "f
"f y "f x $
"f x "f z
%& f = ' z ! y ,
!
,
!
*
"
y
"
z
"
z
"
x
"x "y )
(
= (0 ! 0, 0 ! 0, 1 ! (!1) )
=( 0 , 0 ,
Functions that tend to curl around have large curls
The equations of optics are
Maxwell’s equations.
! !
$%E = ! /"
! !
$%B = 0
!
y
!
! !
#B
$& E = '
#t
!
! !
#E
$ & B = µ"
#t
!
where E is the electric field, B is the magnetic field,
! is the charge density, " is the permittivity, and µ is
the permeability of the medium.
2
)
So this function has a curl of 2z"
Derivation of the Wave Equation
from Maxwell’s Equations
Take
!
!"
of:
!
! !
!B
"# E = $
!t
!
! ! !
!
!B
" # [" # E ] = " # [$ ]
!t
Change the order of differentiation on the RHS:
! ! !
! ! !
" # [" # E ] = $ [" # B]
!t
x
!
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
But:
! ! !
! ! !
" # [" # E ] = $ [" # B] %
!t
!
Taking the 2nd ! " yields:
!
! ! !
"
"E
# $ [# $ E ] = % [ µ!
]
"t
"t
!
! ! !
"2 E
# $ [# $ E ] = % µ! 2
"t
!
!
!
assuming that µ
and ! are constant
in time.
! ! !
!
Lemma (cont’d): ! " [! " f ] = !(! # f ) $ ! 2 f
Proof (cont’d):
Now, look at the RHS:
!
! ! !
!(! " f ) # ! 2 f
2
!2 fx ! f y !2 fz
+
+
,
2
!x
!x!y !x!z
! #2 f y #2 fx
= %
$
% #x#y #y 2
'
y-component:
! #2 fx #2 fz
= %
$ 2
' #x#z #x
2
!2 fx ! f y !2 fz
+ 2 +
,
!x!y !y
!y!z
!
!2 f
!2 f
!2 f
!2 f !2 f !2 f
"# 2 f = ( " 2x " 2x " 2x , " 2y " 2y " 2y ,
!x
!y
!z
!x
!y
!z
!2 fz !2 fz !2 fz
" 2 " 2 )
!x 2
!y
!z
" ! #2 fx #2 fz "
&& $ % 2 $
&
#x#z (
( ' #z
" !
& $ %%
( '
! #2 fz #2 f y " !
= %
$
$
% #z#y #z 2 && %%
'
( '
z-component:
2
#2 fz # f y "
$
&
#y 2 #z#y &(
#2 f y
#x
2
$
#2 fx "
&
#x#y &(
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
becomes:
2
!2 fx ! f y !2 fz
+
+
)
!x!z !z!y !z 2
"
x-component:
Using the lemma,
! ! !
! !f !f
!f
"(" # f ) = " ( x + y + z )
!x !y !z
=(
!
! ! !
! " !f !f y !f x !f z !f y !f x #
$ % [$ % f ] = $ % ' z &
,
&
,
&
(
) !y !z !z !x !x !y *
! !
! " B , we have:
Or:
! ! !
Proof: Look first at the LHS of the above formula:
!
! !
"E
# $ B = µ!
"t
Substituting for
!
!
Lemma: ! " [! " f ] = !(! # f ) $ ! 2 f
!
! ! !
"2 E
# $ [# $ E ] = % µ! 2
"t
!
! ! !
!
"2 E
#(# $ E ) % # 2 E = % µ! 2
"t
If we now assume zero charge density: ! = 0, then
! !
!"E = 0
and we’re left with the Wave Equation!
!
!
"2 E
# E = µ! 2
"t
2
Why light waves are transverse
The magnetic-field direction in a light wave
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:
!E y
!y
=
!Ez !By !Bz
=
=
=0
!z
!y
!z
What is the direction of the magnetic field?
! !
! !
Now, in a charge-free medium, ! " E = 0 and ! " B = 0
!Ex !E y !Ez
+
+
=0
!x
!y
!z
that is,
Substituting the zero
values, we have:
!Ex
=0
!x
!Bx !By !Bz
+
+
=0
!x
!y
!z
and
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)].
!Bx
=0
!x
Use:
$
!
" !E !E !E !E !E !E #
!B ! !
= %& E = ' z $ y , x $ z , y $ x (
!t
!z !z
!x !x
!y *
) !y
So:
!
!E #
!B "
$
= % 0, 0, y &
!t '
!x (
In other words:
"
!Bz !E y
=
!t
!x
So the longitudinal fields are at most constant, and not waves.
And the magnetic field points in the z-direction.
The magnetic-field strength in a light wave
An Electromagnetic Wave
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?
Start with:
"
!Bz !E y
=
!t
!x
and
!
E y (r , t ) = E0 exp %#i (kx " ! t )&$
"
t
We can integrate:
Bz ( x, t ) = Bz ( x, 0) "
Take Bz(x,0) = 0
So:
Bz ( x, t ) = "
But " / k = c:
The electric and magnetic fields are in phase.
#
0
ik
E0 exp [i (kx " ! t ) ]
"i! !
1
Bz ( x, t ) = E y ( x, t )
c
!E y
!x
snapshot of the
wave at one time
dt
Differentiating Ey with
respect to x yields an ik,
and integrating with
respect to t yields a 1/-i".
The electric field, the magnetic field, and the k-vector are all
perpendicular:
! ! !
E!B " k
The Energy Density of a Light Wave
Why we neglect the magnetic field
Using B = E/c, and
we have:
1
c=
!µ
(
!
!
! !
F = qE + q v ! B
, which together imply that
B = E !µ
Taking the ratio of
the magnitudes
of the two forces:
)
11 2
1
UB =
E !µ = ! E 2 = U E
2µ
2
Total energy density:
U = UE +UB = !E
!
!
Felectrical Fmagnetic
The force on a charge, q, is:
1
The energy density of an electric field is: U E = ! E 2
2
11 2
The energy density of a magnetic field is: U B =
B
2µ
Since B = E/c:
2
Fmagnetic
!
Felectrical
Fmagnetic
Felectrical
!
qvB
qE
!
where v is the
charge velocity
! !
v " B = vB sin !
# vB
v
c
So the electrical and magnetic energy densities in light are equal.
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.
The Poynting Vector: S = c2 ! E x B
The Irradiance (often called the Intensity)
The power per unit area in a beam.
A
Justification (but not a proof):
U V
=
c #t
=
U V / ( A #t ) = U A c #t / ( A #t ) = U c = c " E2
=
c2
"EB
!
!
!
And the direction E ! B " k is reasonable.
t +T / 2
"
! !
S (r , t ') dt '
t !T / 2
real amplitudes
U A c #t
So the energy per unit time per unit area:
! !
1
S (r , t ) =
T
Substituting a light wave into the expression for the Poynting vector,
!
! !
S = c 2 ! E " B, yields:
Energy passing through area A in time #t:
=
A light wave’s average power
per unit area is the irradiance.
U = Energy density
! !
! !
!
!
S (r , t ) = c 2 ! E0 $ B0 cos 2 (k % r & " t & # )
The average of cos2 is 1/2:
vector magnitude
! !
S (r , t ) =
! !
= c 2 ! E0 # B0 (1/ 2)
!
" I (r , t ) =
The Irradiance (continued)
Computing the Irradiance (Intensity)
Since the electric and magnetic fields are perpendicular and B0 = E0 / c,
I
=
or:
where:
1
2
! !
c 2 ! E0 " B0
becomes:
I =
=
I
!
1 c ! E0
2
~
2
1
2
! 2
c ! E0
because the real amplitude
squared is the same as the
mag-squared complex one.
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:
! 2
E0 = E0 x E0*x + E0 y E0* y + E0 z E0*z
"
" "
" "
" "
Remember: this formula only works when the wave is of the form:
I = U / (A #t)
or:
! !
! !
!
E (r , t ) = Re E0 exp "i k $ r % ! t #
&
'
"
! !
that is, when all the fields involved have the same k " r # ! t
(
)
Sums of fields: Electromagnetism is linear,
so the principle of Superposition holds.
If E1(x,t) and E2(x,t) are solutions to the wave equation,
then E1(x,t) + E2(x,t) is also a solution.
Proof:
! 2 ( E1 + E2 ) ! 2 E1 ! 2 E2
=
+
! x2
! x2
! x2
and
! 2 ( E1 + E2 ) ! 2 E1 ! 2 E2
=
+
! t2
! t2
! t2
! 2 ( E1 + E2 ) 1 ! 2 ( E1 + E2 ) " ! 2 E1 1 ! 2 E1 # " ! 2 E2 1 ! 2 E2 #
$ 2
=% 2 $ 2
$ 2
&+%
&=0
! x2
c
! t2
c ! t 2 ( ' ! x2
c ! t2 (
'!x
This means that light beams can pass through each other.
It also means that waves can constructively or destructively interfere.
I = P/A
The irradiance of the sum of two waves
! !
If they’re both proportional to exp "&i (k $ r % ! t ) #' , then the irradiance is:
! !
I = 12 c! E0 $ E0* = 12 c! "% E0 x E0 x* + E0 y E0 y* + E0 z E0 z * #&
" "
" "
" "
" "
Different polarizations (say x and y):
I = c! "% E0 x $ E0 x + E0 y $ E0 y #& = I x + I y
!
!
!
!
1
2
*
*
Intensities
add.
Same polarizations (say E0 x = E1 + E2 ):
!
! !
I = 12 c! "% E1 $ E1* + 2 Re {E1 $ E2*}+ E2 $ E2* #&
! !
! !
! !
Therefore:
I = I1 + c! Re {E1 " E2*}+ I 2
! !
Note the
cross term!
The cross term is the origin of interference!
Interference only occurs for beams with the same polarization.
The irradiance of the sum of two waves
of different color
We can’t use the formula because the k’s and "’s are different.
! !
Irradiance of a sum of two waves
Same polarizations
!
!
So we need to go back to the Poynting vector, S ( r , t ) = c 2! E " B
! !
! !
! !
S (r , t ) = c 2! "% E1 + E2 #& $ "% B1 + B2 #&
! !
! !
! !
! !
= c 2! E1 " B1 + E1 " B2 + E2 " B1 + E2 " B2
! !
! !
!
!
E10 cos(k1 $ r % !1 t % "1 ) # B20 cos(k2 $ r % !2 t % " 2 )
! !
This product averages to zero, as does E2 ! B1
{
Different colors:
I = I1 + I 2
}
Different polarizations
Same
colors
Different
colors
I = I1 + I 2 +
c! Re {E1 " E2*}
! !
I = I1 + I 2
I = I1 + I 2
I = I1 + I 2
Intensities add.
Waves of different color (frequency) do not interfere!
Interference only occurs when the waves have the same color and
polarization.