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.