Maxwell’s Equations and Light Waves

! !
Electric field E Magnetic field B y
x
z

Vector derivatives: Div, grad, curl, etc.

Derivation of 3D wave equation from Maxwell’s Equations
Why light waves are transverse waves
Why we neglect the magnetic field
Interference of waves Prof. Rick Trebino
Spherical waves and light bulbs Georgia Tech
www.frog.gatech.edu
The irradiance and intensity
Div, Grad, Curl, and All That

3D Vector Derivatives use the Del operator:

! ⎛∂ ∂ ∂⎞
∇ ≡ ⎜ , , ⎟
⎝ ∂x ∂y ∂z ⎠

For a function of A great book!

The Gradient of a scalar two variables f(x,y):
function f(x, y, z): Direction of
steepest slope
! ⎛ ∂f ∂f ∂f ⎞ at the point (x,y)
∇f ≡ ⎜ , , ⎟
⎝ ∂x ∂y ∂z ⎠ f(x,y)

The gradient points in the !

y ∇f
direction of steepest ascent. x (x,y)
Div, Grad, Curl, and All That
The Divergence of a vector function:

! ! ∂f x ∂f y ∂f z
∇⋅ f ≡ + +
∂x ∂y ∂z

The Divergence is nonzero

if there are sources or sinks.

A 2D source with a y
large divergence:
x

Note that the x-component of this function changes rapidly in the x-

direction, and the same for y and z, the essence of a large divergence.
Div, Grad, Curl, and More All That
The Laplacian of a scalar function is the div of the grad:

2
! ! ! ⎛ ∂f ∂f ∂f ⎞ ¶2 f ¶2 f ¶2 f
∇ f ≡ ∇ ⋅ ∇f = ∇⋅⎜ , , ⎟ = + +
⎝ ∂x ∂y ∂z ⎠ ¶x 2
¶y 2
¶z 2

The Laplacian gives the

curvature of a function
(here the intensity vs. x and y).

The Laplacian of a vector function is the same, but for each

component of f:

2
! ⎛ ∂2 f ∂ 2
f ∂ 2
f ∂ 2
f y
∂ 2
f y
∂ 2
f y ∂ 2
f ∂ 2
f ∂ 2
f ⎞
∇ f = ⎜ x+
⎜ ∂x 2 2
x
+ 2
x
, 2
+ 2
+ 2
, 2
z
+ 2
z
+ z ⎟
2 ⎟
⎝ ∂y ∂z ∂x ∂y ∂z ∂x ∂y ∂z ⎠
Div, Grad, Curl, and Still More All That
!
The Curl of a vector function f :
! ! ⎛ ∂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 :

⎡ ⎤
⎢ x̂ ŷ ẑ ⎥
! ! ⎢ ∂ ∂ ∂ ⎥
∇ × f = Det ⎢ ⎥
⎢ ∂x ∂y ∂z ⎥
⎢ fx fy fz ⎥
⎣ ⎦

Functions that tend to curl around have large curls.

A Function with a Large Curl
!
f (x, y, z) = (− y, x,0) y

!
f (1,0,0) = (0,1,0)
!
f (0,1,0) = (−1,0,0) x
!
f (−1,0,0) = (0,−1,0)
!
f (0,−1,0) = (1,0,0)

! ! ⎛ ∂f ∂f y ∂f ∂f ∂f y ∂f x ⎞
∇ × f = ⎜⎜ z
− , x
− z
, − ⎟
⎝ ∂y ∂z ∂z ∂x ∂x ∂y ⎟⎠

(
= 0 − 0, 0 − 0, 1− (−1) )
=( 0 , 0 , 2 ) So this function has a curl of 2z! .
Lemma: A Result We’ll Need in Order to
Prove a Future Theorem

A lemma is in effect a “mini-theorem.”

The lemma we require involves all of these vector derivatives.

! ! ! ! ! ! 2
!
∇ × (∇ × f ) = ∇(∇ ⋅ f ) − ∇ f

You get to prove this in a homework problem.

The equations of optics are
Maxwell’s equations.
!
! ! ! ! ∂B
∇ ⋅E = 0 ∇ ×E = −
∂t !
! ! ! ! ∂E
∇ ⋅B = 0 ∇ × B = µε
∂t
! !
where E is the electric field, B is the magnetic field, e is the
permittivity, and µ is the permeability of the medium.

As written, they assume a vacuum and no charges.

Derivation of the Wave Equation
from Maxwell’s Equations
! ! !
! ! ∂B ! ! ! ! ∂B
Take ∇× of: ∇ ×E = − yielding: ∇ × ( ∇ ×E ) = ∇ × (− )
∂t ∂t
Change the order of ! ! ! ∂ ! !
differentiation on the RHS: ∇ × ( ∇ ×E ) = − ( ∇ × B )
∂t
!
! ! ∂E
But: ∇ × B = µε
∂t
!
! ! ! ! ! ∂ ∂E
Substituting for ∇ × B : ∇ × (∇ ×E ) = − (µε )
∂t ∂t
!
! ! ! ∂E 2 assuming that µ
⇒ ∇ × (∇ ×E ) = − µε 2 and e are constant
∂t in time.
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
!
! ! ! ∂E 2
Using the lemma, ∇ × (∇ ×E ) = − µε 2
∂t
!
! ! ! 2
! ∂E 2
becomes: ∇(∇ ⋅E ) − ∇ E = − µε 2
∂t
! !
But we’ve assumed zero charge density (r = 0), so: ∇ ⋅E = 0

and we’re left with the Wave Equation:

!
2
! ∂E 2
∇ E = µε 2 where µe = 1/c2
∂t QED
Why Light Waves are Transverse
Suppose a wave propagates in the z-direction. Then it’s a function
of z and t (and not x or y), so all x- and y-derivatives are zero.

¶Ex ¶Ey ¶B x ¶B y
Specifically: = =0 = =0
¶x ¶y ¶x ¶y
! ! ! !
Now, in a charge-free medium, ∇ ⋅E = 0 and ∇ ⋅ B = 0

¶Ex ¶Ey ¶Ez ¶B x ¶B y ¶B z

that is: + + =0 + + =0
¶x ¶y ¶z ¶x ¶y ¶z

So: ¶Ez ¶B z
=0 and =0
¶z ¶z
So the longitudinal fields are at most constant, and not waves.
QED
The Magnetic-Field Direction in a Light Wave
Suppose a wave propagates in the z-direction and has its electric field
along the x-direction.
Þ Ey = Ez = 0 and Ex = Ex(z,t)

What’s the direction of the magnetic field?

!
∂B ! ! ⎛ ∂Ez ∂Ey ∂Ex ∂Ez ∂Ey ∂Ex ⎞
Use: − = ∇ ×E = ⎜⎜ − , − , − ⎟⎟
∂t ⎝ ∂y ∂z ∂z ∂x ∂x ∂y ⎠
!
∂B ⎛ ∂Ex ⎞ B x and B z are at most
So: − = ⎜ 0, ,0 ⎟
∂t ⎝ ∂z ⎠ constants—not waves.

¶B y ¶Ex
and: - =
¶t ¶z
The magnetic field points in the y-direction, perpendicular to the
electric field (x) and the direction of propagation (z).
QED
An Electromagnetic Wave

The electric and magnetic fields are in phase. Snapshot of the

wave at one time
! !
Electric field E Magnetic field B y
x
z

The electric field, the magnetic field, and the k-vector are all
perpendicular:
! ! !
E ×B ∝ k
The Magnetic-Field Strength in a Light Wave
What’s the strength of the magnetic field By?
t

ò
¶B y ¶Ex ¶Ex ( z , t ¢)
- = So: B y ( z , t ) = B y ( z , 0) + - dt ¢
¶t ¶z ¶z
0
Take By(z,0) = 0 (it’s not a wave)

Now: { }
Ex ( z, t ¢) = Re E0 exp éëi ( kz - w t ¢) ùû
¶Ex
So: - = - Re{ikE exp éëi ( kz - w t ¢) ùû}
0
¶z
t t

ò ò
¶Ex ( z, t ¢)
-
¶z
dt ¢ = - { }
Re ikE0 exp éëi ( kz - w t ¢) ùû dt ¢
0 0 t
é ì ik ü ù
= - ê Re í E0 exp éëi ( kz - w t ¢) ùû ýú
ë î -iw þû0
 The Magnetic-Field Strength (continued)
t
From the é ì ik ü ù
previous slide: B y ( z, t ) = - ê Re í E0 exp éëi ( kz - w t ¢) ùû ýú
ë î -iw þû0
t¢ = t t¢ = 0
ì ik ü ì ik ü
= - Re í E0 exp éëi ( kz - w t ) ùû ý + Re í E0 exp éëi ( kz - w ´ 0 ) ùû ý
î -iw þ î -iw þ
Constant in time (not a
k wave), so ignore this term.
=
w
{
Re E0 exp éëi ( kz - w t ) ùû}
Ex ( z, t ) 1
w/k=c So: B y ( z , t ) = Ex ( z , t )
c
The Energy Density of a Light Wave
1
The energy density of an electric field is: U E = eE 2
2
11
The energy density of a magnetic field is: U B = B 2

2µ
1
Using B = E /c and c= , we have: B = E eµ
eµ
So:
11 11 1
UB =
2µ
B =
2

2µ
(E 2
eµ ) =
2
eE 2
= UE

Total energy density: U = U E + U B = eE 2

And the electrical and magnetic energy densities in light are equal.
Why We Neglect the Magnetic Field
! !
The force on a charge, q, is: Felectrical Fmagnetic
! ! ! ! !
where v is the
F = qE + q v × B charge velocity

Taking the ratio of

the magnitudes Fmagnetic qvB ! !
£ v × B = vB sin θ
of the two forces:
Felectrical qE ≤ vB

Fmagnetic v
Since B = E /c: £
Felectrical 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.
The Poynting Vector: S = c2 e E × B
The power per unit area in a beam. U = Energy density

Justification (but not a proof): A

Energy passing through area A in time Dt:

= U V = U A c Dt

So the energy per unit time per unit area is: c Dt

= U V / ( A Dt ) = U A c Dt / ( A Dt ) = U c = c e E 2

= c2 e E B
! ! !
And the direction E × B ∝ k is appropriate.
The Irradiance |Vector magnitude|

A light wave’s average ! ! ! 1

t+T /2
! !
power per unit area is
the irradiance.
I ( r ,t) ≡ S ( r ,t) = ∫
T t−T /2
S ( r , t ʹ) dt ʹ
Time average

Substituting
! 2 ! a light
! wave into the expression for the Poynting vector,
S = c ε E × B yields:
real amplitudes
! ! 2
! ! 2
! !
S ( r ,t) = c ε E0 × B0 cos ( k ⋅ r − ω t − θ )
The average of cos2 is 1/2:
! ! cos2(wt-a)
! 1 1/2
⇒ I ( r ,t) = S ( r ,t) =
2
! !
= c ε E0 × B0 (1/ 2)
t
The Irradiance (continued)
Since the electric and magnetic fields are perpendicular and B0 = E0 /c,
2
! ! !2
I = c ε E0 × B0
1
2 becomes: I = c ε E0
1
2

or: ! 2 because the real amplitude squared

I = c ε E0
1
2
is the same as the mag-squared
complex one, that is, |exp(ij)| = 1.
where:
! 2
! 2 E0 is a vector magnitude
E0 = E0 x E0*x + E0 y E0* y + E0 z E0*z and a complex-number
magnitude

Remember: the above formula only works when the wave is of the
form: ! ! !
! !
E r ,t = Re E0 exp ⎣i k ⋅ r − ω t ⎤⎦
( ) ⎡
( )
! !
that is, when all the fields involved have the same k ⋅ r − ω t
The Irradiance of the Sum of Two Waves
! !
If they’re both proportional to exp ⎣i( k ⋅ r − ω t)⎤⎦ , then the irradiance is:
⎡
! !*
I = cε E0 ⋅ E0 = 12 cε ⎡⎣ E0 x E0*x + E0 y E0*y + E0 z E0*z ⎤⎦
1
2

Different polarizations (say x and y): Irradiances

add.
I = 12 ce éë E0 x E0*x + E0 y E0*y ùû = I x + I y

Same polarizations (say E0 x = E1 + E2 ): Fields add.

{ }
I = 12 ce é E1 E1* + 2 Re E1 E2* + E2 E2* ù
ë û
Therefore: {
I = I1 + ce Re E1 E2* + I 2 } Note the
cross term!

The cross term yields interference!

Interference only occurs for beams with the same polarization.
The Irradiance of the Sum of Two Waves of
the Same Polarization, But Different Color
We can’t use the formula because the k’s and w’s are different.
! ! 2
! !
So we must use the Poynting vector, I ( r ,t) = S( r ,t) = c ε E × B
Identical polarizations allow us to ignore the vector magnitude.
!
I total = S( r ,t) = c 2ε ⎡⎣E1 +E2 ⎤⎦ ⎡⎣B1 +B2 ⎤⎦
I1 I2
= c 2e éë E1 B1 + E1 B2 + E2 B1 + E2 B2 ùû
Consider a cross Because we’re multiplying fields,
term E1 B2 : we can’t use complex fields here.
! ! ! !
E10 cos( k1 ⋅ r − ω1t − θ1 ) B20 cos( k2 ⋅ r − ω 2t − θ 2 )
∞
! ! ! !
∝E10 B20 ∫
−∞
cos ⎡⎣ω1t − ( k1 ⋅ r − θ1 )⎤⎦ cos ⎡⎣ω 2t − ( k2 ⋅ r − θ 2 )⎤⎦ dt
The Irradiance of the Sum of Two Waves of
the Same Polarization, But Different Color
∞
! ! ! !
E10 B20 ∫
−∞
cos ⎡⎣ω1t − ( k1 ⋅ r − θ1 )⎤⎦ cos ⎡⎣ω 2t − ( k2 ⋅ r − θ 2 )⎤⎦ dt

Waves of different frequencies

go in and out of phase. So their
product is as often positive as
negative. t
It integrates (averages) to zero.

Thus the product, E1 B2, averages to zero, as does E2 B1.

Different colors: I total = I1 + I 2 Their irradiances add.

Waves of different color (frequency) do not interfere!

Irradiance of the Sum of Two Waves
Same polarizations Perpendicular polarizations
Coherent
addition
I = I1 + I 2 +
I = I1 + I 2
Same
colors {
ce Re E1 E2 *
}
Incoherent
addition
I = I1 + I 2 I = I1 + I 2
Different
colors

Interference only occurs when the waves have the same color and
polarization.
Irradiance for Many Fields with Random Phases
!
If each field has the same w , k , and polarization, but random phase, qi:
! !
Etotal = (E1 + E2 + ...+ E N ) exp[i( k ⋅ r − ω t)] where Ei = E0 exp(iqi )

I total = I1 + I 2 + ... + I N + ce Re{E1 E2* + E1 E3* + ... + E N -1 E N* }

I1, I2, … In are the irradiances Ei Ej* have the phase factors: exp[i(qi-qj)].
of the various beamlets. When the q’s are random, this sum isn’t 0,
They’re all positive real but it’s small compared to the sum of the
numbers and they add. irradiances.

Itotal ≈ I1 + I2 + … + In All the

relative Im
phases
Re
I1+I2+…+IN
This is the essence of exp[i (qi - q j )] exp[i (q k - ql )]
incoherence.
A spherical wave is also a solution Spherical Waves
to Maxwell's equations.
Note that k and r are
not vectors here.

!
{( )
E ( r ,t) ∝ Re E0 / r exp[i(kr − ω t)] }
where k is a scalar, and
r is the radial coordinate.

Also, the directions of E and B

must still be ┴ to the propagation
direction, so the polarization
varies with angle.

A spherical wave has spherical wave-fronts.

Unlike a plane wave, whose amplitude remains constant as it

propagates, a spherical wave weakens. Its irradiance goes as 1/r2.

A light bulb is not a spherical wave, but its irradiance also goes as 1/r2.
 Argon ion laser: 1 W/cm2
Irradiance Laser pointer: 0.01 mW/cm2
vs. Intensity Sunlight at earth’s surface: 1 kW/m2
Moonlight at earth’s surface: 2 mW/m2

Irradiance (power per unit area) is ideal for describing lasers, which
emit unidirectionally. But the sun and moon, which emit
omnidirectionally, are so far away, the irradiance works well for
them.

Light bulbs and spherical waves emit omnidirectionally, so we

must describe their power per unit solid angle (in steradians),
and not per unit area. This is the Intensity.

Radiance, Radiancy, or Luminance is also used for omnidirec-

tional sources. They’re the power per unit area per unit solid angle.
Intensity Units for Spherical
Waves and Light Bulbs
The radiant power includes all wavelengths.
The luminous power includes only those
the eye can see.

The Candela (cd) is the unit of luminous intensity (power per

steradian). It’s based on a candle: 1 candela = 1/683 W/steradian.

A Lumen (lm) is the unit of luminous power for omnidirectional sources.

It’s the total power emitted into 1 steradian (not 4p steradians). 1 lumen
= (1/4p) candela steradian = 1W/(4p × 683). A 100W (electrical power)
incandescent light bulb emits only about ~2W of total luminous power
(that is, 2W/4p into 1 steradian) and so emits ~1366 lumens.

A Lux is the luminous power per unit area for omnidirectional sources.
1 Lux = 1 Lumen/m2. The lux from a light bulb decreases as 1/r2.
Dynamic Range or Contrast Ratio
When emitting or detecting light, the dynamic range or contrast
ratio is the ratio of the maximum and minimum intensities that the
emitter can emit or the detector can measure.
It’s often expressed in “bits”—the exponent to which 2 must be
raised to yield it.
Dynamic ranges of some common light emitters:
Device Dynamic Range Bits
iPhone/computer display ~1000 ~10
HDTV (LCD & LED) display 500-4000 9-12
HDTV (plasma) display ~8,000 ~13

Dynamic ranges of some common light detectors:

Device Dynamic Range Bits
iPhone camera (jpeg) 256 8
Professional camera (raw) 16,000 14
Human eye >1,000,000 >20