  [  f ]  (  f )   2 f
Proof: Look first at the LHS of the above formula:
 f z f y f x f z f y f x 
  [  f ]    

,

,



y

z

z

x

x

y


Taking the 2nd   yields:
x-component:
 2 f y 2 fx
 

 xy y 2

y-component:
 2 fx 2 fz
 
 2
 xz x
z-component:
 2 fz 2 f y   2 f y 2 fx 
 



 zy z 2   x 2 xy 

 

3/24/2016
  2 fx 2 fz 
   2 


z

x

z

 
2
  2 fz  f y 


  
2
zy 
  y
  [  f ]  (  f )   2 f
Proof (cont’d):
Now, look at the RHS:
(  f )   2 f
f x f y f z
(  f )   (


)
x y z
2
2
2 fx  f y 2 fz
2 fx  f y 2 fz
( 2 

,
 2 
,
x
xy xz xy y
yz
2
2 fx  f y 2 fz

 2 )
xz zy z
2
2
2
2
2
2

f

f

fy

f

f

f
y
y
2
x
x
x
 f  (  2  2  2 ,  2  2  2 ,
x
y
z
x
y
z
3/24/2016
2 fz 2 fz 2 fz
 2  2  2 )
x
y
z
(I) Gauss’ Law
In CGS unit, integrating perpendicular component of E-field over
closed surface  gives total charge within surface.

 
 E (r )  dS  qtot / 
 
E   /
Consider tot(r) = total charge density

qtot    tot (r ) dV
V

 
 
 E (r )  dS     E dV
V
Simple check from a “known” formula
Coulomb force field:
3/24/2016
E
q
4r
2
r
(II) Gauss’ Law for B field
Following Gauss’ Law
  

 Br  ds  qm  0     Bdv
s
v
 
 B  0
There are no magnetic charges or “monopoles”
3/24/2016
(III) Faraday Law
The interplay between B and E
You may know steady current generates magnetic field (or flux B)
I
B
B
I
But also: increasing B generates counter-current because of a ‘rotating’ E field
Or in formula form :
3/24/2016
 
  
  
C E  dl   t S B  dS     E  dS
S

 
B
 E  
t
(IV) Ampere Law
For an entirely free current (quite rare, but possible - see sketch) the
magnetic flux is related to the current through



B

d
l


I




B

d
s
total


c
s
Note last part is after applying Stoke’s theorem.
Defining the total current density Jtot (current per unit area):
I free


  J free  ds
s
We have


  B  J tot
Is this enough?
No. Similar to the way a material influences E due to induced polarization,
materials can also affect B due to induced magnetization current
3/24/2016
(IV) Ampere Law – Continued
The time varying part is missing for the E field
E
dt
This leads to the final result


D
  B  J free  
dt
Displacement current


D  E

E
  B  
dt
3/24/2016
Maxwell’s Equations in Vacuum
E   /
B  0
B
 E  
t
E
  B  
t
where E is the electric field, B is the magnetic field,
 is the charge density,  is the permittivity, and  is
the permeability of the medium.
2

E
2
 E   2  0
t
3/24/2016
2

B
2
 B   2  0
dt
Derivation of the Wave Equation
Take  of:
B
 E  
t
  [  E ]    [
B
]
t
E   /
B  0
B
 E  
t
E
  B  
t
Change the order of differentiation on the RHS:

  [  E ]   [  B]
t

  [  E ]   [  B] 
t
3/24/2016

E
  [  E ]   [ 
]
t
t
2 E
  [  E ]    2
t
assuming that  and 
are constant in time.
The Wave Equations in Vacuum
2 E
  [  E ]    2
t
becomes:
2

E
2
(  E )   E    2
t
If we now assume zero charge density:  = 0, then
E
 0
and we’re left with the Wave Equation in vacuum
3/24/2016
2

E
2
 E   2
t
The Wave Equations in Vacuum
 E
 E   2  0
t
2E 2E 2E
2E

 2   2  0
2
2
x
y
z
t
2
2

i (t  k r  )
E (r , t )  E 0e
3/24/2016
This is really just
three independent
wave equations,
one each for the
x-, y-, and zcomponents of E.
Light is an EM wave

 1  E
2
 E 2 2 0
c dt
2
Why is light a transverse wave?
The strength of E field vs B field?
Why do they perpendicular to each other?
3/24/2016
Light is an EM wave

 1  E
2
 E 2 2 0
c dt
2
E   /
B  0
Why is light a transverse wave?
The strength of E field vs B field?
Why do they perpendicular to each other?
3/24/2016
B
 E  
t
E
  B  
t
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:
E y
Ez By Bz



0
y
z
y
z
Now, in a charge-free medium,
E  0
B  0
that is,
Ex E y Ez


0
x
y
z
Substituting, we have:
3/24/2016
Ex
0
x
Bx By Bz


0
x
y
z
and
Bx
0
x
Correlation between E and B
Suppose a wave propagates in the x-direction and has its electric field
along the y-direction, how to determine the magnetic field?
Ex = Ez= 0, and Ey = Ey(x,t).
What is the direction of the magnetic field?
Use:
 Ez E y Ex Ez E y Ex 
B

  E  

,

,


t

y

z

z

x

x

y


So:
In other words:
E y 
B 

  0, 0,

t 
x 
Bz E y


t
x
And the magnetic field points in the z-direction.
3/24/2016
The Magnetic Field in a Light 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 )  E0ei ( kxt )
t
We can integrate:
Bz ( x, t )  Bz ( x, 0) 
Take Bz(x,0) = 0
ik
Bz ( x, t )  
E0 exp i (kx   t ) 
i
1
Bz ( x, t )  E y ( x, t )
c

0
E y
x
dt
  E (r , t )  i k  E (r , t )
  E (r , t )  i k  E (r , t )
 E (r , t )
 i E (r , t )
t
3/24/2016The electric and magnetic fields are in phase with equal
amplitude.
Why we neglect the magnetic field?
The force on a charge, q, is:
Felectrical Fmagnetic
F  qE  q v  B
so:
Fmagnetic
Felectrical
Since B = E/c:
Fmagnetic
Felectrical
where v is the
charge velocity
qvB

qE
v

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.
3/24/2016
Metamaterials: new waves?
3/24/2016
Light is an Electromagnetic Wave
Electric (E) and magnetic (B) fields are in phase.
The electric field, the magnetic field, and the propagation
direction are all perpendicular.

 1  E
2
 E 2 2 0
c dt
2
3/24/2016
EB  k
The Energy Density of a Light Wave
1 2
The energy density of an electric field is: U E   E
2
11 2
The energy density of a magnetic field is: U B 
B
2
1
c

Using B = E/c, and
, which together imply that B  E 

we have:
11 2
1 2
UB 
E     E  U E

2
2
Total energy density:
U  UE UB   E2
So the electrical and magnetic energy densities in light are equal.
3/24/2016
The Poynting Vector for the Energy Transfer
Rate S  c 2 E  B  k
The power per unit area in a beam.
Justification (but not a proof):
Energy passing through area A in time Dt:
=
U V
=
U A c Dt
So the energy per unit time per unit area:
=
U V / ( A Dt ) = U A c Dt / ( A Dt ) = U c = c  E2
=
c2  E B
And the direction E  B  k is reasonable.
3/24/2016
The Light Intensity (Irradiance)
A light wave’s average power
per unit area is the “irradiance.”
S (r , t )
T
1

T
t T / 2

S (r , t ') dt '
t T / 2
Substituting a light wave into the expression for the Poynting vector,
S  c 2  E  B, yields:
real amplitudes
S (r , t )  c 2  E0  B0 cos 2 (k  r   t   )
 I (r , t )  S (r , t )
T

 c  E0  B0 (1/ 2)
2
3/24/2016
The Light Intensity (Irradiance)
Since the electric and magnetic fields are perpendicular and B0 = E0 / c,
I

1
2
c 2  E0  B0
becomes:
I 
1 c  E0
2
~
2
where
2
E0  E0 x E  E0 y E  E0 z E
3/24/2016
*
0x
*
0y
*
0z
Light is an Electromagnetic Wave
1. The electric field, the magnetic field, and the k-vector are all perpendicular:
EB  k
2. The electric and magnetic fields are in phase.
3. The light density an EM field:
So the electrical and magnetic energy densities in light are equal.
3/24/2016