Transcript 15. Continuity eq and poynting vector and wave solution in free

Continuity Equation
It is about the conservation of charge.
Let charge Q is in volume V.
and the current flowing out through the boundary S is:
 J .da
S
so local conservation of charge says:
So:
dQ
d  (r , t )

d    J .da
dt V dt
S
dQ
d

d   J .da
dt V dt
S
Applying gauss divergence theorem in right hand side:
Since this is true for any volume so:
or

.J  
t
This is known as continuity equation.
Poynting’s Theorem/
energy in electromagneticWaves/
electromagnetic energy
Also known as work-energy theorem of
electrodynamics
or
Energy conservation law in electrodynamics.
Electromagnetic waves carry energy and momentum.
Energy in Electromagnetic waves can be given by
Poynting theorem (vector).
Suppose we have some charge and current configuration
at time t, it produces fields E and B.
Think in the next instant, dt, the charges move around a
bit.
According to the Lorentz force law, the work done dW
in the time dt, done by the electromagnetic forces on a
charge q is
We know
and
q   d
J  v
I
Q
 .V  .r
J  2  2  2 
 v
r
t.r
t.r
t
so the rate at which work is done on all the charges in
a volume V is
J
dW
F.dl
qE.vdt


  (  d ) E.  
dt
dt
dt

v
v
v
Evidently E • J is the work done per unit time, per unit
volume which is, the power delivered per unit volume.
using the Ampere-Maxwell law to eliminate J:
By taking scalar product of above equation with E:
E
E.    B    0 E . J   0 E .
t
After rearranging:
………(1)
From product rule:
Using Faraday’s Law:
Using this in above equation and after rearranging it:
We know,
………(1)
our 1st equation :,
Using calculated terms in our 1st equation:
1  1  2
1  2

E.J   
B  .  E  B     0
E
0  2 t
2 t

 
 
dW
 1
1 2
1
2
   E.J d      0 E  B  d   .  E  B  d
dt
t v 2 
0 
0 v
v
So energy (by applying divergence theorem to second
term):
This is Poynting's theorem.
The first integral on the right is the decrease in the
total energy stored in the fields.
The second term evidently represents the rate at which
energy is carried out of V, across its boundary surface,
by the electromagnetic fields.
Poynting's theorem says, that the work done on the
charges by the electromagnetic force is equal to the
decrease in energy stored in the field, less the energy
that flowed out through the surface.
The energy per unit time, per unit area, transported by
the fields is called the Poynting vector:
Electromagnetic Waves in free space
In free space the Maxwell’s equations are:
.E  0or.D  0
.B  0
B
 E  
t
E
D
D
 B   0 0  0 or H 
t
t
t
  0, J  0,   0,
   0 ,   0
 r  1, r  1
Taking curl of Maxwell’s 3rd equation:
B
 E  
t

    E     B 
t
Using Maxwell’s 4th equation:
  D 
 E
 E  0 
  0 0 2
t  t 
t
2
 E   .E   E
2
2

E
2
  .E    E   0 0 2
t
From equation 1st : .E  0
2

E
2
 E  0 0 2
t
Wave
equations
for E and
H in free
space.
2

H
2
Similarly we can get:  H  0 0
2
t
1 
General wave equation is:    2
v t 2
2
2
1
Comparing: 2   0 0 ; v 
v
1
1
0 0
N .meter
 9 10
4 0
coul 2
;
2
9
0  4 10 N.sec / coul
7
2
2
9 109  N .meter 2  coul 2 
v

7 
2
2 
10  coul  N .sec 
= 3108 meter/sec
Solution of wave equation in free space:
 E
1  E
2
Take equation for E:  E  0 0 2 ;  E  2 2
t
c t
The plane wave solution of this may be written in the
well known form as:
i ( k .r t )
2
2
2
E(r, t )  E0e
Here E0 is complex amplitude of electric field.
k is wave propagation vector.
2
2

k  kn 
n
n n

c
c
n is the unit vector in the direction of propagation of
electromagnetic wave.
Transverse nature of Electromagnetic Waves
Now, take Maxwell’s Ist equation: .E  0
Now
 


i ( k .r t )
.E   i  j  k  .E0e
y
z 
 x
k .r   ik x  jk y  kk z  .  ix  jy  kz   k x x  k y y  k z z
 


i ( k x x  k y y  k z z t )
.E   i  j  k  .  iE0 x  jE0 y  kE0 z  e
y
z 
 x









( k x  k y  k z t )
( k x  k y  k z t )
( k x  k y  k z t )
E0 x ei x y z

E0 y ei x y z

E0 z ei x y z
x
y
z








i ( k x x  k y y  k z z t )
i ( k x x  k y y  k z z t )

E0 x e

E0 y e
x
y

i ( k x x  k y y  k z z t )

E0 z e
z

E
 E0 x e
0z
e
i ( k x x  k y y  k z z t )
i ( k x x  k y y  k z z t )
 ik    E
 ik 
x
0y
e
i ( k x x  k y y  k z z t )
 ik 
y
z

  E0 xik x  E0 y ik y  E0 z ik z  e

i ( k x x  k y y  k z z t )


 i  E0 x k x  E0 y k y  E0 z k z  e

i ( k .r t )
 ik .E0 e
i ( k .r t )

  ik .E
 
.E  i k .E  0  k .E  0
Similarly we can find: k .H  0
This means both E and H are perpendicular to the
direction of motion.
Similarly later two Maxwell’s equations with same
solution can give
k .E  0 H
k .H   0E
Which clearly shows that E, H, k are mutually
perpendicular.
This establishes the transverse nature of light.
Note: Treatment for the electromagnetic radiation in
non-conducting/dielectric medium will be similar to
the free space except the values of permittivity and
permeability will now have certain values for the
particular medium.