Transcript Notes 6 6340 Poynting`s Theoremx

ECE 6340
Intermediate EM Waves
Fall 2016
Prof. David R. Jackson
Dept. of ECE
Notes 6
1
Power Dissipated by Current
Work given to a collection of
electric charges moving in an
electric field:
W   q E   
This could be
conduction current or
impressed current.
   v S  E   
Note: The magnetic field never does any work:
v
E
J  v v
S
  v t
q  v  B     0
q  v  S
Power dissipated per unit volume:
  
W
Pd 
  v
  E   v v   E  J  E
S  t 
t 
Note: We assume no increase in kinetic energy, so the power goes to heat.
2
Power Dissipated by Current (cont.)
Power dissipated per unit volume for ohmic current (we assume here
a simple linear media that obeys Ohm’s law):
Pd  J c  E   E   E   E  E    E  E2
2
Note:
E  E  E E
2
2
Hence, we have
Pd   E 2
3
Power Dissipated by Current (cont.)
Return to the general result:
Pd  J  E
This is the power dissipated per unit volume.
From this we can also write the power generated per unit volume due
to an impressed source current:
Ps   J  E
i
4
Poynting Theorem: Time-Domain
B
 E  M 
t
D
 H  J 
t
From these we obtain
B
H    E   M  H  H 
t
D
E    H   J E  E 
t
Subtract, and use the following vector identity:
H    E   E   H    E  H

5
Poynting Theorem: Time-Domain (cont.)
We then have
B
D
  E  H    M  H  J  E  H 
E
t
t
J  J  E
i
Now let
M M
i
so that
D
B
  E  H    J  E   E  M  H  E 
H 
t
t
i
2
i
6
Poynting Theorem: Time-Domain (cont.)
Next, use
D
 E 
E
  E 

t

t


and
Assume:
D  E
(simple linear media)
E2 
 E 
 E  E   2E  

t
t

t


Hence
 1 E2 
 D 
 E 
E
   E 



t

t
2

t






And similarly,
 1 H 2 
B
H 
 

t
 2 t 
7
Poynting Theorem: Time-Domain (cont.)
S EH
Define
We then have
 1 2 1
2
  S   J E   E  M  H   E   H 
t  2
2

i
i
2
Next, integrate throughout V and use the divergence theorem:


2
ˆ
S

n
dS


J

E

M

H
dV


E


 dV 
S
V
i
i
V
 1 2 1
2

E


H

 dV

t V  2
2

Note: We are assuming the volume to be stationary (not moving) here.
8
Poynting Theorem: Time-Domain (cont.)


2
ˆ
S

n
dS


J

E

M

H
dV


E


 dV 
S
i
i
V
V
 1 2 1
2

E


H

 dV

t V  2
2

Interpretation:
1
We    E 2 dV  stored electric energy
2
V
1
Wm    H 2 dV  stored magnetic energy
2
V
Pd    E 2 dV  dissipated power
V


Ps    J  E  M  H dV  source power
i
i
V
 Pf 
 S  nˆ dS  power flowing out of S
S
(see next figure)
9
Poynting Theorem: Time-Domain (cont.)


2
ˆ
S

n
dS


J

E

M

H
dV


E


 dV 
S
V
i
i
V
 1 2 1
2

E


H

 dV

t V  2
2


Pf  Ps  Pd  We  Wm 
t
or

Ps  Pd  Pf  We  Wm 
t
10
Poynting Theorem: Time-Domain (cont.)

Ps  Pd  Pf  We  Wm 
t
Power flow out
of surface
E
, , 
H
source
11
Poynting Theorem: Note on Interpretation
S  E H
Does the Poynting vector really represent local power flow?
 1 2 1

  S   J E   E  M  H   E   H 2 
t  2
2

i
Consider:
2
i
S   S   A
A  Arbitrary vector function
Note that
S  S
This “new Poynting vector” is equally
valid! They both give same TOTAL
power flowing out of the volume, but
different local power flow.
12
Note on Interpretation (cont.)
Another “dilemma”:
A static point charge is sitting next to a bar magnet.
S  E H  0
N
Is there really power flowing in space?
q
Note: There certainly must be zero net power
out of any closed surface:
S
Ps  Pd  Pf 

We  Wm 
t
13
Note on Interpretation (cont.)
Bottom line: We always get the correct result if we assume
that the Poynting vector represents local power flow.
Because…
In a practical measurement, all we can ever really measure is
the power flowing through a closed surface.
Comment:
At high frequency, where the power flow can be visualized as "photons"
moving in space, it becomes more plausible to think of local power flow. In
such situations, the Poynting vector has always given the correct result that
agrees with measurements.
14
Complex Poynting Theorem
Frequency domain:   E   M i  j   H
  H  J  j  c E
Generalized linear media:
 may be complex.
i
Hence
c    j
H    E   M  H  j   H
i
*

E   H
*

2
*
 E  J  j   c* E
i*


2
Subtract and use the following vector identity:
  A  B   B   A  A   B 

 E H
Hence

 E H
*

*
H
*

   E   E    H
*

  M  H  E  J  j   H  j   c* E
i
*
i*
2
2
15
Complex Poynting Theorem (cont.)
Define
1
*
S  EH
2
Note:
(complex Poynting vector)




1
*
Re S  Re E  H  E  H  S
2
Then

1
1
2
i*
i
*
*
  S   E  J  M  H  j  c E   H
2
2
2

Next use
 c   c  j c   c*   c  j c
collect real and imaginary
     j   Next,
parts in the last term on the RHS.
16
Complex Poynting Theorem (cont.)
S  





1
1
2
i*
i
*
E  J  M  H  j   c E    H
2
2
2


1
2
   c E    H
2
2

or

1
2
1  2 1  2 1
i*
i
*
  S   E  J  M  H  2 j    c E   H     c E    H
2
4
4
 2
2

Next, integrate over a volume V and apply the divergence theorem:
 S  nˆ dS   
S
V


1
1  2 1  2
i*
i
*
E  J  M  H dV  2 j     c E   H  dV
2
4
4

V 
1
2
2
1
    c E    H  dV
2
2

V 
17
Complex Poynting Theorem (cont.)
Final form of complex Poynting theorem:


1
i*
i
*
V  2 E  J  M  H dV 
 S  nˆ dS
S
1  2 1  2
2 j     H   c E  dV
4
4

V
 1  2 1  2 
    c E   H  dV
2
2

V
18
Complex Poynting Theorem (cont.)
Interpretation of Ps :


1
i*
i
*
Ps    E  J  M  H dV
2
V

Re Ps   E  J
i
 M  H
i
 dV
 Ps
V
We therefore identify that
Ps  complex source power  VA
Re Ps  real power (watts) from the sources  W 
Im Ps  imaginary power (vars) from the sources  VAR 
19
Complex Poynting Theorem (cont.)
Interpretation of Pf :

Pf 
S
Re Pf 
1
*
E

H
 nˆ dS


2
  E  H   nˆ dS
 Pf
S
We therefore identify that
Pf  complex power flowing out of S
Re Pf  real power (watts) flowing out of S  W 
Im Pf  imaginary power (vars) flowing out of S  VAR 
20
Complex Poynting Theorem (cont.)
Interpretation of energy terms:


1
1 1
2


 c E   c  E  E * 
4
2 2

1 1

Note: The real-part operator may be
 c  Re E  E * 
added here since it has no effect.
2 2

Note: We know this result represents
1
1
2
stored energy for simple linear media
  c E  E   c E
(  = ), so we assume it is true for
2
2


c
generalized linear media.
1
2
Hence   c E dV 
4
V
1
2
Similarly,    H dV 
4
V
1
2
V 2  c E dV  We
1
V 2   H
2
dV  Wm
Note:
The formulas for
stored energy can
be improved for
dispersive media
(discussed later).
21
Complex Poynting Theorem (cont.)
Interpretation of dissipation terms:




1
2
1
1
* 
* 



 c E   c  E  E    c  Re E  E 
2
2

2

  c E  E   c E
Recall:
2
 
c    j  
 
 is real for simple linear media
For simple linear media:
 

 c      
 
Hence
1
2
2
e


E
dV


E
dV

P
d
V 2 c
V
Note:
This formula gives the correct
time-average power dissipated
due to electric losses for simple
linear media.
We assume the same
interpretation holds for
generalized linear media
( is complex).
22
Complex Poynting Theorem (cont.)
Interpretation of dissipation terms (cont.):
Similarly,
1
2
m


H
dV

P
d
V 2
This is the time-average power dissipated due to magnetic losses.
Note: There is no magnetic conductivity, and hence
no magnetic conduction loss, but there can be
magnetic polarization loss.
23
Complex Poynting Theorem (cont.)
Summary of Final Form


1
i*
i
*

E

J

M

H
dV 
V 2
 S  nˆ dS
S
 1  2 1  2 
    c E   H  dV
2
2

V
1  2 1  2
2 j     H   c E  dV
4
4

V
Ps  Pf  Pd  j  2    Wm  We

24
Complex Poynting Theorem (cont.)
Ps  Pf  Pd  j  2    Wm  We

Pabs
We can write this as Ps  Pf  Pabs
where we have defined a complex power absorbed Pabs :
Re  Pabs   Pd
Im  Pabs    2    Wm  We

25
Complex Poynting Theorem (cont.)
Ps  Pf  Pabs
This is a conservation
statement for complex power.
Im  Pabs    2    Wm  We
Re  Pabs   Pd

VARs
Watts
Pf
E
Complex power
flow out of surface
VARS consumed
Power (watts)
consumed
H
 c,  
Source
26
Example
L
I
+ V -
Ideal (lossless) inductor
Denote:
Pabs  complex power absorbed  Wabs  jQabs
Calculate Pabs using circuit theory, and verify that the result is
consistent with the complex Poynting theorem.
Note: Wabs = 0 (lossless element)
27
Example (cont.)
L
I
+ V -
1 * 1
Pabs  V I   Z I  I *
2
2
1
  j L I  I*
2
1
2
 j L I
2
Ideal inductor
Note:
Pabs  jQabs
(lossless element)
28
Example (cont.)
Qabs
1
2
 Im Pabs   L I
2
  1 * 
  L  Re  I I  

 2
 
L i
2
 2   Wm
 1 2 
 2  L i 
 2


29
Example (cont.)
Since there is no stored electric energy in the inductor, we can write
Qabs  2   Wm  We

Hence, the circuit-theory result is consistent with
the complex Poynting theorem.
Note: The inductor absorbs positive VARS.
30
Example
We use the complex
Poynting theorem to
examine the input
impedance of an antenna.
Pf
Iin
V0
Js
+
-
S
Antenna
Model:
Iin
V0
+
-
Zin=Rin+j Xin
1
1
Pin  V0 I in*   Z in I in  I in*
2
2
1
2
 Z in I in
2
31
Example (cont.)
so
Z in 
2 Pin
I in
Real part:
2
Rin 
2 Re  Pin 
I in
2
Re  Pin   Pin

rad
 P
(no losses in vacuum surrounding antenna)

rad
Note:
The far-field Poynting vector is much easier to
calculate than the near-field Poynting vector.
 Re P
Hence
Rin 
2
I in
2
 Re  S  nˆ  dS
S
32
Example (cont.)
Rin 
2
2
I in
 Re  S  nˆ  dS
S
In the far field (r )
Im S  0
(This follows from plane-wave
properties in the far field.)
Hence
Rin 
2
I in
2
  S  nˆ  dS
S
33
Example (cont.)
Imaginary part:
X in 
2
I in
2
Im  Pin 
where

Im  Pin   Im  Prad
  2  Wm  We 
Hence
X in 
2
I in
2
 Im S  nˆ dS
S
The RHS is zero!
Wm , We
34
Example (cont.)
We can say that
X in 
2
I in
2
 2  W
m
 We 

Hence
4
1
2
2
1
X in 
0 H   0 E  dV
2 
4

I in V  4
However, it would be very difficult to calculate the input impedance
using this formula!
35
Dispersive Material
The permittivity and permeability are now functions of frequency:
    
    
(A lossy dispersive media is assumed here. It is
assumed that the fields are defined over a fairly
narrow band of frequencies.)
The formulas for stored electric and magnetic energy now become:
1 2 
We  E
  
4

1 2 
Wm  H
 
4

Note:
The stored energy should always
be positive, even if the
permittivity or permeability
become negative.
Reference:
J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 263).
36
Momentum Density Vector
The electromagnetic field has a momentum density
(momentum per volume):
p  D B
In free space:
p   0 0  E  H
or
1
p 2 S
c
[(kg m/s)/m3] = [kg/(s m2)]
Reference:
J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 262).
37
Momentum Density Vector (cont.)
Photon:
From physics, we have a relation between the energy E and the
momentum p of a single photon.
E  pc
E  hf
h  6.626068 1034 [J s]
(Planck’s constant)
Photons moving:
Calculation of power flow:
Energy E  A  c t   N p
Sz 

A t
A t
 Ec N p
 pc N p   pN p  c
2
 pz c 2
Hence
2
  ct
v
A
S  Sz zˆ
Np photons per unit volume
1
pz  2 Sz
c
This is consistent with the
previous momentum formula.
38
Momentum Density Vector (cont.)
Example:
Find the time-average force on a 1 [m2] mirror illuminated by
normally incident sunlight, having a power density of 1 [kW/m2].
Note: the fields vary
sinusoidally, but the
frequency is arbitrary.
inc
z
p
Fz 
2

A = 1 m2
z
1
1
inc
 2 Sz  2 (1000)
c
c
pzinc
  Ac t   2
t

inc
z
p

2000
1

 Ac   2  2 (1000)  1 c  
c
c

6
Fz  6.67110 [N]
39
Solar Sail
Sunjammer Project
NASA has awarded $20 million to L’Garde, Inc. (Tustin, CA), a maker of “inflatable space structures,” to
develop a solar sail, which will rely on the pressure of sunlight to move through space when it takes its
first flight as soon as 2014.
A smaller version of L'Garde's solar sail unfurled in a vacuum chamber in Ohio in 2005. This one was about 3,400 square
feet, a quarter the size of the sail the company plans to loft into space as soon as 2014. Photo courtesy NASA and L'Garde.
http://www.lgarde.com
40
Solar Sail (cont.)
http://www.lgarde.com
41
Maxwell Stress Tensor
This gives us the stress (vector force per unit area) on an object,
from knowledge of the fields on the surface of the object.
References:
D. J. Griffiths, Introduction to Electrodynamics, Prentice-Hall, 1989.
J. D. Jackson, Classical Electrodynamics, Wiley, 1998.
J. Schwinger, L. L. DeRaad, Jr., K. A. Milton, and W.-Y. Tsai, Classical
Electrodynamics, Perseus, 1998.
42
Maxwell Stress Tensor
Txx

T  Tyx
Tzx

Txy
Tyy
Tzy
Txz 

Tyz 
Tzz 
1
2
1
Tij   0 EE
0 H
i j  0 H i H j   ij   0 E 
2
2
2



i, j  x, y, z
(for vacuum)
1, i  j
 ij  
0, i  j
(Kronecker delta)
43
Maxwell Stress Tensor (cont.)
Momentum equation:
d
S T  nˆ dS  F  dt V p dV
Total flow rate
of momentum
given to object
from EM field
Total force on object (rate
of change of mechanical
momentum)
Rate of change of
electromagnetic
momentum inside of region
F
p
1
E H
2
c
44
Maxwell Stress Tensor (cont.)
d
S T  nˆ dS  F  dt V p dV
In many practical cases the time-average of the last term (the rate of
change of electromagnetic momentum inside of region) is zero:
 No fields inside the body
 Sinusoidal steady-state fields
In this case we have
F 

Example :
d
sin 2 t    2 sin t  cos t    sin  2t 

dt
d
sin 2 t     sin  2t   0

dt
T  nˆ dS
S
The Maxwell stress tensor (matrix) is then interpreted as the stress
(vector force per unit area) on the surface of the body.
45
Maxwell Stress Tensor (cont.)
Example:
Find the time-average force on a 1 m2 mirror illuminated by
normally incident sunlight, having a power density of 1 [kW/m2].
x
nˆ   zˆ
TEMz wave
A = 1 m2
S
Fz  zˆ  F
Txx

F  T  nˆ  Tyx
 Tzx
Txy
Tyy
Tzy
z
Note:
The fields are assumed to be zero
on the back side of the mirror, and
no fields are inside the mirror.
Txz   0   Txz 

Tyz    0    Tyz 
  

Tzz   1  Tzz 
Fz  Tzz
46
Maxwell Stress Tensor (cont.)
x
nˆ   zˆ
TEMz wave
A = 1 m2
z
1
2
1
Tzz   0 EE


H
H


E

0 H
z z
0
z
z
 0
2
2
1
2
2
1
    0 E  0 H 
2
2
front side
Assume: E  Ex xˆ , H  H y yˆ
2



Exinc / H yinc  0
47
Maxwell Stress Tensor (cont.)
1
1

2
Fz    0Ex  0 H y2 
2
2
front side
1
 0 H y2
2
1
0 H y2
2
1 1

 0  Re  H y H *y  
2 2

Fz 
 
2 
1 1
 0  Re H y 
2 2

2
1
 0 H y
4
(PEC mirror)
We then have:
2
1
Fz  0 H y
4
1
inc 2
 0 2 H y
4
(The tangential magnetic field
doubles at the shorting plate.)
48
Maxwell Stress Tensor (cont.)
1
inc 2
inc 2
Fz  0 2 H y  0 H y
4
Next, use
S zinc  Exinc H yinc 
Fz  0 H
1 inc inc* 1 inc 2
E x H y  H y 0
2
2
E
inc 2
y

 0 2 S zinc / 0
inc
x
/ H yinc  0 

 4  107  2 1000 / 376.7303
so
6
Fz  6.67110 [N]
49
Force on Dielectric Object
Here we calculate the electrostatic force on a
dielectric object in an electric field using dipole moments.
This is an alternative to using the Maxwell stress tensor.
Consider first the force on a single electrostatic dipole:

F  q Er  Er 

q
r  r   r 
r
p  q r
rc
For small dipoles:

F  q  r E  r c 
or

r
q
E
F  p E  r c 
Note that we have a gradient of a vector here. This is called a “dyad.”
50
Force on Dielectric Object (cont.)
Dyadic representation:
E  xˆ



ˆ x  yE
ˆ y  zE
ˆ x  yE
ˆ y  zE
ˆ x  yE
ˆ y  zE
ˆ z  yˆ
ˆ z  zˆ
ˆ z
xE
xE
xE
x
y
z






or
E

E
E
E   xˆ xˆ x  xˆ yˆ y  xˆ zˆ z
x
x
x

Ey
 
Ex
E
ˆ
ˆ
ˆ
ˆ

y
x

y
y
 yˆ zˆ z
 
y
y
y
 
E y
 
Ex
Ez 
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

z
x

z
y

z
z
 


z

z

z
 

Hence we have:


 r E  xˆ  x  yˆ  y  zˆ  z E 
E y
 E
E
 x  xˆ x  yˆ
 zˆ z
x
x
 x
E y

 Ex
Ez
ˆ
ˆ
ˆ


x

y

z

y
y
y

 y
E y

 Ex
Ez 
ˆ
ˆ
ˆ


x

y

z


z

z

z

z



51
Force on Dielectric Object (cont.)
For a dielectric object:
P   0 e E
   0 r
 r  1  e 
V
F  p E  r c   dF  r   P  r  E  r  dV
F   P  r  E  r  dV
V
Since the body does not exert a force on itself, we can use the external electric field E0.
F   P  r  E0  r  dV
V
52
Force on Dielectric Object (cont.)
We then have:
F   0  e  E  r  E0  r  dV
V
We can also write:
F   0 e  E  r  E  r  dV
or
V
1
F   0 e    E  r   E  r   dV
2
V
or


2
1
F   0 e   E  r  dV
2
V
53
Force on Dielectric Object (cont.)
A non-uniform electrical field will generate a net attractive force on a
neutral piece of matter. The force is directed toward the region of
higher field strength.
http://electrogravityphysics.com/html/sec_4.html
54
Force on Magnetic Object
We have (derivation omitted) the magnetostatic force as:
F   M  r  B0  r  dV
V
Recall: M 
m H
Hence, we have
F   m  H  r  B0  r  dV
V
or
or
m
F
B  r  B0  r  dV

 V

  0  r
V

2
m 1
F
 B  r  dV

 V2
55
Foster's Theorem
Consider a lossless system with a port that leads into it:
Zin  jX in
Lossless system
(source free)
Sp
c     
  
PEC enclosure
Coaxial port
z
Foster’s theorem:
dX in
0
d
R. E. Collin, Field Theory of Guided Waves, IEEE Press, Piscataway, NJ, 1991.
56
Foster's Theorem (cont.)
The same holds for the input susceptance:
Z in 
jX in 
X in  
1
Yin
1
jBin
1
Bin
dX in
1 dBin
 2
d
Bin d 
dBin
0
d
57
Foster's Theorem (cont.)
Start with the following vector identity:
  A  B   B   A  A   B 
Hence we have (applying twice, for two different choices of the (vectors)
*
*
*


H   H 
H 
  E 
 
    E   E    








 



*
*
  E*


E   E 
 
 H   H 
 
    H 
    
 


Add these last two equations together:
58
Foster's Theorem (cont.)
*
*
*

  H * 

H
E
H 
 E 

 H  
    E   E    










 



*
*

E   E 
H    
 
    H 
    

Using Maxwell's equations for a source-free region,
*
*
*

  H * 

H
E
H 
  E 

 H  
    j H   E    










 



*
*

E   E 
H    
 
   j E 
    

59
Foster's Theorem (cont.)
*
*
*

  H * 

H
E
H 
  E 

 H  
    j H   E    










 



*
*

E   E 
H    
 
   j E 
    

Using Maxwell's equations again, and the chain rule, we have:
H



E
*
*
*


   H    j E    j   E  j
 



*
*
E



H
*
*


   E     j H   j   H *  j
 



*
*
We then have
60
Foster's Theorem (cont.)
*
*
*

  H * 

H
E

E 
*
 E 

 H  
  E  j
    j H   E    j












 



*
*
 
H   E 
*
H   j
  H  j
 
   j E 
    
 
cancels
Simplifying, we have
*
*


H
E


 
*
  E 

 H    E  j
  E   H   j   H * 


 

 



or
*
*


H
E
  E 

H  




 
2
2 
 
j 
  E      H  
  

 
61
Foster's Theorem (cont.)
*
*


H
E
  E 

H  




 

2
2 
j 
  E      H  
  

 
Applying the divergence theorem,
*
*


H
E
S  E      H   nˆ dS  j 4  We  Wm 
Hypothesis: the total stored energy must be positive.
Therefore,
*
*


H
E
Im   E 

 H   nˆ dS  0


S 

62
Foster's Theorem (cont.)
*
*


H
E
Im   E 

 H   nˆ dS  0


S 

The tangential electric field is only nonzero at the port. Hence we have:
*
*


H
E
Im   E 

 H   nˆ dS  0


Sp 

Assume that the electric field (voltage) at the port is fixed (not changing
with frequency).
Then we have
E
0

*
Zin  jX in
Lossless system
(source free)
c     
  
Sp
Coaxial port
z
63
Foster's Theorem (cont.)
*

H 
Im   E 
  nˆ dS  0
 
Sp 
Then
Zin  jX in
At the coaxial port:
Lossless system
(source free)
E  ˆ E
ˆ  
 c   n
  
H  ˆ H
Hence:
or
Sp
zˆ
Coaxial port
z
 H * 
Im   E
 dS  0

 
Sp 

Im   E H *  dS  0

Sp
(since the electric field is
fixed and not changing
with frequency)
64
Foster's Theorem (cont.)

Im   E H *  dS  0

Sp

Im  2 Pz   0


Im VI z*   0


2
Im   V Yin*   0



Zin  jX in
Lossless system
(source free)
c     
  
Sp
Coaxial port
z
 I z  V Yin 

Im Yin*   0

65
Foster's Theorem (cont.)

Im Yin*   0


Im Yin   0


Bin  0


X in  0

66
Foster's Theorem (cont.)
Example: Short-circuited transmission line
Zin
Z0
ZL = 0
l
Zin  jZ0 tan   l 
   LC
Xin
Inductive
π/2
3π/2
5π/2
l
Capacitive
67