notes 6 6340 Poynting Theorem

Report
ECE 6340
Intermediate EM Waves
Fall 2013
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   
E
This could be
conduction current or
impressed current.
   vS  E   
Note: The magnetic field never does any work:
q  v  B
dr
v
J  vv
S

0
 v t
q   v  S
Power dissipated per unit volume:

 
Pd 
  v
 E   v v  E  J E
S  t
t 

W
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
Note:
E E
2
2
Pd   E
2
 E
2
 E E
2
From this we can also write the power generated per unit volume due
to an impressed source current:
i
Ps   J  E
3
Poynting Theorem: Time-Domain
 E  M 
H  J 
B
t
D
t
From these we obtain
H   E    M  H  H 
E   H

J E  E 
B
t
D
t
Subtract, and use the following vector identity:
H   E   E   H
    E  H 
4
Poynting Theorem: Time-Domain (cont.)
We then have
  E  H
  M
 H  J E  H 
B
t
E
D
t
J  J  E
i
Now let
M  M
i
so that
  E  H
 J
i
E   E  M
2
i
H E
D
t
H 
B
t
5
Poynting Theorem: Time-Domain (cont.)
Next, use
D
E 

E
  E 

t
t 

and
E
D  E
(simple linear media)

 E 

 E  E   2E   
t
t
 t 
2
 1 E 
E 
 D 

E 
   E 


t 
2

t
 t 



2
Hence
And similarly,
 1 H
H 
 
t
 2 t
B
2



6
Poynting Theorem: Time-Domain (cont.)
S EH
Define
We then have
  S   J E   E  M
i
2
i
 1
1
2
H 
E   H

t  2
2
2



Next, integrate throughout V and use the divergence theorem:

S
S  nˆ dS 

V
i
i
 J E  M  H

dV    E dV 
2
V

1
1
2

E

H


t V  2
2
2

 dV

Note: We are assuming the volume to be stationary (not moving) here.
7
Poynting Theorem: Time-Domain (cont.)

S
S  nˆ dS 

i
i
 J E  M  H

dV    E dV 
2
V
V

1
1
2

E

H


t V  2
2
2

 dV

Interpretation:
We 
1
 E dV  sto re d e le ctric e n e rg y
1
H
2
2
V
Wm 
2
2
dV  sto re d m a g n e tic e n e rg y
V
Pd 
  E dV  d issip a te d p o w e r
2
V
Ps 

i
 J E  M
i
H
 dV
 so u rce p o w e r
V
 Pf 
S
 nˆ dS  p o w e r flo w in g o u t o f S
S
(see next figure)
8
Poynting Theorem: Time-Domain (cont.)

S
S  nˆ dS 

V
i
i
 J E  M  H

dV    E dV 
2
V
P f  Ps  Pd 

t

1
1
2

E

H


t V  2
2
2

 dV

 We  Wm 
or
Ps  Pd  Pf 

t
 We  Wm 
9
Poynting Theorem: Time-Domain (cont.)
Ps  Pd  Pf 

t
 We  Wm 
Power flow out
of surface
E
, , 
H
source
10
Poynting Theorem: Note on Interpretation
S EH
Does the Poynting vector really represent local power flow?
  S   J E   E  M
i
Consider:
2
i
2



S  S A
A 
Note that
 1
1
2
H 
E   H

t  2
2
A rb itra ry v e cto r fu n ctio n
 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.
11
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 

t
 We  Wm 
12
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.
13
Complex Poynting Theorem
  E  M  j  H
i
Frequency domain:
  H  J  j c E
i
Hence
(generalized linear media)
H   E   M  H  j   H
i
*

*
E H

 EJ
2
*
i*
 j c E
*
2
Subtract and use the following vector identity:
   A  B   B    A   A    B 

 EH
*


 H    E   E    H
*
*

Hence

 EH
*

i
*
 M  H  E  J
i*
 j  H
2
 j c E
*
2
14
Complex Poynting Theorem (cont.)
S 
Define
1
EH
*
(complex Poynting vector)
2
Note: R e S 
1

Re E  H
2
*

EH
 S
Then
 S  
EJ

2
1
i*
i
 M H
*
 2
1

j 
*
c
E
2
 H
2

Next use
 c   c  j c 
     j  
   c  j c
*
c
Next, collect real and imaginary
parts in the last term on the RHS.
15
Complex Poynting Theorem (cont.)
 S  

2
EJ

EJ
1
i*
i
 M H

*

1
2

j   c E
2
  H
2


1
2

  c E
2
   H
2

or
 S  
1
2
i*
i
 M H
*

1
 2 j    c E
4
2

1
4
 H
2

 1
    c E
 2
2
   H
2

Next, integrate over a volume V and apply the divergence theorem:

S
S  nˆ dS 

V
1
2

E J
1
     c E
2
V 
i*
2
i
 M H

1
2
*

   H
1
dV  2 j     c E
4
V 
2
2

1
4
 H
2

 dV


 dV

16
Complex Poynting Theorem (cont.)
Final form of complex Poynting theorem:
  2 E J
1
V
i*
i
 M H
*
 dV   S  nˆ dS
S
1
2 j      H
4
V 
1
     c E
2
V 
2

2

1
4
1
2
2

 c E  dV
   H

2

 dV

17
Complex Poynting Theorem (cont.)
Interpretation of Ps :
Ps 

V
R e Ps 

2
1

i*
EJ
E  J
i
i
 M H
 M
i
*
H
 dV
 dV
V
 Ps
We therefore identify that
Ps  com plex source pow er
18
Complex Poynting Theorem (cont.)
Interpretation of Pf :
Pf 

S
R e Pf 
1
2
E  H

EH
*
  nˆ dS
  nˆ d S
S
 Pf
We therefore identify that
P f  co m p le x p o w e r flo w in g o u t o f S
19
Complex Poynting Theorem (cont.)
Interpretation of energy terms:
1
4
 c E
2
1
* 
  c 
E E 
2 2

1
2

1
 c 
2
1
2
Hence

1


Re E  E
 c E  E 
1
 4
c
E
2
dV 
V
Similarly,
1
 4  H
V
*
1
2




Note: The real-part operator may be
added here since it has no effect.
 c E
Note: We know this result represents
stored energy for simple linear
media, so we assume it is true for
generalized linear media.
2
1
 2 E
c
2
dV
 We
V
2
dV 
1
 2 
c
H
2
dV
 Wm
V
20
Complex Poynting Theorem (cont.)
Interpretation of dissipation terms:
1
2
  c E
2
1
* 
   c R e E  E 
2



   c E  E    c E
 
c    j  
 
2
real constant 
 

For simple linear media:   c      
 
Hence
1
2
V
2
  c E dV 

 E
2
dV  Pd
e
This is the time-average
power dissipated due to
conducting losses.
We assume the same
interpretation holds for
generalized linear media.
V
21
Complex Poynting Theorem (cont.)
Interpretation of dissipation terms (cont.):
Similarly,
1
2
   H
2
dV  Pd
m
V
This is the time-average power dissipated due to magnetic losses.
22
Complex Poynting Theorem (cont.)
Summary of Final Form

V

2
1
EJ
i*
i
 M H
*
 dV   S  nˆ dS
S
1
     c E
2
V 
2

1
2 j      H
4
V 
1
   H
2
2
Ps  P f  Pd  j  2    Wm  We

1
4
2

 dV

2

 c E  dV


23
Complex Poynting Theorem (cont.)
Ps  P f  Pd  j  2    Wm  We

Pabs
We can write this as Ps  P f  Pabs
where we have defined a complex power absorbed Pabs :
R e  Pabs   Pd
Im  Pabs    2    Wm  We

24
Complex Poynting Theorem (cont.)
Ps  P f  Pabs
This is a conservation
statement for complex power.
Im  Pabs    2    Wm  We
R e  Pabs   Pd

VARs
Watts
Pf
E
Complex power
flow out of surface
VARS consumed
Power (watts)
consumed
H
 c,  
Source
25
Example
L
I
+ V -
Ideal inductor
Denote:
Pabs  com plex pow er absorbed  W abs  jQ abs
Calculate Pabs using circuit theory, and verify that the result is
consistent with the complex Poynting theorem.
Note: Wabs = 0 (lossless element)
26
Example (cont.)
L
I
+ V -
Pabs 
1
V I 
*
2
1
2

1
2

1
Z I  I

*
j L I  I
j L I
Ideal inductor
2
*
Note:
Pabs  jQ abs
2
27
Example (cont.)
Q abs  Im Pabs 
1
L I
2
2

1
* 
  L  Re  I I  
2


L

i
 2

Wm
2

 1

2
 2 
Li 
 2


28
Example (cont.)
Since there is no stored electric energy in the inductor, we can write
Q abs  2 

Wm  We

Hence, the circuit-theory result is consistent with
the complex Poynting theorem.
Note: The inductor absorbs positive VARS.
29
Example
We use the complex Poynting
theorem to examine the input
impedance of an antenna.
Pf
Js
Iin
V0
+
-
S
Antenna
Model:
Iin
V0
+
-
Pin 
1

1
Zin=Rin+j Xin
2
2
V 0 I in 
*
Z in I in
1
2
 Z in I in  I in*
2
30
Example (cont.)
so
Z in 
2 Pin
I in
Real part:
2
R in 
2 R e  Pin 
I in
2
R e  Pin   Pin

 Prad
(no losses in vacuum surrounding antenna)

Note: The far-field Poynting vector is
much easier to calculate than the nearfield Poynting vector.
 R e Prad
Hence
R in 
2
I in
2
 R e  S  nˆ  dS
S
31
Example (cont.)
R in 
2
I in
2
 R e  S  nˆ  dS
S
In the far field (r )
Im S  0
(This follows from plane-wave
properties in the far field.)
Hence
R in 
2
I in
2
  S  nˆ  dS
S
32
Example (cont.)
2
Imaginary part: X in 
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
Wm , We
33
Example (cont.)
X in 
2
I in
2
 2  W
 We 
m

Hence
X in 
4
I in
2
1
  4  0 H
V
2

1
4
2

 0 E  dV

However, it would be very difficult to calculate the input impedance
using this formula!
34
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:
We 
1
Wm 
1
E
4
4
H

2

2
  


  
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).
35
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
p
1
c
2
S
[(kg m/s)/m3] = [kg/(s m2)]
Reference:
J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 263).
36
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:
Sz 
E nergy
A t

E  A c t  N
  c t
p
A t
 Ec N p
 pc N p 
2
 pz c
 pN  c
S  S z zˆ
2
p
Np photons per unit volume
2
Hence
v
A
pz 
1
c
2
Sz
This is consistent with the
previous momentum formula.
37
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].
A = 1 m2
pz
Fz 
inc
2
1

c

inc
pz
2
Sz
c
  Ac t 
t
1

inc
2

2
z
(1000)
inc
z
p
  Ac 
2000
 1

 2  2 (1000)   1  c  
c
c

F z  6.671  10
6
[N ]
38
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
39
Solar Sail (cont.)
http://www.lgarde.com
40
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.
41
Maxwell Stress Tensor
 T xx

T   T yx
T
 zx
T xy
T yy
T zy
T xz 

T yz 
T zz 
1
Tij   0 EE
  0 H i H j   ij   0 E
i j
2
2

1
2
0 H
2



i, j  x, y , z
(for vacuum)
 1, i  j
 ij  
 0, i  j
42
Maxwell Stress Tensor (cont.)
Momentum equation:
 T  nˆ
dS  F 
S
Total flow rate
of momentum
given to object
from EM field
d

dt
p dV
V
Total force on object (rate
of change of mechanical
momentum)
Rate of change of
electromagnetic
momentum inside of region
F
p
1
c
2
EH
43
Maxwell Stress Tensor (cont.)
 T  nˆ
dS  F 
S
d

dt
p dV
V
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


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.
44
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
F z  zˆ   T  nˆ 
 T xx

F  T  nˆ  T yx

 T zx

T xy
T yy
T zy
Note: The fields are assumed to be
zero on the back side of the mirror,
and no fields are inside the mirror.
T xz   0 
  
T yz  0 
  
T zz    1 
F z   T zz
z
  T xz 


 T yz


  T zz 
45
Maxwell Stress Tensor (cont.)
x
nˆ   zˆ
A = 1 m2
TEMz wave
T zz
1
  0 Ez Ez   0 H z H z    0 E
2
1
   0 E
2
2

1
2
0 H
2
2

z
1
2



0 H
2
inc
inc
y

  US

Assume E  Ex xˆ , H  H y yˆ
Ex
/H
 0
46
Maxwell Stress Tensor (cont.)
1
1
2
F z    0Ex   0 H
2
2

1
2
Fz 
1
2
0H
0 H
2
y
2
y
1
* 
 0  Re  H y H y  
2
2


1
 0  Re H
2
2
1

4
0 H y
(PEC mirror)
2
y
1
1



2
y




Fz 
1

1
4
4
0 H y
2
0 2 H
inc
y
2
(The tangential magnetic field
doubles at the shorting plate.)
2
47
Maxwell Stress Tensor (cont.)
Fz 
1
4
0 2 H
inc
y
2
 0 H
inc
y
2
Next, use
inc
Sz
Fz   0 H
inc
y

inc
inc
y

1
2
inc
inc *
Ex H y

1
2
inc
Hy
2
0
2
 0 2 S z
 4   10
 Ex H
inc
7
/ 0

 2 1000  / 376.7303 
F z  6.671  10
6
[N ]
48
Foster's Theorem
Consider a lossless system with a port that leads into it:
Z in  jX in
Lossless system
(source free)
Sp
c     
Coaxial port
  
PEC enclosure
z
dX in
d
0
R. E. Collin, Field Theory of Guided Waves, IEEE Press, Piscataway, NJ, 1991.
49
Foster's Theorem (cont.)
The same holds for the input susceptance:
1
Z in 
Yin
1
jX in 
jB in
X in  
dX in
d

1
B in
1 dB in
B in d 
2
dB in
d
0
50
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
  E 
 
   

*


H 
   E   E   






*
*
 E*


E   E 
 
 H   H  

    H
    
 



Add these last two equations together:
51
Foster's Theorem (cont.)
*
*

  H *
H
E
  E 

 H  



  
*


H 
    E   E    






*
*

E   E 
H  

    H
    


Using Maxwell's equations for a source-free region,
*
*

  H *
H
E
  E 

 H  



  
*


H 
    j  H   E    






*
*

E   E 
H  

   j  E 
    

52
Foster's Theorem (cont.)
*
*

  H *
H
E
  E 

 H  



  
*


H 
    j  H   E    






*
*

E   E 
H  

   j  E 
    

Using Maxwell's equations again, and the chain rule, we have:


H

E
*

*






  H 
  E 
*
*






 j  E 
*
 j  H 
j
*
 j




   E
*
   H
*
 j 
 j 
E
*

H
*

We then have
53
Foster's Theorem (cont.)
*
*

  H *
H
E
  E 

 H  



  
*



E 
*
    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   j 
 


 



E
2
  
   H

  
2



54
Foster's Theorem (cont.)
*
*


H
E
 
  E 

 H   j 
 


 



E
2
  
   H

  
2



Applying the divergence theorem,

S
*
*


H
E

 H   nˆ dS  j 4  We  Wm 
E




Therefore,
Im

S
*
*


H
E

 H   nˆ dS  0
E




55
Foster's Theorem (cont.)
Im

S
*
*


H
E

 H   nˆ dS  0
E




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
Z in  jX in
Lossless system
(source free)
c     
  
Sp
Coaxial port
z
56
Foster's Theorem (cont.)
*

H
Im   E 

Sp 
Then

  nˆ dS  0

At the coaxial port:
Z in  jX in
Lossless system
(source free)
E  ˆ E 
 c   nˆ  
H  ˆ H 
  
Sp
zˆ
Coaxial port
z
Hence:
*

H  
Im   E 
 dS  0

 
Sp 
or


Im

Sp
 E H 
*
 dS  0
(since the electric field is
fixed and not changing
with frequency)
57
Foster's Theorem (cont.)




Im
 E

H
 dS  0
Sp
Im  2 Pz   0

Z in  jX in
Lossless system
(source free)
c     
Im V I   0


*
Im   V


*
z
2
Yin   0

*
  
Iz
Sp
Coaxial port
z
 V Y in 

*
Im  Yin   0

58
Foster's Theorem (cont.)

*
Im  Yin   0


Im  Yin   0





B in  0
X in  0
59
Foster's Theorem (cont.)
Example: Short-circuited transmission line
Zin
Z0
ZL = 0
l
Z in  jZ 0 tan   l 
 
LC
Xin
inductive
π/2
3π/2
5π/2
l
capacitive
60

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