Download Auxiliary Vector Potential Constructing solutions using auxiliary

58
Auxiliary Vector Potential
Constructing solutions using auxiliary vector potentials
• The objective of EM theory is to find possible EM field configurations (modes) for a
given boundary value problem involving wave propagation, radiation, scattering, or
absorption.


E
and
H
) or equally
• This can be done by finding the electric and
magnetic
fields
(


obtaining the auxiliary vector potentials ( A and F )


• In addition to auxiliary vector potentials A and F there are other possible set. For 
example, Hertz vector potentials (  e and  h ). Here, we only concentrate on A and F
• The path for solving EM field configuration is then as follows
Integration
path-1
Sources
 
J,M
Integration
Path-2
Vector
potentials
 
A, F or
e , h
 
E, H
Fields
Differentiation
Path-2
• Depending on the problem at hand, path-2 maybe easier than path-1


B
• Traditionally
E
and
are viewed as physical field quantities, whereas vector

potential ( A ) and its scalar counter part ( e ) are considered as mathematical
constructs. However, there are diverging views on this point!!!
• It is interesting to note that Maxwell
himself derived many of his results by using the

concept of vector potential ( A ) which he called “electromagnetic momentum.”
However this approach was later criticized by other practitioners such as Hertz and
Heaviside.
59
The question of the
 propagation of, not merely the electric potential  but the
vector potential A ... when brought
forward, prove to be one of a metaphysical

nature ... the electric force E and the magnetic force H ... actually represent
the state of the medium everywhere. Heaviside, Philosophical Magazine, 1889.
• Here is what Hertz says about Maxwell’s approach:
I may mention the predominance of the vector potential in [Maxwell’s]
fundamental equations. In the construction of new theory the potential served
as a scaffolding ... it does not appear to me that any ... advantage is attained
by the introduction of the vector potential in the fundamental equations. C.
A. Mead, Collective electrodynamics, 2000.
• Here is different (more modern) point of view:
... the vector potential which appears in quantum mechanics in an explicit form
produces a classical force which depends only on its derivatives. In quantum
mechanics what matters is the interference between nearby paths; it always turns
out that the effects depend only on how much the field
 A changes from point
to point, and therefore only on the derivatives
of A and not on the value

itself. Nevertheless, the vector potential A (together with the scalar potential
 that goes with it) appears to give the most direct description of the physics.
This becomes more and more apparent the more deeply we go into the quantum
theory. In the general theory of quantum electrodynamics, one takes the vector
and scalar potentials as the fundamental quantities in a set of equations that


replace the Maxwell equations: E and B are slowly disappearing from the
modern expression of physical laws, they are being replaced by A and  .
Feynman, Leighton, and Sands, Lectures on Physics, Vol. II, 1984.
• Aharonov- Bohm Effects: What happens to an electron as it passes by an infinitely



long solenoid. The E and B are zero outside the solenoid’s core but A  0 . Despite the
fact that there are no EM forces outside of the
solenoid, electron will experience the presence of

A and its phase will be modified. For the figure

shown, A will introduce a phase shift in the
electrons’ wave functions which can be detected by
interfering the electrons.
60

Equations governing vector potential A

• Since   B  0 


BA    A and


1
HA   A

(1)
(2)
(3)


Subscript A is to remind us that B A and H A are due to vector potential A

• For M  0 (no magnetic source)


  E A   j H A (Faraday’s Law)
(4)
• Use (3) in (4) 



1
  E A   j    A     E A  j A  0




(5)
• Since curl of gradient of any scalar is zero, i.e.,    e   0 , then from (5) we




have E A  jA   e  E A   jA  e where
(6)
e  Scalar potential

A  Vector potential



• Equations (6) and (3) are the expression for E and H in terms of A and e

• We now want to find differential equations governing the behaviors of A and e


1
• We note that from H A    A , for a homogeneous medium we can write






  H A      A 
  H A    A   2 A
(7)



• Using Ampere’s Law   H A  J  jE A in (7) we have




 J  j E A     A   2 A


(8)



• Previously we found the expression for E A to be E A   jA   e . Using this in
(8) we have



J  j  jA  e    A   2 A
(9)


• Recall that  2    2 then (9) can be written as




 2 A   2 A     A  j e   J




• We have defined the curl of A as B A    A , we are at liberty to define the   A .
61




• In light of  2 A   2 A     A  j e   J

A
to be
let us
define
the
divergence
of

  A   je
• Using
 (2)
 in (1)we have
2
2
 A   A   J
and

1
 A
e  
j
(1)
(2)
(3)
(4)


• Finally, our expressions for E A and H A in the last page [Eqs. (6) and (3)] can be
written as
 j



(5)
E A  e  jA 
  A  j A



1
HA   A

(6)



• Now, equations (5) and (6) are expressions for E A and H A in terms of A only
subject to Lorentz gauge.

Equations governing the vector potential F
• Consider a region of space free of charges, i.e. qave  0 , then

 D  0 


DF    F 


1
EF     F



Subscript F is to remind us DF is due to vector potential F

• Recall that Ampere’s Law with J  0 , is given by




1
  H F  j  EF 
  H F  EF
j
• Use (4) in (3) and we have




1
1
  H F     F    H F   j  F 
j



  H F  jF   0
(1)
(2)
(3)
(4)
(5)


• Compare   H F  jF   0 with null identity     m   0 , then it is clear that
62


H F  jF   m 


H F   j  F   m
(1)




1
1
• For homogenous media, from E F     F we have   E F       F 



 1 2
1
  E F     F    F
(2)


• From Faraday’s Law we have



  EF   M  j H F ,
then substitute (3) in (2)


 1

1
 M  j H F     F    2 F


(3)
(4)

• But we already found an expression for H F in (1). Use (1) in (4), and we have




 2 F   2 F  M    F  j m 
(5)
Where again  2   2 



• Once again curl of F is defined by DF    F . We are at liberty to choose the

divergence
of
F
. Let

  F   jm 
(6)

1
(7)
F
m 
j
• Using (6), (5) simplifies to



 2 F   2 F  M
(8)


• Finally, note that H F [Eq. (1)] and E F [Eq. (3) of last page)] can be written in terms

of F according to




j
H F   j F    m   j F 
   F 



1
EF     F

63
Summary



1. Find A from  2 A   2 A   J ,  2   2 



1
2. Find H A from H A    A







1
1
 HA
3. Find E A from E A   j A  j
  A or E A 

j




4. Find F from  2 F   2 F  M



1
5. Find E F from E F     F







1
1
  EF
  F  or H F 
6. Find H F from H F   j F  j

j

7. The total E is given by

 1
 


1
E  E A  E F   j A  j
   A    F


or
 



1
1
E  E A  EF 
 HA   F
j


8. The total H is given by






1
1
H  H A  H F    A  j F  j
   F 


or





1
1
H  HA  HF   A
  EF
j

(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
64


Solutions for A and F


• Recall that governing differential equations for A and F are



 2 A   2 A   J



 2 F   2 F  M
(1)
(2)
• For source located at x , y , z  and observation
point distance R from the source, the solutions to
(1) and (2) are given by



e  jR





A x, y, z  
J
x
,
y
,
z
dv
(3)
4 
R
v



e  j R





F  x, y , z  
M
x
,
y
,
z
dv
(4)
4 
R
v


where J and M have dimensions proportional to 1/m2
z
 x, y , z 

r

R
y

r
x
x, y, z


• For J s and M s dimensions proportional to 1/m we have



e  j R





A x, y, z  
J
x
,
y
,
z
ds 
(5)
s
4 
R
s



e  jR





F  x, y , z  
M
x
,
y
,
z
ds
(6)
s
4 
R
s


• For electric and magnetic current densities I e [Ampere] and I m [volt] we have

 
e  j R





A x, y, z  
I
x
,
y
,
z
dl 
(7)
e
4 c
R

 
e  j R





F  x, y , z  
I
x
,
y
,
z
dl 
(8)
m
4 c
R
65
TEM, TE and TM modes
• The transverse electromagnetic field configuration is a mode for which electric and
magnetic field components are transverse to a given direction. This direction often,
but not always, is the path that wave is traveling.
• For TE mode the electric field is transverse to a given direction and for TM mode
the magnetic field is transverse to a given direction. Again, for TE and TM modes the
aforementioned direction is often, but not always, the direction of propagation.


The conditions on auxiliary vector potentials A and F
for TEM, TE and TM modes




• Recall that E and H in terms of A and F were given by

 1
 


1
E  E A  E F   j A  j
  A    F
(1)






1
1
H  H A  H F    A  j F  j
  F 
(2)
•Let



A  Ax x, y , z a x  Ay  x, y , z a y  Az  x, y, z a z




F  Fx  x, y, z a x  Fy  x, y , z a y  Fz  x, y , z a z
(3)




• Use (3) and (4) in (1) and (2). We get
2
  
1   2 Ax  Ay  2 Az  1  Fz Fy 
 

 

E  a x  jAx  j

z 
  x 2 xy xz    y

2
1   2 Ax  Ay  2 Az  1  Fx Fz 
 


 

a y   j A y  j
 
x 
  xy y 2 yz    z

2
1   2 Ax  Ay  2 Az
 


a z  jAz  j
  xz yz z 2

 1  Fy Fx 
 


   x
y 


(4)
66

• For H we have
2
  
1   2 Fx  Fy  2 Fz  1  Az Ay 
 

 


H  a x  j Fx  j
  x 2 xy xz    y z 

2
1   2 Fx  Fy  2 Fz  1  Ax Az 
 
a y  j Fy  j

 


 
  xy y 2 yz    z x 

(1)
2
1   2 Fx  Fy  2 Fz  1  Ay Ax 
 

 



a z  j Fz  j
  xz yz z 2    x y 





• From expression for E and H in terms of A and F we can see there are at least 3
ways for which we can obtain a TEM mode with respect to z-direction, i.e. TEMz
(HW)
• For example if all the condition listed below are satisfied we have a TEMz mode


 0 , and
Ax  Ay  Az  0 , and Fx  Fy  0 , and
 0 , and
x
y
Fz  Fz  x, y e  j z  Fz  x, y e  j z ,
then
0
o
0
0
 
0 
 
 



0
 2

2
2


j   Ax  Ay  Az  1 Fy Fx 
0
(2)
E z   jAz 


 

  xz yz z 2    x y 








0
0
0
0

 

 



 2

2
2





F
A
A
 Fz  1
1   Fx
y
y
y






H z   j Fz  j
y 
  xz yz z 2    x








 jFz 
j 2
F


z
x, y e  jz  Fz x, y e jz    jFz 
jFz  0
• Note that from (2) and (3) E z  H z  0 .
• We can further calculate the E x , E y , H x , and H y to be
0
0
0
0
 
 
 






2
2
2


1   Ax  Ay  Az  1 Fz Fy 



 

E x   jAx  j
xy xz    y
z 
  x 2








1  
1  
Ex  
Fz  x, y e  jz 
F  x, y e jz  E x  E x
 y
 y z
0
(3)
67
0
0
0
 
0
 
 


0

 2
2
2




A
 Az  1 Fx Fz
1   Ax
y
 

E y   j A y  j


2
x
yz    z
  xy y





1  
1   j z
Ey 
Fz  x, y e  j z 
Fz e  E y  E y
 x
 x
and it can be shown
H x  H x  H x  
H y  H y  H y 

 1 Fz



x



 
 
Ey 
E y (HW)


 
 
Ex 
E , (HW)

 x
(1)
(2)
(3)
(4)
Where expression for E y , E y , E x , E x were given previously (e.g.
E x  
1  
1  
Fz  x, y e  jz and E x  
Fz  x, y e  jz )
 y
 y
(5)
Transverse magnetic wave WRT z-direction (TMz)
• To ensure that wave is a transverse magnetic (TM) field WRT z-direction, it is
sufficient
to ensure the auxiliary vector potential A has only z-component and

F  0.
•For TMz


A  a z Az x, y, z  and F  0
• The field components are then given by
1  2 Az
Ex   j
 xz
1  2 Az
Ey   j
 yz

1  2
 2   2  Az
  z

1 
Hx 
Az
 y
1 Az
Hy  
 x
Hz  0
Ez   j
(6)
(7)
(8)
(9)
(10)
(11)
(12)
• All the field components of the TMz mode can also be expressed in terms of E z
68
Transverse electric field WRT z-direction (TEz)


• To have TEz we require F to have only z-component and A  0 , i.e.,

 
A  0 and F  a z Fz  x, y, z 
(1)
• The field components are given by
1 Fz
Ex  
 y
1 Fz
Ey 
 x
Ez  0
1  2 Fz
 xz
1  2 Fz
Hy  j
 yz
Hx   j
Hz   j

1  2
 2   2  Fz
  z

• All the field components of the TEz mode can also be expressed in terms of H z
Rectangular metallic wave guide
• Rectangular metallic waveguides are routinely used at RF and microwave
frequencies. Their study is not only motivated by their use as RF/microwave
components, but will help us better understand the concept of mode and guided wave
propagation
• In studying the guided wave structures we are usually interested in parameters such as:
field configurations (modes) that are supported by the structure, the structure cutoff
frequency, guided wavelength, wave impedance, phase constant, attenuation
constant, etc.
• For metallic rectangular waveguide, it can be shown that although TEM field
configuration is the lowest order mode, it does not satisfy the boundary conditions
and as such, the waveguide does not support TEM modes
• However, the TE and TM modes satisfy the required boundary conditions and as
such are supported by the structure
69
Transverse Electric Field TEz
• Consider the metallic waveguide of size a  b as shown. The waveguide is infinite in
the z-direction
•From our previous discussion we have seen that TEz modes are obtained if

A  0 and F  aˆ z Fz  x, y, z  which implied
1 Fz
Ex  
 y
1  2 Fz
Hx   j
 xz
1 Fz
Ey 
y
 x
1  2 Fz
Hy  j
 yz
,
b
Ez  0
x

1  2
 2   2  Fz
Hz   j
  z


satisfy
the vector differential equation
• F must


2
2
 F   F  0   2 Fz  x, y, z    2 Fz  x, y, z   0 
a
z
 2 Fz  2 Fz  2 Fz


  2 Fz  0
x 2
y 2
z 2
• Note that Fz  x, y, z  is a scalar function that can be written as (using separation of
variables)
Fz x, y, z   f  x g  y h z 
• Also recall that solutions to  2 Fz   2 Fz  0 are either standing waves (sinusoidal)
or traveling waves (exponential with complex argument)
• The particular form (standing wave or traveling wave) is chosen based on the
boundary conditions to be satisfied
• In the case of our metallic waveguide solutions in x and y must be standing waves
and solution in z-direction (guide is infinite in the z-direction) must be traveling wave
• Hence
70
Fz  x, y, z   f  x g  y h z 


 C1 cos z x   D1 sin  x x   C 2 cos y y   D2 sin  y y   A3 e  j z z  B3 e  j z z

with
 x 2   y 2   z 2   2   2 
• Recall that for e jt time dependency, e  j z z is a positively traveling wave (wave
travels in positive z-direction) and e  j z z is a negatively traveling wave (wave travels in
negative z-direction)
• If source is located such that only positively traveling wave is present, then
B3e  j z z  0  B3  0
• If source is located such that only negatively traveling wave is present, then
y
A3 e  j z z  0  A3  0
yb
• If both positively and negatively traveling waves are
present (a waveguide terminated on a load that is not
matched), then both A3 e  j z z and B3 e  j z z must be included
,
y0
• Here, for simplicity, we assume that only positive traveling
wave exist  B3  0
xa
x0
z
•The Fz is then given by
Fz  x, y , z   C1 cos  x x   D1 sin  x x   C 2 cos  y y   D 2 sin  y y A3 e  j z z
(1)
• We impose the boundary conditions on the top, bottom, left and right walls of the
metallic waveguide,
assuming
a perfect electric conductor (PEC) boundary


condition, i.e. E and H tangential are zero on the walls
• The boundary conditions are:
E x 0  x  a, y  0, z   E x 0  x  a, y  b, z   0 , Bottom and top walls for E x (2)
E z 0  x  a, y  0, z   E z 0  x  a, y  b, z   0 , Bottom and top walls for E z (3)
E y  x  0,0  y  b, z   E y  x  a,0  y  b, z   0 , Left and right walls for E y
(4)
E z  x  0,0  y  b, z   E z x  a,0  y  b, z   0 , Left and right walls for E z
(5)
• Note that the boundary conditions (3) and (5) are not independent and they
represent the same boundary conditions as (2) and (4).
• The necessary and sufficient conditions are to satisfy (2) OR (3) ( E x and E z at
bottom and top walls) AND (4) or (5) ( E y and E z at the left and right walls)
x
71
• Furthermore, note that for TEz modes by definition E z is zero. This means that the
necessary and sufficient B.C.’s for TEz are (2) and (4) of the last page ( E x at the
bottom and the top and E y at the left and right walls)
E x 0  x  a, y  0, z   E x 0  x  a, y  b, z   0
E y  x  0,0  y  b, z   E y  x  a,0  y  b, z   0
• Recall that the vector potential for TEz was given as [Eq. (1), last page]
Fz x, y , z   C1 cos x x   D1 sin  x x  C2 cos  y y   D2 sin  y y A3e  j z z . We then have
Ex  
y
1 Fz
C1 cos x x   D1 sin  x x   C 2 sin  y y   D2 cos y y  A3 e  j z z
  A3
 y



• From E x 0  x  a, y  0, z   0  D2  0
y
• From E x 0  x  a, y  b, z   0  sin  y b   0 
yb
 y b  n n  0,1,2,3 or equally,
n
y 
n  0,1,2,3
b
,
y0
•  y is sometimes referred to as eigenvalue
• If we use our newly found results, we have
 n 
Fz  x, y, z   C1 cos x x   D1 sin  x x C 2 cos
y  A3 e  j z z
 b 
• The E y can be found from
Ey 
1 Fz  x
 C1 sin  x x   D1 cos x x C2 cos n y  A3e  j z x

 x

 b 
• Boundary conditions for E y at left wall is
E y  x  0,0  y  b, z   0  D1  0
• Boundary condition for E y at right wall is
E y  x  a,0  y  b, z   0  sin x a   0 
 x a  m m  0,1,2,3 or equally
m
x 
m  0,1,2,3
a
z
x0
xa
x
72
• Putting it all together, the vector potential Fz is given by
 m 
 n 
 m   n    j  z z
Fz  x, y, z   C1 cos
x C 2 cos
y  A3 e  j z z  Amn cos
x  cos
y e
 a 
 b 
 a   b 
m  0,1,2,3
with
but m  n  0
n  0,1,2,3
Amn is a constant = C1C 2 A3
Propagation constant (wave numbers) and wavelengths
in the x , y and z direction
• From our previous discussion it is clear that propagation constant (or wave number)
along x  x  and along y  y  can be written as
m 2
2a

 x 
; m  0, 1, 2
x
a
m
n 2
2b
y 

 y 
; n  0, 1, 2 and with m  n  0 ,
b
y
n
x 
(1)
(2)
where we have also defined the wavelength along x to be  x and along y to be  y
• Recall that  x   y   z   2   2  or equally well:
2
1
x
2

1
y
2

1
z
2
2

1
2
2
where  is the wavelength in the medium with  and 
(material inside the guide)
• From (1) and (2) note that  x and  y are discrete (one can say they are quantized),
where as  z is a continuous parameter.
• Note that in principle there are infinite numbers of possible  x and  y (eigenvalues)
hence there are infinite number of TEz modes that satisfy the wave equation and the
given boundary condition.
m
n
2
2
2
and  y 
and  x   y   z   2   2  we can see that
a
b
2
2
2
2
 m   n 
 m   n 
2
2
2
2
2
  c ,
 
  z      z  
 

 a   b 
 a   b 
where by definition:  c  cutoff propagation constant or cutoff wave number
• From  x 
73
 m   n 
  
 
 .
 a   b 
2
• Note that  c  
 z 0
  
  c   2f c
 z 0
2
• Hence, the cutoff frequency ( f c ) is given by
 f c mn 
 m   n 

 

 a   b 
2
1
2 
• From the expression  c  
 z 0
m  0,1,2,3
m  n  0
n  0,1,2,3 
2
we can see why  c is called the cutoff propagation
constant. For this wave number,  z  0 and the wave no longer travels along the zdirection.
• The above can be more clearly seen from
  m   2  n  2 
   2   c 2 . Clearly for
 z 2   2   x2   y2   2  

  a   b  


  c   z  0


• The field components for TEmn+z are now given by
y
cos x x sin  y y e  j z



E y   Amn x sin  x x  cos y y e  j z


E x  Amn
z
z

Ez  0

xz
sin  x x  cos y y e  j z

 
 Amn y z cos x x sin  y y e  j z

H x  Amn
HY
Hz


z
z
c 2
  jAmn
cos x x  cos y y e  j z

z
• To appreciate the importance of the cutoff conditions consider the following:
 m  2  n  2 
 z 2   2   c 2   2  
 
 
 a   b  
 z mn
    c
2
2
2

   1   c
 
 
f 
   1   c     1   
 f 
 c 



2
2
for    c  f  f c
74
•  z mn  0
for    c  f  f c
2
 
•  z mn   j  c 2   2   j  c   1
 
2
f 
  j  c   1   j
 f 


 c
2

  1

for  c    f c  f
• If we only consider the positively traveling wave we must choose the sign in front
of the square root appropriately, i.e.
 z mn
 z mn
 z mn
2

  1   c

0
 
f 

   1   c    1   

 f 
 c 

  j  c


  1   j

2
2
for   c  f  f c
for  c  f c  0
2
2
 fc 
   1   j
 f 


 c
2

  1

for  c    f c  f
• Since electric and magnetic fields are proportional to e  j z z e jt then
2
e
 f 
 j 1 c  z
 f 
e jt represents a propagating wave for f  f c
e jt e  j 0 z  e 0 e jt  1e jt represents a standing wave for f c  f


 j   j


e
fc  f
2

 fc 

  1  z

 f 

e j t  e

 z 


2

 fc 
  1  z

 f 

e jt represents an attenuated (evanescent) wave for
75
TEz10 Field components and Field Patterns
, , ,
y
z
,
x