This document discusses vectors and tensors in three dimensions. It defines reciprocal sets of vectors, which satisfy certain orthogonality properties. It introduces the metric tensor or fundamental tensor, which is specified by three non-coplanar vectors. It proves that the determinant of the metric tensor, known as the Gram determinant, is non-zero using properties of the defining vectors.
2. 2
SOLO
TABLE OF CONTENT
References
Dyadics
Vectors & Tensors in a 3D Space
Triple Scalar Product
Reciprocal Sets of Vectors
Vector Decomposition
The Summation Convention
The Metric Tensor or Fundamental Tensor Specified by .3321 ,, Eeee ∈
Change of Vector Base, Coordinate Transformation
Dyadics
Introduction to Dyadics
Dyadics in Reciprocal Coordinates
Identity Dyadic (Unit Dyadic, Idemfactor)
Coordinate Transformation of Dyadics in Reciprocal Coordinates
Dyadics Invariants
Classification of Dyadics
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz)
Differential Equations
3. 3
SOLO
Triple Scalar Product
Vectors & Tensors in a 3D Space
3321 ,, Eeee ∈
are three non-coplanar vectors, i.e.
1e
2e
3e
( ) ( ) 0:,, 321321 ≠×⋅= eeeeee
( ) ( ) ( ) ( )
( ) ( ) ( ) 0,,
,,,,
123123213
132132132321
≠=×⋅=⋅×=
=×⋅=⋅×=
eeeeeeeee
eeeeeeeeeeee
Reciprocal Sets of Vectors
The sets of vectors and are called Reciprocal Sets or Systems
of Vectors if:
321 ,, eee
321
,, eee
DeltaKroneckertheis
ji
ji
ee
j
i
j
i
j
i
δδ
=
≠
==⋅
1
0
Because is orthogonal to and then2e
3e
1
e
( ) ( ) ( )
( )321
321321
1
132
1
,,
1
,,1
eee
keeekeeekeeeeke
=→=×⋅==⋅→×=
and in the same way and are given by:
2
e
3
e
1
e
( )
( )
( )
( )
( )
( )321
213
321
132
321
321
,,,,,, eee
ee
e
eee
ee
e
eee
ee
e
×
=
×
=
×
=
Table of Content
4. 4
SOLO Vectors & Tensors in a 3D Space
Reciprocal Sets of Vectors (continue)
By using the previous equations we get:
( ) ( )
( )
( )[ ] ( )[ ]
( ) ( )321
3
2
321
13323132
2
321
133221
,,,,,, eee
e
eee
eeeeeeee
eee
eeee
ee
=
⋅×−⋅×
=
×××
=×
( )
( )
( )
( )
( )
( )321
213
321
132
321
321
,,,,,, eee
ee
e
eee
ee
e
eee
ee
e
×
=
×
=
×
=
( ) ( )
( ) ( )
0
,,
1
,,
,,
321321
3
3321321
≠=
⋅
==⋅×
eeeeee
ee
eeeeee
Multiplying (scalar product) this equation by we get:
3
e
In the same way we can show that:
Therefore are also non-coplanar, and:
321
,, eee
( ) ( ) 1,,,, 321
321
=eeeeee
( )
( )
( )
( )
( )
( )321
21
3321
13
2321
32
1
,,,,,, eee
ee
e
eee
ee
e
eee
ee
e
×
=
×
=
×
=
1e
2e
3
e
1
e
2
e
3
e
Table of Content
5. 5
SOLO Vectors & Tensors in a 3D Space
Vector Decomposition
Given we want to find the coefficients and such that:3EA∈
321 ,, AAA 321
,, AAA
∑
∑
=
=
=++=
=++=
3
1
3
3
2
2
1
1
3
1
3
3
2
2
1
1
j
j
j
i
i
i
eAeAeAeA
eAeAeAeAA
3,2,1, =iee i
i
are two reciprocal
vector bases
Let multiply the first row of the decomposition by :
j
e
Let multiply the second row of the decomposition by :ie
j
i
j
i
i
i
j
i
ij
AAeeAeA ==⋅=⋅ ∑∑ ==
3
1
3
1
δ
i
j
i
j
j
j
i
j
ji
AAeeAeA ==⋅=⋅ ∑∑ ==
3
1
3
1
δ
Therefore:
ii
jj
eAAeAA
⋅=⋅= &
Then:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )∑
∑
=
=
⋅=⋅+⋅+⋅=
⋅=⋅+⋅+⋅=
3
1
3
3
2
2
1
1
3
1
3
3
2
2
1
1
j
j
j
i
i
i
eeAeeAeeAeeA
eeAeeAeeAeeAA
Table of Content
6. 6
SOLO Vectors & Tensors in a 3D Space
The Summation Convention
j
j
j
j
j eAeAeAeAeA
==++ ∑=
3
1
3
3
2
2
1
1
The last notation is called the summation convention, j is called the dummy
index or the umbral index.
( ) ( ) ( )
( ) ( ) ( ) i
i
i
i
j
j
j
j
j
j
j
j
j
i
i
i
i
i
eAeeAeeAeeA
eAeeAeeAeeAA
=⋅=⋅=⋅=
=⋅=⋅=⋅=
∑
∑
=
=
3
1
3
1
Instead of summation notation we shall use the shorter notation
first adopted by Einstein
∑=
3
1j
j
j
eA
j
j eA
Table of Content
7. 7
SOLO Vectors & Tensors in a 3D Space
Let define:
The Metric Tensor or Fundamental Tensor Specified by .3321 ,, Eeee ∈
jiijjiij geeeeg =⋅=⋅=
3321 ,, Eeee ∈
the metric covariant tensors of
By choosing we get:
( ) ( ) ( ) ( )
j
ijiii
j
jiiiii
egegegeg
eeeeeeeeeeeee
=++=
⋅=⋅+⋅+⋅=
3
3
2
2
1
1
3
3
2
2
1
1
ieA
≡
or: j
iji ege
=
For i = 1, 2, 3 we have:
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
=
=
3
2
1
332313
322212
312111
3
2
1
333231
232221
131211
3
2
1
e
e
e
eeeeee
eeeeee
eeeeee
e
e
e
ggg
ggg
ggg
e
e
e
8. 8
SOLO Vectors & Tensors in a 3D Space
We want to prove that the following determinant (g) is nonzero:
The Metric Tensor or Fundamental Tensor Specified by .3321 ,, Eeee ∈
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
=
=
332313
322212
312111
333231
232221
131211
detdet:
eeeeee
eeeeee
eeeeee
ggg
ggg
ggg
g
g is the Gram determinant of the vectors 321 ,, eee
Jorgen Gram
1850 - 1916
Proof:
Because the vectors are non-coplanars the following
equations:
321 ,, eee
03
3
2
2
1
1
=++ eee
ααα
is true if and only if 0321
=== ααα
Let multiply (scalar product) this equation, consecutively, by :321
,, eee
=
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
→
=⋅+⋅+⋅
=⋅+⋅+⋅
=⋅+⋅+⋅
0
0
0
0
0
0
3
2
1
332313
322212
312111
33
3
23
2
13
1
32
3
22
2
12
1
31
3
21
2
11
1
α
α
α
ααα
ααα
ααα
eeeeee
eeeeee
eeeeee
eeeeee
eeeeee
eeeeee
Therefore α1= α2= α3=0 if and only if g:=det {gij}≠0 q.e.d.
9. 9
SOLO Vectors & Tensors in a 3D Space
Because g ≠ 0 we can take the inverse of gij and fined:
The Metric Tensor or Fundamental Tensor Specified by .3321 ,, Eeee ∈
where Gij
= minor gij having the following property: j
i
kj
ik
ggG δ=
=
=
3
2
1
333231
232221
131211
3
2
1
333231
232221
131211
3
2
1
1
e
e
e
ggg
ggg
ggg
e
e
e
GGG
GGG
GGG
g
e
e
e
and:
g
G
g
gminor
g
ij
ijij
==
Therefore:
g
g
g
g
G
gg
j
i
kj
ik
kj
ik δ
== j
i
kj
ik
gg δ=
Let multiply the equation by gij
and perform the summation on i
j
iji ege
=
jj
ij
ij
i
ij
eeggeg
==
Therefore: i
ijj
ege
=
Let multiply the equation byk
kjj
ege
= i
e
iji
k
kjijji
geegeeee =⋅=⋅=⋅
jiijjiij
geeeeg =⋅=⋅=
i
jkj
ik
ggG δ=
10. 10
SOLO Vectors & Tensors in a 3D Space
Let find the relation between g and
The Metric Tensor or Fundamental Tensor Specified by .3321 ,, Eeee ∈
( ) ( )321321 :,, eeeeee
×⋅=
We shall write the decomposition of in the vector base32 ee
× 321 ,, eee
3
3
2
2
1
1
32 eeeee
λλλ ++=×
Let find λ1
, λ2
, λ3
. Multiply the previous equation (scalar product) by .1e
( ) ( ) i
i
ggggeeeeee 113
3
12
2
11
1
321321 ,, λλλλ =++==×⋅
Multiply this equation by g1i
: ( ) ii
i
ii
ggeeeg λλ ==
1
1
1321
1
,,
Therefore: ( )321
1
,, eeeg ii
=λ
Let compute now:
( ) ( ) ( ) ( ) ( ) ( )321
1
0
323
3
0
322
2
321
1
3232
eeeeeeeeeeeeeeee
×⋅=×⋅+×⋅+×⋅=×⋅× λλλλ
( ) ( )[ ]
( )
( )[ ]
( )
( ) ( )[ ]
( ) ( ) ( ) ( )321
11
321
11
321
2
233322
321
3322332
321
3232
321
32321
,,,,,,,,
,,,,
eee
gg
eee
G
eee
ggg
eee
eeeeeee
eee
eeee
eee
eeee
==
−
=
⋅−⋅⋅
=
××⋅
=
×⋅×
=λ
From those equations we obtain:
( )321
11
1
,, eee
gg
=λ
Finally: ( ) ( ) geeeeee =⇒= 321
2
321
1
,,,,
λ
We can see that if are collinear than and g are zero.321 ,, eee
321
,, eee
Table of Content
11. 11
SOLO Vectors & Tensors in a 3D Space
Let choose another base and its reciprocal
Change of Vector Base, Coordinate Transformation
( )321 ,, fff
( )321
,, fff
[ ]
=
=
→=
3
2
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef j
j
ii
ααα
ααα
ααα
α
where j
i
j
i ef
⋅=α
By tacking the inverse of those equations we obtain:
[ ]
=
=
→=
−
3
2
1
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
f
f
f
L
f
f
f
e
e
e
fe i
j
ij
βββ
βββ
βββ
β
where j
ij
i ef
⋅=β
Because are the coefficients of the inverse matrix with coefficients :
j
i
β j
i
α
i
j
i
k
k
j
δαβ =
12. 12
SOLO Vectors & Tensors in a 3D Space
Let write any vector in those two bases:
Change of Vector Base, Coordinate Transformation (continue – 1)
A
[ ]
=
=
→
⋅=
=
3
2
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef
ef
ji
j
i
j
j
ii
ααα
ααα
ααα
α
α
then:
[ ]
=
=
→
⋅=
= −
3
2
1
1
3
2
1
3
3
3
2
3
1
2
3
2
2
2
1
1
3
1
2
1
1
3
2
1
E
E
E
L
E
E
E
F
F
F
ef
EF T
j
ii
j
i
j
ji
βββ
βββ
βββ
β
β
i
i
j
j
fFeEA
== iijj
fAFeAE
⋅=⋅= &
i
j
ji
i
i
j
j
j
j
i
i
EFfEeEfF ββ =→==
or:
But we remember that:
We can see that the relation between the components F1
, F2
, F3
to E1
, E2
, E3
is
not similar, contravariant, to the relation between the two bases of vectors
to . Therefore we define F1
, F2
, F3
and E1
, E2
, E3
as the contravariant
components of the bases and .
( )321 ,, fff
( )321
,, eee
( )321 ,, fff
( )321 ,, eee
where
13. 13
SOLO Vectors & Tensors in a 3D Space
Let write now the vector in the two bases and
Change of Vector Base, Coordinate Transformation (continue – 2)
A
[ ]
=
=
→=
3
2
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
E
E
E
L
E
E
E
F
F
F
EF j
iji
ααα
ααα
ααα
α
and
i
i
j
j fFeEA
== iijj fAFeAE
⋅=⋅= &
then:
( )321
,, fff
( )321
,, eee
where
j
ij
ef
ijjjjii EfeEeEAfAF j
iji
α
α=⋅
=⋅==⋅=
We can see that the relation between the components F1, F2, F3 to E1, E2, E3 is
similar, covariant, to the relation between the two bases of vectors
to . Therefore wew define F1
, F2
, F3
and E1
, E2
, E3
as the covariant
components of the bases and .
( )321
,, fff
( )321
,, eee
( )321
,, fff
( )321
,, eee
14. 14
SOLO Vectors & Tensors in a 3D Space
We have:
Change of Vector Base, Coordinate Transformation (continue – 3)
( )
( ) j
j
j
j
i
i
i
i
eeAeA
eeAeAA
⋅==
⋅== Ai
contravariant component
Aj covariant component
Let find the relation between covariant and the contravariant components:
j
j
j
ij
i
ege
i
i
eAegAeAA j
iji
===
=
i
i
i
ij
j
ege
j
j eAegAeAA
i
ijj
===
=
Therefore: ij
j
i
ij
i
j gAAgAA == &
Let find the relation between gij and gij
defined in the bases and to
and defined in the bases and .
i
e
i
e
i
f
i
f
ijg ij
g
m
m
kkj
j
ii efef
αα == &
jm
m
k
j
imj
m
k
j
ikiik geeffg αααα =⋅=⋅=
Hence: jm
m
k
j
iik gg αα=
This is a covariant relation of rank two, (similar, two times, to relation
between to .i
f
je
15. 15
SOLO Vectors & Tensors in a 3D Space
Change of Vector Base, Coordinate Transformation (continue – 4)
[ ]
=
=
→
⋅=
=
3
2
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef
ef
ji
j
i
j
j
ii
ααα
ααα
ααα
α
α
Since we have:k
iki fgf
= ( ) m
jm
j
i
ege
j
j
i
ef
k
iki
egefgf m
jmjj
j
ii
αα
α ==
===
and: ( ) ( ) m
jm
j
i
km
kjm
j
i
gg
k
ik egfgfg
mjm
m
k
j
iik
ααα
αα
==
=
Therefore, by equalizing the terms that multiply we obtain:
[ ]
=
=
→=
3
2
1
3
2
1
3
3
3
2
3
1
2
3
2
2
2
1
1
3
1
2
1
1
3
2
1
f
f
f
L
f
f
f
e
e
e
fe
Tkm
k
m
ααα
ααα
ααα
α
We found the relation:
( )jm
j
i gα
16. 16
SOLO Vectors & Tensors in a 3D Space
Change of Vector Base, Coordinate Transformation (continue – 5)
Therefore:
Let take the inverse of the relation by multiplying by and summarize on m:
j
m
β
[ ]
=
=
→= −
3
2
1
1
3
2
1
3
3
3
2
3
1
2
3
2
2
2
1
1
3
1
2
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef
Tmj
m
j
βββ
βββ
βββ
β
km
k
m
fe
α=
jkj
k
km
k
j
m
mj
m fffe
=== δαββ
From the relation: mk
m
kjj
i
i
efef
ββ == &
we have: jmk
m
j
i
mjk
m
j
i
kiik
geeffg ββββ =⋅=⋅=
or: jmk
m
j
i
ik
gg ββ= This is a contravariant relation of rank two.
From the relation:
m
m
kk
jj
i
i
efef
αβ == &
we have: m
j
m
k
i
jm
jm
k
i
j
i
kk
i
eeff δαβαβδ =⋅==⋅
or: This is a relation once covariant and once contravariant of
rank two.
m
j
m
k
i
j
i
k δαβδ =
Table of Content
17. 17
SOLO
conjugate of the dyadic
Dyadics
Introduction to Dyadics
A dyadic has the property that scalar multiplication with a vector produces a vector.
and are two different vectors. A dyadic is a second order tensor
D
V
DV
⋅ VD
⋅
( ) ∑=+++=
i
in BABBBAD
21
A particular dyadic can be obtained by placing two vectors side by side, with neither
a dot nor a cross between them (such as ).BA
General Properties of Dyadics:
( ) ∑=+++=
i
in
BABAAAD
21
( ) ( ) ( ) VDVBAVDBAVBAVDV C
⋅=⋅≠⋅=⋅=⋅=⋅ vector = scalar product with a vector
( ) ( ) ( ) VDVBABAVBAVDV
×=×≠×=×=× dyadic = vector product with a vector
( ) ABBAD
C
C
== :
dyadic = sum of compatible dyadics
( ) ( )( ) ( ) ( )WABVWBAVWBAVWDV
::=⋅⋅=⋅⋅=⋅⋅ scalar = double dot product
( ) ( )( ) WDVWBAVWBAVWDV
⋅×≠×⋅=×⋅=×⋅
( ) ( )( )WBAVWBAVWDV
××=××=×× dyadic = double vector product
vector = double vector scalar product
18. 18
SOLO Dyadics
Introduction to Dyadics
General Properties of Dyadics:
triadic = vector product of two dyads
dyadic = scalar product of two dyads( ) ( ) ( ) 2211221121
BABABABADD
⋅=⋅=⋅
( ) ( ) ( ) 2211221121
BABABABADD
×=×=×
( ) ( ) ( ) ( )( ) 2211221121
BABAVBABAVDDV
⋅×=⋅×=⋅×
dyadic = triple scalar product of
a vector and two dyads( ) ( ) ( ) ( )( )VBABAVBABAVDD
×⋅=×⋅=×⋅ 2211221121
Table of Content
19. 19
SOLO Dyadics
Dyadics in Reciprocal Coordinates
Given:
==
== ∑∑ l
l
li
l
li
j
j
j
ij
j
ii
eAeAeAeAA
==
== ∑∑ m
m
mi
m
mi
k
k
k
ik
k
ii
eBeBeBeBB
====
====
====
=====
∑
∑
∑
∑
→
→
→
→
i
kijikijijk
kj
jk
jl
km
ml
mili
i
k
iji
k
iji
k
jk
jk
j
jl
k
lk
ili
i
ki
j
iki
j
ik
jk
jk
j
km
m
jmi
j
i
i
k
i
j
i
k
i
j
i
jk
kj
jk
kj
k
i
j
iii
BABAdeedeeBA
BABAdeedeeBA
BABAdeedeeBA
BABAdeedeeBABAD
:
:
:
:
lm
kmjlk
l
jl
m
jmkjk
dggdgdgd ===
j
m
kj
mk
k
m
k
m
kmmk
kmmk
k
mkm
ggeeeegeegege δδ ==⋅⋅=⋅== ,,,,
Using we obtain:
jk
mklj
k
lmkm
j
ljlm
dggdggd === φ
mklj
gg
kmjl
gg
20. 20
SOLO Dyadics
Dyadics in Reciprocal Coordinates
Given:
==
== ∑∑ l
l
li
l
li
j
j
j
ij
j
ii
eAeAeAeAA
==
== ∑∑ m
m
mi
m
mi
k
k
k
ik
k
ii
eBeBeBeBB
( ) ( ) [ ]
=
===
3
2
1
321
3
2
1
333231
232221
131211
321
e
e
e
Deee
e
e
e
ddd
ddd
ddd
eeeeedBAD kj
jk
ii
Decomposition of a dyadic in symmetric and anti-symmetric parts:
( ) ( ) [ ]
=
===
3
2
1
321
3
2
1
332313
322212
312111
321
e
e
e
Deee
e
e
e
ddd
ddd
ddd
eeeeedABD
T
kj
kj
ii
C
( ) ( ) ( ) ( )
( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
[ ] [ ]( )
( )
( ) ( )
( ) ( )
( ) ( )
[ ] [ ]( )
−−−−
−−−
−−
+
+++
+++
+++
=
Ω+Φ=−++=−++=
−+
−
3
2
1
2
1
32233113
32232112
31132112
321
3
2
1
2
1
333332233113
322322222112
311321121111
321
0
0
0
2
1
2
1
2
1
2
1
2
1
2
1
e
e
e
dddd
dddd
dddd
eee
e
e
e
dddddd
dddddd
dddddd
eee
eeddeeddDDDDD
TT
DDDD
kj
kjjk
kj
kjjk
symmetricanti
C
symmetric
C
The conjugate dyadic of is:D
Table of Content
21. 21
SOLO Dyadics
Identity Dyadic (Unit Dyadic, Idemfactor)
≠
=
==
kj
kj
k
j
k
j
0
1
δφ ( )
=++=++=
3
2
1
3213
3
2
2
1
13
3
2
2
1
1
100
010
001
e
e
e
eeeeeeeeeeeeeeeI
( ) ( ) ( ) ( ) VeVeVeVeeeeeeeeeeeeeVeVeVVIIV
=++⋅++=++⋅++=⋅=⋅ 3
3
2
2
1
13
3
2
2
1
13
3
2
2
1
1
3
3
2
2
1
1
j
m
kj
mk
k
m
k
m
kmmk
kmmk
k
mkm
ggeeeegeegege δδ ==⋅⋅=⋅== ,,,,
Using we obtain:
j
ij
ij
i
j
i
ji
ijji
ij
j
ij
i
j
ij
i
ggeegeegeegeegI δδ ====== ,
III
=⋅
( ) ( ) ( ) ( ) 3111::
321
=++==⋅⋅==
============ mkjimkjimkji
m
jk
i
m
kj
ij
m
k
i
m
kj
i
m
km
k
j
ij
i
eeeeeeeeII δδδδδδδδ
We also have:
m
j
k
k
jm
l
k
jm
l
k
jm
lm
lk
jk
j
eeddeeeeddeedeedDD
21212121
=⋅=⋅=⋅
DeedeedeedeeeedeeeedID k
jk
jk
jk
m
m
j
mk
m
jm
k
k
jm
l
k
jm
l
k
jm
lm
lk
jk
j
====⋅=⋅=⋅
↔
δδδδ
DeedeedeedeeeedeedeeDI k
lm
lk
lm
j
j
l
jl
m
jm
l
l
jm
l
k
jm
l
k
jm
lm
lk
jk
j
====⋅=⋅=⋅
↔
δδδδ
24. 24
SOLO Dyadics
Identity Dyadic (Unit Dyadic, Idemfactor) (continue – 3)
We found:
( ) ( )
DVeedVg
eedVgeedeeVgDIVDVI
km
ijmk
ji
km
l
j
k
l
ijm
ik
lk
ljm
ijm
i
×==
=⋅=⋅×=⋅×
ε
δεε
km
ijmk
jik
k
jm
ijm
ik
jk
j
i
i
eedVgedegVeedeVDV
εε ==×=×
( ) jm
ijm
ik
k
jm
ijm
ik
jk
j
i
ic
eeVgeegVeeeVIVVIIV
εδεδ ==×=×−=×=×
and:
( ) ( )
VDeedVgeedVg
eedVgeeVgeedIVDVID
mj
kim
k
j
i
ml
lj
ilk
k
j
i
ljm
k
k
jilm
ilm
ilm
i
k
jk
j
kililk
×===
=⋅=×⋅=×⋅
=
→
εε
δεε
εε
mj
kim
k
j
imj
kim
k
j
i
i
i
k
jk
j
eedVgeegdVeVeedVD
εε ==×=×
Using: lm
kmjlk
l
jl
m
jmkjk
dggdgdgd ===
( ) ( ) VDIVDVID
×=×⋅=×⋅
−=
otherwise
ofnpermutatiocyclicakji
ofnpermutatiocyclicakji
kji
0
3,1,2,,1
3,2,1,,1
,,
ε
30. 30
SOLO Dyadics
Coordinate Transformation of Dyadics in Reciprocal Coordinates
kj
jkk
jk
j
k
jk
j
kj
jk
kj
k
i
j
iii eedeedeedeedeeBABAD
======
nm
mnn
mn
m
n
mn
m
nm
mn
ii ffdffdffdffdBAD
=====
where
k
jk
jk
j
n
k
j
mn
mn
mn
m
eedeedffdD
=== βα
Let find the relation between defined in the bases and to
defined in the bases and .
i
e
i
e
i
f
i
f
k
j
d
n
m
d
[ ]
=
=
→
=
= −
3
2
1
1
3
2
1
3
3
3
2
3
1
2
3
2
2
2
1
1
3
1
2
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef T
i
j
i
k
k
j
mj
m
j
βββ
βββ
βββ
δαβ
β
[ ]
=
=
→
⋅=
=
3
2
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef
ef
ji
j
i
j
j
ii
ααα
ααα
ααα
α
α
n
k
j
mn
m
k
j
dd βα=
n
mm
n
n
k
k
j
j
mn
mm
n
k
j
n
k
j
mn
mm
n
k
jk
j
dddd
m
k
k
m
===
δδ
αββααββααβ
m
j
k
nk
j
n
m
dd αβ=
31. 31
SOLO Dyadics
Coordinate Transformation of Dyadics in Reciprocal Coordinates (continuous – 1)
where
[ ]
=
=
→
=
= −
3
2
1
1
3
2
1
3
3
3
2
3
1
2
3
2
2
2
1
1
3
1
2
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef T
i
j
i
k
k
j
mj
m
j
βββ
βββ
βββ
δαβ
β
[ ]
=
=
→
⋅=
=
3
2
1
3
2
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
2
1
e
e
e
L
e
e
e
f
f
f
ef
ef
ji
j
i
j
j
ii
ααα
ααα
ααα
α
α
m
j
k
nk
j
n
m
dd αβ=
We found
[ ]
[ ] [ ] [ ] 1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
:
−
=
=
=
LDL
ddd
ddd
ddd
ddd
ddd
ddd
D
βββ
βββ
βββ
ααα
ααα
ααα
Table of Content
33. 33
SOLO Dyadics
Dyadics Invariants (continue – 1)
We found the following three scalar invariants of a dyadic
Let Compute
( ) ( ) i
k
k
i
m
k
i
m
k
i
m
ki
ji
m
k
j
m
ki
ji
m
k
j
m
i
i
mk
jk
j
ddddddeeeeddeedeedDD ===⋅⋅== δδδ
::
( ) [ ] SDDscalarDtracedddI
===++= 3
3
2
2
1
1
1
( )1
3
3
1
1
2
2
1
2
3
3
2
3
3
2
2
3
3
1
1
2
2
1
1
2
ddddddddddddI −−−++=
[ ]DddddddddddddddddddI det2
3
1
2
3
1
1
3
3
2
2
1
3
3
2
2
1
1
3
3
1
2
2
1
2
2
1
3
3
1
2
3
3
2
1
1
3
=−−−++=
The matrix [ ]
=
333231
232221
131211
ddd
ddd
ddd
D
[ ]
−
−−
−
=
2221
1211
2321
1311
2322
1312
3231
1211
3331
1311
3332
1312
3231
2221
3331
2321
3332
2322
:
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
dd
D adj
The adjoin dyadic is defined as
( ) [ ]
=
3
2
1
321
:
e
e
e
DeeeD adjadj
( ) [ ]
=
3
2
1
321:
e
e
e
DeeeD
The adjoin dyadic is the fourth
dyadic invariant.D
adj
D
34. 34
SOLO Dyadics
Dyadics Invariants (continue – 2)
We have:
If we can define:
and
[ ] [ ] [ ][ ] 33det xadj IDDD =
[ ] 0det 3 ≠= ID [ ] [ ] [ ] [ ] [ ] [ ] 33
11
det/: xadj IDDDDD =→=
−−
( ) [ ]
=
−−
3
2
1
1
321
1
:
e
e
e
DeeeD
( ) [ ] ( ) [ ] ( ) [ ] I
e
e
e
Ieee
e
e
e
Deee
e
e
e
DeeeDD x
=
=
=⋅
−−
3
2
1
33321
3
2
1
1
321
3
2
1
321
1
35. 35
SOLO Dyadics
Dyadics Invariants (continue – 3)
The fifth dyadic invariant is the vector obtained by introducing the cross vector
product between the dyadic vectors
kj
jkk
jk
j
k
jk
j
kj
jk
kj
k
i
j
iii eedeedeedeedeeBABAD
======
kj
jkk
jk
j
k
jk
j
kj
jk
kj
k
i
j
iiiV
eedeedeedeedeeBABAD
×=×=×=×=×=×=
( ) ( )321 ,, eeegee
g
e kj
ijki
=×=
ε
Use
( ) ( ) ( )[ ]321122133113223
eddeddeddgedgD ijk
ijkV
−+−+−== ε
We defined the decomposition of a dyadic in symmetric and anti-symmetric parts:
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )
−−−−
−−−
−−
+
+++
+++
+++
=−++=
−
ΩΦ
3
2
1
32233113
32232112
31132112
321
3
2
1
333332233113
322322222112
311321121111
321
0
0
0
2
1
2
1
2
1
2
1
e
e
e
dddd
dddd
dddd
eee
e
e
e
dddddd
dddddd
dddddd
eeeDDDDD
symmetricanti
C
symmetric
C
Since the matrix representation of the vector cross-product of isV
D
( )
( )
( )
( ) ( )
( ) ( )
( ) ( )
−−−
−−−
−−−
=×
−
−
−
=×
0
0
0
32231331
32232112
13312112
2112
1331
3223
dddd
dddd
dddd
g
dd
dd
dd
gDV
i.e. the
antisymmetric
part of
multiplied by
D
g
Table of Content
36. 36
SOLO
Accordingly we can classify the dyadics as follows:
Dyadics
Classification of Dyadics
Physically, dyadics describe at each point the properties of the field that relate an
input or cause vector to an output or effect vector. If the family of input vectors includes
all magnitudes and directions, then one class of dyadics produces families of output
vectors that also include all magnitudes and directions. Dyadics of this class are called
“complete”. All others are called “incomplete”.
[ ] 0det 3
≠= IDIf
CompleteThe three rows/columns of
[D] are linearly independent
Property Comment Classification
[ ] 0&0det 3
≠== adj
DIDIf PlanarOnly two rows/columns of
[D] are linearly independent
LinearThe three rows/columns of
[D] are linearly dependent
If [ ] 0&0det 3
=== adjDID
A Planar Dyadic can be reduced by a suitable coordinate transformation to the
sum of two dyads (no less) 2211 BABAD
+=
A Linear Dyadic can be reduced by a suitable coordinate transformation to
a single dyad 11BAD
= Table of Content
37. 37
SOLO Dyadics
Differentiation of Dyadics
Define:
( )tD
Suppose we have a dyadics and the vector that are differentiable
functions of the parameter scalar t.
( )tV
( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )[ ] ( )tVtBtAtVtDtWtBtAtVtDtVtU
⋅=⋅=⋅=⋅= :&:
( ) td
Dd
VD
td
Vd
td
Bd
AVB
td
Ad
VBA
td
Vd
td
Ud
⋅+⋅=
⋅+
⋅+⋅=
where:
+
=
td
Bd
AB
td
Ad
td
Dd
:
( ) V
td
Dd
td
Vd
DV
td
Bd
AVB
td
Ad
td
Vd
BA
td
Wd
⋅+⋅=⋅
+⋅
+⋅=
( ) ( ) V
td
Dd
td
Vd
DVD
td
d
td
Dd
VD
td
Vd
DV
td
d
⋅+⋅=⋅⋅+⋅=⋅ &
( ) ( ) V
td
Dd
td
Vd
DVD
td
d
td
Dd
VD
td
Vd
DV
td
d
×+×=××+×=× &
In the same way:
38. 38
SOLO Dyadics
Differentiation of Dyadics
( )zyxV ,,
Gradient of a Vector .
( ) [ ]
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=++
∂
∂
+
∂
∂
+
∂
∂
=∇
z
y
x
z
V
z
V
z
V
y
V
y
V
y
V
x
V
x
V
x
V
zyxzVyVxV
z
z
y
y
x
xV
zyx
zyx
zyx
zyx
1
1
1
111111111
This id a dyadic.
Let compute:
( ) ( ) [ ]
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
++=∇
z
y
x
z
V
y
V
x
V
z
V
y
V
x
V
z
V
y
V
x
V
zyx
z
z
y
y
x
xzVyVxVV
zyx
zyx
zyx
zyx
c
1
1
1
111111111
39. 39
SOLO Dyadics
Differentiation of Dyadics
( )zyxV ,,
Gradient of a Vector (continue – 1) .
( )[ ] ( )[ ]
[ ]
[ ]
∂
∂
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
+
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
+
∂
∂
−
∂
∂
+
+
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∇−∇+∇+∇=∇
z
y
x
z
V
z
V
y
V
z
V
x
V
z
V
y
V
y
V
x
V
z
V
x
V
y
V
x
V
zyx
z
y
x
z
V
z
V
y
V
z
V
x
V
z
V
y
V
y
V
y
V
x
V
z
V
x
V
y
V
x
V
x
V
zyx
VVVVV
zyzxz
yzxy
xzxy
zyzxz
yzyxy
xzxyx
CC
1
1
1
2
1
2
1
2
1
0
2
1
2
1
2
1
0
111
1
1
1
2
1
2
1
2
1
2
1
2
1
2
1
111
2
1
2
1
Let decompose the gradient of the vector in the symmetric and anti-symmetric parts.
40. 40
SOLO Dyadics
Differentiation of Dyadics
( )zyxV ,,
Gradient of a Vector (continue – 2)
Let find the scalar and vector invariants of [ ]
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=∇
z
y
x
z
V
z
V
z
V
y
V
y
V
y
V
x
V
x
V
x
V
zyxV
zyx
zyx
zyx
1
1
1
111
( ) [ ]
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=∇
z
y
x
z
V
y
V
x
V
z
V
y
V
x
V
z
V
y
V
x
V
zyxV
zyx
zyx
zyx
c
1
1
1
111
[ ] ( )[ ] V
z
V
y
V
x
V
VV zyx
S
c
S
⋅∇=
∂
∂
+
∂
∂
+
∂
∂
=∇=∇ Divergence of V
[ ]
V
z
z
V
y
V
y
x
V
z
V
x
y
V
x
V
V
yzzxxy
V
×∇=
∂
∂
−
∂
∂
+
∂
∂
−
∂
∂
+
∂
∂
−
∂
∂
=∇ 111
( )[ ]
∂
∂
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
+
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
+
∂
∂
−
∂
∂
+
=∇−∇
z
V
z
V
y
V
z
V
x
V
z
V
y
V
y
V
x
V
z
V
x
V
y
V
x
V
VV
zyzxz
yzxy
xzxy
c
2
1
2
1
2
1
0
2
1
2
1
2
1
0
2
1
( )[ ]
V
z
z
V
y
V
y
x
V
z
V
x
y
V
x
V
V
yzzxxy
V
c
×∇−=
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−
∂
∂
−=∇ 111
Rotor of V
42. 42
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
ELECTROMAGNETICSSOLO
The Dyadic (Matrix) Green’s function is the solution of the vector equation( )SF rrG
,
( ) ( )SFSS rrIGkG
−=−×∇×∇ δπ42
where is the unit dyadic or the identity matrix.I
( ) ( )
( ) ( ) ( )
( ) ( )GrrIGkG
GGG
rrIGkG
SSSFS
SSSSSS
SFSS
⋅∇∇+−−=+∇⇒
⇒
∇⋅∇−⋅∇∇=×∇×∇
−=−×∇×∇
δπ
δπ
4
4
22
2
( ) ( )
( ) ( )
⇒
=×∇⋅∇×∇=×∇×∇⋅∇→=∇×∇
−=−×∇×∇⋅∇
00
42
GG
rrIGkG
SSSSSSSS
SFSSS
δπ
and
( ) ( )
( ) ⇒−∇−=⋅∇⇒
⇒−∇=−⋅∇=⋅∇−⇒
SFSS
SFSSFSS
rr
k
G
rrrrIGk
δ
π
δπδπ
2
2
4
44
( ) ( )SFSSSS rr
k
G
−∇∇−=⋅∇∇⇒ δ
π
2
4
Therefore ( ) ( )
( ) ( )
( )SFSSS
SFSSSS
SSSFS
rr
k
IGkG
rr
k
G
GrrIGkG
−
∇∇+−=+∇
⇒
−∇∇−=⋅∇∇
⋅∇∇+−−=+∇
δπ
δ
π
δπ
2
22
2
22
1
4
4
4
43. 43
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 1)
ELECTROMAGNETICSSOLO
( )SFSSS
rr
k
IGkG
−
∇∇+−=+∇ δπ 2
22 1
4
The form of the above equation suggests that can be written in terms of a
Scalar Green’s function as
( )SF rrG
,
( )SF rr
,ψ
( ) ( )SFSSSF rr
k
IrrG
,
1
, 2
ψ
∇∇+=
To find let perform the following calculations:( )SF rr
,ψ
( ) ( )
( ) ψ
ψ
ψ
∇∇+−∇−⋅∇∇=
∇∇+−×∇×∇=−×∇×∇
⇒
∇∇+=
SSSSS
SSSSSS
SS
k
Ik
k
IkGkG
k
IG
2
22
2
22
2
1
1
1
( ) ( )
( ) ( )
( ) ( ) ( )ψψψ
ψψ
ψψ
ψψψψψψ
22
0
222
2222
2222
22222
kIkkI
kkI
kkI
kIkIk
SSSSSS
SSSSSS
SSSSSSSS
SSSSSSSSSSSS
SS
+∇−=∇∇×∇×∇++∇−=
∇∇−∇⋅∇∇++∇−=
∇∇∇−∇∇⋅∇∇++∇−=
∇∇−−∇∇∇−∇−∇∇⋅∇∇+∇∇=
=∇×∇
−
−
−
−−
We can see that: ( ) ( ) ( )SFSSS rrIkIGkG
−=+∇−=−×∇×∇ δπψ 4222
44. 44
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 2)
ELECTROMAGNETICSSOLO
We found that the solution of this equation is:
( ) ( ) ( )SFSSS rrIkIGkG
−=+∇−=−×∇×∇ δπψ 4222
Therefore satisfies the scalar wave equation:( )SF rr
,ψ
( ) ( ) ( )SFSFSFS rrrrkrr
−−=+∇ δπψψ 4,, 22
( ) ( )
SFSF rrrwhere
r
rkj
rr
−=
−
=
exp
,ψ
45. 45
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 3)
ELECTROMAGNETICSSOLO
Using the Second Vector Green Identity
( )( )( ) ( ) ( )( )[ ] ( ) ( ) ( )( )[ ]∫∫
→
⋅⋅×∇×−×∇×⋅=×∇×∇⋅⋅−⋅×∇×∇⋅=
S
SS
V
SSSSV dSnaGEEaGdVEaGaGEI 1
where is an arbitrary constant vectora
iS
nS
n
i
iSS
1=
=dV
dSn
→
1
V
Fr
Sr
F
0r
SF rrr
−= iS
nS
dV
dSn
→
1
V
Fr
Sr
F
0r SF rrr
−=
We have
and we get
( ) ( )( )aGaG SS
consta
SS
⋅×∇×∇=⋅×∇×∇
=
( )( )( ) ( ) ( )( )[ ]
( )[ ] ( ) [ ]{ }
( ) ( )[ ]dVJJjGaaE
dVJJjEkGarraaGkE
EaGaGEI
V
mSe
V
mSeSF
V
SSSSV
∫
∫
∫
×∇+⋅⋅+⋅=
×∇−−⋅⋅−−+⋅⋅=
×∇×∇⋅⋅−⋅×∇×∇⋅=
ωµπ
ωµδπ
4
4 22
We used the fact that, since the sources and the observation point are both
in the volume V,
Sr
Fr
( ) aEdVrraE
V
SF
⋅=−⋅∫ δ
( )[ ] ( ) ( ) ( )( )[ ]∫∫ ⋅×∇×−×∇×⋅⋅+
×∇+⋅⋅−=⋅
→
S
SS
V
mSe dSaGEEaGndVJJjGaEa
14 ωµπ
Therefore we obtain
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
46. 46
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 4)
ELECTROMAGNETICSSOLO
Let develop now the expression
( )[ ] ( ) ( ) ( )( )[ ]∫∫ ⋅×∇×−×∇×⋅⋅+
×∇+⋅⋅−=⋅
→
S
SS
V
mSe dSaGEEaGndVJJjGaEa
14 ωµπ
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 1)
( ) ( ) ( )( )[ ]aGEEaGn SS
⋅×∇×−×∇×⋅⋅
→
1
( ) ( ) ( )
a
k
aa
k
aaaG
S
consta
SSSS
consta
SSSS
×∇=
=∇×∇∇⋅+×∇=
∇∇⋅+×∇=⋅×∇
→
→
=
=
ψ
ψψψ 2
0
2
11
and
( )( ) ( ) ( )
( ) aEnaEn
aEnnaEnaGE
SS
SSS
⋅∇×
×=×∇⋅
×=
=×∇×⋅=⋅×∇×=⋅⋅×∇×
→→
→→→
ψψ
ψψ
11
111
47. 47
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 5)
ELECTROMAGNETICSSOLO
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 2)
Since is symmetric andG
k
IG SS
,
1
2
ψ
∇∇+= GaaG
⋅=⋅
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( )Ena
k
Ena
En
k
IaEnGa
nEGanEaGnEaG
SSSS
SSSS
SSS
×∇×⋅∇∇⋅−×∇×⋅−=
×∇×⋅
∇∇+⋅−=
×∇×⋅⋅−=
××∇⋅⋅=××∇⋅⋅=⋅×∇×⋅
→→
→→
→→→
1
1
1
1
1
1
111
2
2
ψψ
ψ
( ) ( ) ( )
( ) ( ) ( ) ( ) addEna
k
Ena
k
subtractEna
k
Ena
SSSSSS
SSSS
←×∇×∇⋅∇⋅+×∇×⋅∇∇⋅−
←×∇×∇⋅∇⋅−×∇×⋅−=
→→
→→
1
1
1
1
1
1
1
22
2
ψψ
ψψ
But since ( ) ( ) ⇒=∇⋅∇=×∇×∇⇒==∇×∇
→
0&0 aaconsta SSSSSS
ψψψ
( ) ( ) ( ) ( ) ( )
( ) ψ
ψψψψψ
SS
SSSSSSSSSS
a
aaaaa
∇∇⋅=
∇⋅∇+∇∇⋅+×∇×∇+∇×∇×=∇⋅∇
we can develop the following expression
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
→→
→→
⋅×∇×∇∇⋅+××∇⋅∇⋅∇=
×∇×∇⋅∇⋅+×∇×⋅∇∇⋅−
nEa
k
nEa
k
Ena
k
Ena
k
SSSSSS
SSSSSS
1
1
1
1
1
1
1
1
22
22
ψψ
ψψ
( ) ( ) ( ) ( )[ ]
( ) ( )[ ]
→
→
⋅×∇∇⋅×∇=
⋅×∇×∇∇⋅+×∇×∇⋅∇=
nEa
k
nEaEa
k
SSS
SSSSSS
1
1
1
1
2
2
ψ
ψψ
48. 48
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 6)
ELECTROMAGNETICSSOLO
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 3)
We get therefore
( ) ( ) =⋅×∇×⋅
→
nEaG S 1
( ) ( ) ( )
( ) ( ) ( ) ( )Ena
k
Ena
k
Ena
k
Ena
SSSSSS
SSSS
×∇×∇⋅∇⋅+×∇×⋅∇∇⋅−
×∇×∇⋅∇⋅−×∇×⋅−=
→→
→→
1
1
1
1
1
1
1
22
2
ψψ
ψψ
( ) ( ) ( )Ena
k
Ena SSSS ×∇×∇⋅∇⋅−×∇×⋅−=
→→
1
1
1 2
ψψ
( ) ( )[ ]
→
⋅×∇∇⋅×∇+ nEa
k
SSS 1
1
2
ψ
( )( ) aEnnaGE SS
⋅∇×
×=⋅⋅×∇×
→→
ψ11We found that
therefore ( ) ( ) ( )( )[ ]aGEEaGn SS
⋅×∇×−×∇×⋅⋅
→
1
( ) ( ) ( )Ena
k
Ena SSSS
×∇×∇⋅∇⋅−×∇×⋅−=
→→
1
1
1 2
ψψ
( ) ( )[ ] aEnnEa
k
SSSS
⋅∇×
×−⋅×∇∇⋅×∇+
→→
ψψ 11
1
2
( ) ( )
×∇×∇⋅∇+∇×
×+×∇×⋅−=
→→→
En
k
EnEna SSSSS 1
1
11 2
ψψψ
( ) ( )[ ]
→
⋅×∇∇⋅×∇+ nEa
k
SSS 1
1
2
ψ
49. 49
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 7)
ELECTROMAGNETICSSOLO
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 4)
Since and we
get
( ) ( ) ( )( )[ ]aGEEaGn SS
⋅×∇×−×∇×⋅⋅
→
1 ( ) ( )
×∇×∇⋅∇+∇×
×+×∇×⋅−=
→→→
En
k
EnEna SSSSS 1
1
11 2
ψψψ
( ) ( )[ ]
→
⋅×∇∇⋅×∇+ nEa
k
SSS 1
1
2
ψ
( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
µεω 22
=k
( ) ( ) ( )( )[ ]
( ) ( )
( ) ( )[ ]
→
→→→→
→
⋅×∇∇⋅×∇+
∇
×∇+⋅−∇
⋅+∇×
×+×∇×⋅−=
⋅×∇×−×∇×⋅⋅
nEa
k
k
JJjnEnEnEna
aGEEaGn
SSS
S
mSeSSS
SS
1
1
1111
1
2
2
ψ
ψ
ωµψψψ
50. 50
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 8)
ELECTROMAGNETICSSOLO
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 5)
Let compute
( ) ( ) ( )( )[ ]∫
→
⋅⋅×∇×−×∇×⋅
S
SS dSnaGEEaG 1
( ) ( )∫
∇
×∇+⋅−∇
⋅+∇×
×+×∇×⋅−=
→→→→
S
S
mSeSSS dS
k
JJjnEnEnEna 2
1111
ψ
ωµψψψ
( ) ( )[ ] dSnEa
k S
SSS∫
→
⋅×∇∇⋅×∇+ 1
1
2
ψ
In our case the integral is performed over a closed surface S and therefore the last
integral is (using Gauss’ 5 Theorem: ):∫∫ ×∇=×
→
VS
dvAdSAn
1
( ) ( )[ ] ( ) ( )[ ] ( ) ( )[ ] 011
0
5
=×∇∇⋅⋅∇×∇=××∇∇⋅⋅∇=⋅×∇∇⋅×∇ ∫∫∫
→→
V
SSSS
Gauss
S
SSS
S
SSS dvEadSnEadSnEa
ψψψ
Compute (using Gauss’ 4 Theorem: ):( )[ ]∫∫ ⋅∇+∇⋅=
⋅
→
VS
dvABBAdSnAB
1
( )
( )
( ) ∫∫
∫∫
∫
∇⋅+
∇∇
×∇+⋅=
×∇⋅∇+⋅∇
∇
⋅+
∇∇
×∇+⋅=
×∇+⋅
∇
⋅
=
−
→
V
S
e
V
SS
mSe
k
V
mSS
j
S
S
V
SS
mSe
Gauss
S
mSe
S
dvadv
k
JJja
dvJJj
k
adv
k
JJja
dSJJjn
k
a
e
ψ
ε
ρψ
ωµ
ωµ
ψψ
ωµ
ωµ
ψ
µεω
ρω
2
0
22
4
2
2
1
51. 51
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 8)
ELECTROMAGNETICSSOLO
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 5)
Let substitute this result in
( ) ( ) ∫∫∫ ∇⋅+
∇∇
×∇+⋅=
×∇+⋅
∇
⋅
→
V
S
e
V
SS
mSe
S
mSe
S
dvadv
k
JJjadSJJjn
k
a ψ
ε
ρψ
ωµωµ
ψ
22
1
( )[ ] ( ) ( ) ( )( )[ ]∫∫
→
⋅⋅×∇×−×∇×⋅+
×∇+⋅⋅−=⋅
S
SS
V
mSe dSnaGEEaGdVJJjGaEa 14
ωµπ
( )
×∇+⋅
∇∇
+⋅−=⋅ ∫ dVJJj
k
IaEa
V
mSe
SS
ωµψπ 2
4
( )∫
∇
⋅+∇×
×+×∇×⋅−
→→→
S
SSS dSEnEnEna ψψψ 111
( ) ∫∫ ∇⋅+
∇∇
×∇+⋅+
V
S
e
V
SS
mSe dvadv
k
JJja ψ
ε
ρψ
ωµ
2
we obtain
Since this is true for all constant vectors , after simplification and rearranging terms,
we obtain
a
( )∫∫
∇
⋅+∇×
×+×∇×−
∇−×∇+−=
→→→
S
SSS
V
S
e
mSe dSEnEnEndVJJjE ψψψ
π
ψ
ε
ρ
ψψωµ
π
111
4
1
4
1
52. 52
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 9)
ELECTROMAGNETICSSOLO
Solution of the equation: ( ) mSeSS JJjEkE ×∇−−=−×∇×∇ ωµ2
(continue – 6)
Using
( )∫∫
∇
⋅+∇×
×+×∇×−
∇−×∇+−=
→→→
S
SSS
V
S
e
mSe
dSEnEnEndVJJjE ψψψ
π
ψ
ε
ρ
ψψωµ
π
111
4
1
4
1
we obtain
We recovered Stratton-Chu solution
Using the duality relations
we can write
( ) ( ) ∫∫∫∫∫ ×∇−×=×∇−×∇=×∇
→
V
mS
S
m
Gauss
V
mS
V
mS
V
mS
dVJdSJndVJdVJdVJ ψψψψψ 1
5
( )∫∫
∇
⋅+∇×
×++×∇×−
∇−×∇−−=
→→→
S
SSmS
V
S
e
mSe
dSEnEnJEndVJJjE ψψψ
π
ψ
ε
ρ
ψψωµ
π
111
4
1
4
1
⇓
⇓
−
⇓
⇓
−
⇓
⇓
−
⇓
⇓
µ
ε
ε
µ
ρ
ρ
ρ
ρ
e
m
m
e
e
m
m
e
J
J
J
J
E
H
H
E
( )∫∫
∇
⋅+∇×
×+−×∇×−
∇−×∇+−=
→→→
S
SSeS
V
S
m
eSm dSHnHnJHndVJJjH ψψψ
π
ψ
µ
ρ
ψωεψ
π
111
4
1
4
1
53. 53
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 10)
ELECTROMAGNETICSSOLO
Discontinuous Surface Distribution
Stratton-Chu equations are valid only if the vectors are continuous and have
continuous derivatives on the S surface. They cannot be applied, therefore, to the
problem of diffraction at a slit.
HE
,
Suppose we have a slit of surface S1 with the
curve C serving as his boundary. Let assume
any surface S2 closed at infinity that
complements the surface S1 and has in
common the curve C. Assume no sources
0,0,0,0 ==== meme JJ ρρ
Assume also that on S2 we have 0,0 22
== HE
ConkHnkEn me
=×−=×
→→
11 1,1
To overcome the discontinuity problem assume that on curve C we have a
distribution of charges such that
54. 54
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 11)
ELECTROMAGNETICSSOLO
Discontinuous Surface Distribution (continue – 1)
Let return to
( ) ( ) ( )( )[ ]
( ) ( ) ( )( )[ ]∫
∫∫
⋅×∇×−×∇×⋅⋅=
⋅×∇×−×∇×⋅⋅+
×∇+⋅⋅−=⋅
→
+
→
1
21
1
14
00
S
SS
SS
SS
V
mSe
dSaGEEaGn
dSaGEEaGndVJJjGaEa
ωµπ
We found ( ) ( ) ( )( )[ ]
( )
( ) ( )[ ]
→
→→→→
→
⋅×∇∇⋅×∇+
∇
×∇+⋅−∇
⋅+∇×
×+×∇×⋅−=
⋅⋅×∇×−×∇×⋅
nEa
k
k
JJjnEnEnEna
naGEEaG
SSS
S
mSeSSS
SS
1
1
1111
1
2
2
00
ψ
ψ
ωµψψψ
Using Stokes’ Theorem: we have∫∫ ⋅=⋅×∇
CS
rdASdA
( ) ( )[ ] ( ) ( )∫∫∫ ⋅×∇∇⋅=⋅×∇∇⋅=⋅×∇∇⋅×∇
→
C
SS
C
SS
Stokes
S
SSS rdEardEadSnEa
ψψψ
1
1
Therefore
( ) ( ) ( )( )[ ]
( ) ( )∫∫
∫
⋅×∇∇⋅+
∇
⋅+∇×
×+×∇×⋅−=
⋅×∇×−×∇×⋅⋅=⋅
→→→
→
C
SS
S
SSS
S
SS
rdEa
k
dSEnEnEna
dSaGEEaGnEa
ψψψψ
π
2
1
111
14
1
1
55. 55
Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations
(continue – 12)
ELECTROMAGNETICSSOLO
Discontinuous Surface Distribution (continue – 2)
Using the duality relations
( ) ( )∫∫ ⋅×∇∇⋅+
∇
⋅+∇×
×+×∇×⋅−=⋅
→→→
C
SS
S
SSS
rdEa
k
dSEnEnEnaEa
ψψψψπ 2
1
1114
1
Since this is true for all constant vectors , we obtaina
( ) ( )∫∫ ⋅×∇∇+
∇
⋅+∇×
×+×∇×−=
→→→
C
SS
S
SSS rdE
k
dSEnEnEnE
ψ
π
ψψψ
π 2
4
1
111
4
1
1
Using and we getµεω 22
=kHjES µω=×∇
∫∫ ⋅∇+
∇
⋅+∇×
×+
×−=
→→→
C
S
S
SS rdH
j
dSEnEnHnjE
ψ
εωπ
ψψψµω
π 4
111
4
1
1
we can write
⇓
⇓
−
⇓
⇓
−
⇓
⇓
−
⇓
⇓
µ
ε
ε
µ
ρ
ρ
ρ
ρ
e
m
m
e
e
m
m
e
J
J
J
J
E
H
H
E
∫∫ ⋅∇−
∇
⋅+∇×
×+
×−−=
→→→
C
S
S
SS rdE
j
dSHnHnEnjH
ψ
µωπ
ψψψεω
π 4
111
4
1
1
56. 56
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References
[1] Vavra, M.H., “Aero-Thermodynamics and Flow Turbomachines”,
John Wiley & Sons, 1960
Appendix B: ”Introduction to Operations Involving Dyadics”, pp.531-557
Dyadics
[2] Reddy, J.N. & Rasmussen, M.L., “Advanced Engineering Analysis”,
John Wiley & Sons, 1982, Ch. 1.5: Dyadics and Tensors, pp.107-152
[3] Chou, P.C., Pagano, N.J., “Elasticity - Tensor, Dyadic and Engineering Approaches”,
Dover, 1992, Ch. 11: Vector and Dyadic Notation in Elasticity, pp.225-244
[4] Chen-To Tai, “Dyadic Green Functions in Electromagnetic Theory”, 2nd
Ed.,
IEEE Press, 1993
57. January 6, 2015 57
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Technion
Israeli Institute of Technology
1964 – 1968 BSc EE
1968 – 1971 MSc EE
Israeli Air Force
1970 – 1974
RAFAEL
Israeli Armament Development Authority
1974 – 2013
Stanford University
1983 – 1986 PhD AA