1. 1 Dyadics SOLO HERMELIN Updated: 9.04.07http://www.solohermelin.com

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 Greens 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 = == 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 i j i j j j i j ji AAeeAeA === == 3 1 3 1 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 is true if and only if 0321 === 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 eeeeee eeeeee eeeeee eeeeee eeeeee eeeeee Therefore 1= 2= 3=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 = = = 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 == j i kj ik gg = 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 = 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 ++= Let find 1 , 2 , 3 . Multiply the previous equation (scalar product) by .1e ( ) ( ) i i ggggeeeeee 113 3 12 2 11 1 321321 ,, =++== Multiply this equation by g1i : ( ) ii i ii ggeeeg == 1 1 1321 1 ,, Therefore: ( )321 1 ,, eeeg ii = Let compute now: ( ) ( ) ( ) ( ) ( ) ( )321 1 0 323 3 0 322 2 321 1 3232 eeeeeeeeeeeeeeee =++= ( ) ( )[ ] ( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( ) ( )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 == = = = = From those equations we obtain: ( )321 11 1 ,, eee gg = Finally: ( ) ( ) geeeeee == 321 2 321 1 ,,,, 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 where j i j i ef = 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 where j ij i ef = Because are the coefficients of the inverse matrix with coefficients : j i j i i j i k k j = 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 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 i i j j fFeEA == iijj fAFeAE == & i j ji i i j j j j i i EFfEeEfF === 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 and i i j j fFeEA == iijj fAFeAE == & then: ( )321 ,, fff ( )321 ,, eee where j ij ef ijjjjii EfeEeEAfAF j iji = ==== 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 == & jm m k j imj m k j ikiik geeffg === Hence: jm m k j iik gg = 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 Since we have:k iki fgf = ( ) m jm j i ege j j i ef k iki egefgf m jmjj j ii == === and: ( ) ( ) m jm j i km kjm j i gg k ik egfgfg mjm m k j iik == = 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 We found the relation: ( )jm j i g 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 [ ] = = = 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 km k m fe = jkj k km k j m mj m fffe === From the relation: mk m kjj i i efef == & we have: jmk m j i mjk m j i kiik geeffg === or: jmk m j i ik gg = This is a contravariant relation of rank two. From the relation: m m kk jj i i efef == & we have: m j m k i jm jm k i j i kk i eeff === or: This is a relation once covariant and once contravariant of rank two. m j m k i j i k = 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 ===== ,,,, Using we obtain: jk mklj k lmkm j ljlm dggdggd === 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 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] [ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] [ ]( ) + +++ +++ +++ = +=++=++= + 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 ( ) =++=++= 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 ===== ,,,, Using we obtain: j ij ij i j i ji ijji ij j ij i j ij i ggeegeegeegeegI ====== , III = ( ) ( ) ( ) ( ) 3111:: 321 =++==== ============ mkjimkjimkji m jk i m kj ij m k i m kj i m km k j ij i eeeeeeeeII 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 ====== DeedeedeedeeeedeedeeDI k lm lk lm j j l jl m jm l l jm l k jm l k jm lm lk jk j ====== 22. 22 SOLO Dyadics Identity Dyadic (Unit Dyadic, Idemfactor) (continue 1) ( ) ( ) ( ) ( ) ( )[ ]211231331232231 3 3 2 2 1 13 3 2 2 1 1 eeeeVeeeeVeeeeVg eeeeeeeVeVeVIV ++= ++++= ( ) ( ) ( ) ( )32121 3 13 2 32 1 ,, 111 eeegee g eee g eee g e ==== We found: Using: ( ) WVWIVWIV ==VVIIV == ( ) ( ) ( ) ( ) ( )[ ]211231331232231 3 3 2 2 1 1 3 3 2 2 1 1 eeeeVeeeeVeeeeVg eVeVeVeeeeeeVI ++= ++++= ( )c IVVIIV == ( ) ( ) ( ) ( )( ) ( ) = ++= ===== 3 2 1 12 13 23 321 211231331232231 0 0 0 e e e VV VV VV eeeg eeeeVeeeeVeeeeVg eeVgeegVeeeVIVVIIV jm ijm ik k jm ijm ik jk j i i c 23. 23 SOLO Dyadics Identity Dyadic (Unit Dyadic, Idemfactor) (continue 2) ( ) ( ) ( ) = = === == 3 2 1 12 13 23 12 13 23 12 13 23 12 13 23 3 2 1 23321331 32231221 31132112 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 e e e VV VV VV WW WW WW WW WW WW VV VV VV g e e e WVWVWVWV WVWVWVWV WVWVWVWV g WVIWVWIVWVI VIVIVVI ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] WVeeWVWVeeWVWVeeWVWV eWVWVeWVWVeWVWVg eWeWeWeeeeVeeeeVeeeeVgWIVWVI k m k m ee =++= ++= ++++== = 21 1221 13 3113 32 2332 312212311312332 3 3 2 2 1 1211231331232231 ( ) ( ) WVWIVWVI == ( ) ( ) ( ) ( )[ ]211231331232231 eeeeVeeeeVeeeeVgIVVIIV c ++=== 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 == === km ijmk jik k jm ijm ik jk j i i eedVgedegVeedeVDV === ( ) jm ijm ik k jm ijm ik jk j i ic eeVgeegVeeeVIVVIIV ===== 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 === === = mj kim k j imj kim k j i i i k jk j eedVgeegdVeVeedVD === Using: lm kmjlk l jl m jmkjk dggdgdgd === ( ) ( ) VDIVDVID == = otherwise ofnpermutatiocyclicakji ofnpermutatiocyclicakji kji 0 3,1,2,,1 3,2,1,,1 ,, 25. 25 SOLO Dyadics Identity Dyadic (Unit Dyadic, Idemfactor) (continue 4) ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = +++ +++ ++ = ++= === 3 2 1 3 3 2 3 1 3 3 2 2 2 1 2 3 1 2 1 1 1 12 13 23 321 3 2 1 3 12 3 21 2 12 2 21 1 12 1 21 3 31 3 13 2 31 2 13 1 31 1 13 3 23 3 32 2 23 2 32 1 23 1 32 321 31 3 22 3 133 3 11 3 322 3 33 3 21 21 2 22 2 133 2 11 2 322 2 33 2 21 11 1 22 1 133 1 11 1 322 1 33 1 21 122133113223321 0 0 0 e e e ddd ddd ddd VV VV VV eeeg e e e dVdVdVdVdVdV dVdVdVdVdVdV dVdVdVdVdVdV eeeg eededVededVededV eededVededVededV eededVededVededV g eededVededVededVg eedVgedegVeedeVDV k kkkkkk km ijmk jik k jm ijm ik jk j i i = otherwise ofnpermutatiocyclicakji ofnpermutatiocyclicakji kji 0 3,1,2,,1 3,2,1,,1 ,, 26. 26 SOLO Dyadics Identity Dyadic (Unit Dyadic, Idemfactor) (continue 5) Let compute: m ijm jim ijm ji j j i i eWVgegWVeWeVWV === and: ( ) ( ) WVVWeeWVWVeeWVeeWVeeWV eeWVeWVgeeWVI ji jiij jnik ji ji injk ij ji for nk kmn ijm ji n lmn k k lijm jim ijm jil k k l ijm ==== == ==== = ,, 1 ( ) ji jiji j j i i j j i i eeVWWVeWeVeVeWWVVW == g e ee nnmlml ,, = ( ) ( ) ( ) ( ) = == 3 2 1 12 13 23 12 13 23 12 13 23 12 13 23 321 3 2 1 23321331 32231221 31132112 321 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 e e e VV VV VV WW WW WW WW WW WW VV VV VV eee e e e WVWVWVWV WVWVWVWV WVWVWVWV eeeeeWVWVWVI ji jiij == otherwise ofnpermutatiocyclicanml ofnpermutatiocyclicanml nml nml 0 3,1,2,,1 3,2,1,,1 :,, ,, m ijmji egee = 27. 27 SOLO Dyadics Identity Dyadic (Unit Dyadic, Idemfactor) (continue 2) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = + + + = ++++== = 3 2 1 12 13 23 12 13 23 321 3 2 1 2 2 1 1 3 2 3 1 2 3 3 3 1 1 2 1 1 3 1 2 3 3 2 2 321 3 3 2 2 1 1 211231331232231 0 0 0 0 0 0 ,, e e e WW WW WW VV VV VV eee e e e WVWVWVWV WVWVWVWV WVWVWVWV eee eWeWeWeeeeVeeeeVeeeeVgWIVWVI g e ee kkjiji ( ) ( ) ( ) ( )[ ]211231331232231 eeeeVeeeeVeeeeVgIVVIIV c ++=== ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = + + + = ++++== = 3 2 1 12 13 23 12 13 23 321 3 2 1 2 2 1 1 2 3 1 3 3 2 3 3 1 1 1 2 3 1 2 1 3 3 2 2 321 2112313312322313 3 2 2 1 1 0 0 0 0 0 0 ,, e e e VV VV VV WW WW WW eee e e e WVWVWVWV WVWVWVWV WVWVWVWV eee eeeeVeeeeVeeeeVgeWeWeWIVWVIW g e ee kkjiji 28. 28 SOLO Dyadics Identity Dyadic (Unit Dyadic, Idemfactor) (continue 2) ( ) ( ) ( ) ( ) ( ) ( ) ( ) VWWVWVI e e e VV VV VV WW WW WW WW WW WW VV VV VV eee e e e WVWVWVWV WVWVWVWV WVWVWVWV eee IVWWIVVIWWVI == = = == 3 2 1 12 13 23 12 13 23 12 13 23 12 13 23 321 3 2 1 23321331 32231221 31132112 321 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ( ) ( ) ( ) == 3 2 1 12 13 23 12 13 23 321 0 0 0 0 0 0 e e e WW WW WW VV VV VV eeeWIVWVI ( ) ( ) ( ) == 3 2 1 12 13 23 12 13 23 321 0 0 0 0 0 0 e e e VV VV VV WW WW WW eeeIVWVIW 29. 29 SOLO Dyadics Identity Dyadic Algebra (Summary) 3 3 2 2 1 13 3 2 2 1 1 eeeeeeeeeeeeI ++=++= VVIIV == III = 3: =II DDIID == ( ) WVWIVWIV == ( )c IVVIIV == ( ) ( ) WVWIVWVI == ( ) ( ) DVDIVDVI == ( ) ( ) VDIVDVID == ( ) WVVWWVI = Table of Content Identity Dyadic (Unit Dyadic, Idemfactor) (continue 6) ( ) ( ) ( ) ( ) ( ) VWWVWVIIVWWIVVIWWVI === 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 === 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 [ ] = = = = 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 n k j mn m k j dd = 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 === m j k nk j n m dd = 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 [ ] = = = = 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 m j k nk j n m dd = 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 Table of Content 32. 32 SOLO Dyadics Dyadic Invariants The invariants of the dyadic are derived from the following invariant equation [ ] [ ] [ ] [ ] [ ] [ ]( ) [ ] [ ] [ ] [ ]( ) [ ]( ) 0detdetdetdet detdetdet 3333 1 1 1 33 1 3333 === == DIDILL LDILLDLIDI xx xxx This is the invariant characteristic equation derived from the dyadic [ ]( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) 0 detdet 2 3 1 2 3 1 2 2 1 3 3 1 1 3 3 2 2 1 3 3 1 2 2 1 2 3 3 2 1 1 3 3 2 2 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 12 3 3 2 2 1 13 2 2 1 3 2 3 1 2 1 3 3 1 1 3 3 2 3 3 1 2 1 2 2 1 2 3 3 2 3 3 2 2 3 3 2 22 1 1 3 3 2 3 1 3 3 2 2 2 1 2 3 1 2 1 1 1 33 =+++ +++++= +++++= = dddddddddddddddddd ddddddddddddddd ddddddddddddddddddd ddd ddd ddd DI x We found the following three scalar invariants of the dyadic ( ) ( ) 2 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 1 3 3 1 1 2 2 1 2 3 3 2 3 3 2 2 3 3 1 1 2 2 1 1 23 3 2 2 1 1 1 ddddddddddddddddddI ddddddddddddIdddI ++= ++=++= [ ] [ ] [ ] [ ] 1 = LDLDWe found that under a coordinate transformation :[ ] [ ] [ ]eLf = Any dyadic has five important invariants associated with: three scalar invariants, one dyadic invariant and one vector invariant. 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 ===== :: ( ) [ ] 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 == Use ( ) ( ) ( )[ ]321122133113223 eddeddeddgedgD ijk ijkV ++== We defined the decomposition of a dyadic in symmetric and anti-symmetric parts: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) + +++ +++ +++ =++= 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 41. 41 Vector AnalysisSOLO Dyadic Identities Summary ( ) ( ) ( ) CbaCabCba == ( ) ( ) ( )CbaCabCba = ( ) CCC += ( ) CCC += ( ) ( ) CCC 2 = 0= C aCCa T = [ ]TT aCCa = ( ) ( ) BCaBaC TT = 42. 42 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations ELECTROMAGNETICSSOLO The Dyadic (Matrix) Greens function is the solution of the vector equation( )SF rrG , ( ) ( )SFSS rrIGkG = 42 where is the unit dyadic or the identity matrix.I ( ) ( ) ( ) ( ) ( ) ( ) ( )GrrIGkG GGG rrIGkG SSSFS SSSSSS SFSS +=+ = = 4 4 22 2 ( ) ( ) ( ) ( ) === = 00 42 GG rrIGkG SSSSSSSS SFSSS and ( ) ( ) ( ) = == SFSS SFSSFSS rr k G rrrrIGk 2 2 4 44 ( ) ( )SFSSSS rr k G = 2 4 Therefore ( ) ( ) ( ) ( ) ( )SFSSS SFSSSS SSSFS rr k IGkG rr k G GrrIGkG +=+ = +=+ 2 22 2 22 1 4 4 4 43. 43 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 1) ELECTROMAGNETICSSOLO ( )SFSSS rr k IGkG +=+ 2 22 1 4 The form of the above equation suggests that can be written in terms of a Scalar Greens function as ( )SF rrG , ( )SF rr , ( ) ( )SFSSSF rr k IrrG , 1 , 2 += To find let perform the following calculations:( )SF rr , ( ) ( ) ( ) += += += SSSSS SSSSSS SS k Ik k IkGkG k IG 2 22 2 22 2 1 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 0 222 2222 2222 22222 kIkkI kkI kkI kIkIk SSSSSS SSSSSS SSSSSSSS SSSSSSSSSSSS SS +=++= ++= ++= += = We can see that: ( ) ( ) ( )SFSSS rrIkIGkG =+= 4222 44. 44 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 2) ELECTROMAGNETICSSOLO We found that the solution of this equation is: ( ) ( ) ( )SFSSS rrIkIGkG =+= 4222 Therefore satisfies the scalar wave equation:( )SF rr , ( ) ( ) ( )SFSFSFS rrrrkrr =+ 4,, 22 ( ) ( ) SFSF rrrwhere r rkj rr = = exp , 45. 45 Dyadic Greens 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 ++= += = 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 = ( )[ ] ( ) ( ) ( )( )[ ] + += S SS V mSe dSaGEEaGndVJJjGaEa 14 Therefore we obtain Solution of the equation: ( ) mSeSS JJjEkE = 2 46. 46 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 4) ELECTROMAGNETICSSOLO Let develop now the expression ( )[ ] ( ) ( ) ( )( )[ ] + += S SS V mSe dSaGEEaGndVJJjGaEa 14 Solution of the equation: ( ) mSeSS JJjEkE = 2 (continue 1) ( ) ( ) ( )( )[ ]aGEEaGn SS 1 ( ) ( ) ( ) a k aa k aaaG S consta SSSS consta SSSS = =+= += = = 2 0 2 11 and ( )( ) ( ) ( ) ( ) aEnaEn aEnnaEnaGE SS SSS = = === 11 111 47. 47 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 5) ELECTROMAGNETICSSOLO Solution of the equation: ( ) mSeSS JJjEkE = 2 (continue 2) Since is symmetric andG k IG SS , 1 2 += GaaG = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )Ena k Ena En k IaEnGa nEGanEaGnEaG SSSS SSSS SSS = += = == 1 1 1 1 1 1 111 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) addEna k Ena k subtractEna k Ena SSSSSS SSSS + = 1 1 1 1 1 1 1 22 2 But since ( ) ( ) ==== 0&0 aaconsta SSSSSS ( ) ( ) ( ) ( ) ( ) ( ) 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 ( ) ( ) ( ) ( )[ ] ( ) ( )[ ] = += nEa k nEaEa k SSS SSSSSS 1 1 1 1 2 2 48. 48 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 6) ELECTROMAGNETICSSOLO Solution of the equation: ( ) mSeSS JJjEkE = 2 (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 ( ) ( ) ( )Ena k Ena SSSS = 1 1 1 2 ( ) ( )[ ] + nEa k SSS 1 1 2 ( )( ) aEnnaGE SS = 11We found that therefore ( ) ( ) ( )( )[ ]aGEEaGn SS 1 ( ) ( ) ( )Ena k Ena SSSS = 1 1 1 2 ( ) ( )[ ] aEnnEa k SSSS + 11 1 2 ( ) ( ) + += En k EnEna SSSSS 1 1 11 2 ( ) ( )[ ] + nEa k SSS 1 1 2 49. 49 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 7) ELECTROMAGNETICSSOLO Solution of the equation: ( ) mSeSS JJjEkE = 2 (continue 4) Since and we get ( ) ( ) ( )( )[ ]aGEEaGn SS 1 ( ) ( ) + += En k EnEna SSSSS 1 1 11 2 ( ) ( )[ ] + nEa k SSS 1 1 2 ( ) mSeSS JJjEkE = 2 22 =k ( ) ( ) ( )( )[ ] ( ) ( ) ( ) ( )[ ] + + + += nEa k k JJjnEnEnEna aGEEaGn SSS S mSeSSS SS 1 1 1111 1 2 2 50. 50 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 8) ELECTROMAGNETICSSOLO Solution of the equation: ( ) mSeSS JJjEkE = 2 (continue 5) Let compute ( ) ( ) ( )( )[ ] S SS dSnaGEEaG 1 ( ) ( ) + + += S S mSeSSS dS k JJjnEnEnEna 2 1111 ( ) ( )[ ] dSnEa k S SSS + 1 1 2 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 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 2 0 22 4 2 2 1 51. 51 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 8) ELECTROMAGNETICSSOLO Solution of the equation: ( ) mSeSS JJjEkE = 2 (continue 5) Let substitute this result in ( ) ( ) + += + V S e V SS mSe S mSe S dvadv k JJjadSJJjn k a 22 1 ( )[ ] ( ) ( ) ( )( )[ ] + += S SS V mSe dSnaGEEaGdVJJjGaEa 14 ( ) + += dVJJj k IaEa V mSe SS 2 4 ( ) + + S SSS dSEnEnEna 111 ( ) + ++ V S e V SS mSe dvadv k JJja 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 111 4 1 4 1 52. 52 Dyadic Greens Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue 9) ELECTROMAGNETICSSOLO Solution of the equation: ( ) mSeSS JJjEkE = 2 (continue 6) Using ( ) + + += S SSS V S e mSe dSEnEnEndVJJjE 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 1 5 ( ) + ++ = S SSmS V S e mSe dSEnEnJEndVJJjE 111 4 1 4 1 e m m e e m m e J J J J E H H E ( ) + + += S SSeS V S m eSm dSHnHnJHndVJJjH 111 4 1 4 1 53. 53 Dyadic Greens 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 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 Greens 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 We found ( ) ( ) ( )( )[ ] ( ) ( ) ( )[ ] + + + += nEa k k JJjnEnEnEna naGEEaG SSS S mSeSSS SS 1 1 1111 1 2 2 00 Using Stokes Theorem: we have = CS rdASdA ( ) ( )[ ] ( ) ( ) == C SS C SS Stokes S SSS rdEardEadSnEa 1 1 Therefore ( ) ( ) ( )( )[ ] ( ) ( ) + + += = C SS S SSS S SS rdEa k dSEnEnEna dSaGEEaGnEa 2 1 111 14 1 1 55. 55 Dyadic Greens 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 2 1 1114 1 Since this is true for all constant vectors , we obtaina ( ) ( ) + + += C SS S SSS rdE k dSEnEnEnE 2 4 1 111 4 1 1 Using and we get 22 =kHjES = + + + = C S S SS rdH j dSEnEnHnjE 4 111 4 1 1 we can write e m m e e m m e J J J J E H H E + + = C S S SS rdE j dSHnHnEnjH 4 111 4 1 1 56. 56 SOLO 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 SOLO 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