Tensor. Syllabus: x x

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1 Tenso Sllabus: Tenso Calculus : Catesan tensos. Smmetc and antsmmetc tensos. Lev Vvta tenso denst. Pseudo tensos. Dual tensos. Dect poduct and contacton. Dads and dadc. Covaant, Contavaant and med tensos. Chstoffel smbols and dffeentaton of tensos. Q1. What do ou mean b a tenso? nswe: - The tensos ae quanttes obeng cetan tansfomaton laws. The tenso analss ma be egaded as an ndspensable pat of the stud whch s athe fo the mathematcal fomulaton of natual laws, whch eman nvaant wth espect to dffeent fame of efeences. We defne a tenso of ank n as follows: n β β βn n µ 1µ... µ n n µ µ µ n n β1β... βn 1 1 Tenso s the genealzaton of the tem vecto and tenso calculus o athe tenso analss s a genealzaton of vecto analss. Tensos ae of mpotant n man aeas of phscs ncludng geneal elatvt and electomagnetc theo. One of the moe polfc souces of tenso quanttes s the ansotopc sold. Hee the elastc, optcal, electcal and magnetc popetes ma well nvolve tensos. Let us consde the flow of cuent. In ths contet Ohm s law can be wtten as, J σe () Whee J Cuent denst E Electc feld. If we have an sotopc medum, σ, the conductvt s a scala and fo the component, J1 σe 1 () Howeve, f the medum s ansotopc, the cuent denst n the -decton ma depend on the electc feld n the - and z-decton as well as on the feld n the -decton. ssumng a lnea elatonshp we can eplace equaton () b, J1 σ 11E1 + σ1e + σ13e3 () In geneal, J σ k Ek (v) Fo odna 3-dmensonal space the scala conductvt, σ, has gven wa to a set of nne elements, σ k to effect the equed change n magntude and decton. σ σ σ σ σ σ σ k 1 3 σ31 σ3 σ 33 Ths aa of 9-elements actuall foms a tenso. (v)

2 Genealzng equaton (), f a elaton BC wth and C as nonpaallel vectos holds n all oentatons of a Catesan sstem, then B s sad to be a tenso. The descpton of electc, optcal and magnetc popetes of ansotopc sold (vaous cstals) ma well nvolve tensos. Q. Eplan dumm suff and eal suff wth eample. nswe:- Dumm suff: When n an ndeed epesson an nde occus once as a lowe nde o subscpt and agan as an uppe nde o supescpt then the nde s called a dumm nde o dumm suff. e.g. In the epesson a, the nde s used as both subscpt and supescpt and hence t s a dumm nde. Fee nde o eal nde: When n an ndeed epesson an nde doesn t occu as a lowe nde as well as an uppe nde, then the nde s called fee nde o eal nde. e.g. n the epesson a, the nde s a dumm nde wheeas the nde s a eal o fee nde snce t s used as supescpt onl. Konecke delta smbol: If a patcula sstem of second ode s consdeed and s denoted b δ ;, 1,,,n and s defned as follows; δ 1 fo 0 fo Such a sstem s called Konecke delta. Some popetes of Konecke delta: () If 1,, 3,, n ae ndependent vaables then, δ () If one deals wth n-dmensons, then, δ δ n Poof: B summaton conventon; 1 n δ δ1 + δ δn n () δ δk δk Poof: B summaton conventon; 1 n 1 n δ δk δ1δk + δδk δδk δn 1δk + δnδk

3 δ k δ k (v) δk k k (v) Konecke delta s nvaant unde co-odnate tansfomaton: Poof: k m k l δ δ l m k m l 1 [ δ m m 1 when l m and 0 when l m] k δ k (v) Konecke delta ank. Poof: k δ has the tansfomaton popet lke that of a med tenso of second Let us consde the tansfomaton of a med tenso ( ν µ ) of ank. ν β ν µ β () µ Konecke delta s defned as; k k k δl () l l Now let us choose, β to δ β Fom equaton (); ν β ν β ν δ δ β µ µ Now eplace ν k and µ l then, ν ν µ β µ k k k k β β l l l β l δ δβ So fom the defnton of the med tenso, t follows that 16 components n 4-dmensonal space. k δ s a med tenso of ode wth

4 Q. Eplan contavaant tenso, covaant and med tenso of ank two. Contavaant tenso: f (n ) quanttes β (Whee 1,, 3 n; β 1,, 3 n) n a coodnate sstem ae elated to (n ) othe quanttes β n anothe coodnate sstem b the followng tansfomaton laws; β β µν () µ ν then the quanttes β ae sad to be the components of a Contavaant tenso of second ank. Equaton () s the Contavaant law fo tenso analss. Covaant tenso: f (n ) quanttes β (Whee 1,, 3 n; β 1,, 3 n) n a coodnate sstem ae elated to (n ) othe quanttes β n anothe coodnate sstem b the followng tansfomaton laws; k m β km β () then the quanttes β ae sad to be the components of a Contavaant tenso of second ank. Equaton () s the Covaant law fo tenso analss. Med tenso: f (n ) quanttes β (Whee 1,, 3 n; β 1,, 3 n) n a coodnate sstem ae elated to (n ) othe quanttes β n anothe coodnate sstem b the followng tansfomaton laws; m km β () k β then the quanttes β ae sad to be the components of a med tenso of second ank. Equaton () s the Contavaant law fo tenso analss. n mpotant eample of med tenso of second ank s Konecke delta ( δ β ). Q. What s the ank of a tenso? The ank of a tenso s detemned b the numbe of suffes o ndces attached to t. s a matte of fact the ank of a tenso when ased as a powe to the numbe of dmensons, elds the total numbe of the components of the tensos and hence the numbe of the components that epesent the tenso. s such a tenso of ank n n fou dmensonal space has 4 n components. Consequentl the ank of a tenso gve the numbe of the mode of changes of the phscal quantt when passng fom one sstem to anothe whch s n otaton elatve to othes. It s clea fom the dscusson that a quantt that doesn t change when the as ae otated s a tenso ank zeo. Snce the numbe of mode of changes s then zeo. These quanttes ae named as tensos of ank zeo and ae scala. Wheeas the tensos of ank one ae vectos. v

5 Q. Show that the veloct of a flud s a Contavaant tenso of ank one. ssumng that () t s the coodnate of a movng patcle wth the tme t. Then we d have v as the veloct of the patcle. In tansfomed coodnates the components of dt d veloct ae v. dt d d β β Now v v dt β dt β Ths shows that the veloct s a Contavaant tenso of ank one. Q. Defne smmetc and ant-smmetc tenso. Smmetc tensos: If a tenso be such that f ts two Contavaant o covaant ndces ae ntechanged wthout alteng the value of the tenso, then the tenso s called smmetc tenso wth espect to these ndces. Thus f p p ; then the tenso s sad to be smmetc wth espect to the ndces and. nt-smmetc tensos: tenso s sad to be ant-smmetc o skew-smmetc wth espect to an to of ts Contavaant o covaant ndces f the components of the tenso changes sgn upon the ntechange of the ndces. Thus f p p ; then the tenso s sad to be ant-smmetc o skew-smmetc wth espect to the ndces and. ddton of tensos: The sum of two o moe tensos of same ank and tpe (.e. same numbe of covaant and Contavaant ndces) s also a tenso of the same ank and tpe. Thus f lm and pq B ae two tensos of ank thee then tenso of ank thee. ddton of tensos s commutatve and assocatve. lm pq s + B C s also a Subtacton of tensos: The subtacton of two tensos of same ank and tpe (.e. same numbe of covaant and Contavaant ndces) s also a tenso of the same ank and tpe. lm Thus f and tenso of ank thee. pq B ae two tensos of ank thee then lm pq s B C s also a Theoem: The sum of two tensos of same ank esults n thd tenso of same ank. v

6 Poof: We have to pove µ + Bµ Cµ. Let us consde µ and B µ and epess them n mat fom as follows: a11 a1 a13 b11 b1 b13 µ a1 a a 3 and Bµ b1 b b 3 a 31 a3 a 33 b31 b3 b 33 a11 a1 a13 b11 b1 b13 µ + Bµ a1 a a 3 + b1 b b 3 a 31 a3 a 33 b31 b3 b 33 a11 + b11 a1 + b1 a13 + b13 a1 b1 a b a3 b a 31 b31 a3 b3 a33 b f the elaton between the coeffcent a s and b s ae such that aµ + bµ cµ c11 c1 c13 then, µ + Bµ c1 c c 3 C µ c 31 c3 c 33 whch s a tenso of same ank as of µ and B µ. Theoem: The dffeence of two tensos of same ank esults n thd tenso of same ank. Poof: We have to pove µ Bµ Cµ. Let us consde µ and B µ and epess them n mat fom as follows: a11 a1 a13 b11 b1 b13 µ a1 a a 3 and Bµ b1 b b 3 a 31 a3 a 33 b31 b3 b 33 a a a B a a a a a a µ µ b b b b b b b b b v

7 a11 b11 a1 b1 a13 b13 a1 b1 a b a3 b 3 a 31 b31 a3 b3 a33 b 33 f the elaton between the coeffcent a s and b s ae such that aµ bµ cµ c11 c1 c13 then, µ Bµ c1 c c 3 C µ c 31 c3 c 33 whch s a tenso of same ank as of µ and B µ. Q. defne oute poduct and show that the oute poduct of two tensos s a tenso whose ode s sum of odes of the two consttuent tensos. The oute poduct of two tensos s a tenso whose ank s the sum of the anks of the gven tensos. Thus f and be the anks of two tensos, then the oute poduct wll be a tenso of ank +. If χ β and Bφ σ ae two tensos of ank 3 and espectvel, then the oute poduct χ φ χφ β Bσ Cβσ s a tenso of ank 5 (3 + ). To pove t, consde the tansfomaton equaton of the gven tensos as follows; µ ν µν kl k l m () m and B β q p B p β q () Multplng () and () we get; µ ν m q µν β k l p β kl m B B µ ν m q µν β k l p β p q klp mq C C () The above (equaton ) s a tansfomaton law fo a tenso of ank 5. hence the oute poduct of two tensos one µν of ank 3 and B β of ank s a tenso Cµν β of ank 5. The oute poduct of tensos s commutatve and assocatve. v

8 Q. Eplan the meanng of contacton wth eample. The algebac opeaton b whch the ank of a tenso ma be loweed b (o b an even numbe) s known as contacton. In the pocess of contacton one covaant and one Contavaant nde of a med tenso ae set equal and the epeated nde assumed ove, the esult s a tenso of ank loweed b than the ognal tenso. Let us consde a med tenso µνσ ρ of ank 5 wth Contavaant ndces µ, ν, σ and covaant ndces, ρ. The tansfomaton low fo the gven tenso s; µ ν σ δ τ µνσ βγ ρ β γ ρ δτ () To appl the pocess of contacton we put ρ σ µ ν σ δ τ µνσ βγ σ β γ σ δτ σ τ τ τ Now δ 1 γ σ γ γ when τ γ 0 when τ γ µ ν δ µνσ βγ σ β δγ () The above s a tansfomaton law fo a med tenso of ank 3. Hence µνσ σ s a med tenso of ank 3, whch ma be denoted b µν. Thus the pocess of contacton enables us to obtan a tenso of ank (-) fom a med tenso of ank. Q. Eplan nne poduct of two tensos wth eample. Inne poduct: The oute poduct of two tensos followed b contacton esults a new tenso called an nne poduct of the two tensos and the pocess s called the nne multplcaton of two tensos. Let us consde two tensos σ µν and B ρ. The oute poduct of these two tensos s, µν µν σ Bρ Cσρ () v

9 Q. Show that eve tenso can be epanded as a sum of smmetc and ant-smmetc tensos wth espect to ntechange of an two same tpe of ndces. Poof: Let... p... s a tenso of ank p. 1 1 p 1 1 [ ] [ ] 1... p 1... p 1... p 1... p 1... p 1... p p + p p p 1... p 1... p C D p 1... p Now 1... p [ p p C ] p + p p [ p p ] p {Intechange of 1 and }. p 1... p C 1... p 1... p C 1... p s smmetc. 1 nd D... p [ p p p ] p p [ p p ] p {Intechange of 1 and }. p 1... p 1... p D 1... p D 1... p s ant-smmetc. Hence eve tenso can be epanded as a sum of smmetc and ant-smmetc tensos wth espect to ntechange of an two same tpes of ndces. Q. Show that the smmet popetes of tensos ae nvaant. Let us consde a tenso µν be smmetc n ts fst two ndces. That s µν µν, then we have to show that µν µν. Now, β γ µν µ ν βγ () and

10 µν β γ µ ν βγ () s the gven tenso s smmetc n fst two ndces, we have, and µν µν βγ βγ Usng ths elaton and compang equatons () and () we fnd that both the equatons ae dentcal. µν µν Whch follows that the gven tenso s n othe (tansfomed) coodnate sstem s also smmetc n fst two ndces. Hence the smmetc popet of a tenso s nvaant. Q. Show that the nne poduct of two tensos each of ank 1 s a tenso of ank zeo. Let us consde two tensos µ and B ν each of ank 1. The oute poduct of these two tensos s; µ µ Bν Cν (sa) we get; µ µ µ Cµ, whch s a scala o tenso of ank zeo. Thus the nne poduct of two tensos each of ank 1 s a tenso of ank zeo. Now applng the contacton pocess b settng ν µ B β Q. If 0 aβ fo all values of vaables 1,, n, then show that aµν + aνµ 0. We have, β aβ 0 () Dffeentatng equaton () wth espect to µ we get, β β aβ + aβ 0 µ µ β β a δ + a δ 0 β µ β µ β µ µβ a + a 0...( ) gan dffeentatng equaton () wth espect to ν we get; β aµ + aµβ 0 ν ν β aβδν + aβδν 0 a + a 0 νµ µν Hence the equed esult.

11 Q. Show that s not a tenso although s a co-vaant tenso of ank one. µ We have; (Covaant law) () Now dffeentatng equaton () wth espect to β we get; + β β β + + µ µ β β µ β µ β The pesence of second tem n equaton () shows that () µ does not tansfom as tenso laws. Hence s not a tenso. µ Q. If and B ae the components of a Contavaant and covaant tensos of ank one, then show that We have So that; C B ae the components of a med tensos of ank two. β and B B β β C B B β Equaton () shows that C tansfoms as a med tenso of ank two. Q. If, -z and z ae the components of a covaant tenso n ectangula coodnates, then fnd ts covaant components n sphecal coodnates. Suppose 1, and z 3 ; then the covaant components of the tenso ae; ()

12 1 1 3 z z Let µ be the covaant components n sphecal coodnates., θ, ϕ and Then, 1 3 () µ µ Now the tansfomaton equatons ae, snθ cosφ ; snθ snφ and z cosθ In the estng case; 1 1sn cos 3, 1sn sn 3, 3 c 1 cos Now fom equaton (); (sn cos 3) 1 + sn sn 3( 3) + cos. 13 (snθ cos φ)( sn θsnφcos φ) + snθsn φ(snθsnφ cos θ) + cos θ( snθcosθcos φ) Smlal; ( cosθ cos φ)( sn θsnφcos φ) + ( cosθsn φ)(snθsnφ cos θ) + ( sn θ( snθcosθcos φ) 3 ( snθ sn φ)( sn θsnφcos φ) + ( snθcos φ)(snθsnφ cos θ) Q. Fnd the components of a vecto n pola co-odnate whose components n Catesan coodnates ae &, & and &&, &&. Lev Cvta tenso: Lev Cvta smbol ε k o ε k s a nvaant component tenso, whch s antsmmetc n ts all ndces. + ε ε + ε ε + ε ε k k k k k k But f I + ε k k k 0 ε ε ε ε k k

13 Thus t s clea that wheneve two ndces ae equal the component s zeo. Thus although a tenso of 3 d ank n 3 dmenson has 7 components but among them onl 6 ae nonvanshng. ll of them have ethe +1 o -1. The onl non-vanshng components of ε k ae those fo whch all the ndces ae dffeent and the ae equal to +1 o -1 accodng to as (k) s an even o odd pemutaton of (13).e ε k ε13 ε31 ε ε ε ε 1 T + 1 f (k) s an even pemutaton of (13) -1 f (k) s an odd pemutaton of (13) 0 othewse (contan two o moe epeated ndces) tenso denst s defned to be the quantt whch tansfoms as; 1 l1 l a w 1,... k1k.., T k k b l l Whee w s a constant, the value of whch chaactezes a tenso and s called the weght of the tenso denst. The Lev Cvta smbol ε k o ε k ae the denstes of weght -1 o +1. The Lev Cvta smbol ε can be used to wte the coss poduct of two vectos. Let D B D1 B3 3B ε B + ε B ε B 1 Smlal; D D ε B ε B 3 3 and Dµ ε µ B Pseudo Tenso: If thee be a tenso ε kl of ank 4, defned such that; + 1 f (kl) s an even pemutaton of (013) ε kl -1 f (kl) s an odd pemutaton of (013) 0 othewse (contan two o moe epeated ndces)

14 These ae temed as components of pseudo tenso of ank 4. * Fom eve antsmmetc tenso µσ of the second ank a pseudo tenso µσ of the same ank can be obtaned b multplng the fome wth a pseudo tenso of ank 4. 3 * 1 µστρ.e. µσ ε β, β 0 Thus the poduct of a tenso wth a pseudo tenso s a pseudo tenso. It s called dual of a gven tenso. Dual of a covaant tenso: dual of a covaant tenso tenso of ank m n such that; 1... m n 1... m n.1... n B ε k Fo eample k of ank n s defned as the full Contavaant 1... n 1... n s a tenso of ank then ts dual of ank one s gven b; B ε B ε 3 + ε If s a poduct of two vectos sa XY actuall the vecto poduct of the two consttuent vectos, then we see that the dual of s X and Y. Dads and Dadc: nothe mpotant epesentaton of a ank tenso s a dadc. dad s a pa of vectos wtten n a defnte ode (. B). The fst s known as the antecedent and the second one s the consequent. The scala poduct of a dadc s a vecto but the poduct can be defned n two was: (. B). C ( B. C) C.(. B) B(. C) In the fst case C s called the postfacto and n the second case, the pefacto. In geneal the two poducts ae not equal. Howeve the magntude and decton of the poduct s geneall dffeent fom that of C. double dot poduct of two dads s defned as; ( C. ).( B. D) C(. B). D Dadc s a lnea polnomal of dads..e.,. B + CD + ˆ + ˆ + zkˆ B B ˆ + B ˆ + B kˆ z v

15 . B B ˆˆ + B ˆˆ + B k ˆ ˆ + B ˆˆ + B ˆˆ + B ˆk ˆ + B k ˆˆ + B k ˆˆ + B kk ˆˆ z Ths s called the nonon fom of the dad because thee ae nne coeffcents. The unt dadc s defned as; ˆ ˆ + ˆˆ + kˆˆ k 1 Theefoe we have; z z z z z v