VEKTORANALYS. GAUSS s THEOREM and STOKES s THEOREM. Kursvecka 3. Kapitel 6-7 Sidor 51-82

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VEKTORANAY Kursvecka 3 GAU s THEOREM and TOKE s THEOREM Kaptel 6-7 dor 51-82

TARGET PROBEM EECTRIC FIED MAGNETIC FIED N + Magnetc monopoles do not est n nature. How can we epress ths nformaton for E and B usng the mathematcal formalsm? 1

TARGET PROBEM et s consder some EECTRIC CHARGE 1 + 2 - + and two closed surfaces, 1 and 2 1 does not contan an charge. It has no sources and no snks: no feld lnes destroed and no feld lnes created nsde 1 1 E d = 2 do contan a charge. It has a snk. Feld lnes are destroed nsde 2 2 E d Q E d = Guass s law (see the 6 th week of ths course for detals or ε Teoretsk elektroteknk ) We want fnd the dfferental form of the Guass s law. (.e. to epress the Guass s law wthout usng ntegrals) to ntroduce the dvergence of a vector feld A, the Gauss s theorem A d = V dvadv dv A 2

THE DIVERGENCE (DIVERGENEN) In cartesan coordnates, the dvergence of a vector feld A s: DEFINITION dva + + (1) It s a measure of how much the feld dverges (or converges) from (to) a pont. EXAMPE: Assume that A s the veloct feld of a gas. If heated, the gas wll epand creatng a veloct feld that wll dverge. The dvergence of A n the heatng pont wll be postve If cooled, the gas wll contract creatng a veloct feld that wll converge on the coolng poston. The dvergence wll be negatve The heatng poston s a source of the veloct feld and the coolng poston s a snk of the veloct The dvergence s a measurement of sources or snks (ths wll be more clear usng the Guass s theorem) 3

A d = V dvadv THE GAU s THEOREM where s a closed surface that forms the boundar of the volume V and A s a contnuousl dfferentable vector feld defned on V. (2) ˆn 2 d 2 V 2 =f 2 (,) ˆn 1 d 1 1 =f 1 (,) d d p dd = d nˆ eˆ = d eˆ 2 2 2 dd = d nˆ eˆ = d eˆ 1 1 1 4

THE GAU s THEOREM PROOF V dvadv = + + ddd = V ddd + ddd + ddd V V V et s calculate the last term: f2 (, ) V f(, ) p 1 p [ (,, (, )) (,, (, ))] ddd = dd d = A f A f dd = 2 1 2 1 dd s the projecton on p of the small element surfaces on d 1 and d 2. dd = eˆ nˆd = eˆ nˆ d Therefore: 1 1 2 2 = A (,, f (, )) eˆ nˆ d + A (,, f (, )) eˆ nˆd = A eˆ nd ˆ 2 2 2 1 1 1 Whch means: V dv = A ˆ ˆ e nd (3) 5

THE GAU s THEOREM PROOF In the same wa we get: V V dv = A ˆ ˆ e nd dv = A ˆ ˆ e nd (4) (5) Addng together equatons (3), (4) and (5) we fnall obtan: = + + = dvadv ddd ddd ddd V V V V A eˆ nd ˆ + A eˆ nd ˆ + A eˆ nd ˆ = A d 6

Rearrange n logc order the steps to prove the Gauss s theorem 8- Add all the three terms together n order to obtan the flu of. 5- Wrte down the volume ntegral of dva 3- Consder the projecton of the surface element on the plane, t wll be dd. The projecton wll dentf a nfntesmal surface element (d 2 ) on the lower surface. 6.4- Re-arrange the ntegrals n d 1 and d 2 n order to have obtan a flu ntegral of (,,A ). 1- Consder a closed surface. 6- plt the volume ntegral nto three terms. 6.1- Consder onl the term whch depends on the -dervatve of A. 7- Repeat the same for the terms whch depend on the -dervatve of A and on the - dervatve of A. 6.3- Epress dd n order to obtan d 1 and d 2. 6.2- Remove the -dervatve b solvng the ntegral n d. What wll reman s just the ntegral n dd. 2- Dvde the surface n two parts, an upper surface and a lower surface and consder an nfntesmal surface element d 1 on the upper surface. 4- Wrte the epresson that relates dd to d 1 and d 2. A

PROOF THE GAU s THEOREM What f we consder a more complcated volume? V V 2 We dvde the volume V n smaller and smpler volumes V = V1+ V2 +... = V V dvadv = dvadv = V V 1 A d = A d 7

PHYICA INTERPRETATION uppose that s the veloct feld of a gas et s appl the Gauss theorem to a volume V of the gas vr ( ) v d = V dv() v dv Ths term s the volume per second [m 3 /s] that flows out (n) from the closed surface If there are no snks and no sources, then no gas flows n and no gas flows out from. Ths mples that the flow v d s ero. Therefore, dv( v ) = dv( v ) = dv( v ) < dv( v ) > No snk and no source flu s destroed and there s a snk flu s created and there s a source 8

TARGET PROBEM Magnetc monopoles do not est n nature. What ths mples, n terms of the magnetc feld? Magnetc monopoles do not ests the flu of B s ero et s appl the Gauss s theorem to the magnetc feld: Gauss B d = V dvbdv B d = dvb = Eercse: appl the Gauss s theorem to the Guass s law: Q E d = ε One of the four Mawell s equatons 9

WHICH TATEMENT I WRONG? 1- The dvergence of a vector feld s a scalar (ellow) 2- The dvergence s related to a measurement of the flu (red) 3- The Gauss theorem translates a surface ntegral nto a volume ntegral (green) 4- The Gauss theorem can be appled also to a non closed surface (blue) 1

VEKTORANAY CUR (ROTATIONEN) and TOKE s THEOREM

TARGET PROBEM The current I s flowng n a conductor How to calculate the magnetc feld? I B We need: Defnton of the curl (or rotor) of a vector feld rot A The tokes theorem A dr = rota d A law that relates the current wth the magnetc feld: the fourth Mawell s equaton (wth statc electrc feld): rotb = µ j (see Teoretsk elektroteknk ) 11

THE CUR (ROTATIONEN) rot A DEFINITION (n cartesan coordnate) rot A eˆ ˆ ˆ e e A = =,, A A A rot stands for rotaton In fact, the curl s a measure of how much the drecton of a vector feld changes n space,.e. how much the feld rotates. In ever pont of the space, rot A s a vector whose length and drecton charactere the rotaton of the feld A. The drecton s the as of rotaton of A The magntude s the magntude of rotaton of A 12

THE CUR rota EXAMPE A (,, ) = (,,) Eercse: calculate the curl of A Drecton: the drecton s the as of rotaton,.e. perpendcular to the plane of the fgure The sgn (negatve, n ths case) s determned b the rght-hand rule Magntude: the amount of rotaton In ths eample, t s constant and ndependent of the poston,.e. the amount of rotaton s the same at an pont. 13

THE CUR rota PHYICA INTERPRETATION Consder the rotaton of a rgd bod around the -as P The coordnates of a pont P on the bod located at the dstance a from the -as and at = changes n tme: t ( ) = acosωt t ( ) = asnωt = a The veloct of the pont P s: v () t = aωsn ωt= ωt () v () t = aωcos ωt= ωt () v= ω, ω, v = Therefore rot v = (,, 2ω ) ( ) ω = 1 2 rot v 14

THE TOKE THEOREM A dr = rota d where A s a vector feld, s a closed curve and s a surface whose boundar s defned b. A must be contnuousl dfferentable on ˆn 15

PROOF THE TOKE THEOREM Fve steps: 1. We dvde n man smaller (nfntesmal) surfaces: 2. We project on: the -plane the -plane the -plane = 1 2 3 3. We prove the tokes theorem on (the onl dffcult part) 4. We add the results for the projectons together and we obtan the tokes theorem on 5. We add the results for together and we obtan the tokes theorem on 16

PROOF THE TOKE THEOREM eˆ et s consder the plane surface located n the -plane (.e. =constant= ) wth boundar defned b the curve et s calculate A dr = A dr A (,, ) d + A (,, ) d + A (,, ) d Term 1 Term 2 Term 3 Term 3 = (=constant! d=) Term 1 2 =g() A (,, ) d b a = A d = + 1 2 1 2 (,, ) A (,, ) d + A (,, ) d = A (, f ( ), ) d + A (, g( ), ) d = a b a 1 b =f() 17

THE TOKE THEOREM PROOF [ ] b b b = A (, f( ), ) d A (, g( ), ) d= A (, f( ), ) A (, g( ), ) d= a a a (,, ) dd = dd = dd b f ( ) b g( ) a g( ) a f ( ) Therefore we get: Term 1 A (,, ) d = dd In a smlar wa: Term 2 A (,, ) d = dd Addng Term 1, Term 2 and Term 3: A dr = dd It s the -component of rota!! 18

THE TOKE THEOREM o can rewrte t as: A dr = ( rota) dd = ( rota) eˆ d In a smlar wa we have: A dr = ( rota) eˆ d A dr = ( rota) eˆ d dd = eˆ nd ˆ = eˆ d Now let s add everthng together: A dr + A dr + A dr = A dr ( rota) eˆ d + ( rota) eˆ d + ( rota) eˆ d = rota d 19

Rearrange n logc order the steps to prove the tokes theorem 3.2 - Consder onl the ntegral n d and prove that 3 - Prove the tokes theorem on : 3.1 - Wrte the lne ntegral of the vector feld along the boundar of and splt the ntegral nto three terms. 2 -Dvde the surface n small areas and consder the projecton of on the,, planes 3.3 -Repeat the same for the ntegral n d and d 5 -Prove the tokes theorem on : add together all the epressons obtaned for 4 -Prove the tokes theorem on : 4.1 -Repeat the same procedure for and A (,, ) d = dd 4.2 - add together the epressons for the ntegrals n to and obtanng: 1 - Consder a closed path and a surface whose boundar s defned b the closed path. 3.5 -Rewrte dd to obtan ( ) ˆ 3.4 -Add the three ntegrals n d, d and d to obtan = ( ) A dr = rota e d A dr rota dd A dr = rota d

PROOF A dr = rota d But we are nterested n the whole. o we add these small contrbutons altogether: THE TOKE THEOREM 1 2 3 rota d = rota d = A dr = A dr A dr = rot A d 2

TARGET PROBEM Now we can calculate the magnetc feld B dstance a from the conductor. Ampere s law rotb = µ j Where j s the current denst: at a I B B dr = rot B d = µ j d = µ j d = µ I tokes Ampere I = j d 21

THE GREEN FORMUA IN THE PANE THEOREM (7.1 n the tetbook) D Q P dd = Pd + Qd ( ) PROOF We can start from tokes theorem A dr = rot A d A dr = ( A ) ( ) d + Ad + Ad = Ad + Ad But we are n a plane, so we can assume A=(A,A,) rot A d = eˆ ˆ edd eˆ eˆ eˆ A A =1 D A dd = A d + A d ( ) whch s the Green formula for P=A and Q=A 22

CUR FREE FIED AND CAAR POTENTIA DEFINITION: A vector feld A s curl free f rot A = ometmes called rrotatonal THEOREM (7.5 n the tetbook) rot A = has a scalar potental φ, A = gradφ PROOF (1) rot A = A dr = rot A d = If the crculaton s ero, then the feld s conservatve and has a scalar potental. ee theorem 4.5 n the tetbook. (2) A = gradφ rot A rot grad rot φ, φ φ, φ = = = eˆ ˆ ˆ e e φ φ φ φ φ =,, = (,,) 23

OENOIDA FIED AND VECTOR POTENTIA DEFINITION: A vector feld B s called solenodal f dvb = DEFINITION: The vector feld B has a vector potental A f, B = rota THEOREM (7.7 n the tetbook) B has a vector potental A, B = rota dvb = PROOF (1) B has a vector potental B rota = dvb = dv( rota) = (2) dvb = et s tr to fnd a soluton A to the equaton B = rota We start lookng for a partcular soluton A* of ths knd: ( (,, ), (,, ), ) * * * A = A A 24

CUR FREE FIED AND CAAR POTENTIA PROOF Assumng B=rotΑ we obtan: * * = B A(,, ) = B( d,, ) + F (, ) * * = B A(,, ) = B( d,, ) + G (, ) * * B B F G = B d + d = B But dvb= B B B + = B F G d + = B = B(,, ) B(,, ) F G = B (,, ) A soluton to ths equaton s: F (, ) = G (, ) = B (,, ) d The general soluton can be found usng : ( ) (,, ) (,, ), (,, ), * A = B d B d B d ( ) B=rotΑ * * * rot A A = B B = A A = gradψ A = A + grad ψ 25

WHICH TATEMENT I WRONG? 1- The curl of a vector feld s a scalar (ellow) 2- The curl s related to the lne ntegral of a feld along a closed surface (red) 3- tokes theorem translates a lne ntegral nto a surface ntegral (green) 4- The tokes theorem can be appled onl to a closed curve (blue) 26