From BiotSavart Law to Divergence of B (1)


 Lorin Moore
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1 From BotSavart Law to Dvergence of B (1) Let s prove that BotSavart gves us B (r ) = 0 for an arbtrary current densty. Frst take the dvergence of both sdes of BotSavart. The dervatve s wth respect to the measurement pont and the ntegral s wth respect to the source pont, so we we can take the dervatve nsde the ntegral: B (r ) = 10 7 λ dτ J ( r ) ˆ Next, apply the dentty from the cover of Grffths: ( a b ) = b ( a ) a ( b ) Usng the dentty, we get λ J ( r ) ˆ λ = ˆ J λ ( ) J ˆ The frst term s zero, because J r ( ) depends only on the source pont, and the dervatve s wth respect to the measurement pont.
2 From BotSavart Law to Dvergence of B (2) The curl n sphercal coordnates s v = 1 r snθ + 1 r + 1 r 1 snθ θ snθ v φ ( ) φ v θ φ v r r r v φ ( ) ( r r v θ ) θ v r φ ˆ θ ˆ We have v = 1 r 2 ˆ r wth only a radal component, so r ˆ v = 1 r 1 snθ φ 1 r 2 θ ˆ + 1 r θ 1 r 2 φ ˆ But 1 2 s ndependent of angle, so r φ 1 r 2 = θ Thus the second term s zero and we have B (r ) = 0 1 r 2 = 0
3 Component Notaton Vector notaton s a convenent shorthand, and s powerful enough for Newtonan mechancs and OK for electrostatcs But you probably notced last lecture when we demonstrated that the BotSavart law for magnetostatcs was equvalent to B = µ 0 J that we had to nvoke some denttes that are hard to remember and hard to nterpret. Component notaton s an alternatve to vector notaton that makes such problems more tractable, and also can be extended to more complcated problems. Bascally, we drop the arrownotaton or boldfacenotaton, and ndcate vector components by subscrpts: a a Thus an equaton lke F = ma becomes F = ma (whch s really 3 equatons for the 3 components of the force and acceleraton n both cases). (Note that not every logcal groupng of 3 numbers s a vector. Temperature, pressure, and humdty are a convenent group, but aren t a vector s the physcs sense. Vector really means somethng that can be changes n a certan way when vewed from a rotated coordnate system).
4 Dot Products, Summaton Notaton A dotproduct sn t a vector at all, but a scalar number wth no drecton. What t means s A B = A x B x + A y B y + A z B z ( ) or any other The subscrpt mght be ( x,y,z), or r,θ,φ three orthogonal drectons (but beware of rght vs left!) If we just use ( 1,2,3) for the ndces, we can make a more compact dot product: A B = 3 =1 A B Such sums come up so often that we ll assume summaton over any repeated ndex unless stated otherwse: A B = A B (The dea of assumed summaton over repeated ndces was nvented by Ensten, and he was as proud of that as he was of relatvty!)
5 Kronecker Delta Symbol It s useful to defne the Kronecker delta symbol δ j whch s unty f = j and zero otherwse: δ j = +1 f = j 0 f j The Kronecker delta symbol does for sumatons what the Drac delta functon does for ntegrals: A δ j = A δ j = A j We often run nto combnatons lke j A B j δ j = A B j δ j By totally conventonal algebra, we can swap the order of the summatons over and j, factor B j out of the sum over, then apply the prevous result: But ths s just A B! A B j δ j = B j A δ j = B j A j j j j
6 Dot Products and Matrx Notaton If we vew A and B as matrces wth 1 column and 3 rows, we could also wrte the dotproduct as a matrx product: A B = A T B = B T A The superscrpt T means matrx transpose. If we wrte the components explctly, [ ] A B = A T B = A A 2 A 3 B 1 B 2 B 3 = A 1 B 1 + A 2 B 2 + A 3 B 3 Remember that wth matrces, the multplcatonorder matters: BA T = AB T s a 3 by 3 matrx, not a scalar! The components of δ j n matrx notaton are δ j = It s just the unt matrx!
7 Cross Products and Alternatng Tensor In Cartesan coordnates, the cross product s A B = x ˆ ( A y B z A y B z )+ y ˆ A z B x A x B z ( ) + ˆ ( ) z A x B y A y B x To wrte ths n component notaton compactly, we nvent a symbol called the alternatng tensor ε jk ε jk = 0 f = j, j = k, or k = +1 f jk =123, 231, f jk = 321, 213, 132 Thnk of the 3 ndces as wrappng around: j k j k... If the order s ncreasng, t s +1. If the order s decreasng, t s 1. If any two ndces are the same, t s zero. The kth component of the cross product s: ( A B ) k = ε jk A B j j = ε jk A B j The frst ndex of ε jk goes wth A, the second wth B, the thrd s the ndex of the resultng crossproduct vector. Let s check t. For the zcomponent of the crossproduct, we want ( k = 3). Only the terms ( = 1, j = 2) and = 2, j =1 are nonzero, and these gve us ( +A 1 B 2 A 2 B 1 ) whch s just ( +A x B y A y B x ), whch s rght. ( )
8 DoubleCrossProducts The doublecrossproduct dentty s A ( B C ) = B ( A C ) C ( A B ) ( ) mean? It says take the What does a term lke B A C vector B, and multply t by the scalar number ( A C ). Usng our new notaton we wrte [ A ( B C )] k = ε jk A ε lmj B l C m j The term n parentheses s the j component of B C, and we use l,m as the summaton ndces to make t clear they are dfferent from the,j ndces of the other summaton. Wth the sums explct lke ths, everythng s just a number and the order of operatons doesn t matter, so we wrte [ A ( B C )] k = ε jk ε lmj A B l C m jlm lm
9 Alternatng Tensor Identty If you can remember ths dentty, you can derve all the doublecross denttes on the fly: k ε jk ε lmk = δ l δ jm δ m δ jl For a gven k value, the two ε s wll have the same sgn f = l and j = m, and have the opposte sgn f = m and j = l, and any other combnaton wll gve zero. Note that the last ndex of the two ε s match. You can always get ths by usng the cyclc property of the ε s: ε jk = ε jk = ε kj The other 4 ndces could be anythng n any order; just transcrbe them to the δ s n the prescrbed pattern: a δ wth both frst ndces tmes a δ wth both second ndces, mnus the δ s wth the frst ndex from one and the second ndex from the other.
10 DoubleCross Revsted Apply ths to the doublecross product: [ A ( B C )] k = ε jk ε lmj A B l C m Frst, make the thrd ndces agree = ε kj ε lmj A B l C m jlm Next, use the sum over j to turn the product of ε s to a product of δ s: jlm = ( δ kl δ m δ km δ l )A B l C m lm Then, evaluate the δ s, rememberng that we expect to end up wth a kndex left over: ( ) = A B k C A B C k Then wrte the result n vector notaton agan: = B ( A C ) C ( A B )
11 Dervatves n Component Notaton The gradent of a scalar (whch s a functon of a vector) s f ( x ) = f x x ˆ + f y y ˆ + f z z ˆ In component notaton, the th component s [ f ( x )] = f x Ths comes up so often that there s a shorthand verson: ( ) x Gradent: ( f ) f Dvergence: A A = A Curl: Note as well that ( A ) k ε jk A j x j = x x j = δ j j = ε jk A j
12 Provng Vector Dervatve Identtes Curl of crossproduct, summatons assumed: [ ( A B )] k = ε jk ( ε lmj A l B m ) The ε s just a coeffcent, so t factors out of the dervatve. Then cycle the ndces on the frst ε: = ( ε kj ε lmj ) ( A l B m ) Now apply the alternatngtensor dentty, and the product rule for the dervatve: Contract the δ s: = ( δ kl δ m δ km δ l )( A l B m + B m A l ) = ( A k B + B A k ) ( A B k + B k A ) We can now wrte t n vector notaton agan = A ( B ) + ( B )A ( A )B B ( A ) I fnd the meanng more clear n the component form!
13 Provng Vector Dervatve Identtes (2) Dvergence of a curl n component notaton: ( A ) = k ( ε jk A j ) Brng the dervatves together: = ε jk ( k A j ) But because k s symmetrc under k, but ε jk s antsymmetrc under k, ths has to be zero. The same argument works for the curl of a gradent: The curl of a curl: ( f ) = ε jk j f = 0 ( ) = ε jk ( ε lmj l A m ) = ε kj ε lmj l A m A ( ) = ( δ kl δ m δ km δ l )( l A m ) = ( k A ) ( A k ) = k ( A ) ( )A k Ths form s clearer than the equvalent vector form: = ( A ) 2 A
14 Vector Dervatves of Radal Vector Functons A radal vector functon ponts n the radal drecton, and s a functon only of radus. Whle we mght wrte t wth a radal unt vector and a functon of the magntude of the radus, we could equally well wrte t as x f ( x x ) In component form, usng two dfferent subscrpts x f j x j x j The vector dervatve operator gves x x f x j x j k j = f x j x j x x + x k j = f ( x 2 )δ k + x x k ( ) x 2 = f ( x 2 f x2 )δ k + 2 j ( ) x x k x k f x j x j Ths s symmetrc wth respect to k j f ( x2 ) ( x 2 ) x j x j
15 BotSavart to Dvergence of B, Revsted Start wth the BotSavart ntegral, usng r for the source coordnate and r for the measurement pont B (r ) = 10 7 J r λ λ d τ λ = r r 2 ( ) ˆ Let s take the dvergence of B n component notaton k B k = 10 7 k ε jk J r λ j ( ) ˆ d τ Move the through the ε, and through the J ( r ), snce the dervatve s wth respect to the measurement pont rather than the source pont: = 10 7 J r λ j ( ) ε jk k ˆ λ d τ 2 Ths dervatve wll be symmetrc under exchange of j and k, but the ε s antsymmetrc, so the result s zero. Wasn t that smpler than the vectornotaton verson?
16 Now the curl of B: Curl of B, Revsted ε jk B j =10 7 ε jk ε lmj J l r ( ) ˆ λ m d τ Move the ε s together, move the through the J ( r ): = 10 7 ε jk ε lmj J l λ ˆ m λ d τ 2 Cycle the ndces and apply the ε δ dentty: Contract the δ s = 10 7 ε kj ε lmj J l λ ˆ m λ d τ 2 = 10 7 ( δ kl δ m δ km δ l )J l λ ˆ m λ d τ 2 = 10 7 J k λ ˆ λ J 2 ˆ λ k d τ
17 Curl of B, Revsted (2) Do the second term frst, usng some trcks. Frst, by the defnton λ = r r, we can change dervatve varables by changng the sgn: J λ ˆ k λ = J λ ˆ 2 k Next, from the dervatve of products rule we can say J J ˆ λ k ˆ λ k = λ ˆ k J + J = λ ˆ k J J ˆ λ k ˆ λ k The left term s zero for magnetostatcs wth steady currents because J = J = 0. The rght term s a pure dvergence, and we are ntegratng over t. We expand the ntegraton volume untl there are no currents at the surface, and apply the dvergence theorem: J λ ˆ k d τ = J λ ˆ k d a = 0 surface volume
18 Curl of B, Revsted (3) Now the remanng term that we skpped over before ε jk B j =10 7 J k λ ˆ ( r ) λ d τ 2 Translatng back nto vector form, we recognze as our frend the 3d Drac delta functon ˆ λ = ˆ λ = 4πδ3 λ ( ) Thus we get ε jk B j =10 7 J k 4πδ ( 3 λ )d τ = 4π 10 7 J k ( r ) Back n vector notaton, ths s B = µ 0 J
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