Appendix A. Appendices. A.1 ɛ ijk and cross products. Vector Operations: δ ij and ɛ ijk

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1 Appendix A Appendices A1 ɛ and coss poducts A11 Vecto Opeations: δ ij and ɛ These ae some notes on the use of the antisymmetic symbol ɛ fo expessing coss poducts This is an extemely poweful tool fo manipulating coss poducts and thei genealizations in highe dimensions, and although many low level couses avoid the use of ɛ, I think this is a mistake and I want you to become poficient with it In a catesian coodinate system a vecto V has components V i along each of the thee othonomal basis vectos ê i,ov = i V i ê i The dot poduct of two vectos, A B, is bilinea and can theefoe be witten as A B = ( A i ê i ) B j ê j (A1) i j = A i B j ê i ê j (A2) i j = A i B j δ ij, (A3) i j whee the Konecke delta δ ij is defined to be 1 if i = j and 0 othewise As the basis vectos ê k ae othonomal, ie othogonal to each othe and of unit length, we have ê i ê j = δ ij Doing a sum ove an index j of an expession involving a δ ij is vey simple, because the only tem in the sum which contibutes is the one with j = i Thus j F (i, j)δ ij = F (i, i), which is to say, one just eplaces j with i in all 263

2 264 APPENDIX A APPENDICES the othe factos, and dops the δ ij and the summation ove j So we have A B = i A i B i, the standad expession fo the dot poduct 1 We now conside the coss poduct of two vectos, A B, which is also a bilinea expession, so we must have A B =( i A i ê i ) ( j B j ê j )= i j A i B j (ê i ê j ) The coss poduct ê i ê j is a vecto, which can theefoe be witten as V = k V k ê k But the vecto esult depends also on the two input vectos, so the coefficients V k eally depend on i and j as well Define them to be ɛ,so ê i ê j = ɛ kij ê k k It is easy to evaluate the 27 coefficients ɛ kij, because the coss poduct of two othogonal unit vectos is a unit vecto othogonal to both of them Thus ê 1 ê 2 =ê 3,soɛ 312 =1andɛ k12 =0ifk = 1 o 2 Applying the same agument to ê 2 ê 3 and ê 3 ê 1, and using the antisymmety of the coss poduct, A B = B A, we see that ɛ 123 = ɛ 231 = ɛ 312 =1; ɛ 132 = ɛ 213 = ɛ 321 = 1, and ɛ = 0 fo all othe values of the indices, ie ɛ = 0 wheneve any two of the indices ae equal Note that ɛ changes sign not only when the last two indices ae intechanged (a consequence of the antisymmety of the coss poduct), but wheneve any two of its indices ae intechanged Thus ɛ is zeo unless (1, 2, 3) (i, j, k) is a pemutation, and is equal to the sign of the pemutation if it exists Now that we have an expession fo ê i ê j, we can evaluate A B = i A i B j (ê i ê j )= j i ɛ kij A i B j ê k j k (A4) Much of the usefulness of expessing coss poducts in tems of ɛ s comes fom the identity ɛ kij ɛ klm = δ il δ jm δ im δ jl, (A5) k which can be shown as follows To get a contibution to the sum, k must be diffeent fom the unequal indices i and j, and also diffeent fom l and m Thus we get 0 unless the pai (i, j) and the pai (l, m) ae the same pai of 1 Note that this only holds because we have expessed ou vectos in tems of othonomal basis vectos

3 A1 ɛ IJK AND CROSS PRODUCTS 265 diffeent indices Thee ae only two ways that can happen, as given by the two tems, and we only need to veify the coefficients If i = l and j = m, the two ɛ s ae equal and the squae is 1, so the fist tem has the pope coefficient of 1 The second tem diffes by one tansposition of two indices on one epsilon, so it must have the opposite sign We now tun to some applications Let us fist evaluate A ( B C)= i A i ɛ B j C k = ɛ A i B j C k jk (A6) Note that A ( B C) is, up to sign, the volume of the paallelopiped fomed by the vectos A, B, and C Fom the fact that the ɛ changes sign unde tanspositions of any two indices, we see that the same is tue fo tansposing the vectos, so that A ( B C)= A ( C B) = B ( C A)= B ( A C) = C ( A B)= C ( B A) Now conside V = A ( B C) Using ou fomulas, V = ɛ kij ê k A i ( B C) j = ɛ kij ê k A i ɛ jlm B l C m lm Notice that the sum on j involves only the two epsilons, and we can use ɛ kij ɛ jlm = j j ɛ jki ɛ jlm = δ kl δ im δ km δ il Thus V k = ( ilm j ɛ kij ɛ jlm )A i B l C m = ilm(δ kl δ im δ km δ il )A i B l C m = δ kl δ im A i B l C m δ km δ il A i B l C m ilm ilm = i A i B k C i i A i B i C k = A C B k A B C k, so A ( B C)= B A C C A B (A7) This is sometimes known as the bac-cab fomula

4 266 APPENDIX A APPENDICES Execise: Using (A5) fo the manipulation of coss poducts, show that ( A B) ( C D)= A C B D A D B C The deteminant of a matix can be defined using the ɛ symbol Fo a 3 3 matix A, det A = ɛ A 1i A 2j A 3k = ɛ A i1 A j2 A k3 Fom the second definition, we see that the deteminant is the volume of the paallelopiped fomed fom the images unde the linea map A of the thee unit vectos ê i,as (Aê 1 ) ((Aê 2 ) (Aê 3 )) = det A In highe dimensions, the coss poduct is not a vecto, but thee is a genealization of ɛ which emains vey useful In an n-dimensional space, ɛ i1 i 2 i n has n indices and is defined as the sign of the pemutation (1, 2,,n) (i 1 i 2 i n ), if the indices ae all unequal, and zeo othewise The analog of (A5) has (n 1)! tems fom all the pemutations of the unsummed indices on the second ɛ The deteminant of an n n matix is defined as det A = n ɛ i1 i 2 i n A p,ip i 1,,i n p=1 A2 The gadient opeato We can define the gadient opeato = i ê i (A8) While this looks like an odinay vecto, the coefficients ae not numbes V i but ae opeatos, which do not commute with functions of the coodinates x i We can still wite out the components staightfowadly, but we must be caeful to keep the ode of the opeatos and the fields coect The gadient of a scala field Φ( ) is simply evaluated by distibuting the gadient opeato Φ =( i ê i )Φ( ) = i ê i Φ (A9)

5 A2 THE GRADIENT OPERATOR 267 Because the individual components obey the Leibnitz ule AB so does the gadient, so if A and B ae scala fields, AB =( A)B + A B = A B+A B, (A10) The geneal application of the gadient opeato to a vecto A gives an object with coefficients with two indices, a tenso Some pats of this tenso, howeve, can be simplified The fist (which is the tace of the tenso) is called the divegence of the vecto, witten and defined by A = ( i ê i ) ( j ê j B j )= ij ê i ê j B j = ij δ ij B j = B i (A11) i In asking about Leibnitz ule, we must emembe to apply the divegence opeato only to vectos One possibility is to apply it to the vecto V =ΦA, with components V i =ΦA i Thus (Φ A) = i (ΦA i ) = i Φ A i +Φ i A i = ( Φ) A +Φ A (A12) We could also apply the divegence to the coss poduct of two vectos, ( A B) = i ( A B) i = i ( jk ɛ A j B k ) = ɛ (A j B k ) = A j ɛ B k + B k ɛ A j (A13) This is expessible in tems of the culs of A and B The cul is like a coss poduct with the fist vecto eplaced by the diffeential opeato, so we may wite the i th component as ( A) i = ɛ A k (A14) jk x j We see that the last expession in (A13) is ( A j ɛ kij )B k B k A j ɛ jik =( ij j x A) B A ( B) (A15) i k ik

6 268 APPENDIX A APPENDICES whee the sign which changed did so due to the tanspositions in the indices on the ɛ, which we have done in ode to put things in the fom of the definition of the cul Thus ( A B)=( A) B A ( B) (A16) Vecto algeba identities apply to the cul as to any odinay vecto, except that one must be caeful not to change, by eodeing, what the diffeential opeatos act on In paticula, Eq A7 is A ( B)= i A i Bi i A i B (A17) A3 Gadient in Spheical Coodinates The tansfomation between Catesian and spheical coodinates is given by = (x 2 + y 2 + z 2 ) 1 2 x= sin θ cos φ θ= cos 1 (z/) y= sin θ sin φ φ= tan 1 (y/x) z= cos θ The basis vectos {ê, ê θ, ê φ } at the point (, θ, φ) ae given in tems of the catesian basis vectos by ê = sin θ cos φ ê x + sin θ sin φ ê y + cos θ ê z ê θ = cos θ cos φ ê x + cos θ sin φ ê y sin θ ê z ê φ = sin φ ê x + cos φ ê y By the chain ule, if we have two sets of coodinates, say s i and c i,andwe know the fom a function f(s i ) and the dependence of s i on c j, we can find f c i = j f s s j c s j c i, whee s means hold the othe s s fixed while vaying s j In ou case, the s j ae the spheical coodinates, θ, φ, while the c i ae x, y, z Thus f = f ê θφ x yz φ x yz θ x x yz + f ê θφ y xz φ y xz y (A18) θ y xz

7 A3 GRADIENT IN SPHERICAL COORDINATES f θφ z xy φ z xy θ z xy ê z We will need all the patial deivatives s j c i Fom 2 = x 2 + y 2 + z 2 we see that = x = y = z x yz y xz z xy Fom cos θ = z/ = z/ x 2 + y 2 + z 2, sin θ = x yz zx (x 2 + y 2 + z 2 ) = 2 cos θ sin θ cos φ 3/2 3 so Similaly, = x yz = y xz cos θ cos φ cos θ sin φ Thee is an exta tem when diffeentiating wt z, fom the numeato, so sin θ = 1 z xy z2 = 1 cos2 θ = 1 sin 2 θ, 3 so = 1 sin θ z xy Finally, the deivatives of φ can easily be found fom diffeentiating tan φ = y/x Using diffeentials, sec 2 φdφ = dy x ydx x 2 = dy dx sin θ sin φ sin θ cos φ sin 2 θ cos 2 φ so = 1 sin φ x yz sin θ = 1 y xz cos φ sin θ =0 z xy

8 270 APPENDIX A APPENDICES Now we ae eady to plug this all into (A18) Gouping togethe the tems involving each of the thee patial deivatives, we find f = f ( x θφ êx + y êy + z êz) ( cos θ cos φ cos θ sin φ ê x + ê y sin θ ) φ êz ( 1 sin φ θ sin θ êx + 1 ) cos φ sin θ êy = f ê + 1 f ê θφ θ + 1 f ê φ sin θ φ θ Thus we have deived the fom fo the gadient in spheical coodinates