Index Notation for Vector Calculus - New Mexico Institute ...

1 Index Notation for Vector CalculusbyIlan Ben-Yaacov and Francesc RoigCopyrightc 2006 Index Notation , also commonly known as subscript Notation or tensor Notation ,is an extremely useful tool for performing Vector algebra. Consider the coordinatesystem illustrated in Figure 1. Instead of using the typical axis labelsx,y, andz,we usex1,x2, andx3, orxii= 1,2,3 The corresponding unit basis vectors are then e1, e2, and e3, or eii= 1,2,3 The basis vectors e1, e2, and e3have the following properties: e1 e1= e2 e2= e3 e3= 1(1) e1 e2= e1 e3= e2 e3= 0(2)x1x2x3a1a2a3ae1e2e3 Figure 1: Reference coordinate NotationWe now introduce theKronecker deltasymbol ij. ijhas the following prop-erties: ij={0i6=j1i=ji, j= 1,2,3(3)Using Eqn 3, Eqns 1 and 2 may be written in Index Notation as follows: ei ej= iji, j= 1,2,3(4)In standard Vector Notation , a Vector ~Amay be written in component form as~A =Ax i+Ay j+Az k(5)Using Index Notation , we can express the Vector ~Aas~A =A1 e1+A2 e2+A3 e3=3 i=1Ai ei(6)Notice that in the expression within the summation, the indexiisrepeated.}

2 Re-peated indices are always contained within summations, or phrased differently arepeated indeximpliesa summation. Therefore, the summation symbol is typi-cally dropped, so that~Acan be expressed as~A =Ai ei 3 i=1Ai ei(7)This repeated Index Notation is known as Einstein s convention. Any repeatedindex is called adummy Index . Since a repeated Index implies a summation overall possible values of the Index , one can always relabel a dummy Index , ~A =Ai ei=Aj ej=Ak eketc. A1 e1+A2 e2+A3 e3(8)Copyrightc 2006 by Ilan Ben-Yaacov and Francesc RoigIndex Notation3 The Scalar Product in Index NotationWe now show how to expressscalar products(also known as inner productsor dot products) using Index Notation . Consider the vectors~aand~b, which can beexpressed using Index Notation as~a=a1 e1+a2 e2+a3 e3=ai ei~b=b1 e1+b2 e2+b3 e3=bj ej(9)Note that we use different indices (iandj) for the two vectors to indicate that theindex for~bis completely independent of that used for~a.

3 We will first write out thescalar product~a ~bin long-hand form, and then express it more compactly usingsome of the properties of Index Notation .~a ~b=(3 i=1ai ei) 3 j=1bj ej =3 i=13 j=1[(ai ei) (bj ej)]=3 i=13 j=1[aibj( ei ej)] (commutative property)=3 i=13 j=1(aibj ij) (from Eqn 3)Summing over all values ofiandj, we get~a ~b=a1b1 11+a1b2 12+a1b3 13+a2b1 21+a2b2 22+a2b3 23+a3b1 31+a3b2 32+a3b3 33=a1b1 11+a2b2 22+a3b3 33=a1b1+a2b2+a3b3=3 i=1aibi=aibi=ajbj=akbkCopyrightc 2006 by Ilan Ben-Yaacov and Francesc Roig4 Index NotationDoing this in a more compact Notation gives us~a ~b= (ai ei) (bj ej)=aibj ei ej=aibj ij=aibi=a1b1+a2b2+a3b3 Notice that when we have an expression containing ij, we simply get rid of the ijand seti=jeverywhere in the 1: Kronecker delta reductionReduce ij jk ki.

4 Ij jk ki= ik ki(remove ij,setj=ieverywhere)= ii(remove ik,setk=ieverywhere)=3 i=1 ii=3 i=11 = 1 + 1 + 1 = 3 Here we can see that ii= 3(Einstein convention implied)(10)Note also that ij jk= ik(11)Example 2:~rand rin Index Notation (a) Express~rusing Index Notation .~r=x1 e1+x2 e2+x3 e3=xi eiCopyrightc 2006 by Ilan Ben-Yaacov and Francesc RoigIndex Notation5(b) Express rusing Index Notation . r=~r|~r|=~r(~r ~r)1/2=xi ei(xjxj)1/2(c) Express~a rusing Index Notation .~a r=~a ~r|~r|=aixi(xjxj)1/2 The Cross Product in Index NotationConsider again the coordinate system in Figure 1. Using the conventional right-hand rule for cross products, we have e1 e1= e2 e2= e3 e3= 0 e1 e2= e3 e2 e1= e3 e2 e3= e1 e3 e2= e1 e3 e1= e2 e1 e3= e2(12)To write the expressions in Eqn 12 using Index Notation , we must introduce thesymbol ijk, which is commonly known as the Levi-Civita tensor, the alternatingunit tensor, or the permutation symbol (in this text it will be referred to as thepermutation symbol).

5 Ijkhas the following properties: ijk= 1if (ijk) is an even (cyclic) permutationof (123), 123= 231= 312= 1 ijk= 1if (ijk) is an odd (noncyclic) permutationof (123), 213= 321= 132= 1 ijk= 0if two or more subscripts are the same, 111= 112= 313= 2006 by Ilan Ben-Yaacov and Francesc Roig6 Index NotationHence, we may rewrite the expressions in Eqn 12 as follows: e1 e2= 123 e3 e2 e1= 213 e3 e2 e3= 231 e1 e3 e2= 321 e1 e3 e1= 312 e2 e1 e3= 132 e2(13)Now, we may write a single generalized expression for all the terms in Eqn 13: ei ej= ijk ek(14)Here ijk ek 3 k=1 ijk ek(kis a dummy Index ). That is, this works because e1 e2= 12k ek=3 k=1 12k ek= 121 e1+ 122 e2+ 123 e3= e3 The same is true for all of the other expressions in Eqn 13. Note that ei ei= iik ek= 0, since iikfor all values ofk.

6 Ijkis also given by the following formula. ijk=12(i j)(j k)(k i)i, j, k= 1,2,3(15)This is a remarkable formula that works for ijkif you do not want to calculate theparity of the permutation(ijk). Also note the following property of ijk: ijk= jik= any two subscripts reverses the sign of the permutation symbol (orin other words ijkisanti-symmetric). Also, ijk= kij= permutations of the subscripts do not change the sign of ijk. TheseCopyrightc 2006 by Ilan Ben-Yaacov and Francesc RoigIndex Notation7properties also follow from the formula in Eqn , let s consider the cross product of two vectors~aand~b, where~a=ai ei~b=bj ejThen~a ~b= (ai ei) (bj ej) =aibj ei ej=aibj ijk ekThus we write for the cross product:~a ~b= ijkaibj ek(16)All indices in Eqn 16 are dummy indices (and are therefore summed over) sincethey are repeated.

7 We can always relabel dummy indices, so Eqn 16 may be writtenequivalently as~a ~b= pqrapbq erReturning to Eqn 16, thekth component of~a ~bis(~a ~b)k= ijkaibjwhere now onlyiandjare dummy indices. Note that the cross product may alsobe written in determinant form as follows:~a ~b= e1 e2 e3a1a2a3b1b2b3 (17)The follwoing is a very important identity involoving the product of two per-mutation symbols. ijk lmn= il im in jl jm jn kl km kn (18)Copyrightc 2006 by Ilan Ben-Yaacov and Francesc Roig8 Index NotationThe proof of this identity is as follows: If any two of the indicesi, j, korl, m, nare the same, then clearly the left-hand side of Eqn 18 must be zero. This condition would also result in twoof the rows or two of the columns in the determinant being the same, sotherefore the right-hand side must also equal zero.

8 If(i, j, k)and(l, m, n)both equal (1,2,3), then both sides of Eqn 18 areequal to one. The left-hand side will be1 1, and the right-hand side willbe the determinant of the identity matrix. If any two of the indicesi, j, korl, m, nare interchanged, the correspondingpermutation symbol on the left-hand side will change signs, thus reversingthe sign of the left-hand side. On the right-hand side, an interchange oftwo indices results in an interchange of two rows or two columns in thedeterminant, thus reversing its , all possible combinations of indices result in the two sides ofEqn 18 being equal. Now consider the special case of Eqn 18 wheren= this case, the repeated indexkimplies a summation over all values ofk. Theproduct of the two permutation symbols is now ijk lmk= il im ik jl jm jk kl km kk (note kk= 3)= 3 il jm 3 im jl+ im jk kl ik jm kl+ ik jl km il jk km= 3 il jm 3 im jl+ im jl il jm+ im jl il jm(from Eqn 11)(19)Or finally ijk lmk= il jm im jl(20)Copyrightc 2006 by Ilan Ben-Yaacov and Francesc RoigIndex Notation9 Eqn 20 is anextremelyuseful property in Vector algebra and Vector calculusapplications.

9 It can also be expressed compactly in determinant form as ijk lmk= il im jl jm (21)The cyclic property of the permutation symbol allows us to write also ijk klm= il jm im jlTo recap: ei ej= ijand~a ~b=aibi ei ej= ijk ekand~a ~b= ijkaibj ekThese relationships, along with Eqn 20, allow us to prove any Vector 3: Vector identity proofShow for the double cross product:~a (~b ~c) = (~a ~c)~b (~a ~b)~cStart with the left-hand side (LHS):~a (~b ~c) = (ai ei) ( jklbjck el)= jklaibjck( ei el)= jklaibjck ilh eh= jkl hilaibjck eh= ( jh ki ji kh)aibjck eh= jh kiaibjck eh ji khaibjck eh=aibjci ej aibick ek= (aici) (bj ej) (aibi) (ck ek)=(~a ~c)~b (~a ~b)~cCopyrightc 2006 by Ilan Ben-Yaacov and Francesc Roig10 Index NotationExample 4: The scalar triple productShow that~a (~b ~c) =~c (~a ~b) =~b (~c ~a)~a (~b ~c) = (ai ei) ( jkmbjck em)= jkmaibjck( ei em)= jkmaibjck im= jkiaibjckor~a (~b ~c) = ijkaibjckFrom our permutation rules, it follows that~a (~b ~c) = ijkaibjck= kijckaibj=~c (~a ~b)= jkibjckai=~b (~c ~a)Copyrightc 2006 by Ilan Ben-Yaacov and Francesc Roig