Index Notation for Vector Calculus

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.

2 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. 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.}

3 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.

4 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.

5 Kronecker delta reductionReduce ij jk ki: 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).

6 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.

7 Note that ei ei= iik ek= 0, since iikfor all values ofk. 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.

8 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.

9 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. 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.

10 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. 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).