Last Latexed: September 8, 2005 at 10:14 1 Last Latexed: September 8, 2005 at 10:14 2 It is easy to evaluate the 27 coefficients kij, because the cross product of two ijk and cross products in 3-D orthogonal unit vectors is a unit vector orthogonal to both of them. Thus Euclidean space eˆ1 eˆ2 =ˆe3,so312 =1andk12 =0ifk = 1 or 2. Applying the same argument× toe ˆ eˆ ande ˆ eˆ , and using the antisymmetry of the cross 2 × 3 3 × 1 These are some notes on the use of the antisymmetric symbol for ex- product, A~ B~ = B~ A~,weseethat ijk × − × pressing cross products. This is an extremely powerful tool for manipulating = = =1; = = = 1, cross products and their generalizations in higher dimensions, and although 123 231 312 132 213 321 − many low level courses avoid the use of ,IthinkthisisamistakeandIwant and = 0 for all other values of the indices, i.e. = 0 whenever any you to become proficient with it. ijk ijk two of the indices are equal. Note that changes sign not only when the last In a cartesian coordinate system a vector ~ has components along each V Vi two indices are interchanged (a consequence of the antisymmetry of the cross of the three orthonormal basis vectorse ˆ ,orV~ = V eˆ . The dot product i Pi i i product), but whenever any two of its indices are interchanged. Thus is of two vectors, A~ B~ , is bilinear and can therefore be written as ijk · zero unless (1, 2, 3) (i, j, k) is a permutation, and is equal to the sign of the permutation if it→ exists. A~ B~ =(X Aieˆi) X Bjeˆj (1) · · Now that we have an expression fore ˆ eˆ , we can evaluate i j i × j = X X AiBjeˆi eˆj (2) ~ ~ i j · A B = X X AiBj(ˆei eˆj)=X X X kijAiBjeˆk. (4) × i j × i j k = X X AiBjδij, (3) i j Much of the usefulness of expressing cross products in terms of ’s comes from the identity where the Kronecker delta δ is defined to be 1 if i = j and 0 otherwise. ij = δ δ δ δ , (5) As the basis vectors ˆ are orthonormal, orthogonal to each other and of X kij k`m i` jm im j` ek i.e. k − unit length, we havee ˆ eˆ = δ . i j ij which can be shown as follows. To get a contribution to the sum, k must be Doing a sum over an· index j of an expression involving a δ is very simple, ij different from the unequal indices i and j, and also different from ` and m. because the only term in the sum which contributes is the one with j = i. Thus we get 0 unless the pair (i, j)andthepair(`, m) are the same pair of Thus F (i, j)δ = F (i, i), which is to say, one just replaces j with i in all Pj ij different indices. There are only two ways that can happen, as given by the the other factors, and drops the δ and the summation over j.Sowehave ij two terms, and we only need to verify the coefficients. If i = ` and j = m, ~ ~ 1 A B = Pi AiBi, the standard expression for the dot product the two ’s are equal and the square is 1, so the first term has the proper · ~ ~ We now consider the cross product of two vectors, A B,whichisalso coefficient of 1. The second term differs by one transposition of two indices a bilinear expression, so we must have A~ B~ =( A ×eˆ ) ( B eˆ )= × Pi i i × Pj j j on one epsilon, so it must have the opposite sign. Pi Pj AiBj(ˆei eˆj ). The cross producte ˆi eˆj is a vector, which can therefore We now turn to some applications. Let us first evaluate ×~ × be written as V = Pk Vkeˆk. But the vector result depends also on the two input vectors, so the coefficients Vk really depend on i and j as well. Define A~ (B~ C~ )=X Ai X ijkBjCk = X ijkAiBjCk. (6) · × them to be ijk,so i jk ijk eˆi eˆj = X kijeˆk. × k Note that A~ (B~ C~ ) is, up to sign, the volume of the parallelopiped formed · × by the vectors ~, ~ ,and~ . From the fact that the changes sign under 1Note that this only holds because we have expressed our vectors in terms of orthonor- A B C mal basis vectors. Last Latexed: September 8, 2005 at 10:14 3 Last Latexed: September 8, 2005 at 10:14 4 transpositions of any two indices, we see that the same is true for transposing In higher dimensions, the cross product is not a vector, but there is a gen- the vectors, so that eralization of which remains very useful. In an n-dimensional space, i1i2...in has n indices and is defined as the sign of the permutation (1, 2,...,n) A~ (B~ C~ )= A~ (C~ B~ )=B~ (C~ A~)= B~ (A~ C~ ) → · × − · × · × − · × (i1i2 ...in), if the indices are all unequal, and zero otherwise. The analog of = C~ (A~ B~ )= C~ (B~ A~). (5) has (n 1)! terms from all the permutations of the unsummed indices on · × − · × − the second . The determinant of an n n matrix is defined as Now consider V~ = A~ (B~ C~ ). Using our formulas, × × × n ~ ~ ~ det A = X i i ...in Y Ap,ip . V = X kijeˆkAi(B C)j = X kijeˆkAi X jlmBlCm. 1 2 i ,...,i p=1 ijk × ijk lm 1 n Notice that the sum on j involves only the two epsilons, and we can use X kijjlm = X jkijlm = δklδim δkmδil. j j − Thus Vk = X(X kijjlm)AiBlCm = X(δklδim δkmδil)AiBlCm ilm j ilm − = X δklδimAiBlCm X δkmδilAiBlCm ilm − ilm = X AiBkCi X AiBiCk = A~ CB~ k A~ BC~ k, i − i · − · so A~ (B~ C~ )=B~ A~ C~ C~ A~ B.~ (7) × × · − · This is sometimes known as the bac-cab formula. Exercise: Using (5) for the manipulation of cross products, show that (A~ B~ ) (C~ D~ )=A~ C~ B~ D~ A~ D~ B~ C.~ × · × · · − · · The determinant of a matrix can be defined using the symbol. For a 3 3 matrix A, × det A = X ijkA1iA2jA3k = X ijkAi1Aj2Ak3. ijk ijk From the second definition, we see that the determinant is the volume of the parallelopiped formed from the images under the linear map A of the three unit vectorse ˆi,as (Aeˆ ) ((Aeˆ ) (Aeˆ )) = det A. 1 · 2 × 3.