16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #2 - Mathematical aside: Vectors, indicial notation and summation convention Indicial notation In 16.21 we’ll work in a an euclidean three-dimensional space R 3 . Free index: A subscript index () i will be denoted a free index if it is not repeated in the same additive term where the index appears. Free means that the index represents all the values in its range. ? Latin indices will range from 1 to, (i,j,k,... = 1, 2, 3), ? greek indices will range from 1 to 2, (α,β,γ,... = 1, 2). Examples: 1. a i1 implies a 11 ,a 21 ,a 31 . (one free index) 2. x α y β implies x 1 y 1 ,x 1 y 2 ,x 2 y 1 ,x 2 y 2 (two free indices). 3. a ij implies a 11 ,a 12 ,a 13 ,a 21 ,a 22 ,a 23 ,a 31 ,a 32 ,a 33 (two free indices implies 9 values). 4. ?σ ij + b i = 0 ?x j 1 ? has a free index (i), therefore it represents three equations: ?σ 1j + b 1 = 0 ?x j ?σ 2j + b 2 = 0 ?x j ?σ 3j + b 3 = 0 ?x j Summation convention: When a repeated index is found in an expression (inside an additive term) the summation of the terms ranging all the possible values of the indices is implied, i.e.: 3 a i b i = a i b i = a 1 b 1 + a 2 b 2 + a 3 b 3 i=1 Note that the choice of index is immaterial: a i b i = a k b k Examples: 1. a ii = a 11 + a 22 + a 33 2. t i = σ ij n j implies the three equations (why?): t 1 = σ 11 n 1 + σ 12 n 2 + σ 13 n 3 t 2 = σ 21 n 1 + σ 22 n 2 + σ 23 n 3 t 3 = σ 31 n 1 + σ 32 n 2 + σ 33 n 3 Other important rules about indicial notation: 1. An index cannot appear more than twice in a single additive term, it’s either free or repeated only once. a i = b ij c j d j is INCORRECT 2. In an equation the lhs and rhs, as well as all the terms on both sides must have the same free indices ? a i b k = c ij d kj free indices i,k, CORRECT ? a i b k = c ij d kj + e i f jj + g k p i q r INCORRECT, second term is missing free index k and third term has extra free index r 2 ? Vectors A basis in R 3 is given by any set of linearly independent vectors e i , (e 1 , e 2 , e 3 ). From now on, we will assume that these basis vectors are orthonormal, i.e., they have a unit length and they are orthogonal with respect to each other. This can be expressed using dot products: e 1 .e 1 = 1, e 2 .e 2 = 1, e 3 .e 3 = 1, e 1 .e 2 = 0, e 1 .e 3 = 0, e 2 .e 3 = 0,... Using indicial notation we can write these expression in very succinct form as follows: e i .e j = δ ij In the last expression the symbol δ ij is de?ned as the Kronecker delta: δ ij = 1 if i = j, 0 if i ?= j Example: a i δ ij =a 1 δ 11 + a 2 δ 21 + a 3 δ 31 , a 1 δ 12 + a 2 δ 22 + a 3 δ 32 , a 1 δ 13 + a 2 δ 23 + a 3 δ 33 =a 1 1 + a 2 0 + a 3 0, a 1 0 + a 2 1 + a 3 0, a 1 0 + a 2 0 + a 3 =a 1 , a 2 , a 3 or more succinctly: a i δ ij = a j , i.e., the Kronecker delta can be thought of an “index replacer”. A vector v will be represented as: v = v i e i = v 1 e 1 + v 2 e 2 + v 3 e 3 3 The v i are the components of v in the basis e i . These components are the projections of the vector on the basis vectors: v = v j e j Taking the dot product with basis vector e i : v.e i = v j (e j .e i ) = v j δ ji = v i Transformation of basis Given two bases e i , ?e k and a vector v whose components in each of these bases are v i and v? k , respectively, we seek to express the components in basis in terms of the components in the other basis. Since the vector is unique: v = ? ?v m e m = v n e n Taking the dot product with ?e i : e i = v? m (? e i ) = v n (e n .?v.? e m .? e i ) But v? m (? e i ) = v? m δ mi = v? i from which we obtain:e m .? v? i = v.? e i )e i = v j (e j .? Note that (e j .?e i ) are the direction cosines of the basis vectors of one basis on the other basis: e j .? e i ? cos ? e i = ?e j ??? e j ?e i 4