16.21 Techniques of Structural Analysis and sig 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 R3 Free index: A subscript index Oi will be denoted a free inder 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,(a, B, y,...= 1, 2) Examples 1. ail implies a11, a21, a31.(one free index) 2. ayB implies 191, T132, 291, 2y2(two free indices) 3. ai; implies a11, a12, a13, a21, a22, a23, a31, a32, a33(two free indices implies 9 valu b=0
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 R3. 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. ai1 implies a11, a21, a31. (one free index) 2. xαyβ implies x1y1, x1y2, x2y1, x2y2 (two free indices). 3. aij implies a11, a12, a13, a21, a22, a23, a31, a32, a33 (two free indices implies 9 values). 4. ∂σij + bi = 0 ∂xj 1
has a free index(i), therefore it represents three equations +b1=0 +b3=0 Summation convention: When a repeated inder 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 aibi aibi=alb1 +a2b2+a3b3 Note that the choice of index is immaterial b i= abk xamples: 1.aa=a11+a22+a33 2. t:=Jiin, implies the three equations(why?) t1 t2=021m1+0222+23 t3=03n1+02n2+0333 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 ai= bijC d, 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 C 了 Lee indices 2 k CORRECT aibk= Ciidki+e;fii+gkPi gr INCORRECT, second term is missing free index k and third term has extra free index r
� has a free index (i), therefore it represents three equations: ∂σ1j + b1 = 0 ∂xj ∂σ2j + b2 = 0 ∂xj ∂σ3j + b3 = 0 ∂xj 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 aibi = aibi = a1b1 + a2b2 + a3b3 i=1 Note that the choice of index is immaterial: aibi = akbk Examples: 1. aii = a11 + a22 + a33 2. ti = σijnj implies the three equations (why?): t1 = σ11n1 + σ12n2 + σ13n3 t2 = σ21n1 + σ22n2 + σ23n3 t3 = σ31n1 + σ32n2 + σ33n3 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. ai = bij cjdj 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 • aibk = cijdkj free indices i, k, CORRECT • aibk = cijdkj +eifjj +gkpiqr INCORRECT, second term is missing free index k and third term has extra free index r 2
Vectors a basis in R is given by any set of linearly independent vectors ei, (e1, e2, e3) 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 1 =0,ele3=0, Using indicial notation we can write these expression in very succinct form as follows In the last expression the symbol diy is defined as the Kronecker delta if i 0ifi≠ Example a10;=a101+a2021+a3531, a1613+a20623+a30 a1+a20+a30 a10+a21+a30, a10+a20+a3 or more succinctly: a, i=a, i.e., the Kronecker delta can be thought of an Index replacer A vector v will be represented as viei=Ue1 Uge2+ U3e3 3
� Vectors A basis in R3 is given by any set of linearly independent vectors ei, (e1, e2, e3). 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: e1.e1 = 1, e2.e2 = 1, e3.e3 = 1, e1.e2 = 0, e1.e3 = 0, e2.e3 = 0, ... Using indicial notation we can write these expression in very succinct form as follows: ei.ej = δij In the last expression the symbol δij is defined as the Kronecker delta: δij = 1 if i = j, 0 if i �= j Example: aiδij =a1δ11 + a2δ21 + a3δ31, a1δ12 + a2δ22 + a3δ32, a1δ13 + a2δ23 + a3δ33 =a11 + a20 + a30, a10 + a21 + a30, a10 + a20 + a3 =a1, a2, a3 or more succinctly: aiδij = aj , i.e., the Kronecker delta can be thought of an “index replacer”. A vector v will be represented as: v = viei = v1e1 + v2e2 + v3e3 3
The vi are the components of v in the basis e;. These components are the projections of the vector on the basis vectors Taking the dot product with basis vector e i Uie,ei=U;S Transformation of basis Given two bases e;, ek and a vector v whose components in each of these bases are vi and ik, respectively, we seek to express the components in basis in terms of the components in the other basis. Since the vector is unique Taking the dot product with e But im(em ei)=immi=i; from which we obtain U,(ej.ei) Note that(ej. ei)are the direction cosines of the basis vectors of one basis on the other basis e,e;=lej‖| eill cose e
The vi are the components of v in the basis ei. These components are the projections of the vector on the basis vectors: v = vjej Taking the dot product with basis vector ei: v.ei = vj (ej .ei) = vjδji = vi Transformation of basis Given two bases ei, ˜ek and a vector v whose components in each of these bases are vi 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˜em = vnen Taking the dot product with ˜ei: v.˜ei = v˜m(˜em.˜ei) = vn(en.˜ei) But v˜m(˜ ei) = v˜mδmi = v˜i em.˜ from which we obtain: v˜i = v.˜ei = vj (ej.˜ei) Note that (ej.˜ei) are the direction cosines of the basis vectors of one basis on the other basis: ej .˜ei = �ej��˜ei� cos e �j˜ei 4
