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Tensors in Generalized Coordinate Systems: Components and Direct Notation

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Tensors in Generalized Coordinate Systems: Components and Direct Notation Tensors in generalized coordinate systems: components and direct notation Math 1550 lecture notes, Prof. Anna Vainchtein 1 Vectors in generalized coordinates Consider a generalized coordinate system (x1, x2, x3) with the local basis {e1, e2, e3}. The basis is not necessarily orthogonal, let alone orthonormal. It comes along with its reciprocal basis {e1, e2, e3}. Recall that we can write a vector a in terms of either basis, using contravariant components ai and covariant components ai, respectively: 1 2 3 i 1 2 3 i a = a e1 + a e2 + a e3 = a ei, a = a1e + a2e + a3e = aie . (1) Recall also that we can find the covariant and contravariant components of the vector by taking dot products with the basis vectors and reciprocal basis vectors, respectively: i i ai = a · ei, a = a · e (2) Consider now another coordinate system (¯x1, x¯2, x¯3), with the basis vec- tors p p p ¯ei = αi ep, αi = ¯ei · e . (3) Recall that the reciprocal basis vectors then transform via the inverse trans- formation (see the first set of notes): i −1 i p −1 i i ¯e = (α )pe , (α )p = ep · ¯e (4) Recall also that covariant and contravariant components of a vector (first order tensor) transform according to p i −1 i p a¯i = αi ap, a¯ = (α )pa . (5) Notice that that the transformation law for the covariant components in- volves the direct transformation matrix α, while the one for contravariant components has the inverse transformation matrix α−1. As we have seen 1 before, this is becausea ¯i = a·¯ei, where ¯ei is related to ei via the direct trans- formation, while ina ¯i = a · ¯ei, the reciprocal vector ¯ei transforms according to (4). In particular, the new and old coordinates of the same point with position vector 1 2 3 1 2 3 r = x e1 + x e2 + x e3 =x ¯ ¯e1 +x ¯ ¯e2 +x ¯ ¯e3 are related by i −1 i p p p i x¯ = (α )px , x = αi x¯ , and thus we can also represent the direct and inverse transformation matrices in terms of partial derivatives of the old and new coordinates with respect to one another: ∂xp ∂x¯i αp = , (α−1)i = (6) i ∂x¯i p ∂xp Using this, we can rewrite (5) as ∂xp ∂x¯i a¯ = a , a¯i = ap (7) i ∂x¯i p ∂xp Finally, recall that covariant and contravariant components are not inde- pendent. They are related by i ik k a = g ak, ai = gika , ik i k where we recall that gik = ei ·ek and g = e ·e are covariant and contravari- ant components of the metric tensor. We called this raising or lowering the index via the metric tensor. Remark. In the case of two rectangular coordinate systems with orthonor- i i i i mal bases that we considered earlier, we have e = ei, ¯e = e , ai = a and α becomes an orthogonal matrix, α−1 = αT . Thus, in this case we have j −1 j αi = Qij and (α )i = Qji, where Qij = ¯ei · ej are components of an orthog- onal matrix. Example 1. Consider the vector a = x1x2i1 + x2x3i2 + x1x3i3, where {i1, i2, i3} is the standard basis in the Cartesian coordinate system (x1, x2, x3). 2 a) Find the covariant components of a in parabolic cylindrical coordinates (v, w, z) defined by v2 − w2 x = , x = vw, x = z. 1 2 2 3 b) Express its contravariant components in terms of covariant ones. Solution. a) The new basis is ∂x ∂x ∂x ¯e = 1 i + 2 i + 3 i = vi + wi 1 ∂u 1 ∂u 2 ∂u 3 1 2 ∂x ∂x ∂x ¯e = 1 i + 2 i + 3 i = −wi + vi 2 ∂v 1 ∂v 2 ∂v 3 1 2 ∂x ∂x ∂x ¯e = 1 i + 2 i + 3 i = i 3 ∂z 1 ∂z 2 ∂z 3 3 The transformation matrix is v w 0 j [αi ] = [¯ei · ij] = −w v 0 0 0 1 j The covariant components of a in the new basis area ¯i = αi aj, or 1 2 2 1 2 2 2 2 a¯1 v w 0 2 (v − w )vw 2 (v − w )v w + vw z 1 2 2 2 2 a¯2 = −w v 0 vwz = − 2 (v − w )vw + v wz . 1 2 2 1 2 2 a¯3 0 0 1 2 (v − w )z 2 (v − w )z √ 2 2 b) Note that the new basis is orthogonal, with h1 = |¯e1| = v + w = |¯e2| = h2 and h3 = |¯e3| = 1. Thus the contravariant components of the 1 metric tensor areg ¯ij = 0 for i 6= j andg ¯11 =g ¯22 = andg ¯33 = 1. v2 + w2 Therefore, 1 1 a¯1 =g ¯11a¯ = a¯ , a¯1 =g ¯22a¯ = a¯ , a¯3 =g ¯33a¯ =a ¯ . 1 v2 + w2 1 2 v2 + w2 2 3 3 Example 2. Let f(x1, x2, x3) be a scalar field. If we change to new coor- dinatesx ¯i =x ¯i(x1, x2, x3), we have, via the chain rule, ∂f ∂f ∂xk = , ∂x¯i ∂xk ∂x¯i 3 which by the first equation in (6) yields ∂f ∂f = αk . ∂x¯i i ∂xk ∂f Thus, v = transform as covariant components of the vector (gradient i ∂xi of f) ∂f v = ei = ∇f, ∂xi where ei are the reciprocal basis vectors associated with coordinates (x1, x2, x3). Of course, the vector v also has contravariant components ∂f ∂f vi = gikv = gik , v = gik e . k ∂xk ∂xk i 2 General higher order tensors By analogy with above transformation laws for covariant and contravariant components of a vector, we can now introduce a more general definition of a second order tensor, no longer considering only rectangular coordinate sys- tems: Definition. A second order tensor in Rd is a quantity uniquely specified 2 by d numbers (its components). These components can be covariant (Aij), ij ·j i contravariant (A ) or mixed (Ai , A·j), and they transform under the change of basis (3) according to ¯ p q Aij = αi αj Apq (8) ¯ij −1 i −1 j pq A = (α )p(α )qA (9) ¯·j p −1 j ·q Ai = αi (α )qAp (10) ¯i −1 i q p A·j = (α )pαj A·q (11) ·j The little dot helps us denote the actual position of the index, e.g. in Ai ·j j j the index j comes second. Since Ai 6= A·i in general, writing Ai is mislead- ing, since we don’t know which of the two is actually meant (Kronecker delta j i δi = δj is an exception). Note that in the transformation laws the covariant indices always require direct transformation matrix α, while the contravari- ant indices come along with inverse transformation matrix α−1, just like 4 it was for the vector transformation laws (5). The indices in the transfor- mation matrices are arranged so that the indices that are not summed over are in the same position as in the new component, while the summation is over the opposite indices. The only exception is Cartesian coordinates with orthonormal bases, where reciprocal vectors coincide with the regular ones, and the covariant, contravariant and mixed components coincide for each pair of indices. Recalling that the transformation matrix can be represented as (6), we can also write (8)-(11) as ∂xp ∂xq A¯ = A ij ∂x¯i ∂x¯j pq i j ¯ij ∂x¯ ∂x¯ pq A = p q A ∂x ∂x (12) ∂xp ∂x¯j A¯·j = A·q i ∂x¯i ∂xq p ∂x¯i ∂xq A¯i = Ap ·j ∂xp ∂x¯j ·q This is how second order tensors are defined in some books. Similar to vectors, the covariant, contravariant and mixed components of a second order tensor are related to one another via the metric tensor, which raises or lowers the corresponding indices. We have mn ·n m Aij = gimgjnA = gjnAi = gimA·j ij im jn im ·j jn i A = g g Amn = g Am = g A·n ·j jn mj (13) Ai = g Ain = gimA i im in A·j = g Amj = gjnA We can now easily generalize this to write transformation laws for higher order tensors. For example, ¯ p q r Aijk = αi αj αkApqr ¯ijk −1 i −1 j −1 k pqr A = (α )p(α )q(α )r A ¯ij −1 i −1 j r pq A··k = (α )p(α )qαkA··r ¯·j· p −1 j r ·q· Ai·k = αi (α )qαkAp·r, and so on. Once again, we define a third order tensor as an object whose various components transform according to these rules - and this is what we 5 need to check to verify these are components of a tensor. The components of a third order tensor are again related by the metric tensor: m mn mnl Aijk = gimA· jk = gimginA·· j = gimgjngklA , etc. ijkl Exercise. Write the transformation laws for the components Aijkl, A , ··kl ·jk· Aij and Ai··l of a fourth order tensor and the relations between these com- ponents. Example 1. Contravariant components of a tensor A in the basis e1 = (0, 1, 1), e2 = (1, 0, 1), e3 = (1, 1, 1) are −1 2 0 ij [A ] = 2 0 3 . 0 3 −2 i ·j Find its mixed components A·j and Ai and covariant components Aij. i im Solution. We have A·j = A gmj. The metric tensor is given by 2 1 2 [gmj] = [em · ej] = 1 2 2 . 2 2 3 Therefore, −1 2 0 2 1 2 0 3 2 i im [A·j] = [A ][gmj] = 2 0 3 1 2 2 = 10 8 13 .
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