Cartesian Tensors Reference: Jeffreys Cartesian Tensors 1 Coordinates and Vectors z x 3 e 3 y x 2 e 2 e 1 x x 1 Coordinates x i, i 123,, Unit vectors: e i, i 123,, General vector (formal definition to follow) denoted by components e.g. u u i Summation convention (Einstein) repeated index means summation: Cartesian Tensors Geoffrey V. Bicknell
u i v i 3 u i v i u ii u ii i 1 3 i 1 2 Orthogonal Transformations of Coordinates (1) x 3 x 3 x2 x 2 x 1 x 1 x i a i x (2) a i Transformation Matrix (3) Position vector r x i e i x e a i x i e x i e i x i ( a i e ) x i e i e i a i e (4) i.e. the transformation of coordinates from the unprimed to the primed frame implies the reverse transformation from the primed to the unprimed frame for the unit vectors. Cartesian Tensors 2/13
Kronecker Delta δ i 1 if i 0 otherwise (5) 2.1 Orthonormal Condition: Now impose the condition that the primed reference is orthonormal e i e δ i and e i e δ i (6) Use the transformation e i a ki e k a l e l e a ki a l e k e l a ki a l δ kl a ki a k (7) NB the last operation is an example of the substitution property of the Kronecker Delta. Since e i e δ i, then the orthonormal condition on a i is a ki a k δ i (8) In matrix notation: a T a I (9) Also have a ik a k aa T δ i (10) Cartesian Tensors 3/13
2.2 Reverse transformations x i a i x a ik x i a ik a i x δ k x x k x k a ik x i x i a i x (11) i.e. the reverse transformation is simply given by the transpose. Similarly, e i a i e (12) 2.3 Interpretation of a i Since e i a i e (13) then the a i are the components of e i wrt the unit vectors in the unprimed system. 3 Scalars, Vectors & Tensors 3.1 Scalar (f): f( x i) f( x i ) (14) Example of a scalar is f r 2 x i x i. Examples from fluid dynamics are the density and temperature. 3.2 Vector (u): Prototype vector: x i General transformation law: Cartesian Tensors 4/13
x i a i x u i a i u (15) 3.2.1 Gradient operator Suppose that f is a scalar. Gradient defined by ( grad f ) i ( f ) i x i (16) Need to show this is a vector by its transformation properties. x i x x x i (17) Since, x a k x k (18) then and x a x k δ ki a i i x i a i x (19) Hence the gradient operator satisfies our definition of a vector. 3.2.2 Scalar Product u v u i v i u 1 v 1 + u 2 v 2 + u 3 v 3 (20) is the scalar product of the vectors and. Exercise: u i v i Show that u v is a scalar. Cartesian Tensors 5/13
3.3 Tensor Prototype second rank tensor x i x General definition: T i a ik a l T kl (21) Exercise: Show that u i v is a second rank tensor if u i and v are vectors. Exercise: u i u i, x (22) is a second rank tensor. (Introduces the comma notation for partial derivatives.) In dyadic form this is written as grad u or u. 3.3.1 Divergence Exercise: Show that the quantity v div v v i x i (23) is a scalar. 4 Products and Contractions of Tensors It is easy to form higher order tensors by multiplication of lower rank tensors, e.g. T ik T i u k is a third rank tensor if T i is a second Cartesian Tensors 6/13
rank tensor and u k is a vector (first rank tensor). It is straightforward to show that T ik has the relevant transformation properties. Similarly, if T ik is a third rank tensor, then T i is a vector. Again the relevant tr4ansformation properties are easy to prove. 5 Differentiation following the motion This involves a common operator occurring in fluid dynamics. Suppose the coordinates of an element of fluid are given as a function of time by x i x i () t (24) v i The velocities of elements of fluid at all spatial locations within a given region constitute a vector field, i.e. v i ( x, t) v i The derivative of a function, fluid is given by: f( x i, t) along the traectoryof a parcel df ----- dt + dx ------ i t x i dt t + v i x i (25) Cartesian Tensors 7/13
Derivative of velocity If we follow the traectory of an element of fluid, then on a particular traectory x i x i () t. The acceleration of an element is then given by: f i dv i d v vi ( x dt dt ()t t, ) i v i dx + t dt x v i v i + v t x (26) Exercise: Show that is a vector. 6 The permutation tensor f i ε ik ε ik 0 if any of i,, k are equal 1 if i,, k unequal and in cyclic order 1 if i,, k unequal and not in cyclic order (27) e.g. ε 112 0 ε 123 1 ε 321 1 (28) Is ε ik a tensor? In order to show this we have to demonstrate that ε ik, when defined the same way in each coordinate system has the correct transformation properties. Cartesian Tensors 8/13
Define ε ik ε lmn a il a m a kn ε 123 a i1 a 2 a k3 + ε 312 a i3 a 1 a k2 + ε 231 a i2 a 1 a k2 + ε 213 a i2 a 1 a k3 + ε 321 a i3 a 2 a k1 + ε 132 a i1 a 3 a k2 a i1 ( a 2 a k3 a 3 a k2 ) a i2 ( a 1 a k3 a 3 a k2 ) + a i3 ( a 1 a k2 a 2 a k1 ) (29) a i1 a i2 a i3 a 1 a 2 a 3 a k1 a k2 a k3 In view of the interpretation of the, the rows of this determinant represent the components of the primed unit vectors in the unprimed system. Hence: a i ε ik e e e i k (30) This is zero if any 2 of i,, k are equal, is +1 for a cyclic permutation of unequal indices and -1 for a non-cyclic permutation of unequal indices. This is ust the definition of ε ik. Thus ε ik transforms as a tensor. 6.1 Uses of the permutation tensor 6.1.1 Cross product Define c i ε ik a b k (31) then Cartesian Tensors 9/13
c 1 ε 123 a 2 b 3 + ε 132 a 3 b 2 a 2 b 3 a 3 b 2 c 2 ε 231 a 3 b 1 + ε 213 a 1 b 3 a 3 b 1 a 1 b 3 c 3 ε 312 a 1 b 2 + ε 321 a 2 b 1 a 1 b 2 a 2 b 1 (32) These are the components of c a b. 6.1.2 Triple Product In dyadic notation the triple product of three vectors is: t u v w (33) In tensor notation this is t u i ε ik v w k ε ik u i v w k (34) 6.1.3 Curl u k ( curl u) i ε ik x (35) e.g. u 3 u 2 ( curl u) 1 ε 123 + ε x 132 2 x 3 u 3 u 2 x 2 x 3 (36) etc. 6.1.4 The tensor ε iks ε mps Define T ikmp ε iks ε mps (37) Properties: If i k or m p then T ikmp 0. Cartesian Tensors 10/13
If i m we only get a contribution from the terms s i and k i, s. Consequently k p. Thus ε iks ± 1 and ε mps ε iks ± 1 and the product ε iks ε iks ( ± 1) 2 1. If i p, similar argument tells us that we must have s i and k m i. Hence, ε iks ± 1, ε mps + 1 ε iks ε mps 1. So, i m, k p 1 unless i i p, k m 1 unless i k 0 k 0 These are the components of the tensor δ im δ kp δ ip δ km. ε iks ε mps δ im δ kp δ ip δ km (38) (39) 6.1.5 Application of ε iks ε mps ( curl ( u v) ) i ε ik ( εklm u x l v m ) ε ik ε klm ( ul v x m ) ( δ il δ m δ im δ l ) u l x v u v m + m l x u i v x u m v i m x u v m v i + i x u x m (40) 7 The Laplacean u i v i v u x x u v u + i x v i x ( v u u v+ u v v u) i 2 φ 2 2 φ φ φ + + x 2 1 x 2 2 x 2 3 2 2 φ -------------- x i x i (41) Cartesian Tensors 11/13
8 Tensor Integrals 8.1 Green s Theorem n i V S In dyadic form: In tensor form: V ( u) dv ( u n) ds S (42) u i dv x i V S u i n i ds Extend this to tensors: Flux of u through S (43) T i V dv T x i n ds S Flux of T i through S (44) Cartesian Tensors 12/13
8.2 Stoke s Theorem t i n C S In dyadic form: In tensor form: S ( curl u) nds u tds C (45) u k n i S ε ik ds u x i t i ds C (46) Cartesian Tensors 13/13