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Cartesian Tensors 1 Coordinates and Vectors Cartesian Tensors Reference: Jeffreys Cartesian Tensors 1 Coordinates and Vectors zx= 3 e3 yx= 2 e2 e1 xx= 1 Coordinates xi, i = 123,, Unit vectors: ei, i = 123,, General vector (formal definition to follow) denoted by compo- nents e.g. u = ui Summation convention (Einstein) repeated index means summa- tion: Cartesian Tensors © Geoffrey V. Bicknell 3 3 uivi = ∑uivi uii = ∑uii (1) i = 1 i = 1 2 Orthogonal Transformations of Coordinates x3 ′ ′ x2 x3 x2 ′ x1 x1 ′ xi = aijx j (2) aij = Transformation Matrix (3) Position vector ′ ′ r ==xiei x j e j ′ ⇒ a jixie j = xiei (4) ′ xi()a jie j = xiei ′ ⇒ ei = a jie j 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 ==1 if ij ij (5) = 0 otherwise 2.1 Orthonormal Condition: Now impose the condition that the primed reference is orthonormal ′ ′ ei ⋅δe j =ij and ei ⋅δe j = ij (6) Use the transformation ′ ′ ei ⋅ e j = akiek ⋅ aljel ′ ′ = akialjek ⋅ el (7) = akialjδkl = akiakj NB the last operation is an example of the substitution property of the Kronecker Delta. Since ei ⋅δe j = ij, then the orthonormal condition on aij is akiakj = δij (8) In matrix notation: aT aI= (9) Also have T aika jk ==aa δij (10) Cartesian Tensors 3/13 2.2 Reverse transformations ′ ′ xi = aijx j ⇒ aikxi ===aikaijx j δkjx j xk (11) ′ ′ ∴xk = aikxi ⇒ xi = a jix j i.e. the reverse transformation is simply given by the transpose. Similarly, ′ ei = aije j (12) 2.3 Interpretation of aij Since ′ ei = aije j (13) ′ then the aij are the components of ei wrt the unit vectors in the unprimed system. 3 Scalars, Vectors & Tensors 3.1 Scalar (f): ′ fx()i = fx()i (14) 2 Example of a scalar is f ==r xixi. Examples from fluid dynam- ics are the density and temperature. 3.2 Vector (u): Prototype vector: xi General transformation law: Cartesian Tensors 4/13 ′ ′ xi = aijx j ⇒ ui = aiju j (15) 3.2.1 Gradient operator Suppose that f is a scalar. Gradient defined by ∂ f ()grad f i ==()∇f i (16) ∂xi Need to show this is a vector by its transformation properties. ∂ f ∂ f ∂x j = (17) ′ ′ ∂xi ∂x j∂xi Since, ′ x j = akjxk (18) then ∂x j ==a δ a ∂x′ kj ki ij i (19) ∂ f ∂ f and = a ′ ij ∂xi ∂x j Hence the gradient operator satisfies our definition of a vector. 3.2.2 Scalar Product uv⋅⋅⋅ ==uivi u1v1 ++u2v2 u3v3 (20) is the scalar product of the vectors ui and vi . Exercise: Show that uv⋅⋅⋅ is a scalar. Cartesian Tensors 5/13 3.3 Tensor Prototype second rank tensor xix j General definition: ′ T ij = aika jlT kl (21) Exercise: Show that uiv j is a second rank tensor if ui and v j are vectors. Exercise: ∂ui uij, = (22) ∂x j 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 ∇ ⋅ v ==div v i (23) ∂xi 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 ijk = T ijuk is a third rank tensor if T ij is a second Cartesian Tensors 6/13 rank tensor and uk is a vector (first rank tensor). It is straightfor- ward to show that T ijkhas the relevant transformation properties. Similarly, if T ijk is a third rank tensor, then T ijj 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. Sup- pose the coordinates of an element of fluid are given as a function of time by xi = xi()t (24) vi The velocities of elements of fluid at all spatial locations within a given region constitute a vector field, i.e. vi = vi()x j, t The derivative of a function, fx()i, t along the trajectoryof a parcel fluid is given by: df ∂ f ∂ f dxi ∂ f ∂ f ----- ==+ ------- + vi (25) dt ∂t ∂xi dt ∂t ∂xi Cartesian Tensors 7/13 Derivative of velocity If we follow the trajectory of an element of fluid, then on a partic- ular trajectory xi = xi()t . The acceleration of an element is then given by: dvi d ∂vi ∂vi dx j ∂vi ∂vi f i ==vi()x j()t , t =+ =+ v j (26) dt dt ∂t ∂x jdt ∂t ∂x j Exercise: Show that f i is a vector. 6 The permutation tensor εijk εijk = 0 if any of ijk,, are equal = 1 if ijk,, unequal and in cyclic order (27) = –1 if ijk,, unequal and not in cyclic order e.g. ε112 = 0 ε123 = 1 ε321 = –1 (28) Is εijk a tensor? In order to show this we have to demonstrate that εijk, when defined the same way in each coordinate system has the correct transforma- tion properties. Cartesian Tensors 8/13 Define ′ εijk = εlmnaila jmakn = ε123ai1a j2ak3 ++ε312ai3a j1ak2 ε231ai2a j1ak2 +++ε213ai2a j1ak3 ε321ai3a j2ak1 ε132ai1a j3ak2 = ai1()a j2ak3 – a j3ak2 – ai2()a j1ak3 – a j3ak2 (29) + ai3()a j1ak2 – a j2ak1 ai1 ai2 ai3 = a j1 a j2 a j3 ak1 ak2 ak3 In view of the interpretation of the aij, the rows of this determinant represent the components of the primed unit vectors in the unprimed system. Hence: ′ ′ ′ ′ εijk = ei ⋅ e j × ek (30) This is zero if any 2 of ijk,, are equal, is +1 for a cyclic permuta- tion of unequal indices and -1 for a non-cyclic permutation of une- ′ qual indices. This is just the definition of εijk . Thus εijk transforms as a tensor. 6.1 Uses of the permutation tensor 6.1.1 Cross product Define ci = εijka jbk (31) then Cartesian Tensors 9/13 c1 ==ε123a2b3 + ε132a3b2 a2b3 – a3b2 c2 ==ε231a3b1 + ε213a1b3 a3b1 – a1b3 (32) c3 ==ε312a1b2 + ε321a2b1 a1b2 – a2b1 These are the components of cab= × . 6.1.2 Triple Product In dyadic notation the triple product of three vectors is: t = uv⋅⋅⋅ × w (33) In tensor notation this is tu==iεijkv jwk εijkuiv jwk (34) 6.1.3 Curl ∂uk ()curl u i = εijk (35) ∂x j e.g. ∂u3 ∂u2 ∂u3 ∂u2 ()curl u 1 ==ε123 + ε132 – (36) ∂x2 ∂x3 ∂x2 ∂x3 etc. 6.1.4 The tensor εiksεmps Define T ikmp = εiksεmps (37) Properties: • If ik= or mp= then T ikmp = 0. Cartesian Tensors 10/13 • If im= we only get a contribution from the terms si≠ and kis≠ , . Consequently kp= . Thus εiks = ±1 and 2 εmps ==εiks ±1 and the product εiksεiks ==()±1 1 . • If ip= , similar argument tells us that we must have si≠ and − km= ≠ i. Hence, εiks = ±1, εmps = +1 ⇒ εiksεmps = –1 . So, imk==, p ⇒ 1 unless i =0k ⇒ (38) i ==p, k m ⇒ –1 unless i =0k ⇒ These are the components of the tensor δimδkp – δipδkm. ∴δεiksεmps = imδkp – δipδkm (39) 6.1.5 Application of εiksεmps ∂ ∂ ()curl ()uv× i ==εijk ()εklmulvm εijkεklm ()ulvm ∂x j ∂x j ∂ul ∂vm = ()δilδ jm – δimδ jl vm + ul ∂x j ∂x j (40) ∂ui ∂u j ∂vm ∂vi = vm – vi + ui – u j ∂xm ∂x j ∂xm ∂x j ∂ui ∂vi ∂v j ∂u j = v j – u j + ui – vi ∂x j ∂x j ∂x j ∂x j = ()v ⋅ ∇uu– ⋅ ∇vu+ ∇ ⋅ vv– ∇ ⋅ u i 7 The Laplacean 2 2 2 ∂φ ∂φ ∂φ ∂2φ 2 ==++ --------------- (41) ∇ φ 2 2 2 ∂x1 ∂x2 ∂x3 ∂xi∂xi Cartesian Tensors 11/13 8 Tensor Integrals 8.1 Green’s Theorem ni V S In dyadic form: ∫()∇ ⋅ u dV = ∫()un⋅⋅⋅ dS (42) V S In tensor form: ∂ui dV ==uinidS Flux of u through S (43) ∫∂xi ∫ V S Extend this to tensors: ∂T ij dV ==T ijn jdS Flux of T ij through S (44) ∫∂x j ∫ V S Cartesian Tensors 12/13 8.2 Stoke’s Theorem n ti C S In dyadic form: ∫()curl u ⋅ ndS = ∫ut⋅ ds (45) S C In tensor form: ∂uk εijk nidS = uitids (46) ∫ ∂x j ∫ S C Cartesian Tensors 13/13.
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