Cartesian Tensors 1 Coordinates and Vectors

Cartesian Tensors

Reference: Jeffreys Cartesian Tensors 1 Coordinates and Vectors

zx= 3

yx= 2

xx= 1

Coordinates xi, i = 123,,

Unit vectors: ei, i = 123,, General vector (formal definition to follow) denoted by components e.g. u = ui Summation convention (Einstein) repeated index means summation:

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

′ 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 dynamics 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 ijk has 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)

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 transformation 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 permutation 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

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

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