Chapter 3 Cartesian Tensors - damtp

ecause (i) this is equivalent to x = Ix; or (ii) we can check for each value of i (e.g.,when i = 1, RHS = δ 1j x j = δ 11 x 1 + δ 12 x 2 + δ 13 x 3 = x 1 = LHS). The Kronecker delta just“selects” entries: e.g., δ ik a jk is equal to a ji .What is δ ii ? It is not 1.The alternating tensor can be used to write down the vector equation z = x × y insuffix notation:z i = [x × y] i = ɛ ijk x j y k .(Check this: e.g., z 1 = ɛ 123 x 2 y 3 + ɛ 132 x 3 y 2 = x 2 y 3 − x 3 y 2 , as required.) There is one veryimportant property of ɛ ijk :ɛ ijk ɛ klm = δ il δ jm − δ im δ jl .This makes many vector identities easy to prove.(The property may be proved by first proving the generalisation⎛⎞δ il δ im δ in⎜⎟ɛ ijk ɛ lmn = det ⎝δ jl δ jm δ jn ⎠ .δ kl δ km δ knBoth sides clearly vanish if any of i, j, k are equal; or if any of l, m, n are. Now takei = l = 1, j = m = 2, k = n = 3: both sides are clearly 1. Finally consider the effectof swapping say i and j. Once we have proved this generalisation, contract k and n andsimplify, noting that for example δ jk δ km = δ jm .)Example: prove that a × (b × c) = (a . c)b − (a . b)c.[a × (b × c)] i = ɛ ijk a j [b × c] k= ɛ ijk a j ɛ klm b l c m= (δ il δ jm − δ im δ jl )a j b l c m= a j b i c j − a j b j c i= (a . c)b i − (a . b)c i= [(a . c)b − (a . b)c] i ,as required.ɛ ijk can also be used to calculate determinants. The determinant of a 3 × 3 matrixA = (a ij ) is given by ɛ ijk a 1i a 2j a 3k (check this by just expanding the product and sum in49 © R. E. Hunt, 2002

full). This can be written in several other ways; for example,det A = ɛ ijk a 1i a 2j a 3k = ɛ jik a 1j a 2i a 3k[swapping i and j]= −ɛ ijk a 2i a 1j a 3k .This proves that swapping two rows of a matrix changes the sign of the determinant.3.3 What is a Vector?A vector is more than just 3 real numbers. It is also a physical entity: if we know its3 components with respect to one set of Cartesian axes then we know its componentswith respect to any other set of Cartesian axes. (The vector stays the same even if itscomponents do not.)For example, suppose that {e 1 , e 2 , e 3 } is a right-handedorthogonal set of unit vectors, and that a vector v has componentsv i relative to axes along those vectors. That is tosay,v = v 1 e 1 + v 2 e 2 + v 3 e 3 = v j e j .What are the components of v with respect to axes whichhave been rotated to align with a different set of unit vectors{e ′ 1, e ′ 2, e ′ 3}? Letv = v 1e ′ ′ 1 + v 2e ′ ′ 2 + v 3e ′ ′ 3 = v je ′ ′ j.Now e ′ i . e ′ j = δ ij , sov . e ′ i = v je ′ ′ j . e ′ i = v jδ ′ ij = v i′but alsov . e ′ i = v j e j . e ′ i = v j l ijwhere we define the matrix L = (l ij ) byl ij = e ′ i . e j .Thenv i ′ = l ij v j(or, in matrix notation, v ′ = Lv where v ′ is the column vector with components v i). ′ Lis called the rotation matrix.This looks like, but is not quite the same as, rotating the vector v round to a different vector v ′ using atransformation matrix L. In the present case, v and v ′ are the same vector, just measured with respect50 © R. E. Hunt, 2002

andE i = l ji E ′ jfrom the results regarding the transformation of vectors in §3.3. Henceσ ′ ijE ′ j = J ′ i= l ip J p= l ip σ pq E q= l ip σ pq l jq E j′=⇒ (σ ij ′ − l ip l jq σ pq )E j ′ = 0.This is true for all vectors E ′ , and hence the bracket must be identically zero; henceσ ij ′ = l ip l jq σ pq . This tells us how σ transforms.Compare this argument with the corresponding argument for the case Ax = 0 where A is a matrix; if itis true for all x then A must be zero, though this is not the case if it is only true for some x’s.σ is a second rank tensor, because it has two suffixes (σ ij ).Definition: In general, a tensor of rank n is a mathematical object with n suffixes,T ijk... , which obeys the transformation lawT ′ijk... = l ip l jq l kr . . . T pqr...where L is the rotation matrix between frames.Note: for second rank tensors such as σ, the transformation lawT ′ij = l ip l jq T pqcan be rewritten in matrix notation as T ′ = LT L T – check this yourself!Examples of Tensors(i) Any vector v (e.g., velocity) is a tensor of rank 1, because v i ′ = l ip v p .(ii) Temperature T is a tensor of rank 0 – known as a scalar – because it is the samein all frames (T ′ = T ).(iii) The inertia tensor. Consider a mass m which is part of a rigid body, at a locationx within the body. If the body is rotating with angular velocity ! then the mass’svelocity is v = ! × x, and its angular momentum is thereforemx × v = mx × (! × x) = m ( |x| 2 ! − (! . x)x ) .53 © R. E. Hunt, 2002

Changing from a single mass m to a continuous mass distribution with densityρ(x), so that an infinitesimal mass element is ρ(x) dV , we see that the total angularmomentum of a rigid body V is given by∫∫∫h = ρ(x) ( |x| 2 ! − (! . x)x ) dV,Vor, in suffix notation,∫∫∫h i =V∫∫∫=Vρ(x)(x k x k ω i − ω j x j x i ) dVρ(x)(x k x k δ ij − x j x i )ω j dVwhere= I ij ω j∫∫∫I ij =ρ(x)(x k x k δ ij − x i x j ) dVVis the inertia tensor of the rigid body. Note that the tensor I does not depend on!, only on properties of the body itself; so it may be calculated once and for allfor any given body. To see that it is indeed a tensor, note that both h and ! arevectors, and apply arguments previously used for the conductivity tensor.(iv) Susceptibility χ. If M is the magnetization (magnetic moment per unit volume)and B is the applied magnetic field, then for a simple medium we have M =χ (m) B where χ (m) is the magnetic susceptibility. This generalises to M i = χ (m)ij B jwhere χ (m)ij is the magnetic susceptibility tensor. Similarly for polarization densityP in a dielectric: P i = χ (e)ij E j where E is the electric field and χ (e)ij is the electricsusceptibility tensor.(v) The Kronecker delta itself. We have defined δ ij without reference to frame; i.e., itscomponents are by definition the same in all frames (δ ′ ij ≡ δ ij ). Surprisingly, then,we can show that it is a tensor:l ip l jq δ pq = l ip l jp = δ ij = δ ij′(from §3.3), which is exactly the right transformation law. We can also show thatɛ ijk is a tensor of rank 3.Both δ ij and ɛ ijk are isotropic tensors: that is, their components are the same inall frames.54 © R. E. Hunt, 2002

(vi) Stress and strain tensors. In an elastic body, stresses (forces) produce displacementsof small volume elements within the body. Let this displacement at a location x beu; then the strain tensor is defined to bee ij = 1 ( ∂ui+ ∂u )j.2 ∂x j ∂x iThe stress tensor p ij is defined as the j th component of the forces within the bodyacting on an imaginary plane perpendicular to the i th axis. Hooke’s law for simplemedia says that stress ∝ strain. We can now generalise this to the tensor formulationp ij = k ijkl e klwhere k ijkl is a fourth rank tensor, which expresses the linear (but possibly anisotropic)relationship between p and e.3.5 Properties of TensorsLinear Combination of TensorsIf A ij and B ij are second rank tensors, and α, β are scalars, then T ij = αA ij + βB ij is atensor.Proof:T ′ij = α ′ A ′ ij + β ′ B ′ ij= αl ip l jq A pq + βl ip l jq B pq= l ip l jq (αA pq + βB pq )as required.= l ip l jq T pqThis result clearly extends to tensors of rank n.Contraction (also known as the Inner Product)If T ij is a tensor then T ii is a scalar. Proof:T ′ii = l ip l iq T pq = δ pq T pq = T pp = T ii ,so T ii has the same value in all frames as required.55 © R. E. Hunt, 2002

We can extend this result: if T ijk...lmn... is a tensor of rank n then S jk...mn... = T ijk...imn...is a tensor of rank n − 2. Proof:S ′ jk...mn... = T ′ijk...imn...= l ip l jq l kr . . . l iα l mβ l nγ . . . T pqr...αβγ...= (l ip l iα )l jq l kr . . . l mβ l nγ . . . T pqr...αβγ...= δ pα l jq l kr . . . l mβ l nγ . . . T pqr...αβγ...= l jq l kr . . . l mβ l nγ . . . S qr...βγ... .Outer ProductIf a and b are vectors then the outer product T ij defined by T ij = a i b j is a tensor of ranktwo. Proof:as required.T ′ij = a ′ ib ′ j = l ip a p l jq b q = l ip l jq a p b q = l ip l jq T pqSimilarly (left as an exercise for the reader) we can show that if A ijk... is a tensor ofrank m and B lmn... is a tensor of rank n, then T ijk...lmn... = A ijk... B lmn... is a tensor of rankm + n.Example: if a and b are vectors then a . b is a scalar. Proof: T ij = a i b j , being anouter product of two vectors, is a tensor of rank two. Then T ii = a i b i , being a contractionof a tensor, is a scalar, as required. Note that |a| 2 = a . a and |b| 2 are also scalars; hencea.b/|a| |b| = cos θ is a scalar, so that the angle between vectors is unaffected by a changeof frame.3.6 Symmetric and Anti-Symmetric TensorsA tensor T ijk... is said to be symmetric in a pair of indices (say i, j) ifT ijk... = T jik...or anti-symmetric in i, j ifT ijk... = −T jik... .For a second rank tensor we need not specify the indices as there are only two tochoose from! For example, δ ij is symmetric; ɛ ijk is anti-symmetric in any pair of indices.56 © R. E. Hunt, 2002

Note: if A ij is a symmetric second rank tensor then the matrix corresponding to A issymmetric, i.e. A = A T . Similarly for an anti-symmetric tensor.Suppose that S ij is a symmetric tensor and A ij an anti-symmetric tensor. ThenS ij A ij = 0. Proof:S ij A ij = −S ij A ji = −S ji A ji= −S ij A ij (swapping dummy i and j)=⇒ 2S ij A ij = 0,as required. Try to work out also how to see this “by inspection”, by considering appropriatepairs of components.Example: for any vector a, a × a = 0 because[a × a] i = ɛ ijk a j a kand ɛ ijk is anti-symmetric in j, k whilst a j a k is symmetric.The properties of symmetry and anti-symmetry are invariant under a change of frame:that is, they are truly tensor properties. For example, suppose that T ij is symmetric.ThenT ′ij = l ip l jq T pqso that T ′ij is also symmetric.= l ip l jq T qp= l jq l ip T qp = T ′ ji,(Alternative, and simpler, proof for second rank tensors:using T T = T .)T ′ = LT L T =⇒ T ′T = (LT L T ) T = LT T L T = LT L T = T ′Symmetry and anti-symmetry occur frequently in practical applications. For example,the strain tensor e ij = 1 2 (∂u i/∂x j + ∂u j /∂x i ) is clearly symmetric. In most situations thestress tensor is also symmetric; but in some circumstances (for instance in crystallographyor geodynamics) it is forced to be anti-symmetric while the strain remains symmetric.Inertia tensors are always symmetric; conductivity and susceptibility tensors usually are.57 © R. E. Hunt, 2002

Decomposition into Symmetric and Anti-Symmetric PartsAny second rank tensor T ij can be uniquely expressed as the sum of a symmetric and ananti-symmetric tensor; forT ij = S ij + A ijwhereS ij = 1 2 (T ij + T ji ), A ij = 1 2 (T ij − T ji )are symmetric and anti-symmetric respectively. Exercise: prove that S and A are tensors.Furthermore, any anti-symmetric tensor A ij can be expressed in terms of a vector !(sometimes known as the dual vector) such thatA ij = ɛ ijk ω k .Proof: define ! byω k = 1 2 ɛ klmA lm .Thenɛ ijk ω k = 1 2 ɛ ijkɛ klm A lm= 1 2 (δ ilδ jm − δ im δ jl )A lm= 1 2 (A ij − A ji ) = A ijas required. ! is a vector as it is a contraction of two tensors.This definition of ! actually corresponds to setting⎛⎞0 ω 3 −ω 2⎜⎟A = ⎝−ω 3 0 ω 1 ⎠ .ω 2 −ω 1 0Example: suppose that two symmetric second rank tensors R ij and S ij are linearlyrelated. Then there must be a relationship between them of the form R ij = c ijkl S kl . Itis clear that c ijkl must be symmetric in i, j (for otherwise, R ij would not be). It is notnecessarily the case that it must also be symmetric in k, l, but without loss of generalitywe may assume that it is, by the following argument. Decompose c ijkl into a part c (s)ijklwhich is symmetric in k, l and a part c (a)ijklwhich is anti-symmetric. ThenR ij = c (s)ijkl S kl + c (a)ijkl S kl = c (s)ijkl S klbecause the second term is the contraction of an anti-symmetric tensor with a symmetricone, which we showed was zero above. Hence we can ignore any anti-symmetric part ofc ijkl .58 © R. E. Hunt, 2002

3.7 Diagonalization of Symmetric Second RankTensorsSuppose T ij is a symmetric second rank tensor. We shall show that there exists a framesuch that, if we transform T to that frame, it has components given by⎛ ⎞λ 1 0 0T ′ ⎜ ⎟= ⎝ 0 λ 2 0 ⎠ .0 0 λ 3This process is known as diagonalization. The values λ 1 , λ 2 and λ 3 are known as theprincipal values of T , and the Cartesian coordinate axes of the corresponding frame areknown as the principal axes. We will see that in fact the principal values are just theeigenvalues of the matrix corresponding to T , and the principal axes are the eigenvectors.Because T is symmetric, we know that there are 3 real eigenvalues and that we canfind 3 corresponding eigenvectors which are orthogonal and of unit length. Let λ 1 , λ 2 ,λ 3 be the eigenvalues and e ′ 1, e ′ 2, e ′ 3 be the eigenvectors (arranged as a right-handed setof orthonormal vectors). Change frame to one in which the coordinate axes are alignedwith {e ′ 1, e ′ 2, e ′ 3}. What is T ′ ?Recall that L T = ( ∣ )e ′ ∣1 e′2 e′3 ; i.e., the three columns of L T are the vectors e ′ 1, e ′ 2and e ′ 3 (measured relative to the first frame). Hence in matrix notation,T L T = T ( e ′ 1= ( λ 1 e ′ 1∣ e′2∣ ∣ e′3))∣ λ2 e ′ ∣2 λ3 e ′ 3 .SoT ′ = LT L T =⎛⎜⎝e ′ T1e ′ T2e ′ T3⎞ ⎛⎛ ⎞λ 1 0 0⎜ ⎟= ⎝ 0 λ 2 0 ⎠0 0 λ 3⎞⎟ ⎜⎠ ⎝λ 1 e ′ 1 λ 2 e ′ 2 λ 3 e ′ ⎟3⎠because, for example, the top LHS entry is given by e ′ 1 . λ 1 e ′ 1, and the top RHS entry ise ′ 1 . λ 3 e ′ 3.⎛There is another way of seeing that T ′ = ⎝ λ ⎞1 0 00 λ 2 0 ⎠. The equation T e ′ 1 = λ 1 e ′ 1 is true in any frame0 0 λ 3(because T is a tensor, e ′ 1 a vector and λ 1 a scalar). In particular it is true in the frame with {e ′ 1, e ′ 2, e ′ 3}59 © R. E. Hunt, 2002

as coordinate axes. But, measured in this frame, e ′ 1 is just (1, 0, 0) T , and T has components T ′ ; so⎛T ′ ⎝ 1 ⎞ ⎛0⎠ = ⎝ λ ⎞10 ⎠0 0which shows that the first column of T ′ is (λ 1 , 0, 0) T . Similarly for the other columns.Note: the three principal values are invariants of T . That is, whatever frame we startfrom, when we diagonalize T we will obtain the same values of λ. The eigenvalues areproperties of the tensor, not of the coordinate system.3.8 Isotropic TensorsAn isotropic tensor is one whose components are the same in all frames, i.e.,T ′ijk... = T ijk... .We can classify isotropic tensors up to rank four as follows:Rank 0: All scalars are isotropic, since the tensor transformation law states that T ′ = Tfor tensors of rank zero.Rank 1: There are no non-zero isotropic vectors.Rank 2: The most general isotropic second rank tensor is λδ ij where λ is any scalar,as proved below.Rank 3: The most general isotropic third rank tensor is λɛ ijk .Rank 4: The most general isotropic fourth rank tensor iswhere λ, µ, ν are scalars.λδ ij δ kl + µδ ik δ jl + νδ il δ jkWhat is the physical significance of an isotropic tensor? Consider the conductivitytensor σ ij in an isotropic medium. As the medium is the same in all directions, we expectthat σ ij will be isotropic too. Hence σ ij = λδ ij andJ i = σ ij E j = λδ ij E j = λE i ,i.e., J = λE. So we recover the “simple version” of the conductivity law, as we mightexpect.60 © R. E. Hunt, 2002

Isotropic Second Rank TensorsConsider a general tensor T of rank two, with components T ijwith respect to some set of axes {e 1 , e 2 , e 3 }. Suppose that T isisotropic. Its components should then be unaltered by a rotationof 90 ◦ about the 3-axis, i.e., with respect to new axese ′ 1 = e 2 , e ′ 2 = −e 1 , e ′ 3 = e 3 .The matrix of this rotation is⎛ ⎞0 1 0⎜ ⎟L = ⎝−1 0 0⎠ .0 0 1Using the matrix formulation of the transformation law for tensors, we see that⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞T 11 ′ T 12 ′ T 13′ 0 1 0 T 11 T 12 T 13 0 −1 0⎜⎝T 21 ′ T 22 ′ T 23′ ⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟⎠ = ⎝−1 0 0⎠⎝T 21 T 22 T 23 ⎠ ⎝1 0 0⎠T 31 ′ T 32 ′ T 33′ 0 0 1 T 31 T 32 T 33 0 0 1⎛⎞T 22 −T 21 T 23⎜⎟= ⎝−T 12 T 11 −T 13 ⎠ .T 32 −T 31 T 33But, because T is isotropic, T ′ij = T ij . Hence, comparing matrix entries, we have:T 11 = T 22 ;T 13 = T 23 = −T 13 so that T 13 = T 23 = 0;T 31 = T 32 = −T 31 so that T 31 = T 32 = 0.Similarly, considering a rotation of 90 ◦ about the 2-axis, we find that T 11 = T 33 and thatT 12 = T 32 = 0, T 21 = T 23 = 0. Therefore all off-diagonal elements of T are zero, and alldiagonal elements are equal, say λ. Thus⎛ ⎞λ 0 0⎜ ⎟T = ⎝0 λ 0⎠ ,0 0 λor in suffix notation, T ij = λδ ij .In summary, we have shown that any isotropic second rank tensor must be equal toλδ ij for some scalar λ.61 © R. E. Hunt, 2002

3.9 Tensor Differential OperatorsA tensor field is a tensor which depends on the location x. For example:(i) Temperature is a scalar field (a tensor field of rank zero), because T = T (x).(ii) Any vector field F(x), such as a gravitational force field, is a tensor field of rank one.In particular, x is itself a vector field, because it is a vector function of position!(iii) In a conducting material where the conductivity varies with location, we have σ ij =σ ij (x), a tensor field of rank two.We are interested here in calculating the derivatives of tensor fields; we start withscalars and vectors.We can rewrite the definitions of grad, div, and curl using suffix notation.Grad:[∇Φ] i = ∂Φ∂x iDiv: ∇ . F = ∂F 1∂x 1+ ∂F 2∂x 2+ ∂F 3∂x 3= ∂F i∂x iCurl:[∇ × F] i = ɛ ijk∂F k∂x jThere is another notation worth knowing: if u, v are vectors then we define the vector()∂ ∂ ∂(u . ∇)v = u 1 + u 2 + u 3 v.∂x 1 ∂x 2 ∂x 3In suffix notation,Laplace’s equation ∇ 2 Φ = 0 becomesin suffix notation. Similarly,[(u . ∇)v] i = u j∂v i∂x j.∂ 2 Φ∂x i ∂x i= 0[∇ 2 F] i = ∂2 F i∂x j ∂x j(note that we only use Cartesian coordinates here).We sometimes find it useful to use the differential operator ∂ i defined by∂ i =∂∂x i.62 © R. E. Hunt, 2002

Then[∇Φ] i = ∂ i Φ; ∇ . F = ∂ i F i ; [∇ × F] i = ɛ ijk ∂ j F k .It turns out that ∂ i is in fact a tensor of rank one. We know that x j = l ij x ′ i (fromx = L T x ′ ) so that∂x j∂x ′ i= ∂ (l∂x ′ kj x ′ ∂x ′ kk) = l kji∂x ′ i= l kj δ ik = l ij .(This looks obvious but has to be proved very carefully!) Now let T be some quantity(perhaps a scalar or a tensor of some rank). Then∂ ′ iT = ∂T∂x ′ i= ∂T∂x j∂x j∂x ′ i= ∂T ∂x 1+ ∂T ∂x 2+ ∂T ∂x 3∂x 1 ∂x ′ i ∂x 2 ∂x ′ i ∂x 3 ∂x ′ i= l ij∂T∂x j= l ij ∂ j T.This is true for any quantity T , so∂ ′ i = l ij ∂ j ,i.e., ∂ i transforms like a vector, and is hence a tensor of rank one.This result allows us to prove that ∇Φ, ∇ . F and ∇ × F are scalars or vectors (asappropriate). For example, to show that if F is a vector field then ∇ × F is a vector field:[∇ × F] ′ i = ɛ ′ ijk∂ ′ jF ′ k= l ip l jq l kr ɛ pqr l js ∂ s l kt F t[ɛ, ∂ and F are tensors]= l ip (l jq l js )(l kr l kt )ɛ pqr ∂ s F t= l ip δ qs δ rt ɛ pqr ∂ s F tas required.= l ip ɛ pqr ∂ q F r= l ip [∇ × F] p ,Alternatively, we can just state that ∇×F is a contraction of the tensor outer product T ijklm = ɛ ijk ∂ l F m(because [∇ × F] i = T ijkjk ).As an example of a tensor field of rank three, consider the derivative of the conductivitytensor, ∂ i σ jk . This cannot be written using ∇.63 © R. E. Hunt, 2002