1. Divergence of a product: Given that φ is a scalar field and v a vector field, show

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1 1. Divergence of a product: Given that φ is a scalar field and v a vector field, show that div(φv) = (gradφ) v + φ div v grad(φv) = (φv i ), j g i g j = φ, j v i g i g j + φv i, j g i g j = v (grad φ) + φ grad v Now, div(φv) = tr(grad(φv)). Taking the trace of the above, we have: div(φv) = v (grad φ) + φ div v 2. Show that grad(u v) = (grad u) T v + (grad v) T u u v = u i v i is a scalar sum of components. grad(u v) = (u i v i ), j g j = u i, j v i g j + u i v i, j g j Now grad u = u i, j g i g j swapping the bases, we have that, (grad u) T = u i, j (g j g i ). Writing v = v k g k, we have that, (grad u) T v = u i, j v k (g j g i )g k = u i, j v k g j δ i k = u i, j v i g j It is easy to similarly show that u i v i, j g j = (grad v) T u. Clearly,

2 As required. grad(u v) = (u i v i ), j g j = u i, j v i g j + u i v i, j g j = (grad u) T v + (grad v) T u 3. Show that grad(u v) = (u )grad v (v )grad u u v = ε ijk u j v k g i Recall that the gradient of this vector is the tensor, grad(u v) = (ε ijk u j v k ), l g i g l = ε ijk u j, l v k g i g l + ε ijk u j v k, l g i g l = ε ikj u j, l v k g i g l + ε ijk u j v k, l g i g l = (v )grad u + (u )grad v 4. Show that div (u v) = v curl u u curl v We already have the expression for grad(u v) above; remember that div (u v) = tr[grad(u v)] = ε ikj u j, l v k g i g l + ε ijk u j v k, l g i g l = ε ikj u j, l v k δ i l + ε ijk u j v k, l δ i l = ε ikj u j, i v k + ε ijk u j v k, i = v curl u u curl v

3 5. Given a scalar point function φ and a vector field v, show that curl (φv) = φ curl v + (grad φ) v. curl (φv) = ε ijk (φv k ), j g i = ε ijk (φ, j v k + φv k, j )g i = ε ijk φ, j v k g i + ε ijk φv k, j g i = (grad φ) v + φ curl v 6. Show that div (u v) = (div v)u + (grad u)v u v is the tensor, u i v j g i g j. The gradient of this is the third order tensor, grad (u v) = (u i v j ), k g i g j g k And by divergence, we mean the contraction of the last basis vector: div (u v) = (u i v j ) k (g i g j )g k = (u i v j ) k g i δ j k = (u i v j ) j g i = u i, j v j g i + u i v j, j g i = (grad u)v + (div v)u

4 7. For a scalar field φ and a tensor field T show that grad (φt) = φgrad T + T gradφ. Also show that div (φt) = φ div T + Tgradφ grad(φt) = (φt ij ), k g i g j g k = (φ, k T ij + φt ij, k )g i g j g k = T gradφ + φgrad T Furthermore, we can contract the last two bases and obtain, div(φt) = (φ, k T ij + φt ij, k )g i g j g k = (φ, k T ij + φt ij, k )g i δ j k = T ik φ, k g i + φt ik, k g i = Tgradφ + φ div T 8. For two arbitrary tensors S and T, show that grad(st) = (grad S T ) T T + S grad T grad(st) = (S ij T jk ), α g i g k g α = (S ij,α T jk + S ij T jk, α )g i g k g α = (T kj S ji,α + S ij T jk, α )g i g k g α = (grad S T ) T T + S grad T 9. For two arbitrary tensors S and T, show that div(st) = (grad S): T + T div S grad(st) = (S ij T jk ), α g i g k g α

5 = (S ij,α T jk + S ij T jk, α )g i g k g α div(st) = (S ij,α T jk + S ij T jk, α )g i (g k g α ) = (S ij,α T jk + S ij T jk, α )g i α δ k = (S ij,k T jk + S ij T jk, k )g i = (grad S): T + S div T 10. For two arbitrary vectors, u and v, show that grad(u v) = (u )gradv (v )gradu grad(u v) = (ε ijk u j v k ), l g i g l = (ε ijk u j, l v k + ε ijk u j v k, l )g i g l = (u j, l ε ijk v k + v k, l ε ijk u j )g i g l = (v )gradu + (u )gradv 11. For a vector field u, show that grad(u ) is a third ranked tensor. Hence or otherwise show that div(u ) = curl u. The second order tensor (u ) is defined as ε ijk u j g i g k. Taking the covariant derivative with an independent base, we have grad(u ) = ε ijk u j, l g i g k g l

6 This gives a third order tensor as we have seen. Contracting on the last two bases, div(u ) = ε ijk u j, l g i g k g l 12. Show that div (φi) = grad φ = ε ijk u j, l g i δ k l = ε ijk u j, k g i = curl u Note that φi = (φg αβ )g α g β. Also note that grad φi = (φg αβ ), i g α g β g i The divergence of this third order tensor is the contraction of the last two bases: div (φi) = tr(grad φi) = (φg αβ ), i (g α g β )g i = (φg αβ ), i g α g βi 13. Show that curl (φi) = ( grad φ) = φ, i g αβ g βi g α = φ, i δ α i g α = φ, i g i = grad φ Note that φi = (φg αβ )g α g β, and that curl T = ε ijk T αk, j g i g α so that, curl (φi) = ε ijk (φg αk ), j g i g α = ε ijk (φ, j g αk )g i g α = ε ijk φ, j g i g k = ( grad φ)

7 14. Show that the dyad u v is NOT, in general symmetric: u v = v u (u v) u v = ε ijk u j v k g i ((u v) ) = ε αiβ ε ijk u j v k g α g β = (δ α j δ β k δ α k δ β j ) u j v k g α g β = ( u α v β + u β v α )g α g β = v u u v 15. Show that curl (v ) = (div v)i grad v so that (v ) = ε αβk v β g α g k curl T = ε ijk T αk, j g i g α curl (v ) = ε ijk ε αβk v β, j g i g α = (g iα g jβ g iβ g jα ) v β, j g i g α = v j, j g α g α v i, j g i g j = (div v)i grad v

8 16. Show that div (u v) = v curl u u curl v div (u v) = (ε ijk u j v k ), i Noting that the tensor ε ijk behaves as a constant under a covariant differentiation, we can write, div (u v) = (ε ijk u j v k ), i = ε ijk u j, i v k + ε ijk u j v k, i = v curl u u curl v 17. Given a scalar point function φ and a vector field v, show that curl (φv) = φ curl v + (grad φ) v. 18. Show that curl (grad φ) = o For any tensor v = v α g α curl (φv) = ε ijk (φv k ), j g i = ε ijk (φ, j v k + φv k, j )g i = ε ijk φ, j v k g i + ε ijk φv k, j g i = (grad φ) v + φ curl v curl v = ε ijk v k, j g i Let v = grad φ. Clearly, in this case, v k = φ, k so that v k, j = φ, kj. It therefore follows

9 that, curl (grad φ) = ε ijk φ, kj g i = o. The contraction of symmetric tensors with anti-symmetric led to this conclusion. Note that this presupposes that the order of differentiation in the scalar field is immaterial. This will be true only if the scalar field is continuous a proposition we have assumed in the above. 19. Show that curl (grad v) = 0 For any tensor T = T αβ g α g β curl T = ε ijk T αk, j g i g α Let T = grad v. Clearly, in this case, T αβ = v α, β so that T αk, j = v α, kj. It therefore follows that, curl (grad v) = ε ijk v α, kj g i g α = 0. The contraction of symmetric tensors with anti-symmetric led to this conclusion. Note that this presupposes that the order of differentiation in the vector field is immaterial. This will be true only if the vector field is continuous a proposition we have assumed in the above.

10 20. Show that curl (grad v) T = grad(curl v) From previous derivation, we can see that, curl T = ε ijk T αk, j g i g α. Clearly, curl T T = ε ijk T kα, j g i g α so that curl (grad v) T = ε ijk v k, αj g i g α. But curl v = ε ijk v k, j g i. The gradient of this is, grad(curl v) = (ε ijk v k, j ), α g i g α = ε ijk v k, jα g i g α = curl (grad v) T 21. Show that div (grad φ grad θ) = 0 The gradient of this vector is the tensor, grad φ grad θ = ε ijk φ, j θ, k g i grad(grad φ grad θ) = (ε ijk φ, j θ, k ), l g i g l = ε ijk φ, jl θ, k g i g l + ε ijk φ, j θ, kl g i g l The trace of the above result is the divergence we are seeking: div (grad φ grad θ) = tr[grad(grad φ grad θ)] = ε ijk φ, jl θ, k g i g l + ε ijk φ, j θ, kl g i g l = ε ijk φ, jl θ, k δ i l + ε ijk φ, j θ, kl δ i l = ε ijk φ, ji θ, k + ε ijk φ, j θ, ki = 0 Each term vanishing on account of the contraction of a symmetric tensor with an

11 antisymmetric. 22. Show that curl curl v = grad(div v) grad 2 v Let w = curl v ε ijk v k, j g i. But curl w ε αβγ w γ, β g α. Upon inspection, we find that w γ = g γi ε ijk v k, j so that curl w ε αβγ (g γi ε ijk v k, j ), β g α = g γi ε αβγ ε ijk v k, jβ g α Now, it can be shown that g γi ε αβγ ε ijk = g αj g βk g αk g βj so that, curl w = (g αj g βk g αk g βj )v k, jβ g α = v β, jβ g j g βj v α, jβ g α = grad(div v) grad 2 v Also recall that the Laplacian (grad 2 ) of a scalar field φ is, grad 2 φ = g ij φ, ij. In Cartesian coordinates, this becomes, grad 2 φ = g ij φ, ij = δ ij φ, ij = φ, ii as the unit (metric) tensor now degenerates to the Kronecker delta in this special case. For a vector field, grad 2 v = g βj v α, jβ g α. Also note that while grad is a vector operator, the Laplacian (grad 2 ) is a scalar operator.

12 23. Show that g γi ε αβγ ε ijk = g αj g βk g αk g βj Note that g iα g iβ g iγ g γi g iα g γi g iβ g γi g iγ g γi ε αβγ ε ijk = g γi g jα g jβ g jγ = g jα g jβ g jγ g kα g kβ g kγ g kα g kβ g kγ δ γ α δ γ β δ γ γ = g jα g jβ g jγ g kα g kβ g kγ = δ γ α gjβ g jγ g kβ g kγ δ γ β gjα g kα g jγ g kγ + δ γ γ gjα g kα g jβ g kβ = gjβ g jα g kβ g kα g jβ gjα g kα g kβ + 3 g jβ gjα g kα g kβ = g jβ gjα g kα g kβ = g αj g βk g αk g βj 24. Given that φ(t) = A(t), Show that φ (t) = A A(t) : A Now, φ 2 A: A d dt (φ2 ) = 2φ dφ dt = da da : A + A: dt dt = 2A: da dt

13 as inner product is commutative. We can therefore write that dφ dt = A φ : da dt = A A(t) : A as required. 25. Given a tensor field T, obtain the vector w T T v and show that its divergence is T: ( v) + v div T The gradient of w is the tensor, (T ji v j ), k g i g k. Therefore, divergence of w (the trace of the gradient) is the scalar sum, T ji v j, k g ik + T ji, k v j g ik. Expanding, we obtain, div (T T v) = T ji v j, k g ik + T ji, k v j g ik = T j k, k v j + T j k v j, k = (div T) v + tr(t T grad v) = (div T) v + T: (grad v) Recall that scalar product of two vectors is commutative so that div (T T v) = T: (grad v) + v div T

14 26. For a second-order tensor T define curl T ε ijk T αk, j g i g α show that for any constant vector a, (curl T) a = curl (T T a) Express vector a in the invariant form with covariant components as a = a β g β. It follows that (curl T) a = ε ijk T αk, j (g i g α )a = ε ijk T αk, j a β (g i g α )g β = ε ijk T αk, j a β g i δ β α = ε ijk (T αk ), j g i a α = ε ijk (T αk a α ), j g i The last equality resulting from the fact that vector a is a constant vector. Clearly, (curl T) a = curl (T T a) 27. For any two vectors u and v, show that curl (u v) = [(grad u)v ] T + (curl v) u where v is the skew tensor ε ikj v k g i g j. Recall that the curl of a tensor T is defined by curl T ε ijk T αk, j g i g α. Clearly therefore, curl (u v) = ε ijk (u α v k ), j g i g α = ε ijk (u α, j v k + u α v k, j ) g i g α

15 = ε ijk u α, j v k g i g α + ε ijk u α v k, j g i g α = (ε ijk v k g i ) (u α, j g α ) + (ε ijk v k, j g i ) (u α g α ) = (ε ijk v k g i g j )(u α, β g β g α ) + (ε ijk v k, j g i ) (u α g α ) = (v )(grad u) T + (curl v) u = [(grad u)v ] T + (curl v) u upon noting that the vector cross is a skew tensor. 28. Show that curl (u v) = div(u v v u) The vector w u v = w k g k = ε kαβ u α v β g k and curl w = ε ijk w k, j g i. Therefore, curl (u v) = ε ijk w k, j g i = ε ijk ε kαβ (u α v β ), j g i = (δ α i δ β j δ β i δ α j ) (u α v β ), j g i = (δ α i δ β j δ β i δ α j ) (u α, j v β + u α v β, j )g i = [u i, j v j + u i v j, j (u j, j v i + u j v i, j )]g i = [(u i v j ), j (u j v i ), j ]g i = div(u v v u) since div(u v) = (u i v j ), α g i g j g α = (u i v j ), j g i.

16 29. Given a scalar point function φ and a second-order tensor field T, show that curl (φt) = φ curl T + ((grad φ) )T T where [(grad φ) ] is the skew tensor ε ijk φ, j g i g k curl (φt) ε ijk (φt αk ), j g i g α = ε ijk (φ, j T αk + φt αk, j ) g i g α = ε ijk φ, j T αk g i g α + φε ijk T αk, j g i g α = (ε ijk φ, j g i g k ) (T αβ g β g α ) + φε ijk T αk, j g i g α = φ curl T + ((grad φ) )T T 30. For a second-order tensor field T, show that div(curl T) = curl(div T T ) Define the second order tensor S as curl T ε ijk T αk, j g i g α = S i.α g i g α The gradient of S is S i.α, β g i g α g β = ε ijk T αk, jβ g i g α g β Clearly, div(curl T) = ε ijk T αk, jβ g i g α g β = ε ijk T αk, jβ g i g αβ = ε ijk T β k, jβ g i = curl(div T T )

17 31. Show that if φ defined in the space spanned by orthogonal coordinates x i, then 2 (x i φ) = 2 φ x i + xi 2 φ. By definition, 2 (x i φ) = g jk (x i φ),jk. Expanding, we have g jk (x i φ),jk = g jk (x i,jφ + x i φ,j ),k = g jk (δ j i φ + x i φ,j ),k = g jk (δ j i φ,k + x i,kφ,j + x i φ,jk ) = g jk (δ j i φ,k + δ k i φ,j + x i φ,jk ) = g ik φ,k + g ij φ,j + x i g jk φ,jk When the coordinates are orthogonal, this becomes, 2 (h i ) 2 Φ x i + xi 2 Φ where we have suspended the summation rule and h i is the square root of the appropriate metric tensor component. 32. In Cartesian coordinates, If the volume V is enclosed by the surface S, the position vector r = x i g i and n is the external unit normal to each surface element, show that 1 6 (r r) nds S equals the volume contained in V. r r = x i x j g i g j = x i x j g ij

18 By the Divergence Theorem, (r r) nds S = [ (r r)]dv V = l [ k (x i x j g ij )] g l g k dv V = l [g ij (x i, k x j + x i x j, k )] g l g k dv V = g ij g lk (δ k i x j + x i δ k j ), l dv V = 2g ik g lk x i, l dv V = 2δ i l δ l i dv V = 6 dv V 33. For any Euclidean coordinate system, show that div u v = v curl u u curl v Given the contravariant vector u i and v i with their associated vectors u i and v i, the contravariant component of the above cross product is ε ijk u j v k.the required divergence is simply the contraction of the covariant x i derivative of this quantity: (ε ijk u j v k ),i = ε ijk u j,i v k + ε ijk u j v k,i where we have treated the tensor ε ijk as a constant under the covariant derivative. Cyclically rearranging the RHS we obtain, (ε ijk u j v k ),i = v k ε kij u j,i + u j ε jki v k,i = v k ε kij u j,i + u j ε jik v k,i where we have used the anti-symmetric property of the tensor ε ijk. The last expression shows clearly that div u v = v curl u u curl v

19 as required. 34. For a general tensor field T show that, curl(curl T) = [grad 2 (tr T) div(div T)]I + grad(div T) + (grad(div T)) T grad(grad (tr T)) grad 2 T T curl T = ε αst T βt, s g α g β = S.β α g α g β curl S = ε ijk S.k α, j g i g α so that curl S = curl(curl T) = ε ijk ε αst T kt, sj g i g α g iα g is g it = g jα g js g jt T kt, sj g i g α g kα g ks g kt = [ giα (g js g kt g jt g ks ) + g is (g jt g kα g jα g kt ) +g it (g jα g ks g js g kα ] T kt, sj g i g α ) = [g js T.t t, sj T.. sj, sj ](g α g α ) + [T αj.., sj g jα T.t t, sj ](g s g α ) + [g jα T.t s., sj g js T.t α., sj ](g t g α ) = [grad 2 (tr T) div(div T)]I + (grad(div T)) T grad(grad (trt)) + (grad(div T)) grad 2 T T

20 35. When T is symmetric, show that tr(curl T) vanishes. curl T = ε ijk T βk, j g i g β tr(curl T) = ε ijk T βk, j g i g β = ε ijk T βk, j δ i β = ε ijk T ik, j which obviously vanishes on account of the symmetry and antisymmetry in i and k. In this case, curl(curl T) = [ 2 (tr T) div(div T)]1 grad(grad (trt)) + 2(grad(div T)) 2 T as (grad(div T)) T = grad(div T) if the order of differentiation is immaterial and T is symmetric. 36. For a scalar function φand a vector v show that the divergence of the vector vφ is equal to, v grad φ + φ div v grad (vφ) = (v i φ), j g i g j = (φv i, j + v i φ, j )g i g j Taking the trace of this equation, div vφ = tr(grad (vφ)) = (φv i, j + v i φ, j )g i g j = (φv i, j + v i φ, j )δ j i

21 Hence the result. = φv i, i + v i φ, i = φ div v + v gradφ 37. Show that curl u v = (v grad u ) + (u div v ) (v div u) (u grad v ) Taking the associated (covariant) vector of the expression for the cross product in the last example, it is straightforward to see that the LHS in indicial notation is, ε lmi (ε ijk u j v k ),m Expanding in the usual way, noting the relation between the alternating tensors and the Kronecker deltas, ε lmi (ε ijk u j v k ) = δ lmi,m jki (u j,mv k u j v k,m) = δ lm jk (u j,mv k u j v k,m) = δ j l δ j m δ k l δ k m (uj,mv k u j v k,m) = (δ j l δ k m δ k l δ j m )(u j,mv k u j v k,m) = δ j l δ k m u j,mv k δ j l δ k m u j v k,m + δ k l δ j m u j,mv k δ k l δ j m u j v k,m = u l,mv m u m,mv l + u l v m,m u m v l,m Which is the result we seek in indicial notation.

22 38.. In Cartesian coordinates let x denote the magnitude of the position vector r = x i e i. Show that (a) x, j = x j x, (b) x, ij = 1 x δ ij x ix j (x) 3, (c) x,ii = 2 x, (d) If U = 1 x, then U, ij = δ ij x 3 + 3x ix j x 5 U, ii = 0 and div ( r x ) = 2 x. (a) x = x i x i x,j = x i x i x j (b) x,ij = = x i x i (x i x i ) (x ix i ) x j = 1 2 x i x i [x i δ ij + x i δ ij ] = x j x. x ( x i x i ) = ( x x i x x i x j x i x j x ) = x i j x j (x) 2 (c) x,ii = 1 x δ ii x ix i (x) 3 = 3 x (x)2 (x) 3 = 2 x. (d) U = 1 so that x U,j = 1 x = 1 x x j x x = 1 1 x j x 2 x x j = x j x 3 Consequently, = xδ x i x j ij x (x) 2 = 1 x δ ij x ix j (x) 3

23 U,ij = (U, x i ) = ( x i j x j x 3) = = U,ii = δ ii x 3 x 3 ( δ ij ) + x i ( (x3 ) x + 3x ix i x 5 x 6 x 3 ( ( x x 2 )) + x i (x j x 3 ) j x x j ) = 3 x 3 + 3x2 x 5 = 0. div ( r x ) = (x j x ), j = 1 x x j, j + ( 1 x ),j x 6 = 3 x + x j [ ( 1 x x 2) j x ] = 3 x x jx j x 3 xj = x3 δ ij + x i (3x2 x ) x 6 = δ ij x 3 = 3 x + x j ( x (1 x ) dx dx j ) = 3 x 1 x = 2 x + 3x ix j x Define the dyadic and squared times products of tensors as, (A B)C = (B: C)A and (A B)C = ACB T Show that (A B) (C D) = ACB T D so that (A B) (C D)E = (A B)(D: E)C = (ACB T )D: E = (ACB T D)E (A B) (C D) = ACB T D

24 40. Define the dyadic and squared times products of tensors as, (A B)C = (B: C)A and (A B)C = ACB T For vectors a, b and tensors A, B show that (A B) (a b) = Aa Bb. (A B) (a b) = A(a b)b T = Aa Bb 41. Define the dyadic and squared times products of tensors as, (A B)C = (B: C)A and (A B)C = ACB T For vectors a, b, c and d show that (a b) (c d) = (a c) (b d) For a tensor E, ((a b) (c d))e = (a b)e(d c) so that (a b) (c d) = (a c) (b d). = (a c)[(e T b) d] = (a c) tr((d b)e) = (a c)[(b d): E] = ((a c) (b d))e

25 42. Define the tensor basis G ij g i g j, observe that unlike the scalar component g ij, the tensor G ij is not symmetrical in its indices; furthermore, show that I g ij G ij g αβ G αβ is the fourth order unit tensor. By the definition of G ij g i g j, It is immediately clear that G ij [G ji ] T. It is therefore not symmetric in its components. We further observe that g ij G ij is the component representation of the second-order unit tensor. Lastly, I is the fourth-order unit tensor. This is evident because, given any secondorder tensor T, IT = T. To show this to be true, take any component representation of T and expand IT: IT = (g ij G ij g αβ G αβ )T = (g ij G ij g αβ G αβ )T kl g i g j = (g ij G ij g αβ G αβ )T kl G kl = g ij g αβ T kl (G ij G αβ )G kl = g ij g αβ T kl G ij G kl G βα = g ij g αβ T kl g jk g lβ G iα = δ i k δ β l T kl G iα = T iα G iα = T

26 Showing that, I = I I 43. Given that I = I I, show that, I = g ij g kl G ij G kl = g ik g jl G ij G kl The first expression is recognizable as I I since I = I I = g ij G ij g αβ G αβ = g ij g αβ G ij G αβ Let us see how the second expression operates on a second-order tensor: g ik g jl (G ij G kl )T = g ik g jl (G ij G kl )T αβ g α g β = g ik g jl T αβ (G ij G kl )g α g β = g ik g jl T αβ G ij (G kl : (g α g β )) = g ik g jl T αβ G ij g kα g lβ = δ i α δ j β T αβ G ij = T ij G ij = T confirming that g ik g jl G ij G kl = I = g ik g jl (g i g j ) (g k g l ). 44. For a second-order tensor A show that AI = IA = A where I is the fourth-order unit tensor. Note that I = I I. Therefore, AI = A(I I) = I T AI = A. Similarly, IA = (I I)A = IAI T = A since the identity tensor is symmetric and hence self-

27 transpose. 45. The transposer tensor T turns a second-order tensor into its transpose: TS = S T = ST; show that T = g il g jk G ij G kl TS = g il g jk (G ij G kl )S = g il g jk (G ij G kl )(S αβ g α g β ) = g il g jk S αβ G ij (G kl : (g α g β )) = g il g jk S αβ G ij (g k g β )(g l g α ) = g il g jk S αβ G ij δ β k δ α l = S ji G ij = S T ST = Sg il g jk (G ij G kl ) = (S αβ g α g β )g il g jk (G ij G kl ) = g il g jk S αβ ((g α g β ): G ij ) G kl = g il g jk S αβ G kl (g i g α )(g j g β ) = g il g jk S αβ G kl δ α i δ β j = S ij G ji = S T

28 46. Define the symmetrizer, S and anti symmetrizer, W tensors as fourth order tensors that return the symmetric and antisymmetric parts of a second-order tensor; show that S = 1 (I + T) and W = 1 (I T). 2 2 Consider a tensor A. Similarly, SA = 1 2 (I + T)A = 1 2 (IA + TA) = 1 2 (A + AT ) = sym A WA = 1 2 (I T)A = 1 2 (IA TA) = 1 2 (A AT ) = skw A 47. For any second-order tensor A Show that SA = AS, and that WA = AW where S is the fourth-order symmetrizer tensor. [Hint: AI = IA, TS = ST] Consider a tensor A. SA = 1 2 (I + T)A = 1 2 (IA + TA) = 1 2 (A + AT ) = sym A AS = A ( 1 2 (I + T)) = 1 2 (AI + AT) = 1 (IA + TA) = sym A 2 so that SA = AS = sym A. Similarly,

29 WA = 1 2 (I T)A = 1 2 (IA TA) = 1 2 (A AT ) = skw A AW = A ( 1 2 (I T)) = 1 2 (AI AT) = 1 2 (IA TA) = 1 2 (A AT ) = skw A showing that WA = AW = skw A 48. For the fourth order tensors S, T, and W show that (a) TT = I, (b) TS = ST, (c) SS = S (d) WW = W and (e) SW = WS = O. (a) An indicial proof TT = I is straightforward. A direct proof is however more illuminating: Consider the double transpose: (b) TTA = TA T = (A T ) T = A = IA showing clearly that TT = I. TS = T ( 1 2 (I + T)) = 1 2 (TI + TT) = 1 (T + I) = S 2 ST = ( 1 2 (I + T)) T = 1 2 (IT + TT) = 1 (T + I) = S 2 so that TS = ST = S (c) For a second-order tensor A

30 SSA = S(sym A) = ( 1 (I + T)) sym A 2 = 1 2 sym A + 1 sym A 2 = sym A = SA so that SS = S. (d) For a second-order tensor A WWA = W(skw A) = ( 1 2 (I T)) skw A = 1 2 skw A + 1 skw A 2 = skw A = WA (e) For a second-order tensor A SWA = S(skw A) = ( 1 (I + T)) skw A 2 Similarly, = 1 2 skw A 1 skw A 2 = OA = O

31 WSA = W sym A = ( 1 (I T)) sym A 2 = 1 2 sym A 1 sym A = O 2 showing that SW = WS = O the fourth-order zero tensor. 49. Given that, in indicial notation, the transposer T = g il g jk G ij G kl, show that TT = I. TT = (g il g jk G ij G kl )(g αγ g βδ G αβ G δγ ) = g il g jk g αγ g βδ G ij G δγ (G kl : G αβ ) = g il g jk g αγ g βδ G ij G δγ (δ α k δ β l ) = g il g jk g kγ g lδ G ij G δγ = g il g jk g kγ g lδ (g i g j ) (g δ g γ ) = g il g jk (g i g j ) (g l g k ) = g ik g jl (g i g j ) (g k g l ) = I

32 50. The position vector in the above example r = x i e i. Show that (a) div r = 3, (b) div (r r) = 4r, (c) div r = 3, and (d) grad r = 1 and (e) curl (r r) = r grad r = x i, j e i e j = δ ij e i e j = 1 div r = x i, j e i e j = δ ij δ ij = δ jj = 3. r r = x i e i x j e j = x i x j e i e j grad(r r) = (x i x j ), k e i e j e k = (x i, k x j + x i x j, k )e i e j e k = (δ ik x j + x i δ jk )δ jk e i = (δ ik x k + x i δ jj )e i = 4x i e i = 4r curl(r r) = ε αβγ (x i x γ ), β e α e i = ε αβγ (x i, β x γ + x i x γ, β )e α e i = ε αβγ (δ iβ x γ + x i δ γβ )e α e i = ε αiγ x γ e α e i + ε αββ x i e α e i = ε αγi x γ e α e i = r 51. Define the magnitude of tensor A as, A = tr(aa T ) Show that A A = A A By definition, given a scalar α, the derivative of a scalar function of a tensor f(a) is f(a) : B = lim f(a + αb) A α 0 α

33 for any arbitrary tensor B. In the case of f(a) = A, A A : B = lim α 0 A + αb α A + αb = tr(a + αb)(a + αb) T = tr(aa T + αba T + αab T + α 2 BB T ) Note that everything under the root sign here is scalar and that the trace operation is linear. Consequently, we can write, lim α 0 α So that, or, A + αb = lim α 0 as required since B is arbitrary. 52. Show that I 3 (S) = det(s) tr (BA T ) + tr (AB T ) + 2αtr (BB T ) 2 tr(aa T + αba T + αab T + α 2 BB T ) A A : B = A A : B A A = A A = S c the cofactor of S. Clearly S c = det(s) S T = I 3 (S) S T. Details of this for the contravariant = 2A: B 2 A: A = A A : B

34 components of a tensor is presented below. Let Differentiating wrt S αβ, we obtain, det(s) S S = 1 3! εijk ε rst S ir S js S kt αβ g α g β = 1 3! εijk ε rst [ ir αβ S js S kt + S ir js αβ S kt + S ir S js kt αβ ] g α g β = 1 3! εijk ε rst [δ i α δ r β S js S kt + S ir δ j α δ s β S kt + S ir S js δ k α δ t β ] g α g β = 1 3! εαjk ε βst [S js S kt + S js S kt + S js S kt ]g α g β Which is the cofactor of [S αβ ] or S = 1 2! εαjk ε βst S js S kt g α g β [S c ] αβ g α g β 53. For a scalar variable α, if the tensor T = T(α) and T dt det(t) tr(t T 1 ) dα d, Show that det(t) = dα Let A T T 1 so that, T = AT. In component form, we have T j i = A m i T j m. Therefore, d dα det(t) = d dα (eijk T i 1 T j 2 T k 3 ) = e ijk (T i 1 T j 2 T k 3 + T i 1 T j 2 T k 3 + T i 1 T j 2 T k 3 )

35 = e ijk (A l 1 T i l T j 2 T k 3 + T i 1 A m 2 T j m T k 3 + T i 1 T j 2 A n 3 T k n ) = e ijk [(A 1 1 T i 1 + A 2 1 T i 2 + A 3 1 T i 3 ) T j 2 T k 3 + T i 1 ( A 1 2 T j 1 + A 2 2 T j 2 + A 3 2 T j 3 ) T k 3 + T i 1 T j 2 ( A 1 3 T k 1 + A 2 3 T k 2 + A 3 3 T k 3 )] All the boxed terms in the above equation vanish on account of the contraction of a symmetric tensor with an antisymmetric one. (For example, the first boxed term yields, e ijk A 2 1 T i 2 T j 2 T k 3 Which is symmetric as well as antisymmetric in i and j. It therefore vanishes. The same is true for all other such terms.) α det(t) = eijk [(A 1 1 T i 1 )T j 2 T k 3 + T i 1 (A 2 2 T j 2 )T k 3 + T i 1 T j 2 (A 3 3 T k 3 )] = A m m e ijk T i 1 T j 2 T k 3 = tr(t T 1 ) det(t) as required. 54. Prove Liouville s Theorem that for a scalar variable α, if the tensor T = T(α) and T dt, d det(t) = det(t) tr(t T 1 ) by direct methods. dα dα We choose three constant, linearly independent vectors a, b and c so that [a, b, c] det(t) = [Ta, Tb, Tc] Differentiating both sides, noting that the RHS is a product,

36 [a, b, c] d dα det(t) = d [Ta, Tb, Tc] dα = [ dt dt dt a, Tb, Tc] + [Ta, b, Tc] + [Ta, Tb, dα dα dα c] = [ dt dα T 1 Ta, Tb, Tc] + [Ta, dt dα T 1 Tb, Tc] + [Ta, Tb, dt dα T 1 Tc] = tr ( dt dα T 1 ) [Ta, Tb, Tc] Clearly, d det(t) = det(t) tr (dt dα dα T 1 ) 55. Without breaking down into components, use Liouville s theorem, α det(t) = det(t) tr(t T 1 ), to establish the fact that det(t) T = T c Start from Liouville s Theorem, given a scalar parameter such that T = T(α), (det(t)) = det(t) tr [( T α α ) T 1 ] = [det(t) T T ]: ( T α ) By the simple rules of multiple derivative, α (det(t)) = [ (det(t))]: ( T T α )

37 It therefore follows that, Hence 56. If T is invertible, show that T [ T (det(t)) [det(t) T T ]]: ( T α ) = 0 T (det(t)) = [det(t) T T ] = T c T (log det(t)) = T T (log det(t)) det(t) (log det(t)) = det(t) T = 1 det(t) Tc = 1 det(t) T T det(t) = T T 57. If T is invertible, show that T (log det(t 1 )) = T T T (log det(t 1 )) = (log det(t 1 )) det(t 1 ) det(t 1 ) T 1 T 1 = det(t 1 ) T c ( T 1 T T ) T 1

38 1 = det(t 1 ) det(t 1 ) T T ( T 1 T T ) = T T T T T T = T T 58. Given that A is a constant tensor, Show that tr(as) = AT In invariant components terms, let A = A ij g i g j and let S = S αβ g α g β. AS = A ij S αβ (g i g j )(g α g β ) = A ij S αβ (g i g β )δ j α = A ij S jβ (g i g β ) tr(as) = A ij S jβ (g i g β ) tr(as) = = A ij S jβ δ i β = A ij S ji αβ tr(as)g α g β = Aij S ji αβ g α g β = A ij δ j α δ i β g α g β = A ij g j g i = A T = (AT : S)

39 as required. 59. Given that A and B are constant tensors, show that tr(asbt ) = A T B First observe that tr(asb T ) = tr(b T AS). If we write, C B T A, it is obvious from the above that tr(cs) = CT. Therefore, tr(asbt ) = (B T A) T = A T B 60. Given that A and B are constant tensors, show that tr(ast B T ) = B T A Observe that tr(as T B T ) = tr(b T AS T ) = tr[s(b T A) T ] = tr[(b T A) T S] [The transposition does not alter trace; neither does a cyclic permutation. Ensure you understand why each equality here is true.] Consequently, tr(ast B T ) = tr[(bt A) T S] = [(B T A) T ] T = B T A

40 61. Let S be a symmetric and positive definite tensor and let I 1 (S), I 2 (S)andI 3 (S) be the three principal invariants of S show that (a) I 1 (S) I 2 (S) = I 1 (S)I S and (c) I 3 (S) I 1 (S) = I 3 (S) S 1 can be written in the invariant component form as, I 1 (S) Recall that I 1 (S) α = tr(s) = S α hence I 1 (S) = I 1(S) g j i g j = α α j i g i g j i = δ α i δ j α g i g j = δ j i g i g j = I which is the identity tensor as expected. I 2 (S) = I 1(S) i j g i g j = I the identity tensor, (b) in a similar way can be written in the invariant component form as, I 2 (S) = 1 I 1 (S) [S α 2 α S β β S α β S β α ] g i g j i j where we have utilized the fact that I 2 (S) = 1 2 [tr2 (S) tr(s 2 )]. Consequently,

41 I 2 (S) = 1 2 j [S α α S β β S α β S β α ] g i g j i = 1 2 [δ α i δ j α S β β + δ β i δ j β S α α δ β i δ j α S α β δ α i δ j β S β α ] g i g j = 1 2 [δ j i S β β + δ j i S α α S i j S i j ] g i g j = (δ j i S α α S i j )g i g j = I 1 (S)1 S Differentiating wrt S αβ, we obtain, det(s) S S = 1 3! εijk ε rst S ir S js S kt αβ g α g β = 1 3! εijk ε rst [ ir αβ S js S kt + S ir js αβ S kt + S ir S js kt αβ ] g α g β = 1 3! εijk ε rst [δ i α δ r β S js S kt + S ir δ j α δ s β S kt + S ir S js δ k α δ t β ] g α g β = 1 3! εαjk ε βst [S js S kt + S js S kt + S js S kt ]g α g β Which is the cofactor of [S αβ ] or S = 1 2! εαjk ε βst S js S kt g α g β [S c ] αβ g α g β

42 62. For a tensor field Ξ, The volume integral in the region Ω E, (grad Ξ) dv = Ω Ω Ξ n ds where n is the outward drawn normal to Ω the boundary of Ω. Show that for a vector field f (div f) Ω Replace Ξ by the vector field f we have, (grad f) Ω dv = f n Ω dv = f n Ω Taking the trace of both sides and noting that both trace and the integral are linear operations, therefore we have, (div f) Ω dv = tr(grad f) Ω = tr(f n) Ω = f n Ω ds ds ds dv ds

43 63. Show that for a scalar function Hence the divergence theorem becomes, (grad φ) dv = φn ds Ω Ω Recall that for a vector field, that for a vector field f (div f) dv = f n ds Ω Ω if we write, f = φa where a is an arbitrary constant vector, we have, (div[φa]) dv = φa n Ω Ω For the LHS, note that, div[φa] = tr(grad[φa]) The trace of which is, ds = a φn Ω grad[φa] = (φa i ), j g i g j = a i φ, j g i g j a i φ, j g i g j = a i φ, j δ i j = a i φ, i = a grad φ For the arbitrary constant vector a, we therefore have that, ds (div[φa]) Ω dv = a grad φ Ω dv = a φn Ω ds grad φ Ω dv = φn Ω ds

44 64. Given tensors A, B and C we define the tensor square product with operator,, in the equation, (A B)C = ACB T, show that the square product of two tensors A and B has the component form, A B = A ik B jl g i g j g k g l Let A = A ij g i g j, B = B kl g k g l, C = C αβ g α g β. If we assume that A B = A ik B jl g i g j g k g l, Then the product, (A B)C = A ik B jl (g i g j g k g l )(C αβ g α g β ) = A ik B jl C αβ δ α k δ β l g i g j = A ik B jl C kl g i g j To complete the proof, we only need to show that the above result equals ACB T. To do so, we change the basis order in B so that, ACB T = (A ij g i g j )(C αβ g α g β )(B kl g l g k ) = A ij C αβ B kl g i g k δ α j δ β l = A ij C jl B kl g i g k = A ik B jl C kl g i g j = (A B)C We can therefore conclude that A B = A ik B jl g i g j g k g l..

45 65. Given tensors A, B and C we define the tensor square product with operator,, in the equation, (A B)C = ACB T, and that A B = A ik B jl g i g j g k g l and show A(B C) = (B T C T )A. By the definition of the tensor square product, the last expression, (B T C T )A = B T AC We therefore need to show that A(B C) = B T AC. Let, A = A αβ g α g β, B = B ij g i g j, C = C kl g k g l A(B C) = (A αβ g α g β )(B ik C jl g i g j g k g l ) = A αβ B ik C jl g k g l δ i j α δ β = A ij B ik C jl g k g l B T AC = B ij A αβ C kl (g j g i )(g α g β )(g k g l ) = B ij A αβ C kl g j g l δ i k α δ β = B ij A ik C kl g j g l = A ij B ik C jl g k g l = A(B C) We therefore conclude that A(B C) = B T AC = (B T C T )A.

46 66. Use the expression (S + T) c = S c + T c + T T S T + S T T T tr(t)s T tr(s)t T + [tr(s)tr(t) tr(st)]i to obtain an expression for the fourth order tensor, c cof S. We observe that the Gateaux differential, cof(s + αds) = ( cof S) ds = [ α cof(s + αds)] α=0 = [ α (Sc + (αds) c + (αds) T S T + S T (αds) T tr((αds))s T tr(s)(αds) T + [tr(s)tr((αds)) tr(s(αds))]i)] α=0 This can be further simplified as, = [ α (Sc + α 2 (ds) c + α(ds) T S T + αs T (ds) T α tr(ds)s T αtr(s)(ds) T + [αtr(s)tr(ds) αtr(sds)]i)] α=0 = [2α(dS) c + (ds) T S T + S T (ds) T tr(ds)s T tr(s)(ds) T + (tr(s)tr(ds) tr(sds))i] α=0

47 = (ds) T S T + S T (ds) T tr(ds)s T tr(s)(ds) T + [tr(s)tr(ds) tr(sds)]i = [(I S)T + (S T I)T S T I tr(s)t + tr(s)(i I) I S T ]ds from which we can conclude that, cof S = ((I S) + (ST I)) T (S T I + I S T ) + tr(s)((i I) T) 67. Use indicial notation to obtain an expression for the fourth order tensor, c cof S. Hint: Sc = 1 δ 2 ijk lmn j S k msn g l g i c cof S = 1 2 δ lmn ijk α (S j ms k n )g l g i g α g β β = 1 2 δ ijk lmn (δ j α δ β m S k n + δ k α δ β n S j m ) g l g i g α g β = 1 2 δ ijk lmn S k n g l g i g j g m δ ijk lmn S j m g l g i g k g n = δ lmn ijk S k n g l g i g j g m

48 68. Use the fact that for any two tensors A and B, S(A B + B A)S = 2S(A B)S and the derivative of the cofactor to show that for the right Cauchy- Green tensor, cof C = ((I C) + (C I))T (C I + I C) + tr(c)((i I) T) C so that S ( cof C) S = S((I C) + (C I))S (C I + I C) + tr(c)((i I) S) C 69. Given any scalar k > 0, for the scalar-valued tensor function, f(s) = tr(s k ), show that, d ds f(s) = k(sk 1 ) T. When k = 1, So that, Df(S, ds) = d dα f(s + αds) α=0 = d dα tr(s + αds) α=0 = I: ds = tr(1 ds)

49 d tr(s) = I. ds When k = 2, The Gateaux differential in this case, Df(S, ds) = d dα f(s + αds) = d α=0 dα tr{(s + αds)2 } α=0 = d When k = 3, dα tr{(s + αds)(s + αds)} α=0 = tr[ds(s + αds) + (S + αds)ds] α=0 = tr[ds S + S ds] = 2(S) T : ds Df(S, ds) = d dα f(s + αds) α=0 = tr [ d dα (S + αds)(s + αds)] α=0 = d dα tr{(s + αds)3 } α=0 = d dα tr{(s + αds)(s + αds)(s + αds)} α=0 = tr [ d dα (S + αds)(s + αds)(s + αds)] α=0 = tr[ds(s + αds)(s + αds) + (S + αds)ds(s + αds) + (S + αds)(s + αds)ds] α=0 = tr[ds S S + S ds S + S S ds] = 3(S 2 ) T : ds It easily follows by induction that, d ds f(s) = k(sk 1 ) T.

50 70. Find the derivative of the second principal invariant of the tensor S d ds I 2(S) = 1 2 d ds [tr2 (S) tr(s 2 )] = 1 2 [2tr(S)1 2 ST ] = tr(s)1 S T 71. Find the derivative of the third principal invariant of the tensor S By Cayley-Hamilton, in terms of traces only, d ds I 3(S) = 1 6 I 3 (S) = 1 6 [tr3 (S) 3tr(S)tr(S 2 ) + 2tr(S 3 )] d ds [tr3 (S) 3tr(S)tr(S 2 ) + 2tr(S 3 )] = 1 6 [3tr2 (S)I 3tr(S 2 )I 3tr(S)2S T + 2 3(S 2 ) T ] = I 2 I I 1 (S)S T + S 2T

51 72. By the representation theorem, every real isotropic tensor function is a function of its principal invariants. Show that for every isotropic function f(s) the derivative f(s) can be represented as a quadratic function of ST. By the representation theorem, consequently, f(s) f(s) = φ(i 1 (S), I 2 (S), I 3 (S)) = f(s) I 1 I 1 + f(s) I 2 I 2 + f(s) I 3 I 3 = f(s) I 1 = ( f(s) I 1 I + f(s) I 2 + f(s) I 2 + f(s) S 2T I 3 = α 0 I + α 1 S T + α 2 S 2T (tr(s)i S T ) + f(s) I 3 (I 2 I I 1 (S)S T + S 2T ) I 1 (S) + f(s) I 3 where α 0 = f(s) + f(s) I I 1 I 1 (S) + f(s) I 2 I 2 (S), α 1 = f(s) 3 I 2 quadratic representation of the derivative. I 2 (S)) I ( f(s) I 2 f(s) I 3 + f(s) I 3 I 1 (S)) S T I 1 (S) and α 2 = f(s) is the desired I 3

52 73. Show that the derivative of S 2 is (S I + I S T ) The Gateaux differential of F(S) = S 2 is, so that, F(S) F(S, ds) = lim F(S + αds) α 0 α = lim α (S2 + αsds + αds(s) + α 2 (ds) 2 ) = SdS + ds(s) = (S I + I S T )ds = ( F(S) ) ds α 0 = S I + I S T. 74. For three tensors A, B, and C, show that A(B C) = (C B)A Let A = A αβ g α g β, B = B ij g i g j, C = C kl g k g l so that, B C = B ik C jl g i g j g k g l A(B C) = (A αβ g α g β )(B ik C jl g i g j g k g l ) = A αβ B ik C jl δ i α δ j β g k g l = A ij B ik C jl g k g l

53 = B ik C jl g k g l δ α i δ β j A αβ = B ik C jl (g k g l (g i g j )) (A αβ g α g β ) 75. For any positive integer n Given that the derivative n = Sn r (S T ) r 1 n r=1 = S n, S, and that A(B C) = B T AC = (B T C T )A show that the derivative, n(s n 1 ) T Applying the rules of differentiation, tr S n tr Sn = ( n ) (n ) n = I ( S n r (S T ) r 1 ) = I(S n r (S T ) r 1 ) r=1 n = (S n r ) T I(S T ) r 1 r=1 = n(s n 1 ) T n n r=1 = (S n r+r 1 ) T r=1 = (S n 1 ) T 76. Show that if F(S) = S 1, then F(S) = S 1 S T = (S S T ) 1. Observe that the differential of the unit tensor is the zero tensor; n r=1 tr Sn =

54 D(SS 1 ) = D(I) = O Now, the Gateaux derivative follows the same product rules as the regular derivative, hence, so that so that, D(SS 1 ) = D(S)S 1 + S D(S 1 ) F(S) D(S 1 ) = S 1 D(S)S 1 = (S 1 S T ) ds = ( F(S) ) ds = S 1 S T = (S S T ) Given that 1 = S 1 S T = (S S T ) 1 Show that T S T ) = (S T S 1 )T = T(S S T ) 1 = T(S 1 The first and the last results are immediately obvious by invoking the rule that the transpose of a derivative is the derivative of the transpose. Hence we operate the transposer on the given result and obtain,

55 implies that 1 T = S 1 S T = (S S T ) 1 = T(S 1 S T ) = T(S S T ) 1 For the middle result, consider that, Replacing S by S T in the result, 1 S 1. T Consequently, = T because if F(S) = ST T = T T T = S 1 S T, we have T T DF(S, ds) = lim (S + αds)t α 0 α = lim α (ST + αds T ) = ds T = TdS = ( F(S) ) ds α 0 = S T

56 T = T T T 78. Show that T S = (I ST )T + (S T I) The differential DF(S, ds) when F(S) = S T S is, so that 79. Show that ST D(S T S) = ds T S + S T ds = (I S) + (S I)T = (S T S 1 )T = ((I S T )T + S T I) ds = ( F(S) ) ds T S = (I ST )T + (S T I) The differential DF(S, ds) when F(S) = SS T is, D(S T S) = ds S T + SdS T = ((I S) + (S I)T)dS = ( F(S) ) ds

57 so that S T = (I S) + (S I)T. 80. For the integers n > 0, if F(S) = S n for any differentiable tensor, show by mathematical induction that the derivative 0 F(S) = S n r (S T ) r 1 = S n, S r= n+1 Case n = 1 as expected. Case n = 2 S 1, S = S 1 S T S 2, S = S 1 S 2T S 2 S T Assume it is true for arbitrary n we use the fact that UV to obtain = (I VT ) U + (U I) V S n 1, S = (S n S 1 ), S = (I S T ) n + (S n I) 1

58 n = (I S T ) S n r 1 (S T ) r + (S n I)(S 1 S T ) r=1 n = S n r 1 (S T ) r 1 + (S n 1 S T ) r=1 n+1 = S r n+1 (S T ) r r=0 81. For the positive integer n, if F(S) = S n for any differentiable tensor, show by mathematical induction that the derivative F(S) = Sn 1 I + S n 2 S T + S n 3 S 2T + + S S (n 2)T + I S (n 1)T n = S n r (S T ) r 1 r=1 Case n = 1 as expected. = S n, S S, S = S 0 (S T ) 0 = I I = I

59 Case n = 2 S 2, S = S I + I S T Assume it is true for arbitrary n we use the fact that UV to obtain And hence the proof. = (I VT ) U S n+1, S = (S n S), S = (I S T ) n + (Sn I) n = (I S T ) S n r (S T ) r 1 + (S n I)I n r=1 = S n r (S T ) r r=1 n+1 = S n r (S T ) r 1 r=1 + (S n I)I + (U I) V 82. Given two differentiable tensor valued functions U(S) and V(S), of the tensor S show that (U:V) = ( U )T V + ( V )T U In component form, let U = U ij g i g j, V = V kl g k g l and

60 so that, (U: V) U: V = (U ij g i g j ): (V kl g k g l ) = U ij V kl g ik g jl = U ij αβ V kl g ik g jl g α g β + U ij V kl αβ g ik g jl g α g β = U ij αβ ((g α g β ) (g i g j )) (V kl g k g l ) + V kl αβ ((g α g β ) (g k g l )) (U ij g i g j ) = ( U ) T V + ( V ) T U 83. Given two differentiable tensor valued functions U(S) and V(S), of the tensor S show that UV = (I VT ) U + (U I) V U = U ij g i g j, V = V kl g k g l and UV = U ij V.l j g i g l so that, UV = U j ijv.l g i g l g α g β αβ

61 = ( U j ij V j V.l + U.l ij ) g i g l g αβ α g β αβ = g jk (V kl U ij αβ + U ij V kl αβ ) g i g l g α g β = (V kl U ij αβ + U ij V kl αβ ) (g i g j )(g k g l ) g α g β = (I(g i g j )(g k g l )V kl ) ( U ij αβ g α g β ) + (U ij g i g j )(g k g l ) ( V kl αβ g α g β ) = (I(g i g j )V) ( U ij ) + U(gk g l ) V kl = (I V T ) U + (U I) V 84. For the positive integer n, if F(S) = S n for any differentiable tensor, show by mathematical induction that the derivative

62 F(S) = Sn 1 I + S n 2 S T + S n 3 S 2T + + S S (n 2)T + I S (n 1)T n = S n r (S T ) r 1 r=1 Case n = 1 as expected. Case n = 2 = S n, S S, S = S 0 (S T ) 0 = I I = I S 2, S = S I + I S T Assume it is true for arbitrary n we use the fact that UV to obtain = (I VT ) U S n+1, S = (S n S), S = (I S T ) n + (Sn I) n = (I S T ) S n r (S T ) r 1 + (S n I)I n r=1 = S n r (S T ) r + (S n I)I r=1 + (U I) V

63 And hence the proof. n+1 = S n r (S T ) r 1 r=1 85. For a scalar valued function φ and a tensor-valued function F of the tensor S, show that the derivative of the product, φ F (φ(s)f(s)) = F + φ In component form, let F(S) = F ij g i g j, and S = S αβ g α g β. The required derivative, in component form is, (φ(s)f(s)) = (φf ij )g i g j g α g β αβ = ( φ αβ F ij + φ F ij αβ ) g i g j g α g β = (F ij g i g j ) φ αβ g α g β + φ F ij αβ g i g j g α g β = F φ + φ F

64 86. By finding the Gateaux differential of the Cayley-Hamilton equation, F(S) = S 3 tr S S (tr2 S tr S 2 )S I det S = 0, establish the Rivlin s identity, ( F(S) ) T = S 2 T + STS + TS 2 S 2 tr T (ST + TS) tr S + S(tr T tr S tr(st)) + T 2 (tr2 S tr S 2 ) ((cof S): T)I Cayley-Hamilton states that a tensor satisfies its own characteristic equation, hence, F(S) = S 3 tr S S (tr2 S tr S 2 )S I det S = 0 The Gateaux differential for the function F(S) in the direction of T is, lim α 0 d F(S + αt) = ( F(S)) T = 0 dα It is possible to take the Gateaux limit directly, however, in this case, the differential can be more easily found indirectly by taking the derivative, F(S) term by term, using two formulae: S n, S = n r=1 S n r (S T ) r 1 ; (UV), S = (I V T ) U + (U I) V

65 so that so that, 3 S 3, S = S n r (S T ) r 1 = S 2 I + S S 2 + I (S T ) 2 r=1 ( 3 ) T = S2 T + STS + TS 2 (I 1S 2 ) = S 2 I 1 + I 2 1 = S 2 I + I 1 (S I + I S T ) ( (I 1S 2 )) T = S 2 tr T + (ST + TS) tr S (I 2S) = S I 2 + I 2 = S 2 (I tr S S T ) (tr2 S tr S 2 )I ( (I 2S)) T = (S (I tr S S T )) T (tr2 S tr S 2 )IT = S ((I tr S S T ): T) + T 2 (tr2 S tr S 2 )

66 = S(tr T tr S tr(st)) + T 2 (tr2 S tr S 2 ) Lastly, (det S) = cof S. We can therefore write, ( F(S) ) T = S2 T + STS + TS 2 S 2 tr T (ST + TS) tr S + S(tr T tr S tr(st)) + T 2 (tr2 S tr S 2 ) ((cof S): T)I which is the Rivlin identity. 87. Given that S n, S = 0 r= n+1 S n r (S T ) r 1 If φ(s) = tr(s 1 S 1 ) show that φ(s) = 2S 3T Clearly, 0 S 2, S = S 2 r (S T ) r 1 r= 2+1 φ(s) = ( 2 tr(s 2 )) ( 2 ) = S 1 S 2T S 2 S T = I( S 1 S 2T S 2 S T ) = S 3T S 3T = 2S 3T

67 88. Given that S n, S = 0 r= n+1 S n r (S T ) r 1 If φ(s) = (det S )(S 1 : S 1 ) show that φ(s) = (cof S)S 1 : S 1 det S (S 1 S 1 S T + S T S 1 S T ) Clearly, 0 S 1, S = S 1 r (S T ) r 1 φ(s) = ( (det S) r= 1+1 = S 1 S T S 1 + det S 1 ) : S 1 + (det S)S 1 : 1 = (cof S)S 1 : S 1 det S (S 1 S T )S 1 (det S)S 1 (S 1 S T ) = (cof S)S 1 : S 1 det S (S 1 S 1 S 1 ) (det S)(S T S 1 S T ) 89. If φ(s) = (cof S ): (cof S ) = 1 2 ((S: S)2 (S T S): (S T S))show that φ(s) = 2(S: S)S + 2S T SS T : S T S = S + S = IS + SI = 2S = T S + ST = TS + ST I = 2S T

68 Consequently, φ(s) = 1 2 ( (S: S)2 : S ((S T S): (S T S)) T S S ) (: ) ( ((ST S): (ST S)) T ) ( T S S ). = ( T S ) ST S + S T S ( T S ) = 2ST S φ(s) = 1 S)2 S ( (S: ) (: 2 : S ) ( ((ST S): (ST S)) T ) ( T S S ) = 2(S: S)S + 2S T SS T 90. By the representation theorem, every real isotropic tensor function is a function of its principal invariants. Show that for every isotropic function f(s) of an invertible tensor S the derivative f(s) S T and S T. By the representation theorem, can be represented as a linear combination I, f(s) = φ(i 1 (S), I 2 (S), I 3 (S)) and, when S is invertible, I 3 = cof S = I 3S T consequently,

69 f(s) = f(s) I 1 I 1 + f(s) I 2 I 2 + f(s) I 3 I 3 = f(s) I 1 = ( f(s) I 1 I + f(s) I 2 + f(s) I 2 = α 0 I + α 1 S T + α 1 S T (tr(s)i S T ) + f(s) I 3 where α 0 = f(s) + f(s) I I 1 I 1 (S) + f(s) I 2 I 2 (S), α 1 = f(s) 3 I 2 desired representation of the derivative. I 3 S T I 1 (S)) I ( f(s) ) S T + f(s) I I 2 I 3 S T 3 f(s) I 3 I 1 (S) and α 1 = f(s) I 3 I 3 is the 91. Find the derivative of a vector point function u(x 1, x 2, x 3 ) using its covariant components, u α (x 1, x 2, x 3 ), α = 1, 2, 3 Let g α be the reciprocal basis vectors, Let u = u α g α, then du = x k (u αg α )dx k Clearly, = ( u α x k gα + gα x k u α) dx k

70 u x k = u α x k gα + gα x k u α And the projection of this quantity on the g i direction is, u x k g i = ( u α x k gα + gα x k u α) g i = u α x k gα g i + gα x k g iu α = u α x k δ i α + gα x k g iu α Now, g i g j = δ i j so that g i x k g j + g i g j = 0. xk g i x k g j = g i g j i { xk jk }. This important quantity, necessary to quantify the derivative of a tensor in general coordinates, is called the Christoffel Symbol of the second kind. Using this, we can now write that, u x k g i = u α x k δ i α + gα x k g iu α

71 = u i x k { α ik } u α The quantity on the RHS is the component of the derivative of vector u along the g i direction using covariant components. It is also called the covariant derivative of u. 92. Find the derivative of a vector field using contravariant components du = ( ui x k g i + g i x k ui ) dx k = ( ui x k g i + { α ik } g αu i ) dx k So that, u uα = xk x k g α + g α uα xk The components of this in the direction of g i can be obtained by taking a dot product as before: u x k gi = ( uα x k g α + g α x k uα ) g i = uα x k δ α i + g α x k gi u α = ui i + { xk αk } uα

72 93. Express the Christoffel symbols in terms of the metric tensor coefficients We observe that the derivative of the covariant basis, g i (= r x i), g i x j = Taking the dot product with g k, 2 r x j x i = g i x j g k = 1 2 ( g j x i g k + g i x j g k) = 1 2 ( x i [g j g k ] + = 1 2 ( x i [g j g k ] + = 1 2 ( x i [g j g k ] + = 1 2 ( g jk x i + g ki x j g ij x k ) 2 r x i x j = g j x i x j [g i g k ] g j g k x i g i g k x j) x j [g i g k ] g j g i x k g i g j x k) x j [g i g k ] x k [g i g j ]) Which is the quantity defined as the Christoffel symbols of the first kind in the textbooks. It is therefore possible for us to write, [ij, k] g i x j g k = g j x i g k = 1 2 ( g jk x i + g ki x j g ij x k )

73 It should be emphasized that the Christoffel symbols, even though the play a critical role in several tensor relationships, are themselves NOT tensor quantities. (Prove this). However, notice their symmetry in the i and j. The extension of this definition to the Christoffel symbols of the second kind is immediate: Contract the above equation with the conjugate metric tensor, we have, g kα [ij, α] g kα g i x j g α = g kα g j x i g α = g i x j gk = { k ij } = gk x j g i Which connects the common definition of the second Christoffel symbol with the one defined in the above derivation. The relationship, g kα [ij, α] = { k ij } apart from defining the relationship between the Christoffel symbols of the first kind and that second kind, also highlights, once more, the index-raising property of the conjugate metric tensor. 94. Demonstrate the index raising and index lowering attributes of the metric coefficients on Christoffel symbols. (despite the fact that these are, strictly speaking, not tensors in themselves). We contract the above equation with g kβ and obtain,

74 g kβ g kα [ij, α] = g kβ { k ij } δ β α [ij, α] = [ij, β] = g kβ { k ij } so that, g kα { α ij } = [ij, k] Showing that the metric tensor can be used to lower the contravariant index of the Christoffel symbol of the second kind to obtain the Christoffel symbol of the first kind. 95. Find the derivative of a second-order tensor field in terms of its contravariant components. For a second-order tensor T, we can express the components in dyadic form along the product basis as follows: T = T ij g i g j = T ij g i g j = T.j i g i g j = T j. i g j g i This is perfectly analogous to our expanding vectors in terms of basis and reciprocal bases. Derivatives of the tensor may therefore be expressible in any of these product bases. Consider the product covariant bases. We have:

75 T Tij = xk x k g i g j + T ij g i x k g j + T ij g i g j x k Recall that, g i x k gj = { j }. It follows therefore that, ik g i x k gj { α ik } δ j α = g i x k gj { α ik } g α g j = ( g i x k { α ik } g α) g j = 0. Clearly, g i x k = { α ik } g α (Obviously since g j is a basis vector it cannot vanish) Where T Tij = xk x k g i g j + T ij g i x k g j + T ij g i g j x k = Tij x k g i g j + T ij ({ α ik } g α) g j + T ij g i ({ α jk } g α) = Tij x k g i g j + T αj ({ i αk } g i) g j + T iα g i ({ j αk } g j) = ( Tij x k +Tαj { i αk } + Tiα { j αk }) g i g j = T ij,kg i g j

76 T ij = Tij,k x k +Tαj { i αk } + Tiα { j Tij } or αk x k +Tαj i Γ αk + T iα j Γ αk are the components of the covariant derivative of the tensor T in terms of contravariant components on the product covariant bases as shown. 96. Find the derivative of a second-order tensor field in terms of its covariant components. In the same way, by taking the tensor expression in the dyadic form of its contravariant product bases, we can write, T x k = T ij x k gi g j + T ij g i x k gj + T ij g i gj x k = T ij x k gi g j i + T ij Γ αk g j + T ij g i gj x k Again, notice from previous derivation above, { i jk } = gi g x k j so that, gi = x k { i αk } gα i = Γ αk g α Therefore, T x k = T ij x k gi g j g i + T ij x k gj + T ij g i gj x k = T ij x k gi g j i T ij Γ αk g α g j + T ij g i Γ j αk g α

77 So that = ( T ij x k T αjγ ik α T iα Γ jk α ) g i g j = T ij,k g i g j T ij,k = T ij x k T αjγ ik α T iα Γ jk α 97. Show that metric tensor components behave like constants under a covariant differentiation. The proof that this is so is due to Ricci: g ij, k = g ij x k { α ik } g αj { β kj } g iβ = g ij [ik, j] [kj, i] xk = g ij x k 1 2 ( g jk x i + g ji x k g ik x j ) 1 2 ( g kj x j + g ik x j g jk ) = 0. xi The conjugate metric tensor behaves the same way as can be seen from the relationship, g il g lj = δ i j The above can be differentiated covariantly with respect to x k to obtain

78 g il, k g lj + g il g lj, k = δ j i, k 0 + g il g lj, k = δ j i x k { α ik } δ j α + { j αk } δ i α g il g lj, k = 0 { j ik } + { j ik } = 0 The contraction of g il with g lj, k vanishes. Since we know that the metric tensor cannot vanish in general, we can only conclude that g ij, k = Show that the basis vectors are constants under covariant differentiation. In other words, their covariant derivatives vanish Recall that we can express the covariant basis vectors in terms of their dual, g i = g ij g j By Ricci s theorem, the derivatives g ij, j vanish, hence for the vectors g i themselves, the covariant derivatives vanish so that, g i, j = 0. In the same way, g i, j = Show that the alternating tensor, ε ijk, l = 0 that is, the Einstein tensor component is a constant under covariant differentiation. Recall that

79 Taking the covariant derivatives, we have, ε ijk = g i g j g k ε ijk, l = g i, l g j g k + g i g j, l g k + g i g j g k, l = 0 Hence the alternating tensor is a constant under covariant differentiation Let h 1 = g 11, h 2 = g 22 and h 3 = g 33 in an orthogonal curvilinear system 1 ε of coordinates, show that 123 = 1 h 1 h 2 h 3 = { i ε 123 x j h 1 h 2 h 3 x j ij } The covariant derivative, ε ijk,l = ε ijk x l { α il } ε αjk { β jl } ε iβk { γ kl } ε ijγ = 0 is valid for all admissible values of i, j and k. In particular, setting i = 1, j = 2 and k = 3, we obtain, Hence, ε 123,l = ε 123 x l { α 1l } ε α23 { β 2l } ε 1β3 { γ 3l } ε 12γ = 0 = ε 123 x l { 1 1l } ε 123 { 2 2l } ε 123 { 3 3l } ε 123 = ε 123,l = ε 123 x l { i il } ε 123 = 0