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

Transcription

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

2 It is easy to similarly show that u i v i, j g j = (grad v) T u. Clearly, 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

3 = 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 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 = ( 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

4 = (grad u)v + (div v)u 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 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

5 = (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 9. 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 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 10. Show that div (φ1) = grad φ = ε ijk u j, l g i δ k l = ε ijk u j, k g i = curl u Note that φ1 = (φg αβ )g α g β. Also note that

6 grad φ1 = (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 (φ1) = tr(grad φ1) = (φg αβ ), i (g α g β )g i = (φg αβ ), i g α g βi = φ, i g αβ g βi g α 11. Show that curl (φ1) = ( grad φ) = φ, i δ α i g α = φ, i g i = grad φ Note that φ1 = (φg αβ )g α g β, and that curl T = ε ijk T αk, j g i g α so that, curl (φ1) = 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) 12. Show that curl (v ) = (div v)1 grad v (v ) = ε αβk v β g α g k so that curl T = ε ijk T αk, j g i g α curl (v ) = ε ijk ε αβk v β, j g i g α

7 = (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)1 grad v 13. 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 14. Given a scalar point function φ and a vector field v, show that curl (φv) = φ curl v + ( 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

8 15. Show that curl (grad φ) = o For any tensor v = v α g α = ( 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 that, curl (grad φ) = ε ijk φ, kj g i = 0. The contraction of symmetric tensors with unsymmetric 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. 16. 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

9 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 unsymmetric 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. 17. 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 18. Show that div (grad φ grad θ) = 0 grad φ grad θ = ε ijk φ, j θ, k g i

10 The gradient of this vector is the tensor, 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 antisymmetric. 19. 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

11 Now, it can be shown (see #20 below) 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. 20. Show that g γi ε αβγ ε ijk = g αj g βk g αk g βj Note that

12 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 21. 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 as required. dφ dt = A φ : da dt = A A(t) : A 22. 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 23. 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) 24. 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,

15 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 = 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. 25. 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 i α δ j β δ i β δ j α ) (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

16 = 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. 26. Given a scalar point function φ and a second-order tensor field T, show that curl (φt) = φ curl T + (( φ) )T T where [( 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 + (( φ) )T T 27. 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,

17 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 ) 28. Show that if φ defined in the space spanned by orthonormal 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 orthonormal, 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.

18 29. 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. By the Divergence Theorem, r r = x i x j g i g j = x i x j g ij (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 = 6 dv V = 2g ik g lk x i, l dv V = 2δ i l δ l i dv V 30. 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

19 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 as required. 31. For a general tensor field T show that, curl(curl T) = [ 2 (tr T) div(div T)]I + grad(div T) + (grad(div T)) T grad(grad (trt)) 2 T T curl T = ε αst T βt, s g α g β = S.β α g α g β

20 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 ) = [ 2 (tr T) div(div T)]I + (grad(div T)) T grad(grad (trt)) + (grad(div T)) 2 T T 32. When T is symmetric, show that tr (curl T) vanishes. curl T = ε ijk T βk, j g i g β

21 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. 33. For a scalar function Φ and a vector v i show that the divergence of the vector v i Φ is equal to, v Φ + Φ div v (v i Φ),i = Φv i,i + v i Φ,i Hence the result.

22 34. Show that curl u v = (v u) + (u div v) (v div u) (u 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 ),m = δ lmi 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 k l δ j m = (δ j l δ k m δ k l δ j m )(u j,mv k u j v k,m) δ k m (uj,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.

23 35.. In Cartesian coordinates let x denote the magnitude of the position vector r = x i e i. Show that (a) x, j = x j, (b) x, x ij = 1 δ x ij x ix j (x) 3, (c) x,ii = 2, (d) If U = 1, then U, x x 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 = 1 x δ ij x ix j (x) 3 (c) x,ii = 1 x δ ii x ix i (x) 3 = 3 x (x)2 (x) 3 = 2 x. (d) U = 1 x so that = xδ x i x j ij x (x) 2

24 U,j = 1 x = 1 x x j x x = 1 1 x j x 2 x x j = x j Consequently, 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 3 ( x j ( x 2 )) + x i 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 x j (x 3 ) = x3 δ ij + x i (3x2 xj x ) x 6 = 3 x + x j [ ( 1 x x 2) j x ] = 3 x x jx j x 3 = δ ij x 3 = 3 x + x j ( x (1 x ) dx dx j ) = 3 x 1 x = 2 x + 3x ix j x 5

25 36. For vectors u, v and w, show that (u )(v )(w ) = u (v w) (u v)w. The tensor (u ) = ε lmn u n g l g m similarly, (v ) = ε αβγ v γ g α g β and (w ) = ε ijk w k g i g j. Clearly, (u )(v )(w ) = ε lmn ε αβγ ε ijk u n v γ w k (g α g β )(g l g m )(g i g j ) = ε αβγ ε lmn ε ijk u n v γ w k (g α g j )δ β l δ i m = ε αlγ ε lin ε ijk u n v γ w k (g α g j ) = ε lαγ ε lni ε ijk u n v γ w k (g α g j ) = (δ n α δ i γ δ i α δ n γ )ε ijk u n v γ w k (g α g j ) = ε ijk u α v i w k (g α g j ) + ε ijk u γ v γ w k (g i g j ) = [u (v w) (u v)w ] 37. Show that [u, v, w] = tr[(u )(v )(w )] In the above we have shown that (u )(v )(w ) = [u (v w) (u v)w ]

26 Because the vector cross is traceless, the trace of [(u v)w ] = 0. The trace of the first term, u (v w) is obviously the same as [u, v, w] which completes the proof. 38. Show that (u )(v ) = (u v)1 u v and that tr[(u )(v )] = 2(u v) (u )(v ) = ε lmn ε αβγ u n v γ (g α g β )(g l g m ) = ε lmn ε αβγ u n v γ (g α g m )δ β l = ε βmn ε βγα u n v γ (g α g m ) = [δ n γ δ m α δ m γ δ n α ]u n v γ (g α g m ) = u n v n (g α g α ) u n v m (g n g m ) = (u v)1 u v Obviously, the trace of this tensor is 2(u v) 39. 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

27 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 40. 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 for any arbitrary tensor B. In the case of f(a) = A,

28 A A : B = lim A + αb α 0 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 So that, or, A + αb = lim α α 0 = A A : B as required since B is arbitrary. 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 = 2A: B 2 A: A

29 41. Show that I 3 (S) S = det(s) S = S c the cofactor of S. Clearly S c = det(s) S T = I 3 (S) S T. Details of this for the contravariant 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 S S αβ g α g β = 1 3! εijk ε rst [ S ir S αβ S js S kt + S ir S js S αβ S kt + S ir S js S kt S 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

30 42. For a scalar variable α, if the tensor T = T(α) and T dt d dα det(t) = det(t) tr(t T 1 ) 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 d det(t) = dα dα (εijk T 1 i T 2 j T 3 k ) = ε ijk (T i 1 T 2 j T 3 k + T 1 i T j 2 T 3 k + T 1 i T 2 j T k 3 ) = ε ijk (A 1 l T l i T 2 j T 3 k + T 1 i A 2 m T m j T 3 k + T 1 i T 2 j A 3 n T n k ) = ε 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 )] dα, Show that 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, ε 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.)

31 as required. d dα det(t) = εijk [(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 ε ijk T i 1 T j 2 T k 3 = tr(t T 1 ) det(t) 43. Without breaking down into components, establish the fact that det(t) 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 ) It therefore follows that, Hence [ T (det(t)) [det(t) T T ]]: ( T α ) = 0 T = T c

32 T (det(t)) = [det(t) T T ] = T c 44. [Gurtin 3.4.2a] If T is invertible, show that T 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 45. [Gurtin 3.4.2a] 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 2 ) 1 = det(t 1 ) det(t 1 ) T T ( T 2 ) = T T T 1

33 46. Given that A is a constant tensor, Show that tr(as) = AT S 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 ) S tr(as) = = A ij S jβ δ i β = A ij S ji S αβ tr(as)g α g β = Aij S ji S αβ g α g β as required. = A ij δ j α δ i β g α g β = A ij g j g i = A T = S (AT : S)

34 47. Given that A and B are constant tensors, show that S 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 S tr(cs) = CT. Therefore, S tr(asbt ) = (B T A) T = A T B 48. Given that A and B are constant tensors, show that S 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, S tr(ast B T ) = S tr[(bt A) T S] = [(B T A) T ] T = B T A

35 49. 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) S I 1 (S) S = I 1 (S)1 S and (c) I 3 (S) S = I 3 (S) S 1 can be written in the invariant component form as, I 1 (S) S = I 1(S) S i j Recall that I 1 (S) α = tr(s) = S α hence I 1 (S) = I 1(S) g S j i g j = S α α j S i S g i g j i = δ α i δ j α g i g j = δ j i g i g j = 1 which is the identity tensor as expected. I 2 (S) S g i g j S = 1 the identity tensor, (b) in a similar way can be written in the invariant component form as,

36 I 2 (S) = 1 I 1 (S) [S α S 2 α S β β S α β S β α ] g i g j S i j where we have utilized the fact that I 2 (S) = 1 2 [tr2 (S) tr(s 2 )]. Consequently, I 2 (S) S = 1 2 j S [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 S S αβ g α g β = 1 3! εijk ε rst [ S ir S αβ S js S kt + S ir S js S αβ S kt + S ir S js S kt S 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

37 = 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 50. For a tensor field Ξ, The volume integral in the region Ω E, (grad Ξ) Ω Ω dv = 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) dv = f n Ω Ω Replace Ξ by the vector field f we have, (grad f) 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, ds ds

38 (div f) dv = tr(grad f) dv Ω Ω = tr(f n) Ω = f n Ω ds ds 51. Show that for a scalar function Hence the divergence theorem becomes, (grad φ) dv = Ω φn ds 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 ds = a φn Ω Ω Ω For the LHS, note that, div[φa] = tr(grad[φa]) grad[φa] = (φa i ), j g i g j = a i φ, j g i g j ds

39 The trace of which is, 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, (div[φa]) Ω dv = a grad φ Ω dv = a φn Ω ds grad φ Ω dv = φn Ω ds