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- 1. Tensor CalculusDifferentials & Directional Derivatives
- 2. The Gateaux Differential We are presently concerned with Inner Product spaces in our treatment of the Mechanics of Continua. Consider a map, 𝑭: V → W This maps from the domainV to W – both of which are Euclidean vector spaces. The concepts of limit and continuity carries naturally from the real space to any Euclidean vector space.Dept of Systems Engineering, University of Lagos 2 oafak@unilag.edu.ng 12/27/2012
- 3. The Gateaux Differential Let 𝒗0 ∈ V and 𝒘0 ∈ W, as usual we can say that the limit lim 𝑭 𝒗 = 𝒘0 𝒗→ 𝒗0 if for any pre-assigned real number 𝜖 > 0, no matter how small, we can always find a real number 𝛿 > 0 such that 𝑭 𝒗 − 𝒘0 ≤ 𝜖 whenever 𝒗 − 𝒗0 < 𝛿. The function is said to be continuous at 𝒗0 if 𝑭 𝒗0 exists and 𝑭 𝒗0 = 𝒘0Dept of Systems Engineering, University of Lagos 3 oafak@unilag.edu.ng 12/27/2012
- 4. The Gateaux Differential Specifically, for 𝛼 ∈ ℛ let this map be: 𝑭 𝒙 + 𝛼𝒉 − 𝑭 𝒙 𝑑 𝐷𝑭 𝒙, 𝒉 ≡ lim = 𝑭 𝒙 + 𝛼𝒉 𝛼→0 𝛼 𝑑𝛼 𝛼=0 We focus attention on the second variable ℎ while we allow the dependency on 𝒙 to be as general as possible. We shall show that while the above function can be any given function of 𝒙 (linear or nonlinear), the above map is always linear in 𝒉 irrespective of what kind of Euclidean space we are mapping from or into. It is called the Gateaux Differential.Dept of Systems Engineering, University of Lagos 4 oafak@unilag.edu.ng 12/27/2012
- 5. The Gateaux Differential Let us make the Gateaux differential a little more familiar in real space in two steps: First, we move to the real space and allow ℎ → 𝑑𝑥 and we obtain, 𝐹 𝑥 + 𝛼𝑑𝑥 − 𝐹 𝑥 𝑑 𝐷𝐹(𝑥, 𝑑𝑥) = lim = 𝐹 𝑥 + 𝛼𝑑𝑥 𝛼→0 𝛼 𝑑𝛼 𝛼=0 And let 𝛼𝑑𝑥 → Δ𝑥, the middle term becomes, 𝐹 𝑥 + Δ𝑥 − 𝐹 𝑥 𝑑𝐹 lim 𝑑𝑥 = 𝑑𝑥 Δ𝑥→0 Δ𝑥 𝑑𝑥 from which it is obvious that the Gateaux derivative is a generalization of the well-known differential from elementary calculus. The Gateaux differential helps to compute a local linear approximation of any function (linear or nonlinear).Dept of Systems Engineering, University of Lagos 5 oafak@unilag.edu.ng 12/27/2012
- 6. The Gateaux Differential It is easily shown that the Gateaux differential is linear in its second argument, ie, for 𝛼 ∈ R 𝐷𝑭 𝒙, 𝛼𝒉 = 𝛼𝐷𝑭 𝒙, 𝒉 Furthermore, 𝐷𝑭 𝒙, 𝒈 + 𝒉 = 𝐷𝑭 𝒙, 𝒈 + 𝐷𝑭 𝒙, 𝒉 and that for 𝛼, 𝛽 ∈ R 𝐷𝑭 𝒙, 𝛼𝒈 + 𝛽𝒉 = 𝛼𝐷𝑭 𝒙, 𝒈 + 𝛽𝐷𝑭 𝒙, 𝒉Dept of Systems Engineering, University of Lagos 6 oafak@unilag.edu.ng 12/27/2012
- 7. The Frechét Derivative The Frechét derivative or gradient of a differentiable function is the linear operator, grad 𝑭(𝒗) such that, for 𝑑𝒗 ∈ V, 𝑑𝑭 𝒗 𝑑𝒗 ≡ grad 𝑭 𝒗 𝑑𝒗 ≡ 𝐷𝑭 𝒗, 𝑑𝒗 𝑑𝒗 Obviously, grad 𝑭(𝒗) is a tensor because it is a linear transformation from one Euclidean vector space to another. Its linearity derives from the second argument of the RHS.Dept of Systems Engineering, University of Lagos 7 oafak@unilag.edu.ng 12/27/2012
- 8. Differentiable Real-Valued Function The Gradient of a Differentiable Real-Valued Function For a real vector function, we have the map: 𝑓: V → R The Gateaux differential in this case takes two vector arguments and maps to a real value: 𝐷𝑓: V × V → R This is the classical definition of the inner product so that we can define the differential as, 𝑑𝑓(𝒗) ⋅ 𝑑𝒗 ≡ 𝐷𝑓 𝒗, 𝑑𝒗 𝑑𝒗 This quantity, 𝑑𝑓(𝒗) 𝑑𝒗 defined by the above equation, is clearly a vector. oafak@unilag.edu.ng 12/27/2012Dept of Systems Engineering, University of Lagos 8
- 9. Real-Valued Function It is the gradient of the scalar valued function of the vector variable. We now proceed to find its components in general coordinates. To do this, we choose a basis 𝐠 𝑖 ⊂ V. On such a basis, the function, 𝒗 = 𝑣𝑖 𝐠𝑖 and, 𝑓 𝒗 = 𝑓 𝑣𝑖 𝐠𝑖 we may also express the independent vector 𝑑𝒗 = 𝑑𝑣 𝑖 𝐠 𝑖 on this basis.Dept of Systems Engineering, University of Lagos 9 oafak@unilag.edu.ng 12/27/2012
- 10. Real-Valued Function The Gateaux differential in the direction of the basis vector 𝐠 𝑖 is, 𝑓 𝒗 + 𝛼𝐠 𝑖 − 𝑓 𝒗 𝑓 𝑣 𝑗 𝐠 𝑗 + 𝛼𝐠 𝑖 − 𝑓 𝑣 𝑖 𝐠 𝑖 𝐷𝑓 𝒗, 𝐠 𝑖 = lim = lim 𝛼→0 𝛼 𝛼→0 𝛼 𝑗 𝑓 𝑣 𝑗 + 𝛼𝛿 𝑖 𝐠 𝑗 − 𝑓 𝑣 𝑖 𝐠 𝑖 = lim 𝛼→0 𝛼 As the function 𝑓 does not depend on the vector basis, we can substitute the vector function by the real function of the components 𝑣 1 , 𝑣 2 , 𝑣 3 so that, 𝑓 𝑣 𝑖 𝐠 𝑖 = 𝑓 𝑣1, 𝑣 2, 𝑣 3 in which case, the above differential, similar to the real case discussed earlier becomes, 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝐷𝑓 𝒗, 𝐠 𝑖 = 𝜕𝑣 𝑖Dept of Systems Engineering, University of Lagos 10 oafak@unilag.edu.ng 12/27/2012
- 11. 𝑑𝑓(𝒗) Of course, it is now obvious that the gradient of 𝑑𝒗 the scalar valued vector function, expressed in its dual basis must be, 𝑑𝑓(𝒗) 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑖 = 𝑖 𝐠 𝑑𝒗 𝜕𝑣 so that we can recover, 𝑑𝑓 𝒗 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗 𝐷𝑓 𝒗, 𝐠 𝑖 = ⋅ 𝑑𝒗 = 𝑗 𝐠 ⋅ 𝐠𝑖 𝑑𝒗 𝜕𝑣 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝑗 𝛿𝑖 = 𝜕𝑣 𝜕𝑣 𝑖 Hence we obtain the well-known result that 𝑑𝑓(𝒗) 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑖 grad 𝑓 𝒗 = = 𝑖 𝐠 𝑑𝒗 𝜕𝑣 which defines the Frechét derivative of a scalar valued function with respect to its vector argument.Dept of Systems Engineering, University of Lagos 11 oafak@unilag.edu.ng 12/27/2012
- 12. Vector valued Function The Gradient of a Differentiable Vector valued Function The definition of the Frechét clearly shows that the gradient of a vector-valued function is itself a second order tensor. We now find the components of this tensor in general coordinates. To do this, we choose a basis 𝐠 𝑖 ⊂ V. On such a basis, the function, 𝑭 𝒗 = 𝐹𝑘 𝒗 𝐠𝑘 The functional dependency on the basis vectors are ignorable on account of the fact that the components themselves are fixed with respect to the basis. We can therefore write, 𝑭 𝒗 = 𝐹 𝑘 𝑣1 , 𝑣 2 , 𝑣 3 𝐠 𝑘Dept of Systems Engineering, University of Lagos 12 oafak@unilag.edu.ng 12/27/2012
- 13. Vector Derivatives 𝐷𝑭 𝒗, 𝑑𝒗 = 𝐷𝑭 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖 = 𝐷𝑭 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 𝑑𝑣 𝑖 = 𝐷𝐹 𝑘 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖 Again, upon noting that the functions 𝐷𝐹 𝑘 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 , 𝑘 = 1,2,3 are not functions of the vector basis, they can be written as functions of the scalar components alone so that we have, as before, 𝐷𝑭 𝒗, 𝑑𝒗 = 𝐷𝐹 𝑘 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖 = 𝐷𝐹 𝑘 𝒗, 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖Dept of Systems Engineering, University of Lagos 13 oafak@unilag.edu.ng 12/27/2012
- 14. Which we can compare to the earlier case of scalar valued function and easily obtain, 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗 𝐷𝑭 𝒗, 𝑑𝒗 = 𝐠 ⋅ 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖 𝜕𝑣 𝑗 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝐠 𝑘 ⊗ 𝐠 𝑗 𝐠 𝑖 𝑑𝑣 𝑖 𝜕𝑣 𝑗 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝐠 𝑘 ⊗ 𝐠 𝑗 𝑑𝒗 𝜕𝑣 𝑗 𝑑𝑭 𝒗 = 𝑑𝒗 ≡ grad 𝑭 𝒗 𝑑𝒗 𝑑𝒗 Clearly, the tensor gradient of the vector-valued vector function is, 𝑑𝑭 𝒗 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 = grad 𝑭 𝒗 = 𝐠 𝑘 ⊗ 𝐠𝑗 𝑑𝒗 𝜕𝑣 𝑗 Where due attention should be paid to the covariance of the quotient indices in contrast to the contravariance of the non quotient indices.Dept of Systems Engineering, University of Lagos 14 oafak@unilag.edu.ng 12/27/2012
- 15. The Trace & Divergence The trace of this gradient is called the divergence of the function: 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 div 𝑭 𝒗 = tr 𝐠 𝑘 ⊗ 𝐠𝑗 𝜕𝑣 𝑗 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝐠 𝑘 ⋅ 𝐠𝑗 𝜕𝑣 𝑗 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗 = 𝑗 𝛿𝑘 𝜕𝑣 𝜕𝐹 𝑗 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝜕𝑣 𝑗Dept of Systems Engineering, University of Lagos 15 oafak@unilag.edu.ng 12/27/2012
- 16. Tensor-Valued Vector Function Consider the mapping, 𝑻: V → T Which maps from an inner product space to a tensor space. The Gateaux differential in this case now takes two vectors and produces a tensor: 𝐷𝑻: V × V → T With the usual notation, we may write, 𝐷𝑻 𝒗, 𝑑𝒗 = 𝐷𝑻 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖 = 𝐷𝑻 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖 𝑑𝑻 𝒗 = 𝑑𝒗 ≡ grad 𝑻 𝒗 𝑑𝒗 𝑑𝒗 = 𝐷𝑇 𝛼𝛽 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖 𝐠 𝛼 ⊗ 𝐠 𝛽 = 𝐷𝑇 𝛼𝛽 𝒗, 𝐠 𝑖 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑑𝑣 𝑖 Each of these nine functions look like the real differential of several variables we considered earlier.Dept of Systems Engineering, University of Lagos 16 oafak@unilag.edu.ng 12/27/2012
- 17. The independence of the basis vectors imply, as usual, that 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 𝐷𝑇 𝛼𝛽 𝒗, 𝐠 𝑖 = 𝜕𝑣 𝑖 so that, 𝐷𝑻 𝒗, 𝑑𝒗 = 𝐷𝑇 𝛼𝛽 𝒗, 𝐠 𝑖 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑑𝑣 𝑖 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑑𝑣 𝑖 𝜕𝑣 𝑖 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝐠 𝛼 ⊗ 𝐠 𝛽 𝐠 𝑖 ⋅ 𝑑𝒗 𝜕𝑣 𝑖 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 𝑑𝑻 𝒗 = 𝐠 𝛼 ⊗ 𝐠 𝛽 ⊗ 𝐠 𝑖 𝑑𝒗 = 𝑑𝒗 𝜕𝑣 𝑖 𝑑𝒗 ≡ grad 𝑻 𝒗 𝑑𝒗 Which defines the third-order tensor, 𝑑𝑻 𝒗 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 = grad 𝑻 𝒗 = 𝐠 𝛼 ⊗ 𝐠 𝛽 ⊗ 𝐠𝑖 𝑑𝒗 𝜕𝑣 𝑖 and with no further ado, we can see that a third-order tensor transforms a vector into a second order tensor.Dept of Systems Engineering, University of Lagos 17 oafak@unilag.edu.ng 12/27/2012
- 18. The Divergence of a Tensor Function The divergence operation can be defined in several ways. Most common is achieved by the contraction of the last two basis so that, 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 div 𝑻 𝒗 = 𝐠 𝛼 ⊗ 𝐠 𝛽 ⋅ 𝐠𝑖 𝜕𝑣 𝑖 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝐠 𝛼 𝐠 𝛽 ⋅ 𝐠𝑖 𝜕𝑣 𝑖 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 𝑖 𝜕𝑇 𝛼𝑖 𝑣 1 , 𝑣 2 , 𝑣 3 = 𝑖 𝐠 𝛼 𝛿𝛽 = 𝑖 𝐠𝛼 𝜕𝑣 𝜕𝑣 It can easily be shown that this is the particular vector that gives, for all constant vectors 𝒂, div 𝑻 𝒂 ≡ div 𝑻T 𝒂Dept of Systems Engineering, University of Lagos 18 oafak@unilag.edu.ng 12/27/2012
- 19. Real-Valued Tensor Functions Two other kinds of functions are critically important in our study. These are real valued functions of tensors and tensor-valued functions of tensors. Examples in the first case are the invariants of the tensor function that we have already seen. We can express stress in terms of strains and vice versa. These are tensor-valued functions of tensors. The derivatives of such real and tensor functions arise in our analysis of continua. In this section these are shown to result from the appropriate Gateaux differentials. The gradients or Frechét derivatives will be extracted once we can obtain the Gateaux differentials.Dept of Systems Engineering, University of Lagos 19 oafak@unilag.edu.ng 12/27/2012
- 20. Frechét Derivative Consider the Map: 𝑓: T → R The Gateaux differential in this case takes two vector arguments and maps to a real value: 𝐷𝑓: T × T → R Which is, as usual, linear in the second argument. The Frechét derivative can be expressed as the first component of the following scalar product: 𝑑𝑓(𝑻) 𝐷𝑓 𝑻, 𝑑𝑻 = : 𝑑𝑻 𝑑𝑻 which, we recall, is the trace of the contraction of one tensor with the transpose of the other second-order tensors. This is a scalar quantity.Dept of Systems Engineering, University of Lagos 20 oafak@unilag.edu.ng 12/27/2012
- 21. Guided by the previous result of the gradient of the scalar valued function of a vector, it is not difficult to see that, 𝑑𝑓(𝑻) 𝜕𝑓(𝑇11 , 𝑇12 , … , 𝑇 33 ) 𝛼 = 𝐠 ⊗ 𝐠𝛽 𝑑𝑻 𝜕𝑇 𝛼𝛽 In the dual to the same basis, we can write, 𝑑𝑻 = 𝑑𝑇 𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 Clearly, 𝑑𝑓(𝑻) 𝜕𝑓(𝑇11 , 𝑇12 , … , 𝑇 33 ) 𝛼 : 𝑑𝑻 = 𝐠 ⊗ 𝐠 𝛽 : 𝑑𝑇 𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 𝑑𝑻 𝜕𝑇 𝛼𝛽 𝜕𝑓(𝑇11 , 𝑇12 , … , 𝑇 33 ) = 𝑑𝑇 𝑖𝑗 𝜕𝑇 𝑖𝑗 Again, note the covariance of the quotient indices.Dept of Systems Engineering, University of Lagos 21 oafak@unilag.edu.ng 12/27/2012
- 22. Examples Let 𝐒 be a symmetric and positive definite tensor and let 𝐼1 𝑺 , 𝐼2 𝑺 and 𝐼3 𝑺 be the three principal invariants of 𝜕𝑰1 𝐒 𝐒 show that (a) = 𝟏 the identity tensor, (b) 𝜕𝐒 𝜕𝑰2 𝐒 𝜕𝐼3 𝐒 = 𝐼1 𝐒 𝟏 − 𝐒 and (c) = 𝐼3 𝐒 𝐒 −1 𝜕𝐒 𝜕𝐒 𝜕𝑰1 𝐒 can be written in the invariant component form 𝜕𝐒 as, 𝜕𝐼1 𝐒 𝜕𝐼1 𝐒 = 𝑗 𝐠𝑖 ⊗ 𝐠𝑗 𝜕𝐒 𝜕𝑆 𝑖Dept of Systems Engineering, University of Lagos 22 oafak@unilag.edu.ng 12/27/2012
- 23. (a) Continued α Recall that 𝐼1 𝐒 = tr 𝐒 = 𝑆α hence 𝜕𝐼1 𝐒 𝜕𝐼1 𝐒 = 𝑗 𝐠𝑖 ⊗ 𝐠𝑗 𝜕𝐒 𝜕𝑆 𝑖 α 𝜕𝑆α = 𝑗 𝐠𝑖 ⊗ 𝐠𝑗 𝜕𝑆 𝑖 = 𝛿 𝑖𝛼 𝛿 𝑗 𝛼 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝛿 𝑗𝑖 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝟏 which is the identity tensor as expected.Dept of Systems Engineering, University of Lagos 23 oafak@unilag.edu.ng 12/27/2012
- 24. 𝜕𝐼2 𝐒 (b) = 𝐼1 𝐒 𝟏 − 𝐒 𝜕𝐒 𝜕𝐼2 𝐒 in a similar way can be written in the invariant component form 𝜕𝐒 as, 𝜕𝐼2 𝐒 1 𝜕𝐼1 𝐒 α 𝛽 𝛽 = 𝑆α 𝑆 𝛽 − 𝑆 α 𝑆α 𝐠 𝑖 ⊗ 𝐠 𝑗 𝛽 𝜕𝐒 2 𝜕𝑆 𝑗 𝑖 1 where we have utilized the fact that 𝐼2 𝐒 = 2 tr 2 𝐒 − tr(𝐒 2 ) . Consequently, 𝜕𝐼2 𝐒 1 𝜕 α 𝛽 𝛽 = 𝑆α 𝑆 𝛽 − 𝑆 α 𝑆α 𝐠 𝑖 ⊗ 𝐠 𝑗 𝛽 𝜕𝐒 2 𝜕𝑆 𝑗 𝑖 1 𝑖 𝛼 𝛽 𝑖 𝛽 α 𝛽 𝛽 = 𝛿 𝛼 𝛿 𝑗 𝑆 𝛽 + 𝛿 𝛽 𝛿 𝑗 𝑆α − 𝛿 𝛽 𝛿 𝑗 𝛼 𝑆α − 𝛿 𝑖𝛼 𝛿 𝑗 𝑆 α 𝐠 𝑖 ⊗ 𝐠 𝑗 𝑖 𝛽 2 1 𝑖 𝛽 𝑗 𝑗 = 𝛿 𝑗 𝑆 𝛽 + 𝛿 𝑗𝑖 𝑆α − 𝑆 𝑖 − 𝑆 𝑖 𝐠 𝑖 ⊗ 𝐠 𝑗 α 2 𝑗 = 𝛿 𝑗𝑖 𝑆α − 𝑆 𝑖 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝐼1 𝐒 𝟏 − 𝐒 αDept of Systems Engineering, University of Lagos 24 oafak@unilag.edu.ng 12/27/2012
- 25. 𝜕𝑰3 𝐒 𝜕det 𝐒 = = 𝐒𝐜 𝜕𝐒 𝜕𝐒 the cofactor of 𝐒. Clearly 𝐒 𝐜 = det 𝐒 𝐒−1 = 𝑰3 𝐒 𝐒−1 as required. Details of this for the contravariant components of a tensor is presented below. Let 1 𝑖𝑗𝑘 𝑟𝑠𝑡 𝐒 = 𝑆= 𝑒 𝑒 𝑆 𝑖𝑟 𝑆 𝑗𝑠 𝑆 𝑘𝑡 3!Differentiating wrt 𝑆 𝛼𝛽 , we obtain, 𝜕𝑆 1 𝑖𝑗𝑘 𝑟𝑠𝑡 𝜕𝑆 𝑖𝑟 𝜕𝑆 𝑗𝑠 𝜕𝑆 𝑘𝑡 𝐠 ⊗ 𝐠 𝛽= 𝑒 𝑒 𝑆 𝑆 + 𝑆 𝑖𝑟 𝑆 + 𝑆 𝑖𝑟 𝑆 𝑗𝑠 𝐠 ⊗ 𝐠𝛽 𝜕𝑆 𝛼𝛽 𝛼 3! 𝜕𝑆 𝛼𝛽 𝑗𝑠 𝑘𝑡 𝜕𝑆 𝛼𝛽 𝑘𝑡 𝜕𝑆 𝛼𝛽 𝛼 1 𝑖𝑗𝑘 𝑟𝑠𝑡 𝛼 𝛽 𝛽 𝛽 = 𝑒 𝑒 𝛿 𝑖 𝛿 𝑟 𝑆 𝑗𝑠 𝑆 𝑘𝑡 + 𝑆 𝑖𝑟 𝛿 𝑗 𝛼 𝛿 𝑠 𝑆 𝑘𝑡 + 𝑆 𝑖𝑟 𝑆 𝑗𝑠 𝛿 𝑘𝛼 𝛿 𝑡 𝐠 𝛼 ⊗ 𝐠 𝛽 3! 1 𝛼𝑗𝑘 𝛽𝑠𝑡 = 𝑒 𝑒 𝑆 𝑗𝑠 𝑆 𝑘𝑡 + 𝑆 𝑗𝑠 𝑆 𝑘𝑡 + 𝑆 𝑗𝑠 𝑆 𝑘𝑡 𝐠 𝛼 ⊗ 𝐠 𝛽 3! 1 𝛼𝑗𝑘 𝛽𝑠𝑡 = 𝑒 𝑒 𝑆 𝑗𝑠 𝑆 𝑘𝑡 𝐠 𝛼 ⊗ 𝐠 𝛽 ≡ 𝑆 c 𝛼𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽 2!Which is the cofactor of 𝑆 𝛼𝛽Dept of Systems Engineering, University of Lagos 25 oafak@unilag.edu.ng 12/27/2012
- 26. Real Tensor Functions Consider the function 𝑓(𝑺) = tr 𝑺 𝑘 where 𝑘 ∈ ℛ, and 𝑺 is a tensor. 𝑓 𝑺 = tr 𝑺 𝑘 = 𝑺 𝑘 : 𝟏 To be specific, let 𝑘 = 3.Dept of Systems Engineering, University of Lagos 26 oafak@unilag.edu.ng 12/27/2012
- 27. Real Tensor Functions The Gateaux differential in this case, 𝑑 𝑑 3 𝐷𝑓 𝑺, 𝑑𝑺 = 𝑓 𝑺 + 𝛼𝑑𝑺 = tr 𝑺 + 𝛼𝑑𝑺 𝑑𝛼 𝛼=0 𝑑𝛼 𝛼=0 𝑑 = tr 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑑𝛼 𝛼=0 𝑑 = tr 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑑𝛼 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 𝑺2 T : 𝑑𝑺Dept of Systems Engineering, University of Lagos 27 oafak@unilag.edu.ng 12/27/2012
- 28. Real Tensor Functions With the last equality coming from the definition of the Inner product while noting that a circular permutation does not alter the value of the trace. It is easy to establish inductively that in the most general case, for 𝑘 > 0, we have, 𝑘−1 T 𝐷𝑓 𝑺, 𝑑𝑺 = 𝑘 𝑺 : 𝑑𝑺 Clearly, 𝑑 T tr 𝑺 𝑘 = 𝑘 𝑺 𝑘−1 𝑑𝑺Dept of Systems Engineering, University of Lagos 28 oafak@unilag.edu.ng 12/27/2012
- 29. Real Tensor Functions When 𝑘 = 1, 𝑑 𝐷𝑓 𝑺, 𝑑𝑺 = 𝑓 𝑺 + 𝛼𝑑𝑺 𝑑𝛼 𝛼=0 𝑑 = tr 𝑺 + 𝛼𝑑𝑺 𝑑𝛼 𝛼=0 = tr 𝟏 𝑑𝑺 = 𝟏: 𝑑𝑺 Or that, 𝑑 tr 𝑺 = 𝟏. 𝑑𝑺Dept of Systems Engineering, University of Lagos 29 oafak@unilag.edu.ng 12/27/2012
- 30. Real Tensor Functions Derivatives of the other two invariants of the tensor 𝑺 can be found as follows: 𝑑 𝐼2 (𝑺) 𝑑𝑺 1 𝑑 = tr 2 𝑺 − tr 𝑺2 2 𝑑𝑺 1 = 2tr 𝑺 𝟏 − 2 𝑺T = tr 𝑺 𝟏 − 𝑺T 2 .Dept of Systems Engineering, University of Lagos 30 oafak@unilag.edu.ng 12/27/2012
- 31. Real Tensor Functions To Determine the derivative of the third invariant, we begin with the trace of the Cayley-Hamilton for 𝑺: tr 𝑺3 − 𝐼1 𝑺2 + 𝐼2 𝑺 − 𝐼3 𝑰 = tr(𝑺3 ) − 𝐼1 tr 𝑺2 + 𝐼2 tr 𝑺 − 3𝐼3 = 0 Therefore, 3𝐼3 = tr(𝑺3 ) − 𝐼1 tr 𝑺2 + 𝐼2 tr 𝑺 1 2 𝐼2 𝑺 = tr 𝑺 − tr 𝑺2 2 .Dept of Systems Engineering, University of Lagos 31 oafak@unilag.edu.ng 12/27/2012
- 32. Differentiating the Invariants We can therefore write, 3𝐼3 𝑺 = tr(𝑺3 ) − tr 𝑺 tr 𝑺2 1 2 + tr 𝑺 − tr 𝑺2 tr 𝑺 2 so that, in terms of traces only, 1 3 𝐼3 𝑺 = tr 𝑺 − 3tr 𝑺 tr 𝑺2 + 2tr(𝑺3 ) 6Dept of Systems Engineering, University of Lagos 32 oafak@unilag.edu.ng 12/27/2012
- 33. Real Tensor Functions Clearly, 𝑑𝐼3 (𝑺) 1 = 3tr 2 𝑺 𝟏 − 3tr 𝑺2 − 3tr 𝑺 2𝑺 𝑇 + 2 × 3 𝑺2 𝑇 𝑑𝑺 6 = 𝐼2 𝟏 − tr 𝑺 𝑺 𝑇 + 𝑺2𝑇Dept of Systems Engineering, University of Lagos 33 oafak@unilag.edu.ng 12/27/2012
- 34. Real Tensor Functions .Dept of Systems Engineering, University of Lagos 34 oafak@unilag.edu.ng 12/27/2012
- 35. The Euclidean Point Space In Classical Theory, the world is a Euclidean Point Space of dimension three. We shall define this concept now and consequently give specific meanings to related concepts such as Frames of Reference, Coordinate Systems and Global ChartsDept of Systems Engineering, University of Lagos 35 oafak@unilag.edu.ng 12/27/2012
- 36. The Euclidean Point Space Scalars, Vectors and Tensors we will deal with in general vary from point to point in the material. They are therefore to be regarded as functions of position in the physical space occupied. Such functions, associated with positions in the Euclidean point space, are called fields. We will therefore be dealing with scalar, vector and tensor fields.Dept of Systems Engineering, University of Lagos 36 oafak@unilag.edu.ng 12/27/2012
- 37. The Euclidean Point Space The Euclidean Point Space ℰ is a set of elements called points. For each pair of points 𝒙, 𝒚 ∈ ℰ, ∃ 𝒖 𝒙, 𝒚 ∈ E with the following two properties: 1. 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒚 ∀𝒙, 𝒚, 𝒛 ∈ ℰ 2. 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝒛 ⇔ 𝒚 = 𝒛 Based on these two, we proceed to show that, 𝒖 𝒙, 𝒙 = 𝟎 And that, 𝒖 𝒙, 𝒛 = −𝒖 𝒛, 𝒙Dept of Systems Engineering, University of Lagos 37 oafak@unilag.edu.ng 12/27/2012
- 38. The Euclidean Point Space From property 1, let 𝒚 → 𝒙 it is clear that, 𝒖 𝒙, 𝒙 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒙 And it we further allow 𝒛 → 𝒙, we find that, 𝒖 𝒙, 𝒙 = 𝒖 𝒙, 𝒙 + 𝒖 𝒙, 𝒙 = 2𝒖 𝒙, 𝒙 Which clearly shows that 𝒖 𝒙, 𝒙 = 𝒐 the zero vector. Similarly, from the above, we find that, 𝒖 𝒙, 𝒙 = 𝒐 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒙 So that 𝒖 𝒙, 𝒛 = −𝒖 𝒛, 𝒙Dept of Systems Engineering, University of Lagos 38 oafak@unilag.edu.ng 12/27/2012
- 39. Dept of Systems Engineering, University of Lagos 39 oafak@unilag.edu.ng 12/27/2012
- 40. Coordinate SystemDept of Systems Engineering, University of Lagos 40 oafak@unilag.edu.ng 12/27/2012
- 41. Coordinate SystemDept of Systems Engineering, University of Lagos 41 oafak@unilag.edu.ng 12/27/2012
- 42. Position Vectors Note that ℰ is NOT a vector space. In our discussion, the vector space E to which 𝒖 belongs, is associated as we have shown. It is customary to oscillate between these two spaces. When we are talking about the vectors, they are in E while the points are in ℰ. We adopt the convention that 𝒙 𝒚 ≡ 𝒖 𝒙, 𝒚 referring to the vector 𝒖. If therefore we choose an arbitrarily fixed point 𝟎 ∈ ℰ, we are associating 𝒙 𝟎 , 𝒚 𝟎 and 𝒛 𝟎 respectively with 𝒖 𝒙, 𝟎 , 𝒖 𝒚, 𝟎 and 𝒖 𝒛, 𝟎 . These are vectors based on the points 𝒙, 𝒚 and 𝒛 with reference to the origin chosen. To emphasize the association with both points of ℰ as well as members of E they are called Position Vectors.Dept of Systems Engineering, University of Lagos 42 oafak@unilag.edu.ng 12/27/2012
- 43. Length in the Point Space Recall that by property 1, 𝒙 𝒚 ≡ 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝟎 + 𝒖 𝟎, 𝒚 Furthermore, we have deduced that 𝒖 𝟎, 𝒚 = −𝒖(𝒚, 𝟎) We may therefore write that, 𝒙 𝒚 = 𝒙 𝟎 − 𝒚(𝟎) Which, when there is no ambiguity concerning the chosen origin, we can write as, 𝒙 𝒚 = 𝒙− 𝒚 And the distance between the two is, 𝒅 𝒙 𝒚 = 𝒅 𝒙− 𝒚 = 𝒙− 𝒚Dept of Systems Engineering, University of Lagos 43 oafak@unilag.edu.ng 12/27/2012
- 44. Metric Properties The norm of a vector is the square root of the inner product of the vector with itself. If the coordinates of 𝒙 and 𝒚 on a set of independent vectors are 𝑥 𝑖 and 𝑦 𝑖 , then the distance we seek is, 𝒅 𝒙− 𝒚 = 𝒙− 𝒚 = 𝑔 𝑖𝑗 (𝑥 𝑖 − 𝑦 𝑖 )(𝑥 𝑗 − 𝑦 𝑗 ) The more familiar Pythagorean form occurring only when 𝑔 𝑖𝑗 = 1.Dept of Systems Engineering, University of Lagos 44 oafak@unilag.edu.ng 12/27/2012
- 45. Coordinate Transformations Consider a Cartesian coordinate system 𝑥, 𝑦, 𝑧 or 𝑦1 , 𝑦 2 and 𝑦 3 with an orthogonal basis. Let us now have the possibility of transforming to another coordinate system of an arbitrary nature: 𝑥 1 , 𝑥 2 , 𝑥 3 . We can represent the transformation and its inverse in the equations: 𝑦 𝑖 = 𝑦 𝑖 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑥 𝑖 = 𝑥 𝑖 (𝑦1 , 𝑦 2 , 𝑦 3 ) And if the Jacobian of transformation,Dept of Systems Engineering, University of Lagos 45 oafak@unilag.edu.ng 12/27/2012
- 46. Jacobian Determinant 𝜕𝑦1 𝜕𝑦1 𝜕𝑦1 𝜕𝑥 1 𝜕𝑥 2 𝜕𝑥 3 𝜕𝑦 𝑖 𝜕 𝑦1 , 𝑦 2 , 𝑦 3 𝜕𝑦 2 𝜕𝑦 2 𝜕𝑦 2 𝑗 ≡ 1, 𝑥2, 𝑥3 ≡ 𝜕𝑥 𝜕 𝑥 𝜕𝑥 1 𝜕𝑥 2 𝜕𝑥 3 𝜕𝑦 3 𝜕𝑦 3 𝜕𝑦 3 𝜕𝑥 1 𝜕𝑥 2 𝜕𝑥 3 does not vanish, then the inverse transformation will exist. So that, 𝑥 𝑖 = 𝑥 𝑖 (𝑦1 , 𝑦 2 , 𝑦 3 )Dept of Systems Engineering, University of Lagos 46 oafak@unilag.edu.ng 12/27/2012
- 47. Given a position vector 𝐫 = 𝐫(𝑥 1 , 𝑥 2 , 𝑥 3 ) In the new coordinate system, we can form a set of three vectors, 𝜕𝐫 𝐠𝑖 = 𝑖 , 𝑖 = 1,2,3 𝜕𝑥 Which represent vectors along the tangents to the coordinate lines. (This is easily established for the Cartesian system and it is true in all systems. 𝐫 = 𝐫 𝑥 1 , 𝑥 2 , 𝑥 3 = 𝑦1 𝐢 + 𝑦2 𝐣 + 𝑦3 𝐤 So that 𝜕𝐫 𝜕𝐫 𝜕𝐫 𝐢 = 1 , 𝐣 = 2 and 𝐤 = 3 𝜕𝑦 𝜕𝑦 𝜕𝑦 The fact that this is true in the general case is easily seen when we consider that along those lines, only the variable we are differentiating with respect to varies. Consider the general case where, 𝐫 = 𝐫 𝑥1 , 𝑥 2 , 𝑥 3 The total differential of 𝐫 is simply,Dept of Systems Engineering, University of Lagos 47 oafak@unilag.edu.ng 12/27/2012
- 48. Natural Dual Bases 𝜕𝐫 1+ 𝜕𝐫 2+ 𝜕𝐫 𝑑𝐫 = 1 𝑑𝑥 𝑑𝑥 𝑑𝑥 3 𝜕𝑥 𝜕𝑥 2 𝜕𝑥 3 ≡ 𝑑𝑥 1 𝐠1 + 𝑑𝑥 2 𝐠 2 + 𝑑𝑥 3 𝐠 3 With 𝐠 𝑖 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑖 = 1,2,3 now depending in general on 𝑥 1 , 𝑥 2 and 𝑥 3 now forming a basis on which we can describe other vectors in the coordinate system. We have no guarantees that this vectors are unit in length nor that they are orthogonal to one another. In the Cartesian case, 𝐠 𝑖 , 𝑖 = 1,2,3 are constants, normalized and orthogonal. They are our familiar 𝜕𝐫 𝜕𝐫 𝜕𝐫 𝐢 = 1 , 𝐣 = 2 and 𝐤 = 3 . 𝜕𝑦 𝜕𝑦 𝜕𝑦Dept of Systems Engineering, University of Lagos 48 oafak@unilag.edu.ng 12/27/2012
- 49. We now proceed to show that the set of basis vectors, 𝐠 𝑖 ≡ 𝛻𝑥 𝑖 is reciprocal to 𝐠 𝑖 . The total differential 𝑑𝐫 can be written in terms of partial differentials as, 𝜕𝐫 𝑑𝐫 = 𝑑𝑦 𝑖 𝜕𝑦 𝑖 Which when expressed in component form yields, 𝜕𝑥 1 1 𝜕𝑥 1 2 𝜕𝑥 1 3 𝑑𝑥 1 = 1 𝑑𝑦 + 2 𝑑𝑦 + 3 𝑑𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑥 2 1 𝜕𝑥 2 2 𝜕𝑥 2 3 𝑑𝑥 2 = 1 𝑑𝑦 + 2 𝑑𝑦 + 3 𝑑𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑥 3 1 𝜕𝑥 3 2 𝜕𝑥 3 3 𝑑𝑥 3 = 1 𝑑𝑦 + 2 𝑑𝑦 + 3 𝑑𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦 Or, more compactly that, 𝑗 = 𝜕𝑥 𝑗 𝑑𝑥 𝑑𝑦 𝑖 = 𝛻𝑥 𝑗 ⋅ 𝑑𝐫 𝜕𝑦 𝑖Dept of Systems Engineering, University of Lagos 49 oafak@unilag.edu.ng 12/27/2012
- 50. In Cartesian coordinates, 𝑦 𝑖 , 𝑖 = 1, . . , 3 for any scalar field 𝜙 (𝑦1 , 𝑦 2 , 𝑦 3 ), 𝜕𝜙 𝑖 = 𝜕𝜙 𝜕𝜙 𝜕𝜙 𝑑𝑦 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 = grad 𝜙 ⋅ 𝑑𝐫 𝜕𝑦 𝑖 𝜕𝑥 𝜕𝑦 𝜕𝑧 We are treating each curvilinear coordinate as a scalar field, 𝑥 𝑗 = 𝑥 𝑗 (𝑦1 , 𝑦 2 , 𝑦 3 ). 𝑗 = 𝛻𝑥 𝑗 ⋅ 𝜕𝐫 𝑑𝑥 𝑑𝑥 𝑚 𝜕𝑥 𝑚 = 𝐠 𝑗 ⋅ 𝐠 𝑚 𝑑𝑥 𝑚 𝑗 = 𝛿 𝑚 𝑑𝑥 𝑚 . The last equality arises from the fact that this is the only way one component of the coordinate differential can equal another. 𝑗 ⇒ 𝐠𝑗 ⋅ 𝐠 𝑚 = 𝛿 𝑚 Which recovers for us the reciprocity relationship and shows that 𝐠 𝑗 and 𝐠 𝑚 are reciprocal systems.Dept of Systems Engineering, University of Lagos 50 oafak@unilag.edu.ng 12/27/2012
- 51. The position vector in Cartesian coordinates is 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 ⊗ 𝑥𝑗 𝒆𝑗 = 𝑥 𝑖 𝑥𝑗 𝒆 𝑖 ⊗ 𝒆 𝒋 grad 𝒓 ⊗ 𝒓 = 𝑥 𝑖 𝑥 𝑗 , 𝑘 𝒆 𝑖 ⊗ 𝒆 𝒋 ⊗ 𝒆 𝒌 div 𝒓 ⊗ 𝒓 = 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 ×Dept of Systems Engineering, University of Lagos 51 oafak@unilag.edu.ng 12/27/2012
- 52. 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 𝐓Dept of Systems Engineering, University of Lagos 52 oafak@unilag.edu.ng 12/27/2012
- 53. 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 × grad𝒗Dept of Systems Engineering, University of Lagos 53 oafak@unilag.edu.ng 12/27/2012
- 54. For any two tensor fields 𝐮 and 𝐯, Show that, 𝐮 × : grad 𝐯 = 𝐮 ⋅ curl 𝐯 𝐮 × : grad 𝐯 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝐠 𝑖 ⊗ 𝐠 𝑘 : 𝑣 𝛼 , 𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝛼 , 𝛽 𝐠 𝑖 ⋅ 𝐠 𝛼 𝐠 𝑘 ⋅ 𝐠 𝛽 𝛽 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝛼 , 𝛽 𝛿 𝑖 𝛼 𝛿 𝑘 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑖 , 𝑘 = 𝐮 ⋅ curl 𝐯Dept of Systems Engineering, University of Lagos 54 oafak@unilag.edu.ng 12/27/2012
- 55. 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 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑙 𝐠 𝑖 𝛿 𝑙𝑘 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑘 𝐠 𝑖 = −curl 𝒖Dept of Systems Engineering, University of Lagos 55 oafak@unilag.edu.ng 12/27/2012
- 56. Show that 𝐜𝐮𝐫𝐥 𝜙𝟏 = 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 × Show that curl 𝒗 × = div 𝒗 𝟏 − grad 𝒗 𝒗 × = 𝜖 𝛼𝛽𝑘 𝑣 𝛽 𝐠 𝛼 ⊗ 𝐠 𝑘 curl 𝑻 = 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼 so that 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 𝟏 − grad 𝒗Dept of Systems Engineering, University of Lagos 56 oafak@unilag.edu.ng 12/27/2012
- 57. 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 Given a scalar point function ϕ and a vector field 𝐯, show that 𝐜𝐮𝐫𝐥 𝜙𝐯 = 𝜙 curl 𝐯 + grad𝜙 × 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 = grad𝜙 × 𝐯 + 𝜙 curl 𝐯Dept of Systems Engineering, University of Lagos 57 oafak@unilag.edu.ng 12/27/2012
- 58. 𝐀 Given that 𝜑 𝑡 = 𝐀(𝑡) , Show that 𝜑 𝑡 = : 𝐀 𝐀 𝑡 𝜑2 ≡ 𝐀: 𝐀 Now, 𝑑 2 = 2𝜑 𝑑𝜑 𝑑𝐀 𝑑𝐀 𝑑𝐀 𝜑 = : 𝐀 + 𝐀: = 2𝐀: 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 as inner product is commutative. We can therefore write that 𝑑𝜑 𝐀 𝑑𝐀 𝐀 = : = : 𝐀 𝑑𝑡 𝜑 𝑑𝑡 𝐀 𝑡 as required.Dept of Systems Engineering, University of Lagos 58 oafak@unilag.edu.ng 12/27/2012
- 59. Given a tensor field 𝑻, obtain the vector 𝒘 ≡ 𝑻 𝐓 𝐯 and show that its divergence is 𝑻: 𝛁𝐯 + 𝐯 ⋅ 𝐝𝐢𝐯 𝑻 The divergence of 𝒘 is the scalar sum , 𝑇𝑗𝑖 𝑣 𝑗 , 𝑖 . Expanding the product covariant derivative we obtain, div 𝑻T 𝐯 = 𝑇𝑗𝑖 𝑣 𝑗 , 𝑖 = 𝑇𝑗𝑖 , 𝑖 𝑣 𝑗 + 𝑇𝑗𝑖 𝑣 𝑗 , 𝑖 = div 𝑻 ⋅ 𝐯 + tr 𝑻T grad𝐯 = div 𝑻 ⋅ 𝐯 + 𝑻: grad 𝐯 Recall that scalar product of two vectors is commutative so that div 𝑻T 𝐯 = 𝑻: grad𝐯 + 𝐯 ⋅ div 𝑻Dept of Systems Engineering, University of Lagos 59 oafak@unilag.edu.ng 12/27/2012
- 60. 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 contravariant 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 𝛽𝛼 = 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 𝒂Dept of Systems Engineering, University of Lagos 60 oafak@unilag.edu.ng 12/27/2012
- 61. 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 = 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.Dept of Systems Engineering, University of Lagos 61 oafak@unilag.edu.ng 12/27/2012
- 62. 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 𝑗 𝑖 𝑗 = 𝛿 𝑖𝛼 𝛿 𝛽 − 𝛿 𝛽 𝛿 𝛼 𝑢𝛼𝑣 𝛽 ,𝑗 𝐠𝑖 𝑗 𝑖 𝑗 = 𝛿 𝑖𝛼 𝛿 𝛽 − 𝛿 𝛽 𝛿 𝛼 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 𝐠𝑖 = 𝑢𝑖 𝑣 𝑗 ,𝑗 − 𝑢 𝑗 𝑣𝑖 ,𝑗 𝐠𝑖 = div 𝒖 ⊗ 𝒗 − 𝒗 ⊗ 𝒖 since div 𝒖 ⊗ 𝒗 = 𝑢𝑖 𝑣 𝑗 , 𝛼 𝐠𝑖 ⊗ 𝐠 𝑗 ⋅ 𝐠 𝛼 = 𝑢 𝑖 𝑣 𝑗 , 𝑗 𝐠 𝑖.Dept of Systems Engineering, University of Lagos 62 oafak@unilag.edu.ng 12/27/2012
- 63. Given a scalar point function 𝜙 and a second-order tensor field 𝑻,show that 𝐜𝐮𝐫𝐥 𝜙𝑻 = 𝜙 𝐜𝐮𝐫𝐥 𝑻 + grad𝜙 × 𝑻 𝑻 where grad𝜙 × is the skew tensor 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝐠 𝒊 ⊗ 𝐠 𝒌curl 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 + grad𝜙 × 𝑻 𝑇 Dept of Systems Engineering, University of Lagos 63 oafak@unilag.edu.ng 12/27/2012
- 64. 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.𝛼 𝐠 𝑖 ⊗ 𝐠 𝛼 𝑖 The gradient of 𝑺 is 𝑆.𝛼 , 𝛽 𝐠 𝑖 ⊗ 𝐠 𝛼 ⊗ 𝐠 𝛽 = 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗𝛽 𝐠 𝑖 ⊗ 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑖 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 𝑻TDept of Systems Engineering, University of Lagos 64 oafak@unilag.edu.ng 12/27/2012
- 65. 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 𝟏 element, show that 𝑺 𝛁 𝒓 ⋅ 𝒓 ⋅ 𝒏𝒅𝑺 equals the volume 𝟔 contained in 𝑽. 𝒓 ⋅ 𝒓 = 𝑥 𝑖 𝑥 𝑗 𝐠 𝑖 ⋅ 𝐠 𝑗 = 𝑥 𝑖 𝑥 𝑗 𝑔 𝑖𝑗 By the Divergence Theorem, 𝛻 𝒓 ⋅ 𝒓 ⋅ 𝒏𝑑𝑆 = 𝛻⋅ 𝛻 𝒓⋅ 𝒓 𝑑𝑉 𝑆 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 𝑗 = 𝑔 𝑖𝑗 𝑔 𝑙𝑘 𝛿 𝑖𝑘 𝑥 𝑗 + 𝑥 𝑖 𝛿 𝑘 , 𝑙 𝑑𝑉 = 2𝑔 𝑖𝑘 𝑔 𝑙𝑘 𝑥 𝑖 , 𝑙 𝑑𝑉 𝑉 𝑉 = 2𝛿 𝑖𝑙 𝛿 𝑙𝑖 𝑑𝑉 = 6 𝑑𝑉 𝑉 𝑉Dept of Systems Engineering, University of Lagos 65 oafak@unilag.edu.ng 12/27/2012
- 66. 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 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗,𝑖 𝑣 𝑘 + 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘,𝑖 ,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 = 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 ,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.Dept of Systems Engineering, University of Lagos 66 oafak@unilag.edu.ng 12/27/2012
- 67. Liouville Formula 𝑑𝐓 𝑑 For a scalar variable 𝛼, if the tensor 𝐓 = 𝐓(𝛼) and 𝐓 ≡ 𝑑𝛼, Show that 𝑑𝛼 det 𝐓 = det 𝐓 tr(𝐓 𝐓 −𝟏 ) Proof: Let 𝐀 ≡ 𝐓 𝐓 −1 so that, 𝐓 = 𝐀𝐓. In component form, we have 𝑇𝑗 𝑖 = 𝐴 𝑖 𝑚 𝑇𝑗 𝑚 . Therefore, 𝑑 𝑑 det 𝐓 = 𝜖 𝑖𝑗𝑘 𝑇𝑖1 𝑇𝑗2 𝑇 3 = 𝜖 𝑖𝑗𝑘 𝑇𝑖1 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝑇 3 𝑘 𝑘 𝑘 𝑘 𝑑𝛼 𝑑𝛼 = 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖 𝑙 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝐴2𝑚 𝑇𝑗 𝑚 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝐴3𝑛 𝑇 𝑘𝑛 𝑙 𝑘 𝑘 = 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖1 + 𝐴1 𝑇𝑖2 + 𝐴1 𝑇𝑖3 1 2 3 𝑇𝑗2 𝑇 3 𝑘 + 𝑇𝑖1 𝐴1 𝑇𝑗1 + 𝐴2 𝑇𝑗2 + 𝐴2 𝑇𝑗3 2 2 3 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝑘 𝐴1 𝑇 1 + 𝐴3 𝑇 2 + 𝐴3 𝑇 3 3 𝑘 2 𝑘 3 𝑘 All the boxed terms in the above equation vanish on account of the contraction of a symmetric tensor with an antisymmetric one.Dept of Systems Engineering, University of Lagos 67 oafak@unilag.edu.ng 12/27/2012
- 68. Liouville Theorem Cont’d (For example, the first boxed term yields, 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖2 𝑇𝑗2 𝑇 3 2 𝑘 Which is symmetric as well as antisymmetric in 𝑖 and 𝑗. It therefore vanishes. The same is true for all other such terms.) 𝑑 det 𝐓 = 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖1 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝐴2 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝐴3 𝑇 3 1 𝑘 2 𝑘 3 𝑘 𝑑𝛼 = 𝐴 𝑚 𝜖 𝑖𝑗𝑘 𝑇𝑖1 𝑇𝑗2 𝑇 3 𝑚 𝑘 = tr 𝐓 𝐓 −1 det 𝐓 as required.Dept of Systems Engineering, University of Lagos 68 oafak@unilag.edu.ng 12/27/2012
- 69. For a general tensor field 𝑻 show that, 𝐜𝐮𝐫𝐥 𝐜𝐮𝐫𝐥 𝑻 = 𝟐 𝐭𝐫 𝑻 − 𝐝𝐢𝐯 𝐝𝐢𝐯 𝑻 𝑰 + 𝐠𝐫𝐚𝐝 𝐝𝐢𝐯 𝑻 + 𝐓 𝛁 𝐠𝐫𝐚𝐝 𝐝𝐢𝐯 𝑻 − 𝐠𝐫𝐚𝐝 𝐠𝐫𝐚𝐝 𝐭𝐫𝑻 − 𝛁 𝟐 𝑻 𝐓 curl 𝑻 = 𝝐 𝛼𝑠𝑡 𝑇 𝛽𝑡 , 𝑠 𝐠 𝛼 ⊗ 𝐠 𝛽 𝛼 = 𝑆 .𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽 𝛼 curl 𝑺 = 𝜖 𝑖𝑗𝑘 𝑆 .𝑘 , 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 𝑔 𝑖𝛼 𝑔 𝑗𝑠 𝑔 𝑘𝑡 − 𝑔 𝑗𝑡 𝑔 𝑘𝑠 + 𝑔 𝑖𝑠 𝑔 𝑗𝑡 𝑔 𝑘𝛼 − 𝑔 𝑗𝛼 𝑔 𝑘𝑡 = 𝑖𝑡 𝑗𝛼 𝑘𝑠 𝑗𝑠 𝑘𝛼 𝑇 𝑘𝑡 , 𝑠𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼 +𝑔 𝑔 𝑔 − 𝑔 𝑔 𝑡 𝑡 = 𝑔 𝑗𝑠 𝑇 .𝑡 , 𝑠𝑗 −𝑇..𝑠𝑗 , 𝑠𝑗 𝐠 𝛼 ⊗ 𝐠 𝛼 + 𝑇 ..𝛼𝑗 , 𝑠𝑗 −𝑔 𝑗𝛼 𝑇 .𝑡 , 𝑠𝑗 𝐠 𝑠 ⊗ 𝐠 𝛼 𝑠. 𝛼. + 𝑔 𝑗𝛼 𝑇 .𝑡 , 𝑠𝑗 −𝑔 𝑗𝑠 𝑇 .𝑡 , 𝑠𝑗 𝐠 𝑡 ⊗ 𝐠 𝛼 T = 𝛻 2 tr 𝑻 − div div 𝑻 𝑰 + grad div 𝑻 − grad grad tr𝑻 + grad div 𝑻 − 𝛻 2 𝑻TDept of Systems Engineering, University of Lagos 69 oafak@unilag.edu.ng 12/27/2012
- 70. 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 𝛽 = 𝜖 𝑖𝑗𝑘 𝑇 𝛽𝑘 , 𝑗 𝛿 𝑖 = 𝜖 𝑖𝑗𝑘 𝑇𝑖𝑘 , 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 𝑰 − grad grad tr𝑻 + 2 grad div 𝑻 − 𝛻 2 𝑻 T as grad div 𝑻 = grad div 𝑻 if the order of differentiation is immaterial and T is symmetric.Dept of Systems Engineering, University of Lagos 70 oafak@unilag.edu.ng 12/27/2012
- 71. For a scalar function Φ and a vector 𝑣 𝑖 show that the divergence of the vector 𝑣 𝑖 Φ is equal to, 𝐯 ⋅ gradΦ + Φ div 𝐯 𝑣 𝑖 Φ ,𝑖 = Φ𝑣 𝑖 ,𝑖 + 𝑣 𝑖 Φ,i Hence the result.Dept of Systems Engineering, University of Lagos 71 oafak@unilag.edu.ng 12/27/2012
- 72. 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, 𝑙𝑚𝑖 𝜖 𝑙𝑚𝑖 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 = 𝛿 𝑗𝑘𝑖 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚 ,𝑚 𝑙𝑚 𝛿 𝑗𝑙 𝛿𝑗 𝑚 = 𝛿 𝑗𝑘 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚 = 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚 𝛿 𝑙𝑘 𝛿 𝑘𝑚 = 𝛿 𝑗𝑙 𝛿 𝑘𝑚 − 𝛿 𝑙𝑘 𝛿 𝑗 𝑚 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚 = 𝛿 𝑗𝑙 𝛿 𝑘𝑚 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝛿 𝑗𝑙 𝛿 𝑘𝑚 𝑢 𝑗 𝑣 𝑘 ,𝑚 + 𝛿 𝑙𝑘 𝛿 𝑗 𝑚 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝛿 𝑙𝑘 𝛿 𝑗 𝑚 𝑢 𝑗 𝑣 𝑘 ,𝑚 = 𝑢 𝑙 ,𝑚 𝑣 𝑚 − 𝑢 𝑚 ,𝑚 𝑣 𝑙 + 𝑢 𝑙 𝑣 𝑚 ,𝑚 − 𝑢 𝑚 𝑣 𝑙 ,𝑚 Which is the result we seek in indicial notation.Dept of Systems Engineering, University of Lagos 72 oafak@unilag.edu.ng 12/27/2012
- 73. Differentiation of Fields Given a vector point function 𝒖 𝑥 1 , 𝑥 2 , 𝑥 3 and its covariant components, 𝑢 𝛼 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝛼 = 1,2,3, with 𝐠 𝛼 as the reciprocal basis vectors, Let 𝜕 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 𝒖 = 𝑢 𝛼 𝐠 𝛼 , then 𝑑𝒖 = 𝑢 𝛼 𝐠 𝛼 𝑑𝑥 𝑘 = 𝐠 + 𝑢 𝛼 𝑑𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 Clearly, 𝜕𝒖 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 𝑘 = 𝑘 𝐠 + 𝑢 𝑘 𝛼 𝜕𝑥 𝜕𝑥 𝜕𝑥 And the projection of this quantity on the 𝐠 𝑖 direction is, 𝜕𝒖 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 ∙ 𝐠 = 𝐠 + 𝑢 ∙ 𝐠𝑖 = 𝐠 ∙ 𝐠𝑖 + ∙ 𝐠 𝑢 𝜕𝑥 𝑘 𝑖 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝛼 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑖 𝛼 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 = 𝛿 + 𝑘 𝑖 𝑘 ∙ 𝐠𝑖 𝑢 𝛼 𝜕𝑥 𝜕𝑥Dept of Systems Engineering, University of Lagos 73 oafak@unilag.edu.ng 12/27/2012
- 74. Christoffel Symbols 𝑗 Now, 𝐠 𝑖 ∙ 𝐠 𝑗 = 𝛿 𝑖 so that 𝜕𝐠 𝑖 𝑖∙ 𝜕𝐠 𝑗 𝑘 ∙ 𝐠𝑗 + 𝐠 𝑘 = 0. 𝜕𝑥 𝜕𝑥 𝜕𝐠 𝑖 𝑖 𝜕𝐠 𝑗 𝑖 ∙ 𝐠 𝑗 = −𝐠 ∙ ≡− . 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑗𝑘 This important quantity, necessary to quantify the derivative of a tensor in general coordinates, is called the Christoffel Symbol of the second kind.Dept of Systems Engineering, University of Lagos 74 oafak@unilag.edu.ng 12/27/2012
- 75. Derivatives in General Coordinates Using this, we can now write that, 𝜕𝒖 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 𝜕𝑢 𝑖 𝛼 ∙ 𝐠𝑖 = 𝛿𝑖 + ∙ 𝐠𝑖 𝑢 𝛼 = − 𝑢 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑖𝑘 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 the covariant derivative of 𝒖. Using contravariant components, we could write, 𝜕𝑢 𝑖 𝜕𝐠 𝑖 𝑖 𝜕𝑢 𝑖 𝛼 𝑑𝒖 = 𝐠𝑖 + 𝑢 𝑑𝑥 𝑘 = 𝐠𝑖 + 𝐠 𝑢 𝑖 𝑑𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑖𝑘 𝛼 So that, 𝜕𝒖 𝜕𝑢 𝛼 𝜕𝐠 𝛼 𝛼 = 𝐠𝛼+ 𝑢 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 The components of this in the direction of 𝐠 𝑖 can be obtained by taking a dot product as before: 𝜕𝒖 𝑖 = 𝜕𝑢 𝛼 𝜕𝐠 𝛼 𝛼 𝑖 = 𝜕𝑢 𝛼 𝑖 𝜕𝐠 𝛼 𝑖 𝑢𝛼 = 𝜕𝑢 𝑖 𝑖 ∙ 𝐠 𝐠𝛼+ 𝑢 ∙ 𝐠 𝛿𝛼 + ∙ 𝐠 + 𝑢𝛼 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝛼𝑘Dept of Systems Engineering, University of Lagos 75 oafak@unilag.edu.ng 12/27/2012
- 76. Covariant Derivatives The two results above are represented symbolically as, 𝜕𝑢 𝑖 𝛼 𝑖 𝜕𝑢 𝛼 𝑖 𝑢 𝑖,𝑘 = − 𝑢 𝛼 and 𝑢,𝑘 = + 𝑢𝛼 𝜕𝑥 𝑘 𝑖𝑘 𝜕𝑥 𝑘 𝛼𝑘 Which are the components of the covariant derivatives in terms of the covariant and contravariant components respectively. It now becomes useful to establish the fact that our definition of the Christoffel symbols here conforms to the definition you find in the books using the transformation rules to define the tensor quantities.Dept of Systems Engineering, University of Lagos 76 oafak@unilag.edu.ng 12/27/2012
- 77. Christoffel Symbols 𝜕𝐫 We observe that the derivative of the covariant basis, 𝐠 𝑖 = 𝑖 , 𝜕𝑥 𝜕𝐠 𝑖 𝜕 2𝐫 𝜕 2𝐫 𝜕𝐠 𝑗 = = = 𝜕𝑥 𝑗 𝜕𝑥 𝑗 𝜕𝑥 𝑖 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝜕𝑥 𝑖 Taking the dot product with 𝐠 𝑘 , 𝜕𝐠 𝑖 1 𝜕𝐠 𝑗 𝜕𝐠 𝑖 𝑗 ∙ 𝐠𝑘= 𝑖 ∙ 𝐠𝑘+ 𝑗 ∙ 𝐠𝑘 𝜕𝑥 2 𝜕𝑥 𝜕𝑥 1 𝜕 𝜕 𝜕𝐠 𝑘 𝜕𝐠 𝑘 = 𝐠 ∙ 𝐠𝑘 + 𝐠 ∙ 𝐠𝑘 − 𝐠𝑗 ∙ − 𝐠𝑖 ∙ 2 𝜕𝑥 𝑖 𝑗 𝜕𝑥 𝑗 𝑖 𝜕𝑥 𝑖 𝜕𝑥 𝑗 1 𝜕 𝜕 𝜕𝐠 𝑖 𝜕𝐠 𝑗 = 𝐠𝑗 ∙ 𝐠𝑘 + 𝐠𝑖 ∙ 𝐠 𝑘 − 𝐠𝑗 ∙ 𝑘 − 𝐠𝑖 ∙ 2 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝜕𝑥 𝜕𝑥 𝑘 1 𝜕 𝜕 𝜕 = 𝐠 ∙ 𝐠𝑘 + 𝐠 ∙ 𝐠𝑘 − 𝐠 ∙ 𝐠𝑗 2 𝜕𝑥 𝑖 𝑗 𝜕𝑥 𝑗 𝑖 𝜕𝑥 𝑘 𝑖 1 𝜕𝑔 𝑗𝑘 𝜕𝑔 𝑘𝑖 𝜕𝑔 𝑖𝑗 = + −Dept of Systems Engineering, 𝜕𝑥 𝑖 2 University of Lagos 𝜕𝑥 𝑗 𝜕𝑥 𝑘 77 oafak@unilag.edu.ng 12/27/2012
- 78. The First Kind Which is the quantity defined as the Christoffel symbols of the first kind in the textbooks. It is therefore possible for us to write, 𝜕𝐠 𝑖 𝜕𝐠 𝑗 1 𝜕𝑔 𝑗𝑘 𝜕𝑔 𝑘𝑖 𝜕𝑔 𝑖𝑗 𝑖𝑗, 𝑘 ≡ ∙ 𝐠𝑘 = ∙ 𝐠𝑘 = + − 𝜕𝑥 𝑗 𝜕𝑥 𝑖 2 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝜕𝑥 𝑘 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:Dept of Systems Engineering, University of Lagos 78 oafak@unilag.edu.ng 12/27/2012
- 79. The Second Kind Contract the above equation with the conjugate metric tensor, we have, 𝑘𝛼 𝑘𝛼 𝜕𝐠 𝑖 𝑘𝛼 𝜕𝐠 𝑗 𝜕𝐠 𝑖 𝑘 𝑔 𝑖𝑗, 𝛼 ≡ 𝑔 ∙ 𝐠𝛼= 𝑔 ∙ 𝐠𝛼= ∙ 𝐠𝑘= 𝜕𝑥 𝑗 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝑖𝑗 𝜕𝐠 𝑘 =− 𝑗 ∙ 𝐠𝑖 𝜕𝑥 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.Dept of Systems Engineering, University of Lagos 79 oafak@unilag.edu.ng 12/27/2012
- 80. Two Christoffel Symbols Related We contract the above equation with 𝑔 𝑘𝛽 and obtain, 𝑘𝛼 𝑖𝑗, 𝛼 = 𝑔 𝑘 𝑔 𝑘𝛽 𝑔 𝑘𝛽 𝑖𝑗 𝛼 𝑘 𝛿 𝛽 𝑖𝑗, 𝛼 = 𝑖𝑗, 𝛽 = 𝑔 𝑘𝛽 𝑖𝑗 so that, 𝛼 𝑔 𝑘𝛼 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.Dept of Systems Engineering, University of Lagos 80 oafak@unilag.edu.ng 12/27/2012
- 81. Higher Order Tensors We are now in a position to express the derivatives of higher order tensor fields in terms of the Christoffel symbols. For a second-order tensor 𝐓, we can express the components in dyadic form along the product basis as follows: 𝐓 = 𝑇𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝑇 𝑖𝑗 𝐠 𝑖 ⨂𝐠 𝑗 = 𝑇.𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝑇𝑗.𝑖 𝐠 𝑗 ⨂𝐠 𝑖 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. As an example, take the product covariant bases.Dept of Systems Engineering, University of Lagos 81 oafak@unilag.edu.ng 12/27/2012
- 82. Higher Order Tensors We have: 𝜕𝐓 𝜕𝑇 𝑖𝑗 𝑖𝑗 𝜕𝐠 𝑖 𝑖𝑗 𝜕𝐠 𝑗 𝑘 = 𝑘 𝑖 𝐠 ⨂𝐠 𝑗 + 𝑇 𝑘 ⨂𝐠 𝑗 + 𝑇 𝐠 𝑖 ⨂ 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑘 𝜕𝐠 𝑗 Recall that, 𝑖𝑘 ∙ 𝐠 𝑗 = . It follows therefore that, 𝜕𝑥 𝑖𝑘 𝜕𝐠 𝑖 𝑗− 𝛼 𝛿𝑗 = 𝜕𝐠 𝑖 𝛼 ∙ 𝐠 ∙ 𝐠𝑗 − 𝐠 𝛼 ∙ 𝐠𝑗 𝜕𝑥 𝑘 𝑖𝑘 𝛼 𝜕𝑥 𝑘 𝑖𝑘 𝜕𝐠 𝑖 𝛼 = − 𝐠 𝛼 ∙ 𝐠 𝑗 = 0. 𝜕𝑥 𝑘 𝑖𝑘 𝜕𝐠 𝑖 𝛼 Clearly, 𝑘 = 𝐠𝛼 𝜕𝑥 𝑖𝑘 (Obviously since 𝐠 𝑗 is a basis vector it cannot vanish)Dept of Systems Engineering, University of Lagos 82 oafak@unilag.edu.ng 12/27/2012
- 83. Higher Order Tensors 𝜕𝐓 𝜕𝑇 𝑖𝑗 𝑖𝑗 𝜕𝐠 𝑖 𝑖𝑗 𝐠 ⨂ 𝜕𝐠 𝑗 𝜕𝑥 𝑘 = 𝜕𝑥 𝑘 𝐠 𝑖 ⨂𝐠 𝑗 + 𝑇 𝜕𝑥 𝑘 ⨂𝐠 𝑗 + 𝑇 𝑖 𝜕𝑥 𝑘 𝜕𝑇 𝑖𝑗 𝑖𝑗 𝛼 𝑖𝑗 𝐠 ⨂ 𝛼 = 𝐠 ⨂𝐠 𝑗 + 𝑇 𝐠 ⨂𝐠 𝑗 + 𝑇 𝑖 𝑗𝑘 𝐠 𝛼 𝜕𝑥 𝑘 𝑖 𝑖𝑘 𝛼 𝜕𝑇 𝑖𝑗 𝑖 𝑗 = 𝑘 𝐠 𝑖 ⨂𝐠 𝑗 + 𝑇 𝛼𝑗 𝐠 𝑖 ⨂𝐠 𝑗 + 𝑇 𝑖𝛼 𝐠 𝑖 ⨂ 𝐠𝑗 𝜕𝑥 𝛼𝑘 𝛼𝑘 𝜕𝑇 𝑖𝑗 𝑖 𝑗 𝑖𝑗 = +𝑇 𝛼𝑗 + 𝑇 𝑖𝛼 𝐠 𝑖 ⨂𝐠 𝑗 = 𝑇 ,𝑘 𝐠 𝑖 ⨂𝐠 𝑗 𝜕𝑥 𝑘 𝛼𝑘 𝛼𝑘 Where 𝜕𝑇 𝑖𝑗 𝑖𝑗 𝑖 𝑗 𝜕𝑇 𝑖𝑗 𝑖 𝑗 = 𝑇,𝑘 +𝑇 𝛼𝑗 + 𝑇 𝑖𝛼 or +𝑇 𝛼𝑗 Γ 𝛼𝑘 + 𝑇 𝑖𝛼 Γ 𝛼𝑘 𝜕𝑥 𝑘 𝛼𝑘 𝛼𝑘 𝜕𝑥 𝑘 are the components of the covariant derivative of the tensor 𝐓 in terms of contravariant components on the product covariant bases as shown.Dept of Systems Engineering, University of Lagos 83 oafak@unilag.edu.ng 12/27/2012

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