new tensor stuff 10_12

New Tensor Stuff PhL 10.2.12 After several hours, I finally finished reviewing this document 10.19.12. Notice that somehow 17 days have snuck by since I first started it! While reading about the Ft type tensors in Lai and elsewhere, I have found a "hole" in my tensor doc, and that hole is the element I need to understand the Ft world. A related subject is "the meaning of the word tensor". I think maybe R and S are tensors in a mixed basis, something like that. One foot in each world is what I vaguely said at the time. This has a complicated answer. First, in App E (i) ending example, I have shown a tensor expansion of the identity tensor which has Rij as coefficients. I had some other expansion in App J but I decided just now to get rid of it because I could not really justify it fully. I know that R and S are really basis change matrices, not tensors. // But then I restored that App J expansion because I see it quoted from the web below right in this document. I changed the language to be softer in what is claimed for the expansion. It was not until Appendix E on Tensor Expansions that I ever really did the direct product stuff, sort of an afterthought in tensor doc. I am now reading through that appendix. In the first part, I have arbitrary coefficients αijk and in the two special cases, these become the Aijk and A'ijk contra and covariant components of tensor A. I never refer to αijk as a tensor, just as coefficients. The tensor is A. And when I say tensor, I mean tensorial tensor since I then talk about Aijk as contravariant coefficients. Here I am reviewing the existing Appendix E on tensor expansions: I say nothing about "mixed basis" tensors. This is a hole in the discussion, but let's read on. The dialog is pretty good I think. Orthonormal basis OK. Tensor density OK. I then have a section on "tensor like" objects and the point is that you still have components but they are not tensorial tensor components. (c) on polyadic notation. All OK except for part on A (id jd kd ... ) . Where did that even come from? OK, now I see, it is OK. (d) on dyadic products OK, "tensorial vector" phrase used here. (e) OK, going to the matrix form from the cross product form. (f) OK on the small dot. (g) on operator concept. Still no mixed basis, but this is getting closer to the issue. Bra-ket section seems OK. I have now opened the door for mixed basis, but I have not gone in that room yet, since A is an "operator". The section on "bases related by transformation" seems less solid. I show A(b) = BABT where everything is a matrix, so this relates the matrix A to the matrix A(b) which I guess is OK. I show that for the special case bn = en, we get B = R. There is still no mixed basis stuff. I then have "more on bra-ket" which seems OK, relating dyadic notion to A as operator notion. I am next comparing to M&F on dyadic notation, still all OK. Section (h) gets into the unit vector stuff. I then define matrices N and M. I then have a format anomaly with "Orthogonal application". This is my main application, g=1, for curvilinear coordinates. I do a lot here with the M and N which apply here. I show that M and N are rotations. I have now completed reading Appendix E. I think there is some connection with x" space and what I am confused about. Idea: In App E where I have subsection Orthogonal curvilinear coordinates application , maybe I am going too early to the g = 1 curvilinear case. I have M and N as general objects. I have [A()]ijk... = Mii'Mjj'Mkk'...... A i'j'k'... where at this general point M are not rotations, they are just scaled R, as in Mab ≡ h'a Rab . Can I define my x" space right at this point? It would be a space whose basis vectors are n I guess. Consider x = xiui = x'iei = x'ih'ii = x"ii = x"i"i i In any event we definitely have x"i = h'i x'i and nothing is orthogonal, we are fully general! I could then define a brand new matrix H = diag(h'i) and which is not related to g'ij which is not diagonal. Then I can say x" = FH(x') = H x' RH, SH = H, H-1 and then the connection between x'-space and x"-space is a linear transformation. I am trying to get to a picture here that is fully general. Question: Which vsd file has my Picture drawings? Seems to be lost! Answer is tensor paper.vsd in a subfolder. x" = H F(x) ≡ F"(x) x' = F(x) R",S" = M,N = HR, SH-1 R" = M _________________________ Practice Mixed Tensor Expansion _______________________ Here is that web-quoted (I think one of the Kelly PDF's) expansion of F Part A. dx = F(X,t) dX = F dX dxi = FijdXj = (∂xi/∂Xj) dXj Use en as orthonormal basis in x-space, but En in X-space. Consider this mixed expansion F = Σij [F(e,E)]ij ei Ej = Σij [F(e,E)]ij ei EjT then we have F dX = Σij [F(e,E)]ij ei EjT dX = Σij [F(e,E)]ij ei (EjdX) = Σij [F(e,E)]ij ei dXj = dx = dxi ei => dxi = [F(e,E)]ij dXj = (∂xi/∂Xj) dXj Therefore I have shown that [F(e,E)]ij = (∂xi/∂Xj) = ((X)x)ij So this is the first time ever than I have identified these two objects, and this agrees with what I have seen in a PDF, the one called section 2_3, Part B. Now consider a similar mapping, but we will have dx' = Ft(x,τ) dx where τ is a reference time. Suppose now that at both time t and time τ we use the same set of basis vectors, en. I will now just repeat the steps shown above in this context, dx' = Ft(x,τ) dx = Ft dx dx'i = (Ft)ijdxj = (∂x'i/∂xj) dxj Use en as orthonormal basis in x-space AND in x'-space. Consider this non-mixed expansion Ft = Σij [Ft(e,e)]ij ei ej = Σij [Ft(e,e)]ij ei ejT then we have Ft dx = Σij [Ft(e,e)]ij ei ejT dx = Σij [Ft(e,e)]ij ei (ejdx) = Σij [Ft(e,e)]ij ei dxj = dx' = dx'i ei => dx'i = [Ft(e,e)]ij dxj = (∂x'i/∂xj) dxj Therefore I have shown that [Ft(e,e)]ij = (∂x'i/∂xj) = ((x)x')ij So I have verified this result from the 2_3 paper, where he uses where I use x'. The τ argument has not done anything yet. The spatial argument matches the variables superscripted onto the gradient, that seems clear. I was hoping this more symmetrical expansion would somehow explain why Ft is better than F, but so far I see no reason for that. I think he wants to use τ as a probing variable in the neighborhood of t, perhaps for doing derivatives. How can the above expansion possibly be justified? I want to show that dx' = F dx { Σij Fij [i(S)] [j(S)]T } dx = Σij Fij [i(S)] [j(S)]T { Σkdxkk(S)} = Σij Fij Σkdxk [i(S)] [j(S)]T [ k(S)] = Σij Fij Σkdxk [i(S)] δj,k = [ Σij Fij dxj] i(S) = [ dx'i ] i(S) // using the fact that F dX = dx = dx' . Well it does work after all. The same basis vectors are used for dx and dx' which is something we are always allowed to do. OK, I jammed both these expansions into my appendix J. The following was replaced by a similar discussion with a different picture. Part C: Consider this sequential pair of deformation flows. We start with the bottom flow, then discuss the upper flow. The lower arrow marked by F represents the deforming flow of a particle of continuous matter which starts at location X at time t0 and ends up at location x at time t. The particle might have started out at time t0 as a perfect tiny cube, but then by time t that cube has "deformed" into a rotated and stretched parallelepiped. It is assumed that the flow is reasonable and smooth, we are not considering some kind of "explosion" here. We use the words flow and fluid, but the deformation concept applies to elastic solids as well as fluids since these deform in some way when they are stressed (think jello or even steel). Rather than think of the flow in terms of the edges of this tiny cube, one can instead consider two very closely spaced points in the fluid close to X which are separated by spacing dX at time t0, which we think of as a little stick or "a little dumbbell". At time t, if one carefully tracks the "pathline" of the ends of the stick, one finds that the stick may tumble and stretch and end up as dx at time and location x, as shown on the right above. The relationship between dx and dX is given by dx = F(X,t) dX dxi = FijdXj // Lai p 105 (3.18.13) where the matrix Fij is called "the deformation gradient". It is also known as "the deformation gradient tensor" even though it is not a "tensorial tensor" with respect to any identifiable transformation. This is an example of tensor definition (3) mentioned in Appendix E (i). In Section 5 (o) the above equation was identified with dx' = R(x)dx with R(x) here being F(x,t). The picture on the right above is then the second figure shown in Section 2. Thus, the deformation gradient F is the linearized version (at point x) of some fancy non-linear (and unknown) "flow transformation" F (to avoid confusion with F). Since one can write dxi = (∂xi/∂Xj)dXj , one finds that Fij = (∂xi/∂Xj) or F = (x) // Lai p 105 (3.18.4) where the gradient is with respect to X, so it is really = (X). Thus the name "deformation gradient". [ Notice that (x) is a matrix. In Appendix G the form of (v) for arbitrary vector v is found in arbitrary curvilinear coordinates. ] The deformation F(X,t) depends implicitly on the time t0. At t = t0+ε (with a very small ε) no flow has yet taken place, so dxi = dXi and then F(X,t0) = 1. Time t0 is called the reference time and one could display it by writing F(X,t) = Ft0(X,t). Rto(X,t) = Ft0(X,t) = F(X,t) = F Ft0(X,t) dx = F(X,t) dX\\ Ft(X,τ) Rt(x,τ) = Ft(x,τ) dx' = Ft(x,τ) dx Rt(x, ************************************************* Here is someone else confirming the general formula T' = RTTR for the corotational stress, and confirming the use of my phrase "corotational stress". Yes, it is the Cauchy stress.