oldroyd + CM mods to tensor

Why Oldroyd upper and lower? PhL 10.16.12 CM Mods to Tensor Doc This document contains three repairs I have to make to tensor doc, each marked by ***. // These three repairs were installed into tensor doc at 11:10AM 12.19.12 Meaning of Oldroyd upper and lower. 1 Question: What is going on in Example 8.19.1. 4 Guess as to how the Oldroyd up and down works. 5 Appendix J is starting to wobble. 7 Tensor doc facts which differ for continuum mechanics: review of Section (5) 8 Proposed change to Section 5 (i) **** 12 Finish Scan of Section 5 for CM differences. 13 Proposed Replacement for Section 5 (o) *** 14 (o) Continuum Mechanics and its Metric Tensors 14 Proposed change to Section 8 (c) 7 *** 18 7. Cartesian-View Magnitude Ratios. 18 Meaning of Oldroyd upper and lower. For the lower thing we use JL(τ) ≡ FtT(τ) T(τ) Ft(τ) // Lai p 484 (8.19.12) which I guess we can write as JL ≡ FtT T Ft Remember that Ft is like R in earlier stuff. So maybe write as JL ≡ RT T R => (RT)-1JL(R-1) = T => STJLS = T Now look at the upper thing JU(τ) ≡ Ft-1(τ) T(τ) Ft-1,T(τ) JU ≡ Ft-1 T Ft-1,T JU ≡ R-1T (R-1)T JU ≡ ST ST S-1JU(ST)-1 = T R JU RT = T Here then is where we stand so far JL = RT T R T = STJLS JU = ST ST T = R JU RT Now recall from Section 5 (f) M' = R M RT // contravariant rank-2 tensor ' = ST S // covariant rank-2 tensor Suppose I now make this identification JU = J JL= Then "where we stand so far" becomes = RT T R T = STS ' = ST S // lower J = ST ST T = R J RT M' = R M RT // upper Now maybe in the lower case I need to replace T by Then we have = RT R = STS ' = ST S // lower J = ST ST T = R J RT M' = R M RT // upper Now we have things aligning pretty well. We have to regard now T as being in the x'-space. T → T' J → T Start over. Start with these two equations, which are "tensor correct" T' = R T RT // contravariant rank-2 tensor R = Ft forward flow ' = ST S // covariant rank-2 tensor and above I found that, in Lai notation, T = R JU RT T = STJLS According to my Forward Flow picture, we associate T' with the end of the flow That is , x'-space is the end of the flow, while x-space is the start of the flow, as in Question: Suppose we have some stress tensor T(x,t) viewed in frame S above. Suppose we then define another frame of reference that "flows with the flow". What does that mean? I would be included just to say that it means T' = Ft T FtT based on M' = R M RT // contravariant rank-2 tensor If Ft were a rotation, then T' would be the T seen in a "rotated frame of reference". So based on this idea, I want to say that this J think is T' and then J = Ft T FtT // not true But that is NOT how he defines it. The transposes are swapped. Here then is what we actually have J = FtT T Ft So in fact, J is stress tensor observed in a frame which is related to the original frame by FtT. We have that extra transpose T present, no way around it. Question: How does this work in the simpler corotational situation? T' = Ft T FtT based on M' = R M RT // contravariant rank-2 tensor J = RtT T Rt (8.19.3) We have this exact same reversal problem with the transpose. It is as if J were T measured in a frame of reference that rotates backwards. But in the other case remember that transpose does not mean in inverse, so "backwards" does not solve the issue. But let's try again: J = (RtT) T (RtT)T We can interpret J as being T observed in a frame that is rotated by RtT relative to the start frame. Question: What is going on in Example 8.19.1. We start with a nice static Tij and we note that dtTij = 0. We then go to a rotating frame Rz and compute T' in that frame and then dtT' and we find that dtT' ≠ 0 as expected. But Lai stays nothing about quantity J in either frame. I think he is going to use this formula = dtT + [T,W] (8.18.10) What is this quantity in the starting frame (no prime)? What is W? It is the spin tensor in the sense (v) = D + W But what is v in our given stress situation? Do we have some kind of static material with v = 0? Maybe this situation is a solid bar being stretched along the 1 axis, so there is only a T11 force. Then yes, it is a static situation, so yes, v ≡ 0. This means (v)= 0 and that means D = 0 and W = 0. Therefore = dtT + [T,W] = 0 + [T,0] = 0 + 0 = 0 Then the equation from page 336 is just a frame of reference change thing and I guess if you go to the rotating frame, you will find the solid object to have (v)' = QT which we calculate for our Q(t) as shown. And from that we get W' and from that we get [T',W'] and then we can compute ' = dtT' + [T',W'] and he claims this is 0. But now we have some kind of typo issue because those two things don't add up to 0! // I found and fixed his typos, continue on. So he finds that ' = 0. OK, now this object is supposed to be a rank-2 tensor, which means ' = Q QT and then the fact that both and ' are zero is consistent with them being tensors. In contrast, dtT is not covariant because it is 0 in one frame and not the other. Notice that the rotation Q(t) used in defining J as J = QTRQ is the same Q used in this example. But the rotation in J is in fact the rotation of the polar decomposition of Ft. So in this example, we imagine that we have some Ft = RtUt = VtRt and then we set Q in the example to be this Rt. Guess as to how the Oldroyd up and down works. The two Oldroyd equations are these JU ≡ Ft-1 T Ft-1T JL ≡ FtT T Ft If I now set Ft-1 to "R" we then get JU ≡ RT RT M' = R M RT R = Ft-1 => Ft = S JL ≡ ST T S ' = ST S Now let's assume JU = J JL= and T in the second line should really be Then we have J ≡ RT RT M' = R M RT R = Ft-1 => Ft = S ≡ ST S ' = ST S Finally we have something that makes at least some sense! Now, if T is in Cartesian space, then T = . Then we are even closer! Question: How do I translate all the above into standard notation? (a) first, do this one J = RTRT // dev notation Jij = RiaTabRTbj // dev notation Ria = (Ft-1)ia Jij = RiaTabRjb // dev notation Jij = RiaRjbTab // dev notation Now go into standard notation Jij = RiaRjbTab Ria = (Ft-1)ia (b) second, do this one ≡ ST S ij ≡ (ST)ia ab S bj ij ≡ Sai ab S bj ij ≡ Sai S bj ab Now apply the translation rules from Section 7. Jij = SaiSbj Tab Now use the general idea that Sai = Ria and we then have Jij = Ria Rjb Tab Ria = (Ft-1)ia and this is then in the correct form for a covariant transformation. So in standard notation, we have JL ≡ RT T R → Jij = Ria Rjb Tab JU ≡ ST ST → Jij = RiaRjbTab which we can restate as JL ≡ (Ft-1)T T Ft-1 → Jij = (Ft-1)ia (Ft-1)jb Tab JU ≡ FtT (Ft)T → Jij = (Ft-1)ia (Ft-1)jb Tab and in Cartesian space Tab = Tab or T = . Now comes the interpretation: If T is the stress tensor in some initial frame of reference, then JU is the stress tensor one would measure in a frame of reference which is deforming by Ft-1 relative to the initial frame. Suppose we denote this new frame by ". Then what we really have above is T" ≡ FtT (Ft)T and T" is the stress in the new frame. In our flow we had dx' = Ftdx and I guess dx" = Ft-1dx ?? ******************************************************** Appendix J is starting to wobble. Go back to this point in Appendix J. I seem to have ignored the time argument of Ft(τ) in this discussion. I will add those arguments right now Do any of the usual relative derived tensors transform as tensors with respect to Q(t) ? Again we think of a relative tensor Wt as being a property of the continuous material at time t, a measure of the state of deformation. Such a tensor is objective only if Wt* = Q(t)WtQ(t)T, where everything is at time t. The process is very similar to that carried out above. We start with Bt ≡ FtFtT : Bt*(τ) = Ft*(τ)Ft*(τ)T = [Q(τ) Ft(τ) QT(t)] [Q(τ) Ft(τ) QT(t)]T = Q(τ) Ft(τ) QT(t) Q(t) Ft(τ)T Q(τ)T = Q(τ) Ft(τ) Ft(τ)T Q(τ)T = Q(τ) Bt(τ) Q(τ)T // not a rank-2 tensor since t ≠ τ Next comes Ct ≡ FtTFt : Ct* = Ft*TFt* = [Q(τ) Ft QT(t)]T [Q(τ) Ft QT(t)] = Q(t) FtT Q(τ)T Q(τ) Ft QT(t) = Q(t) FtT Ft QT(t) = Q(t) Ct(τ)QT(t) // yes a rank-2 tensor with respect to Q(t) How can you justify this claim with τ sitting in there. I have shown that Ct*(τ) = Q(t) Ct(τ)QT(t) If course in the limit τ→t this just says 1 = Q(t) QT(t) which is at least correct. Lai fudges this on the bottom of page 472. He suppresses the τ argument so you don't see it, just the way I did above. So I now have to face up to this big problem, it just makes no sense right now. What about Ut and Vt? Ft= RtUt = VtRt Ft* = Rt*Ut* = Vt*Rt* Consider, Ft* = Q(τ) Ft QT(t) = Q(τ) RtUtQT(t) = [Q(τ) RtQT(t)] [Q(t)UtQT(t)] = Rt*Ut* . Since [Q(τ) RtQT(t)] is a rotation and since [Q(t)UtQT(t)] is a symmetric positive definite matrix by the argument given in the previous section, and since the polar decomposition is unique, it must be that Rt* = Q(τ) RtQT(t) and Ut* = Q(t)UtQT(t) // Ut is a rank-2 tensor Finally, write Ft* = Q(τ) FtQT(t) = Q(τ)VtRtQT(t) = [Q(τ)VtQT(τ)] [Q(τ) RtQT(t)] = Vt* Rt* By the same argument used several times above, we conclude that Rt* = Q(τ)RtQT(t) and Vt* = Q(τ)VtQT(τ) // Vt is not a rank-2 tensor, τ ≠ t The rule for transforming Rt is the same as found a few lines above. Here then are the conclusions, with references to Lai page 472: Ft* = Q(τ)FtQT(t) // Lai (8.13.6) Bt* = Q(τ)BtQ(τ)T // Lai (8.13.12) Ct* = Q(t)CtQT(t) // rank-2 tensor with respect to Q(t) so objective // Lai (8.13.10) Ut* = Q(t)UtQT(t) // rank-2 tensor with respect to Q(t) so objective // Lai (8.13.9) Vt* = Q(τ)VtQT(τ) // Lai (8.13.12) Rt* = Q(τ)RtQT(t) // Lai (8.13.8) Notice that among the "normal" tensors, B and V are objective, whereas among the "relative tensors" it is Ct and Ut that are objective. All the other tensors are "non-objective". ****************************************************************** Tensor doc facts which differ for continuum mechanics: review of Section (5) My first comments on this subject appear at the end Section 5 (a), g' = R g RT (Picture A) is only true when it is required that (ds)2 be a scalar. Therefore, in continuum mechanics, this is not true, and "the metric tensor does not transform as a rank-2 tensor" !!! Later I show that in fact G and ' are 1 for two spaces related by rotation, so this says S'T S' = 1 which is true for a rotation. But that is not the real point. In a flow transformation, dx = FdX and F is not a rotation and ds2 ≠ dS2. Pause: Can we have "tensors" in continuum mechanics? A tensor under F would have this rule T 'abcde = Raa' Rbb' Rcc' Rdd' Ree' Ta'b'c'd'e' where R is the linearization of F. But now what is the situation with raising and lowering indices on this sample T thing? If g = 1 and g' = 1, it seems you can raise/lower any index for free at will. Something seems wrong about that idea! Let's go to the simplest vector case V'a = Raa'Va Let x-space be that at the start of the flow, so it is X-space. This takes us to my Section 5 (o) and the Forward Flow table of items. I never said one peep about transformations like the above. I do however talk about two different g objects called ' = 1and ' = RRT which I call the wannabe metric tensor you would get from the rule ' = R g RT = RRT. Now what do we mean by raising and lowering indices? In which sections of Section 5 do we have issues with continuum mechanics? Let's look at section (g) which I paste below: We continue in Picture A. Suppose V is a contravariant vector so V' = RV. [ ok to here ] Construct a new vector W with the following properties ( see Section 7 (u) concerning "covariant equations") Now maybe I need to use ' in what follows. I will try it. Here ' ≡ R g RT . W = V x-space W' = ' V' x'-space Is vector W one of our two vector types, or is it neither? One must examine how it transforms under F: W' = ' V' = (ST S) (RV) = ST (SR)V = ST V = ST W Therefore this new vector W is a covariant vector under F [ correct ] , so it should have an overbar, ≡ V This covariant vector can be regarded as the covariant partner of contravariant vector V. This shows the general idea that applying to any contravariant vector produces a covariant vector! So this is one way to construct covariant vectors if we have a supply of contravariant ones. Conversely, starting with a known covariant vector , one can construct a contravariant vector V ≡ g . Thus, every vector of either type can be thought of as having a partner vector of the other type. An obvious notation is to write as so no extra letter is needed. Then one has = V V = g i = ij Vj Vi= gijj So even though the CM metric tensor is ' = 1, we can still use ' = R g RT = RRT to raise and lower indices over in x'-space and in doing so, convert vectors from one type to the other. Wow, that is new to me. There is a decoupling between the two g's and you have to see which one does what. So, suppose we have in a CM situation V'a = Raa'Va Then over in x-space we can raise and lower at will, but in x'-space we have to use ' . Then I would do this ' = ' V' = RRT V' = B V' 'a = 'ab V'b So amazing, now raising and lowering is done by the Cauchy B, it is no longer wannabe. Is this consistent? Use Forward Flow: ( F here is my usual R) V'a = Faa'Va' => V'a = Faa'Va' Faa' = (F-1)a'a // SN V'a = Faa'Va' => 'a = (F-1)a'aa' = (F-1,T)aa'a' // DN I never thought of things this way! The covariant goes with F-1,T if the contravariant goes with F, but of course that is what I have always said, for example here: V' = R V contravariant Rik(x) ≡ (∂x'i/∂xk) R = S-1 ' = ST covariant Sik(x') ≡ (∂xi/∂x'k) = STki(x') How did I get from the first line to the second line in the pair above with SN and DN labels? Well, I guess I applied ' to both sides! Maybe I should swap my symbols ' ↔ ' ? One is a raising/lowering tensor, the other is the distance metric tensor, and they are different!!! This is going to take a lot of work to clean up, I guess a new doc is in order. Raising/lowering and contravariant/covariant are one world whereas distance metric is another. So let's take another look at tensor doc. In the first sections, there is no real mention of metric tensor, and that discussion really starts in Section 5 (with a few exceptions). section "intro": just describes the Picture B and Picture D, I think it is OK. section (a): I think the text down to (ds)2 = 'km dx'k dx'm is OK for CM. This defines distance ds in x-space in terms of the dx' components and ', some matrix. The name metric tensor seems good. So far there is no mention of covariant vector transformations etc. I then say: "Since (ds)2 is a number which is the same in all three systems (that number is the distance between two points in x-space), the quantity 'km dxk' dxm' is a tensorial scalar. " so this claim needs a CM exemption which is that (ds)2 is NOT a scalar in CM flow. Right after that I have my long comment on the CM exemption so it is in the right place. But maybe I say it wrong here. Go back to this line, "km ≡ Σi GiiS"ikS"im => " = S"TG S" Why should this be NOT true if (ds)2 is NOT a scalar ? The ingredients are these: (ds)2 = ΣiGiidxidxi // distance in x-space dx = S dx' // dx transforms as a vector under F (ds)2 = Σi Gii (ΣkS'ik dx'k) (ΣmS'im dx'm) // follows from above = ΣkΣm { Σi Gii S'ikS'im } dx'k dx'm // rearrange = ΣkΣm 'km dx'k dx'm // definition ' = S'TG S' This is all true for CM !!! So I have made a mistake. I have not said anything yet about distance in x'-space. So in CM we can still have ' = S'TG S' . How would I change my comment? (1) in CM it is still true that ' = S'Tg S' for a flow. (2) in CM it is still true that (ds)2 = Σigiidxidxi in x-space with G = 1 (3) in CM we can define a distance in x'-space (ds')2 = Σi 'dx'idx'j . There is not requirement that this distance be the same as (ds)2, so no requirement that (ds)2 be a scalar with respect to F. section (b). All seems OK as is. Inverse defined as stated. section (c) All g's are symmetric. Seems OK section (d): seems OK for CM section (e): two kinds of rank-2 tensors. Just giving the definition of a rank-2 tensor by the way it transforms. Seems OK for CM. section (f): proof that g is a rank-2 tensor. I think all OK for CM, no mention of scalarity of anything. No requirement for example that VV = V'V' . In fact we don't even have a dot product yet. So far so good. So apart from my one sentence about (ds)2 being scalar, I think Section 5 is OK to this point for CM. section (g): conversion of types. Seems completely OK for CM !!! section (h): in Cartesian space = V, this is OK in CM. section (i): Here we are going to have issues with CM !! Opening comments are all OK I think, Then I define AB for the first time. Now something fishy happens. Consider: A' B' = 'abA'aB'b = 'ab(Raa'Aa') (Rbb'Bb') = 'ab Raa' Rbb' Aa' Bb' = [ (RT)a'a 'ab Rbb' ] Aa' Bb' = [RT ' R]a'b' Aa' Bb' = a'b' Aa' Bb' = ab Aa Bb = A B Whoa!!! This seems OK for CM, but then if I use A' = B' = dx' and A = B = dx, I get a contradiction and I get dx'dx' = dxdx . The only way out is to claim dx'dx' ≠ (ds')2. SO, this points out a potential ambiguity in my use of the symbol. Here I have defined it as the covariant dot product, so perhaps write it as (g) . In Lai CM, however, we write (ds')2 = dx' dx' = dx' g dx' = dx' 1 dx' = dx'idx'i So we really have two different dot products in CM. I have defined to be the covariant one. I might put in a little comment at this point. How about this change: __________________________________________________________________________ Proposed change to Section 5 (i) **** Replace "Going back" through end with the following : reviewed on 10.19.12 and ready to install: DONE In applications in which (ds)2 is regarded as a scalar with respect to transformation F we have (ds')2 = dx' dx' = (ds)2 = dx dx and ds = ds' is called "the invariant distance". Such applications include curvilinear coordinate transformations and relativity transformations. In special relativity, using the Bjorken and Drell notation noted above where g'μν = diag(1,-1,-,1,-1) and c = 1, one writes (standard notation) (dτ)2 = g'μν dx'μdx'ν = dxμdxμ = dx'• dx' = dx • dx = a Lorentz scalar = (dt)2 - dx dx , xμ = (t,x) and dτ is called "the proper time", a particular case of the invariant distance ds. Notice that (dτ)2 < 0 for a spacelike 4-vector dxμ, meaning one that lies outside the future and past lightcones (|dx| > |dt| ). [We now restore to our covariant definition after temporarily using it above for a 3-space Cartesian dot product. ] Going back to Section 3 and the vectors e'n and en, a claim made there can now be verified: |e'n|2 = e'n e'n = en en = |en|2 => |e'n| = |en| . In applications in which (ds)2 is NOT regarded as a scalar with respect to transformation F, which includes continuum mechanics flows, things are a little different. We have here (ds)2 ≡ dx dx = dx' dx' ≠ (ds')2 if (ds') ≡ physical distance in x'-space If x'-space has Cartesian coordinates, we have (ds')2 = dx'idx'i = δi,j dx'i dx'j and this is different from the quantity dx' dx' = 'ij dx'i dx'j . One might go so far as to define two different symbols: dx' g dx' ≡ 'ij dx'i dx'j ≠ (ds')2 dx' 1 dx' ≡ δi,j dx'i dx'j = dx'i dx'i = (ds')2 = physical distance squared By default, we have = g, so then dx' dx' ≠ (ds')2. In this situation, x'-space in effect has two different "metric tensors". The first is the 'km that appears in the entire discussion of this Section 5 and which is involved in raising and lowering indices on a tensor and satisfies ' = ST S (the fact that g transforms as a rank-2 tensor), and so on. The second we might call 'km and this is what determines physical distance in x'-space, (ds')2 = 'km dx'k dx'm . In continuum mechanics with Cartesian x'-space, we then have 'km = δk,m. There are two different "metric tensors" because there are two different "metrics" which are of interest in x'-space : dg2(x,y) = (x-y) g (x-y) = km (x-y)k(x-y)m // raising/lowering metric tensor dg2(x,y) = (x-y) g (x-y) = km (x-y)k(x-y)m // physical distance metric tensor For Cartesian coordinates in x'-space, 'km = δk,m, and our two metric tensors may be identified with those of "Cartesian View x'-space" and "Curvilinear View x-space" as discussed in Section 8 and Appendix C (e). __________________________________________________________________________ Finish Scan of Section 5 for CM differences. Let's continue now with our scan of Section 5 to see if there are other CM issues. section (j): seems OK, talks about relation between g' and en etc. section (k): Jacobian, seems OK section (l) : I think is all OK, no use of section (m) : special relativity OK section (n) : general relativity OK Now I need to completely rewrite section (o) in light of the comment above. Here we go: __________________________________________________________________________ Proposed Replacement for Section 5 (o) *** This is ready to install as of 10.19.12, a complete subsection replacement, check pagination! DONE (o) Continuum Mechanics and its Metric Tensors One can describe (Lai) the forward "flow" of a continuous blob of matter by x = x(X,t) where X = x(X,t0). A "particle" of matter (imagine a tiny cube) that starts at location X at time t0 ends up at x at time t. Two points in the flow separated by dX at t0 end up separated by some dx at t. The relation between them is given by dx = F dX where F is called the deformation gradient. F describes how a particle starting say with a cubic shape at t0 gets deformed into some parallelepiped (3-piped) shape at t (picture below). If we examine dx = F dX we find that | dx | ≠ | dX | since the vector dX typically gets rotated and stretched as dX → dx during the flow. The finite-time flow x = x(X,t) from time t0 to time t can be thought of as a (generally non-linear) transformation of the form x = F(X) as in Section 1 above. Recall from Section 1 that a general transformation was x' = F(x) and the linearized transformation was dx' = R dx. To be compatible with Lai notation which uses symbol F for the deformation gradient, we have renamed the Section 1 transformation F to be F, and we shall see below that R will in fact become deformation gradient F. In this flow we assume Cartesian coordinates in both x-space and X-space, so the metric tensors which determine physical distance in these spaces are both 1. As discussed at the end of Section 5 (i), there are really two metric tensors in this situation called 'km and 'km. The former is the metric tensor which raises and lowers tensor indices in the general tensor formalism, while the latter is the metric tensor which determines physical distance in x'-space. For the usual Cartesian coordinates in x'-space, 'km = δk,m which is to say ' = 1. The other metric tensor is given by ' = ST S where is the metric tensor in x-space. Using Cartesian coordinates there means = 1 and then ' = ST S. In order to put this flow into the notation of this document, let X → x and x → x' so that continuum mechanics this document ( Forward Flow X → x [ x → x'] ) x, X ↔ x', x x = x(X,t) = F(X) ↔ x' = F(x) // Lai p70 (3.1.4) dx = F dX ↔ dx' = R dx // as in Section 2 // Lai p86 (3.7.6), p105 (3.18.3) F ↔ R F-1 ↔ S X = Cartesian ↔ g = 1 x = Cartesian ↔ g' = 1 B = FFT ↔ g' = RRT // as in Section 5 (l) // Lai p121 (3.25.2) B-1 = (F-1)T (F-1) ↔ ' = STS // since ' = g'-1 Thus, the deformation gradient F is just the R matrix of the forward transformation x = x(X,t) = F(X). The raising/lowering metric tensor g' = RRT appears as B = FFT which is known as the left Cauchy-Green deformation tensor (manifestly symmetric, so a viable metric tensor). Alternately, we can consider the above flow going backwards in time and then x = x is the starting position and x' = X is the ending position. For this inverse flow, we let F have the same meaning as in the forward flow, dx = F dX, and thus end up with this translation table where we now x → x and X → x' : continuum mechanics this document ( Inverse Flow x → X [x → x'] ) X, x ↔ x', x X = X(x,t) = F(x) ↔ x' = F(x) // this F is the inverse of the forward flow F dX = F-1 dx ↔ dx' = R dx // as in Section 2, same F as in forward flow F-1 ↔ R F ↔ S // S = R-1 and Sik = (∂xi/∂x'k) ↔ Fik = (∂xi/∂Xk) X = Cartesian ↔ g = 1 x = Cartesian ↔ g' = 1 C = FTF ↔ ' = STS // as in Section 5 (l) // Lai p114 (3.23.2) C-1 = F-1(F-1)T ↔ g' = RRT // since g' = (')-1 To avoid confusion, in both tables g goes with flow X-space and g' and g' go with flow x-space. For the inverse flow, the raising/lowering metric tensor ' = STS appears as C = FTF which is the right Cauchy-Green deformation tensor (again manifestly symmetric, so a viable metric tensor). Given the above flow situation, it is then possible to add two more transformations F1 and F2 which take X-space and x-space to independent sets of curvilinear coordinates X' and x': and we then have an interesting triple application of the notions of Section 1 to a real-world situation. This drawing is the implicit subject of Section 3.29 (p131) of Lai. In (reverse) dyadic notation the deformation gradient is written F = (x) where means (X)so that dx = F dX = (x) dX Fij = (x)ij = ∂j(X)xi = ∂xi/∂Xj The (x) notation is explained in Appendix E, and in Appendix G the object (v) for an arbitrary vector field v(x) is expressed in general curvilinear coordinates. Consider now this picture taken from Section 8 below, Flow X-space Flow x-space We can identify the mapping shown in this picture with our Inverse Flow situation (table above). Section 8 discusses in much detail how length, volume and area transform under a general transformation. The length, area and volume magnitudes on the left are called dL'n = dx'(n), dA'n and dV', while the corresponding quantities on the right are called dx(n), (n) and dV, where the first two items are vectors. Cribbing the results of Section 8 (c) 7 and converting them from standard notation to developmental notation, we have | dx(n)|/ dL'n = h'n = ['nn]1/2 = the scale factor for edge dx(n) | (n)|/ dA'n = (1/h'n) |J| = (1/h'n) g'1/2 = [g'nn g']1/2 = [cof('nn)]1/2 |dV| / dV' = |J| = g'1/2 // where g' ≡ det('ij) = J2 , ' = STS We can then translate these three lines into our Inverse Flow context: | dx(n)| / | dX(n)| = h'n = ['nn]1/2 = [(FTF)nn]1/2 = [Cnn]1/2 // Lai p114 (3.23.6-8) | dAn| / |dA0n| = [g'nng']1/2 = [cof('nn)]1/2 = [cof(FTF)nn)]1/2 = [cof Cnn)]1/2 // Lai p129 (3.27.11) |dV| / |dV0| = |J| = g'1/2= [det('ij)]1/2 = [det(FTF)]1/2 = |det(F)| // Lai p 130 (3.28.3) where edge area volume X-space : dX(n) dA0n dV0 time t0 x-space : dx(n) dAn dV time t Thus, for example, the volume change of a "flowing" particle of continuous matter is given by the Jacobian |J| = |detF| associated with the deformation gradient tensor F. We put quotes on "flowing" only because this might be a particle of solid steel that is momentarily moving and deforming a very small amount during an oscillation or in response to an applied stress. In the middle line above we state that | dAn| / | dA0n| = [cof(FTF)nn)]1/2 and quote Lai p 129 (3.27.11) for verification. However, what Lai (3.27.11) actually says (slightly translated to our notation) is this: dA(n)/dA(n)0 = det(F) | (F-1)T un | un = unit base vector, (un)i = δn,i which seems a far cry from our result [cof(FTF)nn)]1/2. But consider, using the Inverse Flow table, | (F-1)T un |2 = | RT un |2 = [RTun]i[RTun]i = Rni Rni = (RRT)nn = g'nn so det(F) | (F-1)T un | = g'1/2 g'nn1/2 = [cof('nn)]1/2 = [cof(FTF)nn)]1/2 where we use the Section 8 (c) 6 theorem converted to developmental notation, stating that g' g'nn = [cof('nn)] . // more generally, (detA) A-1 = cof(A) if A = AT It might be noted that the Lai book does in fact use our "developmental notation" in that all indices are written "down" (when indices are shown), but no overbars mark covariant objects. Here are a few examples: Lai notation Developmental notation Standard Notation dA0 = dX(1)x dX(2) (3.27.1) 0 = dX(1)x dX(2) (dA0)i= εijk [dX(1)]j [dX(1)]k [divT]i = ∂jTij (4.7.3) [divT]i = jTij [divT]i = ∂jTij Lai writes tensors in bold face such as F for the deformation gradient noted above, or T for the stress tensor. Perhaps this is done to emphasize the notion of a tensor as an operator as in our Appendix E (g). Lai writes a specific matrix as [T], but a matrix element is Tij. Notation is an ongoing burden. __________________________________________________________________________ Since the above ties in with Section 8. There in section (c) 7 I suddenly have some script g's, where did they come from?? Aha! I need to fix that now. Here we go: OK, I fixed the below, and then edited the above. It is all exceedingly ugly! But right now I am just trying to get the Oldroyd stuff understood. ___________________________________________________________________________ Ready to install DONE Proposed change to Section 8 (c) 7 *** 7. Cartesian-View Magnitude Ratios. In the Cartesian View of x'-space one can write the Cartesian x'-space magnitudes as | dx'(n)|c = dL'n | dA'(n)|c = dA'n | dV'|c = dV' Then from the three x-space equations in the above table (just above Two Theorems) one obtains the following three ratios of x-space objects divided by their corresponding Cartesian-View x'-space objects: | dx(n)|/ dL'n = h'n = [g'nn]1/2 = the scale factor for edge dx(n) | dA(n)|/ dA'n = (1/h'n) |J| = (1/h'n) g'1/2 = [ g'nn g']1/2 = [cof(g'nn)]1/2 // Theorem 1 above |dV| / dV' = |J| = g'1/2 // g' ≡ det(g'ij) = J2 It is convenient to make the definition dAn ≡ | dA(n)| and then the above area magnitude ratio relation may be written dAn = dA'n and dA'n = Πi≠ndx'i is just a product of the appropriate curvilinear coordinate variations. ___________________________________________________________________________ Now the script g's are gone and I need make no comments whatsoever. In other words, I had to make an ugly patch to the above little section, and now I can pull out that page and restore it as it was. Let's then see how that makes Section 5 (o) look, so I now continue editing above in the previous section. –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––