RE tensors An

The Rivlin-Eriksen tensors PhL 12.15.12 Things here are much more complicated that Lai lets on. Below I had to do 6 Lemmas and 6 Theorems to be happy with the derivation of (8.11.1) and (8.11.3) !! This is a complicated subject because we are dealing with two "pictures" and one must be clear which is which at any given time. Recall this from Appendix J, Picture 1, where X is a FIXED reference: (dX is a dumbbell near X) dx = F(X,t) dX dxi = FijdXj dxi = (∂xi/∂Xj)dXj , Fij = (∂xi/∂Xj) or F = (x) // Lai p 105 (3.18.4) Picture 2, where x is a FIXED reference: (dx is a dumbbell near x) dx' = Ft(x,τ) dx dx'i = (Ft)ijdxj // Lai p 457 (8.7.2) dx'i = (∂x'i/∂xj)dxj , (Ft)ij = (∂x'i/∂xj) or Ft = (x') // Lai p 457 (8.7.3) Lai's Derivation Path. Part A. We start off with a batch of Lemmas, some collected from earlier in the book: Lemma 1. v = dx/dt = Dx/Dt = (X,t) = (x,t) You can parameterize the velocity of a fluid either way. These are the Lagrangian and Eulerian pictures. The two functions are different, and we normally assume that v means (x,t) which write as v(x,t) . Lemma 2. dv = d(X,t) = (X+dX,t) – (X,t) = ((X) (X,t))dX dv = d(x,t) = (x+dx,t) – (x,t) = ((x) (x,t))dx Here dv is the velocity differential between two close locations in a fluid at time t. You can write dv in the two ways shown, based on Lemma 1. This is a "dumbbell based concept", where the dumbbells marks the two points in space at which you measure your two velocities to get dv. They are dX and dx. The expressions on the far right arise from writing dv using normal calculus of several variables and realizing that dt = 0 so there are no (∂tv)dt contributions. The dt=0 idea is because the two ends of the dumbbell exist at the same time t. Corollary 2. ((X) (X,t))dX = ((x) (x,t))dx Lemma 3. dv = dt(dx) Proof: (dx) = x(X+dX,t) – x(X,t) = dumbbell at time t, so then dt(dx) = dtx(X+dX,t) – dtx(X,t) = (X+dX,t) – (X,t) = dv by Lemma 2. Lemma 4. dt(dx) = ((X) (X,t))dX = ((x) (x,t))dx // Lai p 95 (3.12.5) This follows at once from Lemma 3 and Lemma 2. Corollary 4. dt(dx) = (v)dx // Lai p 95 (3.12.6) This is just a compact statement of the Lemma 4, where (v) = ((x) (x,t)). Lemma 5. (1/2) dt[(ds)2] = dx D dx // page 97, 3.13.11 Proof: Watch carefully: dx D dx = dx (D+W) dx // because W is antisymmetric because: dx W dx = dxiWijdxj = 0 Therefore, according to the p 95 3.13.1 definitions of D and W, dx D dx = dx (v) dx But Corollary 4 says dt(dx) = (v)dx so we have dx D dx = dx dt(dx) Now we can write (ds)2 = dx dx so dt(ds)2 = 2 dx dt(dx) Therefore dx D dx = dx dt(dx) = (1/2) dt(ds)2 QED. Lemma 6: If T is symmetric, and if xTTx = 0 for all x, then T = 0. Let D be the diagonalized version of T T = ST D S D = STST Then xTTx = 0 => xT(ST D S)x = (Sx)T D (Sx) = yTDy = 0 . y = Sx Since x is an arbitrary vector, we can regard y as an arbitrary vector for fixed S. We then have yTDy = 0 for all y, where D is diagonal. This says ΣiyiDiiyi = 0 => Σi yi2Dii = 0 . If we take yi = δi,n so that y = un, we get Σi δi,n δi,n Dii = 0 => Dnn = 0 Thus, D = 0 as a matrix, and it follows that T = ST D S = 0 as well. QED If T is not symmetric, the conclusion does not follow. Part B. We have here a tricky series of steps. Theorem 1. dx ∂τnCt(x,τ) dx = ∂τn ( [ds']2) Here we are in Picture 2 described above, the upper half of the picture where x, dx = fixed. Proof: dx ∂τnCt(x,τ) dx = ∂τn (dx Ct(x,τ) dx ) = ∂τn (dx FtTFt dx ) = ∂τn [ Ft(x,τ)dx Ft(x,τ)dx] = ∂τn [ dx' dx' ] = ∂τn ( [ds']2) QED Notice that x and dx are not functions of τ, they are fixed like X and dX are fixed in Picture 1. That is why the ∂τ operator treats the dx dumbbells as constants. Theorem 2. One can replace ∂t by dt anywhere and everywhere in Theorem 1. In Picture 2, x is fixed just the way X is fixed in Picture 1. This means dt f(x,t) = ∂tf(x,t) for any function f in the Picture 2 context. We can then repeat the entire proof of Theorem 1 dx dτnCt(x,τ) dx = dτn (dx Ct(x,τ) dx ) = dτn (dx FtTFt dx ) = dτn [ Ft(x,τ)dx Ft(x,τ)dx] = dτn [ dx' dx' ] = dτn ( [ds']2) QED Summary of Theorem 1 and Theorem 2: dx ∂τnCt(x,τ) dx = dx dτnCt(x,τ) dx = ∂τn ( [ds']2) = dτn ( [ds']2) (*) Theorem 3. Limit [ dτn ( [ds']2) ]τ=t = dtn( [ds]2). In Picture 2, as we do this limit, we know that Ft→1 and ds' → ds and dτ = dt when τ = t. The expression on the right involves [ds]2(x,t) = dx(x,t) dx(x,t). Theorem 3. dtn( [ds]2) = ∂tn( [ds]2) In Picture 2, x is a fixed reference point, so we know that dt[dx(x,t)] = ∂t[dx(x,t)], and it then follows that dt[ds]2 = ∂t[ds]2 . Recall that x is fixed in Pic 2 the way X is fixed in Pic 1. This is the same idea as Theorem 2 and applies really to any statement in Picture 1. Theorem 4: dtn( [ds]2) = dx An(x,t) dx Proof: From (*) we have dx ∂τnCt(x,τ) dx = dτn ( [ds']2) We take the limit of both sides as τ → t. The LHS is of course dx An(x,t) dx . The RHS according to Theorem 3 is ∂tn( [ds]2) . QED Theorem 5 (recursion) An+1 = (v)TAn + DtAn + An(v) // Lai p 468 (8.11.3) Proof: Start with the result of Theorem 4, dtn( [ds(x,t)]2) = dx An(x,t) dx We now switch to Picture 1 where there is no τ and where X and t0 are the references. Now dx is no longer a constant as it was in Picture 2. We then apply dt to the above equation dtn+1( [ds(x,t)]2) = dt[dx An(x,t) dx] = dt(dx) An dx + dx dtAn dx + dx An dt(dx) Next, we apply the result of Corollary 4 that dt(dx) = (v)dx to write this as dtn+1( [ds(x,t)]2) = (v)dx An dx + dx dtAn dx + dx An (v)dx We then move (v) in the first term to the right of the where it becomes a transpose matrix, dtn+1( [ds(x,t)]2) = dx (v)An dx + dx dtAn dx + dx An (v)dx = dx [ (v)TAn + dtAn + An(v)] dx But of course from theorem 4 we have dtn+1( [ds(x,t)]2) = dx An+1(x,t) dx (*) Comparison of the last two equations says dx [ (v)TAn + dtAn + An(v)] dx = dx An+1(x,t) dx or dx [ (v)TAn + dtAn + An(v) – An+1(x,t)] dx = 0 The object in [...] is known to be symmetric. The two terms added are of course symmetric. We know that since Ct is symmetric, so is An and then so is dtAn. Therefore we can apply Lemma 6 to conclude that An+1(x,t) = (v)TAn + dtAn + An(v) Lai makes no mention of Lemma 6. Theorem 6: A1 = 2D Proof: From Lemma 5 we know that (1/2) dt[(ds)2] = dx D dx And from (*) in the last theorem we know that, with n = 0, dtn( [ds(x,t)]2) = dx A1(x,t) dx Comparison shows that dx A1(x,t) dx = dx 2D dx or dx [A1 – 2D] dx = 0 Since D is the symmetric part of (v) and since A1 is symmetric, we can apply our Lemma 6 to conclude that A1 = 2D QED.