try mixed basis expansion for R

Try Mixed Basis expansion for R PhL 10.19.12 This resulted in an Example added to the end of Appendix E (i) . I also added two small dot products to Section 6 (a) and Section 7 (s) of tensor doc. Example: Mixed basis expansions of operators R and S. Plan A. (wrong) Consider this mixed-basis expansion of object R shown in various notations, R = Σij Rij eiuj = Σij Rij ei(uj)T = Σij Rij eiuj = Σij Rij | ei><uj| direct product matrix dyadic bra-ket The expansion can be solved for Rij to get Rij = (ei)TRuj = ei • R • uj = <ei | R | uj> matrix dyadic bra-ket as we now verify using the matrix notation, Rij = (ei)TRuj = (ei)T ( Σab Rab ea(ub)T ) uj = Σab Rab (ei)Tea(ub)T uj = Σab Rab (ei ea) (ub uj) = Σab Rab δiaδbj = Rij Raising and lowering indices, one can cause all four forms of the R matrix elements to appear R = Σij Rij eiuj = Σij Rij eiuj = Σij Rij eiuj = Σij Rij eiuj Now, I claim that this expansion fails to satisfy this basic equation from Section 3 un = e'n = Ren Here is the proof that it fails, Ren = Σij Rij ei(uj)Ten = Σij Rij ei (uj en) Write ei = Σk (ei uk) uk Then = Σij Rij [Σk (ei uk) uk] (uj en) = Σij Rij Σk (ei uk) (uj en) uk = Σij Rij Σk (ei)k(en)j uk = Σij Rij Σk Ski Sjn uk = Rij Ski Sjn uk = Ski Rij Sjn uk = Ski δi,n uk = Skn uk ≠ un So what is the conclusion here? We try to write a tensor R which we can expand and whose expansion coefficients are our Rij friends. But when we do this in what seems a reasonable manner, we find that the basic rule e'n = Ren = un is not fulfilled. Therefore R is not a tensor. Plan B. (wrong) Try it the other way: R = Σij Rij uiej = Σij Rij ui(ej)T = Σij Rij uiej = Σij Rij | ui><ej| direct product matrix dyadic bra-ket The expansion can be solved for Rij to get Rij = (ui)TRej = ui • R • ej = <ui | R | ej> matrix dyadic bra-ket as we now verify using the matrix notation, Rij = (ui)TRej = (ui)T ( Σab Rab ua(eb)T ) ej = Σab Rab (ui)Tua(eb)T ej = Σab Rab (ui ua) (eb ej) = Σab Rab δiaδbj = Rij Now consider Ren = Σij Rij ui(ej)Ten = Rijuiδjn = Rinui = ????? no good this way either Plan C. (right) Consider this mixed-basis expansion of object R shown in various notations, 1 = Σij Rij eiuj = Σij Rij ei(uj)T = Σij Rij eiuj = Σij Rij | ei><uj| direct product matrix dyadic bra-ket The expansion can be solved for Rij to get Rij = (ei)Tuj = ei • 1 • uj = <ei | 1 | uj> = <ei | uj> matrix dyadic bra-ket as we now verify using the matrix notation, Rij = (ei)T1uj = (ei)T ( Σab Rab ea(ub)T ) uj = Σab Rab (ei)Tea(ub)T uj = Σab Rab (ei ea) (ub uj) = Σab Rab δiaδbj = Rij This is reconsidered below. Plan D. Here is one thing I am pretty sure of (in standard notation only) Rij = <ei | uj> = a change of basis matrix = (ei uj) = (ei)j = Rij so at least something works! Yes, you can think of it this way Rij = <ei | 1 | uj> 1 = Σk |uk><uk| = Σk uk ukT = an expansion Problem. What happens if I lower the i index on the first line to get Rij = <ei | uj> = a change of basis matrix = (ei uj) = (ei)j = Sji The result is a contraction because we don't have R = ST. Try the last line more carefully Rij = <ei | uj> = a change of basis matrix = (ei uj) = (ei)j = Rij g'ai Rij = g'ai <ei | uj> = a change of basis matrix = ??? Raj = <ea | uj> = a change of basis matrix = (ea uj) = (ea)j = Sja (says p 102 td) ≠ Raj The contradiction continues. Let's make sure that (ea)j = Sja as tensor doc claims p 102. Can I trace this in tensor doc? Well, (ea)j = Sja is a statement in DN, not in SN! In SN it stays (ea)j = Sja . Then I can say in SN (ea)j = (ea)j = Sja = Raj = Raj and my bug is fixed!!! free free Don't mix the two notations together please! Question: Where in tensor doc do I state the simple fact that Rij = <ei | uj> ? (DONE) In standard notation this would translate into Rij = <Ei | uj> = Ei uj = (Ei)j These simple facts are just plain missing from tensor doc!! So let's try to wedge it in the right places. It has to be sometime after the E vectors are brought up in Section 6. But at that point we don't have Rij yet. Is there anything to be said? Rij = <ei | uj> = ei uj SN Rij = <Ei | uj> = Ei uj DN = (Ei)j if g = 1 So if g = 1 I find that Rij = (Ei)j or Rnj = (En)j or Rni = (En)i which agrees with what I already have. But if g ≠1, is it still true that Rij = Ei uj ? From my definition Ei uj = g'ni ei uj = g'ni ab (ei)a(uj)b = g'ni ab Sai δjb = g'ni aj Sai = ja Sai g'in = (Sg')ji = (RT)ji = Rij // using td p 40. So maybe there IS something to be said here. Then that translates directly into ei uj = Rij OK, I added Ei uj = Rij to the end of Section 6 (b) with a one line proof simpler than the one I show above here. Right in the next section all the OTHER dit products are done. But What about ei uj ? ei uj = ab(ei)a(uj)b = ab (Sai)(δj,b) = aj Sai If that is true in DN, then in SN it becomes ei uj = gajSai = Sji // but not very interesting I think. Now that I added this item to Section 6, I want to add the corresponding item later in Section 7. It is done, just after the other dot products, so all dot products are kept together. Conclusion: R is NOT a tensor you can expand in this way R = Σij Rij eiuj = Σij Rij ei(uj)T = Σij Rij eiuj = Σij Rij | ei><uj| direct product matrix dyadic bra-ket with Rij = (ei)TRuj = ei • R • uj = <ei | R | uj> matrix dyadic bra-ket Yes, the "verification" works, but in order to get it to work, you have to set operator R = 1: 1 = Σij Rij eiuj = Σij Rij ei(uj)T = Σij Rij eiuj = Σij Rij | ei><uj| direct product matrix dyadic bra-ket with Rij = (ei)Tuj = ei uj = <ei | uj> matrix dyadic bra-ket This IS a valid expansion! It must be true that 1 = Σj ujuj = Σj | uj>< uj| Then you can install ΣiRij ei = uj (*) to get 1 = Σij Rij eiuj So where does this fact come from? Rij ei = uj Verify it in this way Rij (ei)a = (uj)a Rij Ria = δja In DN this says Rij ei = uj = e'j RTji ei = e'j OK, this has been installed at the end of Appendix E (i) in some form. Comment: I started out with a completely wrong expansion that made no sense, then I realized that the mixed expansion I wanted was of the identity operator 1, not some mysterious operator R. But now what about the expansion I have in Appendix J ? R = Σij Rij i(S) j(S0) = Σij Rij [i(S)] [j(S0)]T Rij = (∂x'i/∂xj) and this from section 1 to confirm: Rik(x) ≡ (∂x'i/∂xk) . I think S0 = x-space and S = x'-space so this could be written R = Σij Rij 'i uj 'i = Mik uk there must be some rotation Mik So then we have R = Σij Rij Mik uk uj = Σij (MTki Rij) uk uj = Σij (MTR)kj uk uj But then I think you would want to say MTR = Σij (MTR)kj uk uj R = Σij [R(S,S0)]ij i(S) j(S0) I just don't like this one bit. I am now going to remove it from my Appendix J and just park it here, it was never needed anyway. I threw it in as an example and because I saw someone do exactly this in one of those PDF's somewhere. Leave it out, then you don't have to worry about it. Tensor expansion of the deformation gradient Using the basis vectors as shown above, we claim that the deformation gradient F can be expanded in this manner: F = Σij Fij i(S) j(S0) = Σij Fij [i(S)] [j(S0)]T Fij = (∂xi/∂Xj) where we write the basis in both direct product and matrix form as in Appendix E. This is a "mixed basis expansion" as discussed in Appendix E (i). To verify the above expansion, consider F dX = Σij Fij [i(S)] [j(S0)]T dX = Σij Fij [i(S)] [j(S0)]T { ΣkdXkk(S0)} = Σij Fij ΣkdXk [i(S)] [j(S0)]T [ k(S0)] = Σij Fij ΣkdXk [i(S)] δj,k = [ Σij Fij dXj] i(S) = [ dxi ] i(S) = dx which is the correct result. In other words, the mixed-basis expansion shown gives dx = FdX. Well, I now found that PDF claim, so I am putting it back in , rewritten in a softer mannner.