up and down indices in the A as operator section

Delete this doc in a few days after 4.12.12 Debug: up and down indices in the A as operator section PhL 4.8.12 This concerns Appendix E (g). As I start out, it seems that bi is an arbitrary basis vector, it could be either ei or i for example. Things seem OK until the start of bra-ket notation. In this section things seem OK. I write Anm = <un | A | um > = the x-space components of tensor A A'nm = <en | A | em > = the x'-space components of tensor A In these two special cases, yes you can raise and lower indices because g for the first and g' for the second, no problem. Maybe I should say "in these last two equations the covariant indices can be moved up and down in the usual manner". The reason is that you can apply g to both sides . But in this case αnm = <bn | A | bm > is that still true? I guess it is true because certainly bn and bn are related by g?? [ wrong !! ] Have I even defined bi and bi anywhere?? I have to go way back earlier in the appendix. Well, I know it is true that given any set of basis vectors bk you can FIND bk that make bi bj = δij be true. This does not say that g' raises or lowers one into the other!!! That only works for the full en ones, so please be careful! I really should state this more clearly: en = g'niei and vice versa. ****** I then make the claim that if bi are orthonormal, then bi= bi with a proof that I like. In one application that would say that n = n . I have never thought much about n. I guess in general it would be this: n = en/|en| |en|2 = en en = g'nn => |en| = => n = (g'nn)-1/2 en Also, n = (g'nm/ )em = some weird linear combination of the em This is nothing very useful in the general non-orthogonal case. For orthogonal it becomes n = h'n-2/ h'n-1 en = h'n-1 en = h'n-1 h'nn = n in agreement with the prediction above. OK, I think the "orthonormal basis" section is OK. The rest seems OK down to (c). It seems OK. And OK, all is well down to section (g). It never hurts to review. Operator concept: OK, This brings me again to the bra-ket section,. The vectors bi and bi are in general different, so of course the kets | bi > and | bi > are different. As a reminder, no, g does not raise and lower things in general. BUT, one can always talk about all these objects, αnm = <bn | A | bm > αnm = <bn | A | bm > αnm = <bn | A | bm > αnm = <bn | A | bm > They all exist for any bn, but g does not raise and lower indices on αnm and αnm is in general not a tensor. I don't make this point very clear. OK, I have tuned up the bra-ket section a b it. I made it a lot clearer I think. Now we come to the critical section: "bases are related by a transformation". This is where I get involved in transpose with tilted matrices. Everything prior to this in Appendix E is A-OK. So I will continue right here manana. Manana has arrived. Let's add more detail. Bases are related by a transformation. Consider again ( note that |i> ≡ |ui> on the next line ) [A(b)]nm = <bn | A | bm > = (bn)T A bm = [bn]i Aij [bm]j // = <bn|i><i|A|j><j|bm> where the subscripts i and j are those associated with the un basis. Since x-space is assumed Cartesian, up and down position on these indices does not matter. Comment: all my tilting seems irrelevant therefore. So what is the context of this section? Maybe I am doing the general Picture A and then up and down DOES matter. In that case I really do need tilted indices to be covariant. Yes, I do show this Picture A at the start of Appendix E and state that the entire Appendix is that way! So how do I arrive at my index positions in the opening line? [A(b)]nm = <bn | A | bm > = bn (A bm) = [bn]i(A bm)i where then (A bm)i is a vector component in x-space since A is a tensor. Then (A bm)i = Aij (bm)j ?? At least this would transform as a vector, so that must be right. Then [A(b)]nm = <bn | A | bm > =[bn]i Aij (bm)j But then I am quietly assuming this <bn | A | bm > = <bn|ui><ui|A|uj><uj|bm> and that properly shows the completenesses! So my simple form <bn|i><i|A|j><j|bm> is ambiguous because I was assuming Cartesian and I should not be doing so! So maybe change to read [A(b)]nm = (bn)T A bm = bn (A bm) = [bn]i(A bm)i = [bn]i Aij (bm)j and then show this same line in bra-ket notation (completeness 1 = | ui><ui| ) [A(b)]nm = <bn | A | bm > = <bn|ui><ui|A|uj><uj|bm> What facts do we have on these bn vectors? The only one is bmbn = δmn . I don't really know how to raise or lower a label. You cannot do that with any g object. Wrong! bn is a vector in x-space! Therefore at least I can say this: (bn)i = gij(bn)j So although I cannot move the label, I can move this vector index. Then we have [A(b)]nm = [bn]i Aij (bm)j = [bn]i Aij (bm)j Since A is a tensor in x-space and bm is a vector there, this is allowed. Then can write Bjm ≡ [bm]j and [A(b)]nm = Bin Aij Bjm Now Bjm is a non-tensor object but the first index goes up and down with g. There must be some "metric tensor" object call it wij which raises on the other index. but OK, leave that as is. We then have [A(b)]nm = Bin Aij Bjm = Bin Aij Bjm Maybe I am going to need more on the wij. I just reread my duality section in Section 6, it is OK, uses a instead of b. Let's try Maybe I can add this item. 7. The w matrix. Suppose the bm are a complete set of basis vectors in x-space. It must then be possible to express the dual vectors with some wnm coefficients as Bn = wnmbm. [ In standard notation. bn = wnm bm ]. Now there surely must exist some underlying transformation (call it x' = Fb(x) ) for which the bm are the tangent base vectors. Such a transformation would have linearized matrices Sb and Rb and then (bn)i = (Sb)in. For this particular transformation Fb, wnm would be the symmetric contravariant metric tensor whose components we could write out as , as in the Section 5 (b) Comment, w = RbGRbT or wnm = (Rb)nkGkk'(Rb)Tk'm = (Rb)nkGkk(Rb)mk But for some general F, although wnm remains the same symmetric matrix with the same values, wnm is no longer the metric tensor, nor is it even a tensor, but one still has An = wnmam and an = w-1nmAm so w and w-1 provide conversion between the base vectors and the dual vectors. [ In standard notation, wnm and wnm are then raising and lowering operators for the labels on the base vectors: an = wnm am and an = wnm am where the relation is wnm = (w-1)nm, just as for a metric tensor. If an = en, then wnm = gnm. OK, let's imagine that this item has been added to Section 6 (b) and I can then refer to it. Then I can add this new "fact" to my list: bmbn = δmn bm = wmnbn bm = wmnbn Is this wmn a tensor? It would be the metric tensor for a certain Fb , but how about for F ? I think it is NOT a tensor. If it were, you could argue that bm = wmnbn bm = wmnbn = gmm' wm'nbn = gmm' bm' but I know this is not true. Now let's go back to where we were: [A(b)]nm = Bin Aij Bjm = Bin Aij Bjm If we define Bjm ≡ [bm]j Then Bjm = [bm]j = gjk [bm]k = gjk Bkm Bjm = [bm]j = wmk [bk]j = wmk Bjk = Bjk wkm Bjm = [bm]j = gjn [bm]n = gjn Bnk wkm So the object B is similar to R and S in that its indices are raised and lowered by different operators. Bjm ≡ [bm]j Bjm = [bm]j = gjk Bkm Bjm = [bm]j= Bjk wkm Bjm = [bm]j = gjn Bnk wkm Maybe I need to add something to Section 7 (s) : no let's not. So now try to resume flow: [A(b)]nm = Bin Aij Bjm = Bin Aij Bjm = Bin Aij Bjm I could now lower the m index using w to get [A(b)]nm = Bin Aij Bjm but this is NOT in any kind of matrix form! Now how would you deal with transpose on this strange B object? Can we say Bin = (BT)ni ? Bjm = [bm]j = gjk Bkm (BT)mj = [bj]m = gmk Bkj Then lower the second index (BT)mj = [bj]m = gmk Bkj' wj'j But we know that Bjm ≡ [bm]j . I do NOT conclude that (BT)mj = Bjm . How does that work in the special case that b = e ??? (BT)mj = [ej]m = gmk Bkj' g'j'j It does not apply here either. The reason is that this B matrix is NEVER a tensor, it has indices in different spaces like R. So I am having a lot of trouble getting any kind of standard notation matrix concept here! I can get this [A(b)]nm = Bin Aij Bjm then I don't know what to do with it! Suppose we define in = Bin ij = Aij (b)ij = [A(b)]ij Then we can say that (b)nm = inijjm = Tniijjm = [T]nm and then (b) = T Now we have these facts : (T)ni = in = Bin ni = Bni but we do not say anything about BT here, only T . Maybe I should completely nix everything about BT in "standard notation". It is a can of worms. Suppose I have nixed that stuff then. What do I now say? I am at this point, [A(b)]nm = Bin Aij Bjm (b) = T I want to say something about linear algebra. Now let's try to "get to" the dev notation. bmbn = δmn bm = wmnbn bm = wmnbn becomes Bmbn = δmn Bm = wmnbn bm = (w-1)mnBn I arrived at this point in standard notation, [A(b)]nm = [bn]i Aij (bm)j which I could then convert to [A(b)]nm = [Bn]i Aij (Bm)j Then I could define a matrix B according to Bni ≡ [Bn]i and then we have in standard notation [A(b)]nm = Bni Aij Bmj = Bni Aij BTjm = (BABT)nm and finally A(b) = BABT Now how did this relate to my need for such a relationship? I was doing the unit vector expansion stuff. I guess I have to revamp that whole writeup now and see where it sits. ______________________________ start again _____________________________________ [A(b)]nm = (bn)T A bm = bn (A bm) = [bn]i(A bm)i = [bn]i Aij (bm)j [A(b)]nm = <bn | A | bm > = <bn|ui><ui|A|uj><uj|bm> Suppose I start instead writing things this way [A(b)]nm = (bn)T A bm = bn (A bm) = [bn]i(A bm)i = [bn]i Aij (bm)j [A(b)]nm = [bn]i Aij [bm]j Then suppose at this point I define a different B Bni ≡ [bn]i As before, g does the i index, w does the n index. Now so far we have [A(b)]nm = Bni Aij Bmj Since the second index of B goes with g, and A also goes with g, we can tilt i and j at will. so [A(b)]nm = Bni Aij Bmj Now we can make this fairly reasonable definition (BT)jm ≡ Bmj Then we have [A(b)]nm = Bni Aij (BT)jm Now lower the m index with w to get [A(b)]nm = Bni Aij (BT)jm => [ A(b) = BABT ]SN,dt Now what happens in the special case that bn = en ?? Bni ≡ [en]i = Sin = Rni Bni ≡ [en]i = Sin = Rni => B = R Then we have [ A(e) = RART ]SN,dt // which makes eminent sense. And if bn = un, then Bni ≡ [un]i = δni and then [ A(u) = 1A1T ]SN,dt which is again reasonable. Now what happens if bn = n ? I think this is the case that got all this rolling. From above n = (g'nn)-1/2 en Bni ≡ [n]i = (g'nn)-1/2 [en]i = (g'nn)-1/2 Rni (BT)in ≡ Bni = (g'nn)-1/2 Rni (BT)jm ≡ Bmj = (g'mm)-1/2 Rmj Now we have a problem. The m on Rmj lowers with g', but that on (BT)jm lowers with w. So what is w for this case: n = wnm m Start with en = g'nkek Then write en = (g'nn)1/2n ek = h'k k Then we have en = g'nkek (g'nn)1/2n = g'nk h'k k n = (g'nn)-1/2g'nk h'k k = wnk k => wnk = (g'nn)-1/2g'nk h'k => (w-1)nk = wnk = (g'nn)1/2 g'nk / h'k Then we can say (BT)jm = wmk(BT)jk = wmk (g'kk)-1/2 Rkj = (g'mm)1/2 g'mk / h'k * (g'kk)-1/2 Rkj = {(g'mm)1/2 (g'kk)-1/2 / h'k } g'mk Rkj This is a horrible mess in the general case. Of course one only wants to use this when the x'-space has orthogonal coordinates, in which case = {h'm-1 h'k / h'k } g'mk Rkj = h'm-1 g'mk Rkj = h'm-1 Rmj So go back now to [A(b)]nm = Bni Aij (BT)jm => [ A(b) = BABT ]SN,dt We have shown that Bni ≡ [n]i = (g'nn)-1/2 [en]i = (g'nn)-1/2 Rni we know that i is a g index, so this becomes Bni ≡ [n]i = (g'nn)-1/2 Rni = h'n Rni and we also showed that (BT)jm = h'm-1 Rmj Therefore we get [A()]nm = Bni Aij (BT)jm => [ A(b) = BABT ]SN,dt [A()]nm = h'n Rni Aij h'm-1 Rmj = (h'n/h'm) Rni Aij Rmj = (h'n/h'm) Rni Aij (RT)jm = (h'n/h'm) (RART)nm Here is another derivation of this result: first, assume orthogonal so that \ en = g'nmem = h'n-2en n = (g'nn)-1/2 en = h'n en = h'n h'n-2en = h'n-1en n = h'n-1 en Therefore n = n and this is a special case of bn = bn when orthonormal. Then we have A()nm = <n | A | m > = h'n h'm-1 <en | A | em > = h'n h'm-1 A(e)nm = (h'n /h'm) A'nm = (h'n /h'm) [RART]nm and we have again arrived at the same result. Now if I want to put this into my B context where we have [A(b)]nm = Bni Aij (BT)jm => [ A(b) = BABT ]SN,dt Then the pieces are Bni ≡ [n]i = h'n [en]i = h'nRni Bni ≡ [n]i = h'n [en]i = h'nRni // can lower n on B for free since n = n (BT)in = Bni = h'nRni (BT)in = Bni = h'nRni (BT)jm = Bmj = h'mRmj but we know that Rmj = g'mn Rnj = g'mm Rmj = h'm-2 Rmj = h'm-2(RT)jm so that (BT)jm = Bmj = h'mRmj = h'm-1(RT)jm Then we get [A(b)]nm = Bni Aij (BT)jm = h'nRni Aij h'm-1(RT)jm = (h'n/h'm) (RART)nm or [A()]nm = (h'n/h'm) A'nm = (h'n/h'm) [A(e)]nm So we must only do orthogonal when talking n stuff. What about other index positions? They are all the same!!! [A()]nm = <n | A | m > = (h'n/h'm) [A(e)]nm [A()]nm = <n | A | m > = <n | A | m > = (h'n/h'm) [A(e)]nm etc This is true for any tensor done on orthogonal unit vectors. Now of course we know that [A(e)]nm = g'mk[A(e)]nk = h'm2[A(e)]mk and then we get [A()]nm = all other positions = (h'nh'm) [A(e)]nm and this is in agreement with my general unit vector expansion writeup.