More on Basis Vector Paradoxes REVIEWED

More on Basis Vector Paradoxes PhL 3.15.16 5.16.16: Lots of painful stuff here, working on how to annotate vector components. I used to have ei as the default basis and this really messed things up, such as Paradox far below. I later came to accept axis-aligned as the default basis in both x-space and x'-space, but that is in a later doc I think. It was a very jumbled confused world before I put sticks in the sand. I attempted to shed light on this topic in wedge Chapter 2, but I am still being confused, so here I will try harder to clean this mess up. I will try to do this by asking and then answering a set of questions. I think this method worked and has resulted in a new Section 7.19 in tensor doc where things are in fact cleaned up. The work below was the basis for that new section, but I did have a separate Section 7.19 working doc as well which is now frozen. So this doc is fully understood, no need to peruse it any more. [ 3.17.16] Question #1. In (7.18.1) I see lots of component things like (en)i = Sin. Are all these components in the (u) basis? I think they are, but how would I prove that fact? Answer is now installed into tensor doc: Fact: un and en components are by default presented in the u-basis (7.19.10) e'n and u'n components are by default presented in the e'-basis I leave my original text below, you can see it is totally confused. Go back to the equation in wedge doc below (2.3.3) en ≡ ∂x/∂x'n (en)i = ∂xi/∂x'n = Rni . // from (2.1.5) To which basis is this (en)i component referred? The index i is associated with xi . But xi has no prime, so we know that x = xiui would be the associated expansion, so xi is really x(u)i and this (en)i = (en(u))i. So that is one complicated way "to tell". So I have just shown that (en)i = (en(u))i = Rni and of course I know that (en(e))i = δni but this item does not appear in (7.18.1). So I think we can say that Here is a whole batch of stuff from (7.18.1), (en)i = Sin = Rni (en)i = gijRnj = g'njSij (en')i = δni (e'n)i = g'ni (en)i = gijSjn = Rjig'jn (en)i = Rni (en')i = g'ni (e'n)i = δni I guess I am happy now that all the left four are (u) type, so I would say (en)i = Sin = Rni = (en(u))i (en)i = Rni = (en(u))i (en)i = Sin = Rni = (en(u))i (en)i = Rni = (en(u))i Now what about the primed ones on the right? Do we have any definition like (en)i = ∂xi/∂x'n = Rni for the primed vectors? What is the expansion that goes with (e'n)i ? This is a vector in x'-space so I have to call up this data, V = Vn un = Vn un = V'n en = V'n en = V'n n // x-space expansions, V'n = h'nV'n V' = V'n e'n = V'n e'n = Vn u'n = Vn u'n . // x'-space expansions (7.13.12) So for an x'-space vector here is more detail from tensor doc, V' = V'1 e'1 + V'2 e'2 +... = ΣnV'n e'n where e'n V' = V'n e'n = g'ni e'i V' = V'1 e'1 + V'2 e'2 +... = ΣnV'n e'n where e'n V' = V'n V' = V1u'1 + V2u'2 +... = Σn Vn u'n where u'n V' = Vn u'n = gni u'i V' = V1u'1 + V2u'2 +... = Σn Vn u'n where u'n V' = Vn (7.13.11) Then take the second line and write V' = V'i e'i where e'i V' = V'i How would you apply this to the vector V' = (e'n)? I don't think I have ever pondered this. For one thing, I guess you would say V'i = (V')i= (e'n)i so there are no double primes going on. Then we have e'i V' = V'i e'i e'n = (e'n)i V' = V'i e'i e'n = (e'n)i e'i From this last we see clearly that the answer is this (e'n)i = (e'n(e'))i (e'n)i = (e'n(e'))i and so on. So these components are " e' type components", neither e type nor u type. So the last four equations could be written (en')i = δni = (en'(e'))i (e'n)i = g'ni = (e'n(e'))i (en')i = g'ni = (en'(e'))i (e'n)i = δni = (e'n(e'))i Now let's look at some corresponding u' stuff. Try this on V' = u'n . We read off, V' = Vi u'i where u'i V' = Vi What is the meaning of Vi in this case? Vi = u'i V' = u'i u'n = δin So then the expansion is V' = Vi u'i u'n = δin u'i // which seems correct So one conclusion I guess is that (u'n)i = δni = (u'n(u'))i But what I show is this, (u'n)i = Rin So where does that come from? It is a massive copy and paste thing, and this surely must be (u'n)i = Rin = (u'n(e'))i So we get this parallel stuff going on: (en)i = (en(u))i = Rni (en(e))i = δni (u'n)i = (u'n(e'))i = Sni (u'n(u'))i = δni and recall those rules, g'↔ g R ↔ S en → u'n e'n → un en → u'n e'n → un (7.18.2) This simple rule set really explains everything. Question #2. Consider these equations from tensor doc (3.2.4), en ≡ Se'n . (3.2.4) e'n = Ren I often write this out as follows ("component equations") (en)i = Sij(e'n)j (*) (e'n)i = Rij(en)j Are these (e) type vector components, or are they (u) type, or some other type? As shown in the answer below, the (en)i are u-type while the (e'n)i are e'-type. In more detail one would write (en)(u)i = [S(u,e')]ij(e'n)(e')j = Sij (e'n)(e')j (e'n)(e')i = [R(e',u)]ij(en)(u)j = Rij (en)(u)j Everything is in its natural basis, so my (*) is in fact correct. Answer: In Question 1's answer we had (en)i = (en(u))i = Rni (en')i = (en'(e'))i = δni First of all, let's just try these and see if they "work" : (en)i = Sij(e'n)j (en(u))i = Sij(e'n(e'))j Rni = Sij δnj = Sin works! (e'n)i = Rij(en)j (en'(e'))i = Rij(en(u))j δni = RijRnj = δin works! So one way to do interpretation is this: en ≡ Se'n . (3.2.4) e'n = Ren (en)i = Sij(e'n)j means (en(u))i = Sij(e'n(e'))j (e'n)i = Rij(en)j means (en'(e'))i = Rij(en(u))j Notice that you do NOT use the same indicator on both vectors in either equation. Rule: Wherever you see a δij result, you know the type matches the vector. For examples: (un)i = δni (un)i is u-type (un)i = ui un = <ui | un > and this implies that the other three combinations are u-type (en')i = δni (en')i is e'-type (en')i = e'i en' = <e'i | en' > and this implies that the other three combinations are e'-type On the other hand, (en)i = Rni (en)i is u-type (en)i = ui en = <ui | en > and this implies that the other three combinations are u-type (u'n)i = Sni (u'n)i is e'-type (un')i = e'i un' = <e'i | un' > and this implies that the other three combinations are e'-type Summary of Summary: un and en components are presented in the u-basis e'n and u'n components are presented in the e'-basis I think this is a good summary of basis vectors and their types. Restate again: (un)i = ui un = <ui | un > = gni // = δni um = gmn un (en')i = e'i en' = < e'i | en' > = g'ni // = δni e' m = g' mn e'n (en)i = ui en = <ui | en > = Rni em = g' mn en (un')i = e'i un' = <e'i | un'> = Sni u' m = gmn u'n This tableau and the R and S tableaus verify to me that you can raise and lower the label n in each equation to get a new valid equation. What about the index i? You raise and lower this index with g for a non-primed case, and with g' for a primed case. Fact: In the above, you can raise and lower either index on both sides at will. Ignoring label positions, there are a total of 4 x 4 dot products of interest, and 4 are listed above. What are the other 12? ui e'n = <ui | e'n > = dot product of vectors in different spaces Does this even make sense? A scalar product is defined within a Hilbert Space to make it be a H.S. So we have a scalar product within x-space and within x'-space, so I think the above does NOT make sense. If you try to compute it using say 1 = |ei><ei| you always end up somewhere with a cross-space scalar product, so you cannot compute it. Which metric tensor do you use! Let's list off the kinds of dot products u e u' e' u uu ue uu' ue' e eu ee eu' ee' u' u'u u'e u'u' u'e' e' e'u e'e e'u' e'e' Now first put redundant ones in blue u e u' e' u uu ue uu' ue' e eu ee eu' ee' u' u'u u'e u'u' u'e' e' e'u e'e e'u' e'e' So upper right triangle is redundant. Now put illegal ones in red u e u' e' u uu -- -- -- e eu ee -- -- u' u'u u'e u'u' -- e' e'u e'e e'u' e'e' There are then only 6 distinct legal dot products! We have listed four of them above, so two are missing: Make the ones appearing above green u e u' e' u uu -- -- -- e eu ee -- -- u' u'u u'e u'u' -- e' e'u e'e e'u' e'e' The two missing ones are shown in black. But we know you can remove both primes. But let's just add to our table naively (e'n)i = ei en = <ei | en > = g'in (un)i = u'i u'n = < u'i | u'n > = gin I know that (e'n)i is in the e'-basis. So (e'n)i = <e'i | e'n >, and that agrees. I know that (un)i is in the u-basis. So (un)i = <ui | un >, and that agrees. So how should the full table be presented? (un)i = ui un = <ui | un > = gni = u'i u'n = < u'i | u'n > (en')i = e'i en' = < e'i | en' > = g'ni = ei en = <ei | en > (en)i = ui en = <ui | en > = Rni (un')i = e'i un' = <e'i | un'> = Sni This then is my first complete table of meaningful scalar products, keeping in mind (a,b) = (b,a) and keeping in mind that the label n and the index i can be raised or lowered at will on both sides of any equation. Next, I want to think about matrix elements or R and S operators, it that is meaningful! We start with these two facts: e'n = R en Tensor (3.5.2) u'n = R un Tensor (3.5.3) These are just normal V' = RV type equations, don't need tensor doc to know these two equations. Write as |e'n> = R |en> |u'n> = R |un> We can close in various ways. For example, <e'm |e'n> = <e'm |R |en> = δmn <u'm |e'n> = <u'm |R |en> = Smn = Rnm // (um')n = e'n um' = <e'n | um'> = Smn Now start instead with the second equation and close two ways <e'm |u'n> = <e'm | R |un> = Snm = Rmn // <e'm |u'n> = <u'n |e'm> = Rmn = Snm <u'm |u'n> = <u'm | R |un> = δmn So I now know these matrix elements of R <e'm | R | en> = gmn = <en | S | e'm> <u'm | R | en> = Rnm = <en |S | u'm> <e'm | R | un> = Rmn = <un | S | e'm> <u'm | R | un> = gmn = <un | S | u'm> No other matrix elements of R make any sense. Recall that S = RT covariant notation We can then reverse the above items to get what is shown on the right. So the above table shows the only meaningful matrix elements of R and S. Paradox Resolution Paradox: Now armed with all of the above, let's take a look at our Paradox in Picture A notation: On the one hand, according to Argument 1: < u'i | R ua> // vector on the right exists in x'-space 1 = [R ua]i // idea that < u'i | q > = qi for any vector in x'-space (in u'i expansion) [ but [R ua]i is in fact < e'i | R ua> since R ua exists in x'-space, so the above step is wrong! ] 2 = Σj=1n Rij(ua)j // matrix * vector multiplication 3 = Σj=1n Rij δaj // definition of the basis vectors in x-space 4 = Ria // use up the delta function On the other hand, according to Argument 2 Trying to concentrate the paradox more < u'i | R ua> // vector on the right exists in x-space 5 = < u'i | u'a> // idea that R |ua> = | u'a> for a = 1..n, definition of the | u'a> // this is (10.6.4) of Section 10_6 v.3.doc, but up instead of down 6 = δia // property of regular with dual vectors (see Task A below) Let's now do the two arguments in Dirac notation. First, just "look up" the matrix element: < u'i | R | ua > = δia // <u'i | R | ua> = gia This it would seem that argument 2 is the one that is correct. So what is wrong with Argument 1 ? Example steps one at a time < u'i | R ua> // vector on the right exists in x'-space 1 = [R ua]i // idea that < u'i | q > = qi for any vector in x'-space (in u'i expansion) Since u'a = R ua, we know that R ua is a vector of type u'a and for such a vector, (u'a)i is a component of type e'. So we conclude that: [R ua]i is an e'-type component. Nothing wrong yet. Next step : 1 = [R ua]i // idea that < u'i | q > = qi for any vector in x'-space (in u'i expansion) 2 = Σj=1n Rij(ua)j // matrix * vector multiplication Well, the idea here is that we are thinking about the vector equation u'a = R ua where R is a matrix. The problem I am sure is that the matrix elements of R are going to depend on what basis you use for your vectors. Consider a generic matrix equation in x-space which has axis-aligned basis vectors un. Suppose this transformation takes an x-space vector a and creates a new x-space vector b, a = Mb |a> = M |b> <un| a> = <un|M |b> = <un|M |um> <um| b> an = [M(u)]nm bn OK, now suppose we select a different basis in x-space, perhaps the en . The above becomes a = Mb |a> = M |b> <en| a> = <en|M |b> = <en|M |em> <em| b> an = [M(e)]nm bn The matrix equation looks the same in that it is still a = Mb, but the M matrix elements are completely different! Now let's repeat the generic example with this difference: in a = Mb you start with a vector b in x-space and this generates a vector a' in x'-space. You then have a' = Mb |a'>x' = M |b>x Assume that un is a basis in x-space, and u'n is a basis in x'-space. Then we get x'<u'n |a'>x' = x'<u'n | M | b>x = x'<u'n | M | um>x x< um| b>x which we write as a'n = [M(u',u)]nm bm This is exactly the situation with our operator R, so we have in that case: We have the form a' = Rb, but the matrix is R(u',u): [R(u',u)]nm = <u'n | R | um> = δnm // <u'n | R | um> = gnm Thus we have [R(u',u)]nm = δnm The correct steps in Argument 1 would then be 1 = [R ua]i // idea that < u'i | q > = qi for any vector in x'-space (in u'i expansion) 2 = Σj=1n [R(u',u)]ij(ua)j // matrix * vector multiplication 3 = Σj=1n δij δaj = δia and now Argument 1 gives the same result as Argument 2, Paradox goes away.