Sec 8 issues

An erroneous Section 8 (d) rewrite attempt PhL 2.24.12 This doc treats a few different subjects: (this added 2.28.12) (1) the notion that the finite N-pipeds don't map into each other. (2) a "bug" in an area equation that I fixed by taking |J| → J in an Appendix D rewrite. (3) the notion that dV can be positive or negative in x-space, but dV' is always positive in x'-space. (4) the subject of absolute values in a valid tensor equation. (5) confusion about the theorem RARB = AB. Overview of the first subject of this doc: At one point along the way, in "comments on Section 8", I wrote a reasonable-sounding version of section 8 (d) which is quoted between the lines far below. In that section, I start with the finite N-piped in x-space which has face areas An and volume V. I write out each of these quantities in terms of the ε tensor and the en: (An)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x // en missing V = εabc..x(e1)a(e2)b.... (eN)x I then more or less "conjecture" that my two equations are "covariant" and therefore in x'-space you can just prime everything (including the ε) to get these new equations, (A'n)i = σ (-1)n-1 ε'iabc..x (e'1)a(e'2)b.... (e'N)x // en missing V' = ε'abc..x(e'1)a(e'2)b.... (e'N)x I then install the e'n and ε' objects, do the math, and conclude that (A'n)i = σ |J|2 δni A'n = σ |J|2 e'n V' = |J|2 I then claim that A and V are tensor densities of weight -1 and transform like this A'n = |J| RAn meaning (A'n)i = |J| Rij(An)j V' = |J| V I show that everything is consistent by directly computing A'n and V' from both sets of equations. But it is all WRONG, and I want now to explain why the above "conjecture" is incorrect. First, suppose someone rolls out the vector equation x3 = x1+ x2 in x-space and then conjectures that the equation is covariant and therefore one should have x3' = x'1+ x'2 in x'-space. To check this, we apply R to the first equation to get Rx3 = Rx1+ Rx2 . If it were true that xi' = Rxi , then we would end up with x3' = x'1+ x'2 and the conjecture would be valid. But in general xi' ≠ Rxi so the conjecture is "no good". The "mapping" xi' = Rxi conflicts with the mapping xi' = F(xi). The problem of course is that the position "vector" of a point in space, which is to say x, is not a tensorial vector. This is what is happening above. You can prime the (An)i = ... equation, but the (A'n)i which results is NOT the mapping into x'-space of the finite N-piped face n!! We know that a finite N-piped face in x-space maps into some horrible twisted surface in x'-space under the full mapping F. Thus, if you look at the individual points on area An, you get a conflict between the R mapping and the F mapping! It is not just a matter of vector versus vector density, the whole thing is wrong! Those individual points are points in space, and for these points we are having the exact same problem as in the previous paragraph. The same is true for V. The finite N-piped V maps into some complicated shape in x'-space and the points in the volume again have conflicting mappings between R and R. So V' = J| V is just plain wrong. This has been a somewhat subtle point for me to absorb. The problem arises because we are mapping actual coordinates of space, not things like velocities which might be true vectors. On the other hand, the differential version of the above argument is correct! The reason is that for the differential case, there is no difference between R mapping and F mapping. Notice that for the differential situation we have all these things working in our favor: (1) the R and F mapping of points in configuration space agree since restricted to a tiny region (2) dA and dV are both tensor densities of weight -1. (3) therefore both equations (ie, for dA and dV) are valid tensor density equations and are covariant (4) therefore you CAN put primes on everything as advertised. No need to verify. [ the above four items are all correct.] Here then is the erroneous section I wrote for the finite objects: _____________________________________________________________________________ (d) Objects in x'-space Note that x-space is treated as Cartesian in this section, so g = 1. Objects A'n and V'. In Section (b) it was stated that, with regard to the N-piped in x-space, An = σ (-1)n-1 e1 x e2 ... x eN V = | det [ e1, e2, e3 ... eN] | = | det(Sab) | = g'1/2 = |J| As a reminder, these are the nth far face area and volume of a finite N-piped drawn with the tails of the spanning en vectors located at a point x in x-space corresponding to point x' in x'-space. One can and usually does think of en and therefore An and V as functions of the curvilinear coordinates x'. One wonders what these two objects look like in x'-space. To find out, write both objects in terms of an ε tensor as follows: (An)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x // en missing V = |εabc..x(e1)a(e2)b.... (eN)x| In x'-space one will then have [ put primes on everything and recall that σ ≡ sign[det(Sab)] = ±1 has to do with the handedness of the transformation F and is just a constant, as is (-1)n-1 ] (A'n)i = σ (-1)n-1 ε'iabc..x (e'1)a(e'2)b.... (e'N)x // en missing V' = | ε'abc..x(e'1)a(e'2)b.... (e'N)x | Appendix D (e) item 3 discusses the nature of the ε tensor with regard to transformations, and shows that if one assumes that ε'abc... = εabc... = εabc...= the usual permutation tensor, then one must have ε'abc... = |J|2 εabc... |J|2 = g' = det(g'ij) Section 7 (s) states that (e'n)i = δni. Installing these two facts, the above equations become (A'n)i = σ (-1)n-1 |J|2 εiabc..x δ1a δ2b.... δNx // δn* missing V' = |J|2 εabc..x δ1a δ2b.... δNx so that (A'n)i = σ (-1)n-1 |J|2 εi123..N // index n missing on εi123..N V' = |J|2 ε123..N = |J|2 One the first line, if i ≠ n, then εi123..N = 0 because two indices must be the same (since n missing). On the other hand, if i = n, one finds εi123..N = εn123..N = (-1)n-1 ε123..n...N = (-1)n-1. Therefore (A'n)i = σ |J|2 δni σ ≡ sign[det(Sab)] A'n = σ |J|2 e'n // since (e'n)i = δni (*) V' = |J|2 |J|2 = g' = det(g'ij) = | det(Sab) |2 These objects are then the (vector area of far face n) and volume of our x'-space N-piped. [ wrong !!!] It must be said that the physical interpretation of "area" and "volume" of objects in x'-space with g' ≠ 1 is a mental challenge. Making some kind of reasonable Cartesian-space drawing ( even in 2D or 3D) of an object sitting in non-Cartesian space is frustrating, as demonstrated in Appendix C (e) for a simple 2D curvilinear system, where one has difficulty even drawing a simple vector, much less an area or volume. Despite this difficulty of geometric interpretation, the objects A'n and V' are completely well defined by the various equations stated above. [ this paragraph has some truth but is misdirected here. It is true that the picture is hard to imagine, but that is not why this section is wrong.] Going back to the second pair of equations above, (An)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x // en missing V = | εabc..x(e1)a(e2)b.... (eN)x| since the various en are true tensorial vectors (that is, weight 0), and since it is known that εabc... is a tensor density with weight -1 ( as shown in App. D (d) ), then using the rule of App. D (b) item 3 concerning adding weights, one quickly concludes that An is a vector density of weight -1 and V is a scalar density of weight -1. This implies at once that A'n = |J| RAn meaning (A'n)i = |J| Rij(An)j (***) V' = |J| V Since V = |J| (2nd equation of this section) the second line gives V' = |J|2, consistent with what was just found above. Then using the fact from section (b) that An = |det(Sab)| en = |J| en One finds that (An)i = |J| (en)i = |J| Rni and therefore (A'n)i = |J| Rij(An)j = |J| |J| Rij Rnj = |J|2 δin (**) Bug: why is there no σ here but there was above? Fix it later!!! ( See note added below line below) which again agrees with the previous result above. The covariant magnitude of A'n is determined by | A'n |2 = A'n A'n = g'ab(A'n)a(A'n)b = |J|4 g'ab δan δbn = |J|4g'nn = |J|4h'n-2 so that | A'n | = |J|2 h'n-1 ________________________________________________________________________________ Concerning the above bug noted in red. The "bug" is that I first get (A'n)i = σ |J|2 δni but then later I get that (A'n)i = |J|2 δin with no σ. After I repaired Appendix D replacing |J| → J in all the tensor density stuff, look how things now work starting at (***) above which is the first place we have J not squared: A'n = J RAn meaning (A'n)i = J Rij(An)j (***) V' = J V // corrected tensor density transform equations Since V = |J| (2nd equation of this section) the second line gives V' = J |J| = σ |J|2, consistent with what was just found above. Then using the fact from section (b) that An = |det(Sab)| en = |J| en One finds that (An)i = |J| (en)i = |J| Rni and therefore (A'n)i = J Rij(An)j = J |J| Rij Rnj = σ |J|2 δin (**) Equation (**) now DOES have the σ, so this bug has gone away! New Bug with Volume. When I write V' = J V for the tensor density transformation, we can end up with V' < 0 and that seems sort of reasonable if we have a parity transformation. But look at the starting equation, V = | det [ e1, e2, e3 ... eN] | = | det(Sab) | = g'1/2 = |J| If we regard V = | det [ e1, e2, e3 ... eN] | as covariant, we get V' = | det [ e'1, e'2, e'3 ... e'N] | which would then say V' > 0. So look carefully at this equation V = |εabc..x(e1)a(e2)b.... (eN)x| Is this a valid tensor density equation? The weights match, the indices match, but I don't think that absolute value is "allowed" in a valid tensor equation! Consider a simpler case V(x) = | a(x) b(x) | = | g'ij ai(x)bj(x) | Is this RHS really a scalar field? In the neighborhood of a place where ab = 0 , it seems there will be something discontinuous. Just as in f(x) = |x-2|2 there is a discontinuity at x = 2. I think vector fields have to be continuous but I have said nothing about this. Could I show that the above is NOT a scalar. I don't know what to do about the | | signs when I prime things. In the case of | v | = (vv)1/2, I am happy to covariant this into | v' | = (v'v')1/2 . But in the above case, I don't have a power anywhere. [ probably | | signs just pass through as is, like (-1)n-1] Question: suppose S is a real scalar. Then define Q = | S | I think you have to agree that Q is also a scalar. So I cannot really argue that my physical volume element given by V = |εabc..x(e1)a(e2)b.... (eN)x| is not a "scalar". Well, I know that if we have a right handed g system, V will be V > 0 and I don't need the |...|. But if I want to allow left-handed systems, then V itself could be negative. For purposes of covariance of tensor densities, maybe I should use this definition [ yes ] V = det(S) = J Then we get from the tensor density thing, V' = JV = J2 > 0 This says that V' > 0 no matter what. Maybe this would be our cube volume in x'-space. [ yes ] So let's summarize our volume situation: dV = det [ e1, e2, e3 ... eN] Πidx'i = det(Sab) Πidx'i = J Πidx'i = J dV' Notice that the x'-space volume dV' = Πidx'i is always positive since I assumed that all dx'i > 0. This is the volume of a right-handed N-piped in x'-space. The dV sign in x-space of the en differential N-piped depends on whether there has been a change in coordinate system. Now J2 = g'/g for sure. For a left handed system, J < 0 for sure (my section on this), and dV < 0 in this case. so J = σ(g'/g)1/2 seems a good way to solve J2 = g'/g. Then we continue the above = σ(g'/g)1/2 Πidx'i Now let's go look at the Weight Changing Theorem where I think this idea caused a problem (separate doc). This theorem considers this equation Uabcde = k (sg)w/2 Tabcde = k |g|w/2 Tabcde Since g' = J2g, we say g has weight -2. We also can write (sg') = J2(sg) so (sg) has weight -2. The problem arises when you try to take a square root. We cannot have abs value in a tensor density equation, so we have to make a decision. We first write (sg')1/2 = J (sg)1/2 . What is the meaning of J here? If J < 0, that means our en form a left handed system in x-space. What is going on in x'-space? det [ e'1, e'2, e'3 ... e'N] = ε123.. = +1 So x'-space I discover is always right handed! [ that is correct ] Note: I still need to review why my Weight Changing Theorem math was not working, another doc. Start Over on Volume and consider the Two Ways Way #1: [rejected] use physical volume for the N-piped in x-space. Then dV = |detS| Πidx'i > 0 . Then argue that dV = | dAn dx(n) | = | |det(Sab)| en Πi≠ndx'i en dx'n | = | det(S) | Πidx'i = as above! What then is dV'? Well, we know that dA'n = J R dAn dx'(n) = R dx(n) Therefore dA'n dx'(n) = J R dAn R dx(n) = J R {|det(Sab)| en Πi≠ndx'i } R { en dx'n } = J |J| Πidx'i Ren Ren What is the meaning of this dot? It is surely the dot in x'-space so Ren Ren = g'ab [Ren]a[Ren]b = g'abRac(en)c Rbd(en)d = g'abRac(en)c Rbd(en)d = g'abRacRncRbd Rnd = g'abRacRncδbn = g'anRacRnc = g'anRacRnc = g'anRacRnc = g'an δan = g'nn = 1 Did I know this already REn Ren = E'n e'n = En en = δn,n = 1 yes I did know that. So moving right along, dA'n dx'(n) = J |J| Πidx'i Ren Ren = J |J| Πidx'i But I am trying to determine what dV' is. Of I make this definition dV' ≡ dA'n dx'(n) = J |J| Πidx'i and if I use my result above which says dV ≡ | dAn dx(n) | = |J | Πidx'i Then I find that dV' = J dV and I can then state that dV is a scalar density of weight -1. So to make this work, I have to define dV with the abs and dV' without the abs. Way #2: [ correct ] Use non-physical volume for the N-piped in x'-space. Then dV = detS Πidx'i = could have either sign Then try dV = dAn dx(n) = |det(Sab)| en Πi≠ndx'i en dx'n = | det(S) | Πidx'i > 0 Already we have a problem! This second dV > 0 but the first is not! I think if you go this path, you have to remove abs from the dA as well. A3 = σ e1 x e2 σ = sign [ detS ] A3 = + e1 x e2 A3 = – e1 x e2 Comment: This entire doc was written before I updated the handedness section on 2/26. That made clearer that dV' must always be > 0 and then sign (dV) = sign(detS). _________________________________________________________________ Appendix 1. Aside: is this a theorem? RARB = AB Answer: If A and B are vectors, then it is a theorem. The reason is that RA = A' is a vector in x'-space which has metric tensor g'. I did that all wrong below!!! The proof is then trivial: RARB = A'B'= AB since AB is a scalar. Question: If one writes X = RA, is X a vector in x-space or in x'-space? Argument #1: The equation X = RA means that Xi = RijAj. The components of X are then just a linear combination of the components of Aj . But this does not say that X is a linear combination of vectors which lie in x-space! That was going to be my argument #1, but I have already invalidated it. Argument #2: We know that A'(x') ≡ RA(x) is a vector in x'-space, which means the components A'i are those of a vector in x'-space. At point x, R is the same as transformation F, and so this is what we mean by something being in x'-space. Then since X = A', X must also be a vector in x'-space. Therefore, in a dot product RARB it is the g' metric tensor that should be used, which gives the above result. You could write things this way if you wanted: RARB = AB = A(R-1R)B and then this justifies moving the left side R to the right side where it is written as R-1. RARB = A(R-1R) Meanwhile, here is my bad proof where I used g in the left side dot product: Try to prove it in a generic non-Cartesian x-space with some general g: RARB = gab(RA)a(RB)b = gabRaa'Aa'Rbb'Bb' = (gabRaa'Rbb') Aa' Bb' If it were true that gabRaa'Rbb' = ga'b' that is to say abRaa'Rbb' = a'b' Then yes, our theorem would be true. Is this ever true? Well, consider further abRaa'Rbb' = a'b' Raa'abRbb' = a'b' RTa'aabRbb' = a'b' RT R = I know that g' = R g RT and = RT ' R So this would be true if ' = meaning both spaces have the same metric tensor. That in turn means it is true of R is real orthogonal, ie, a rotation. So NO, this theorem is not in general true.