Alt derivation of Forms Integration REVIEWED

Alt Derivation of Forms Integration PhL 4.28.16 This is my "alternate explanation" using bras and kets as to why you can replace dxi by dxi in the second definition of the forms integration. I do some simple examples here too. This is now installed in Section 10.11 // 5.17.16 A. General Theory. 1. First, use R to pull back the x'-space form <∫S' α' | which I regard as a sum of forms over the x'-space manifold. The result is then a form in x-space. <∫S' αx' | R = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) λ^J = ∫S Σ'I fI(F(x)) Σ'M det(RIM) <u^M | // ordered M sum = ∫S Σ'I fI(F(x)) ΣM RIM <u^M | // full M sum 2. Use the following x-space ket for closure | q > ≡ Σ'J | dxJ> = Σ'J [| dxj1> | dxj2> ..... | dxjk>] = Σ'J | dxj1, dxj2, .... dxjk> where dxi = dxiui Note that <∫S' α' | R = a dual rank-k tensor in Λk(Rn) = a bra in Λk(Rn) V*k(Rn) | q > = Σ'J | dxJ> = a rank-k tensor in Vk(Rn) = a ket in Vk(Rn) The tensor product scalar product is then <∫S' α' | R | q > = ∫S Σ'I fI(F(x)) ΣM RIM <u^M | Σ'J | dxJ> = Σ'I Σ'J ∫S fI(F(x)) ΣM RIM <u^M | dxJ> = Σ'I Σ'M Σ'J ∫S fI(F(x)) ΣM RIM dxjdxj....dxj <u^M | uJ> The last scalar product is given by <u^M | uJ> = λ^M(uJ) = (λm1 ^ λm2 ^ ... ^ λmk) (uj,uj, ...uj) = (1/k)! det [ δjm ] // non-Spivak normalization for wedge product = det [ δjm ] // Spivak normalization for wedge product = det [ δJM ] Using the Spivak normalization we then have <∫S' α' | R | q > = Σ'I Σ'J ∫S fI(F(x)) ΣM RIM det(δJM ) dxjdxj....dxj Now consider ΣM RIM det(δJM ) = ΣM RIM ΣP(-1)P δP(J)M = ΣP(-1)P ΣM RIM δMP(J) = ΣP(-1)P RIP(J) = det(RIJ) The final result is then <∫S' α' | R | q > = Σ'I Σ'J ∫S fI(F(x)) det(RIJ) dxjdxj....dxj B. Apply to a 1-form situation with x-space = R2 and x'-space = R3 as the left and right spaces. The R-matrix has 3 rows and 2 columns so is 3x2. α' = Σi=13 fi(x') λ'i = Σi=13 fi(x') dx'i x'-space is R3 <∫S' αx' | R = ∫S Σi=13 fi(F(x)) Σm=12 Rim <um | The closing ket is this: | q > = Σ'J | dxJ> = ( | dx1> + | dx2> ) We then need this scalar product <um | { dx1 | u1> + dx2 | u2> } = (dx1 δm1 + dx2 δm2) Then <∫S' α' | R | q > = ∫S Σi=13 fi(F(x)) Σm=12 Rim (dx1 δm1 + dx2 δm2) = ∫S Σi=13 fi(F(x)) (dx1 Σm=12Rimδm1 + dx2 Σm=12Rimδm2) = ∫S Σi=13 fi(F(x)) (dx1 Ri1 + dx2 Ri2) How can I interpret this result? Rab ≡ (∂x'a/∂xb) so dx1 Ri1 + dx2 Ri2 = dx1 (∂x'i/∂x1) + dx2 (∂x'i/∂x2) ≈ dx'i So write result as <∫S' α' | R | q > = ∫S Σi=13 fi(F(x)) dx'i = ∫S f(x') dx' = line integral. which seems to be a reasonable result. Note that x' = F(x) is arbitrary as long as smooth, and the curve in x-space called S is an arbitrary smooth curve in a plane positioned as you like, and S' is the resulting curve in R3. At some point along curve S' at some x' you will have this tangent base vector (u'1)i = Ri1(x') The other two are chosen to be orthogonal to this one. This u'1 points along the curve in x'-space and so is like a velocity vector. I think if you were to compute '1 ≡ u'1 / |u'1| then if you choose f(x') = '1. the integral will give you the arc length of the curve. |u'1|2 = Σi=13(u'1)i (u'1)i = Σi=13(Ri1)2 = (R11)2 + (R21)2 + (R31)2 |u'1| = '1 = u'1 / fi(F(x)) = fi(x') = ( '1)i = ( '1)i = Ri1(x') / Then <∫S' α' | R | q > = ∫S Σi=13 fi(F(x)) (dx1 Ri1 + dx2 Ri2) = ∫S Σi=13 Ri1(x') / * (dx1 Ri1 + dx2 Ri2) and I think this would be the arc length of the curve S'. This example highlights several things : 1. It is an example of doing an integral where k < n. In this case we have k = 1 and n = 2. This causes there to be more than one term in the J sum, here there are two terms. I have a drawing which illustrates this type of mapping left to right with red curves. 2. The two terms in the J sum have a simple interpretation. One picks up the contribution due to the horizontal extent of the integration and the other to the vertical extent, in the R2 plane. You cannot omit either term unless it happens that S is vertical or horizontal. Example 1: Suppose S = [0,1] in R2 so dx2 = 0. Want S' to be a circle of radius a in R3 in the x'1-x'2 plane. Then x'1 = acos(2πx1) x'2 = asin(2πx1) x'3 = 0 Rab ≡ (∂x'a/∂xb) R11 = (∂x'1/∂x1) = -2πasin(2πx1) R12 = (∂x'1/∂x2) = 0 R21 = (∂x'2/∂x1) = 2πacos(2πx1) R22 = (∂x'2/∂x1) = 0 R31 = (∂x'3/∂x1) = 0 R32 = (∂x'3/∂x2) = 0 (R11)2 + (R21)2 + (R31)2 = (2πa)2sin2(2πx1)+ (2πa)2cos2(2πx1) + 0 = (2πa)2 = 2πa <∫S' α' | R | q > = ∫S Σi=13 ( Ri1(x') / 2πa) dx1 Ri1 = (1/2πa) ∫S (Ri1)2 dx1 = (1/2πa) !Syntax Error, IΣi=13(Ri1)2 dx1 = (1/2πa) !Syntax Error, I(2πa)2 dx1 = 2πa !Syntax Error, Idx1 = 2πa which is the correct answer. Compare two interpretations of "functional" for the general 1-form case <∫S' α' | R | q > = ∫S Σi=13 fi(F(x)) (dx1 Ri1 + dx2 Ri2) For each S' we get this result as a real number, so ∫S' α' α' : S' → R Here the functional α' acts on curves S' in a space of curves to create a real number. It is not a linear functional because somehow doubling S' does not double the resulting integral because the function fi(x') is arbitrary. But we can consider the contribution to the above mapping of a little piece of the curve S' or S and we get ∫S' α' = Σi=13 fi(F(x)) (dx1 Ri1 + dx2 Ri2) due to little piece dx1,dx2. For a small curve piece, the mapping is linear because you double dx1 and dx2.