Section 10_13

Section 10.13 Integration of Differential Forms PhL 4.13.16 Installed on 5.2.16 do not edit. 5.2.16 final 10.13 Integration of 2-forms We wish now to look in more detail at the integration of 2-forms. The general k-form integration result is stated in (10.12.1). Once again, there is much repetition below intended to reinforce the meaning of various objects. For the moment we set m = 3 and consider this 2-form in x-space R3 , αx = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I = Σ1≤i<i≤3 fii(x) dxi ^ dxi = f12(x) dx1 ^ dx2 + f13(x) dx1 ^ dx3 + f23(x) dx2 ^ dx3 . (10.13.1) We wish to define a meaning for the integration of this 2-form α over a piece of the surface x = φ(t), ∫φ αx = ∫φ Σ'I fI(x) dxi ^ dxi = integral of a 2-form over the surface φ in R3 . (10.13.2) The transformation x = φ(t) is a mapping φ: (t1,t2) → R3 where t1 and t2 are "parameters". Comment: Officially it is the mapping φ which is the "surface", but we loosely refer to the image (trace) of this mapping in R3 as "the surface". The distinction is necessary because many mappings can have the same image surface, such as φ(t1,t2) and φ(t12,t22) where the parameters are "respeeded" (reparametrized). If the integral of the 2-form αx is the same over surfaces φ and ψ which have the same image surface, the two surfaces are called smoothly equivalent surfaces, see Buck p 386 Theorem 2 (ii). So imagine that we have a 2D surface hanging in R3 and as t1 and t2 vary (each perhaps from 0 to 1 in t-space), we move around on the image surface in R3. The problem is how to integrate a 2-form over this surface. We can define the calculational meaning of the above integral in two steps, each being a definition, as outlined in Section 10.11. First definition: ∫φ αx ≡ ∫[0,1]2 φ*(αx) = the integral in t-space of the pullback of αx over the 2-cube [0,1]2 (10.13.3) On the left is an integral of the 2-form αx over a surface φ in R3. On the right is an integral of a different 2-form φ*(αx) (the pullback of αx) over a 2-cube [0,1]2 in R2. Note that αx lies in xΛ2(R3) while φ*(αx) lies in tΛ2(R2). Since our usual mapping is φ* : xΛ1(Rm) → tΛ1(Rn) we have m = 3 and n = 2 (see (10.7.18)). The "tall" m x n R-matrix for this problem is then a 3 x 2 matrix, Rij = (Dφ)ij = ∂φi/∂tj = ∂jφi i = 1,2,3 j = 1,2 R = . (10.13.4) We then compute the pullback φ*(αx) of αx : αx = Σ1≤i<i≤3 fii(x) xλ^I = Σ1≤i<i≤3 fii(x) dxi ^ dxi φ*(αx) = Σ1≤i<i≤3 φ*(fii(x)) φ*(dxi ^ dxi) // (10.9.5) item 3 = Σ1≤i<i≤3 fii(φ(t)) Σ1≤j<j≤2 det dtj ^ dtj // (10.9.16) = Σ1≤i<i≤3 fii(φ(t)) det dt1 ^ dt2 // only one term in Σ1≤j<j≤2 = Σ1≤i<i≤3 fii(φ(t)) det dt1 ^ dt2 // more compact notation // these determinants are the 2x2 minors of the matrix R shown above = Σ1≤i<i≤3 fii(φ(t)) dt1 ^ dt2 // Jacobian notation for determinants = Σ1≤i<i≤3 gii(t) dt1 ^ dt2 where gii(t) = fii(φ(t)) . (10.13.5) Since in this example there are so few terms (three) in the sum Σ1≤i<i≤3 , we just write them out φ*(αx) = [ f12(φ(t)) + f13(φ(t)) + f23(φ(t)) ] dt1 ^ dt2 = G(t) dt1 ^ dt2 where G(t) ≡ [ f12(φ(t)) + f13(φ(t)) + f23(φ(t)) ] . (10.13.6) The pullback φ*(αx) = G(t) dt1 ^ dt2 is a 2-form in dual t-space tΛ2(R2). Using the definition given above, one then has, ∫φ αx ≡ ∫[0,1]2 φ*(α) = ∫[0,1]2 G(t) dt1 ^ dt2 . (10.13.7) Thus the integral of the 2-form αx over the surface φ in x-space is defined to be equal to the integral of the 2-form G(t) dt1 ^ dt2 over a 2-cube in t-space. So far no regular calculus integrals have appeared. Second definition: ∫[0,1]2 G(t) tλ1 ^tλ2 = ∫[0,1]2 G(t) dt1 ^ dt2 ≡ !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 . (10.13.8) On the left is the integral of a 2-form on a 2-cube, on the right is an ordinary calculus integral of a function over the 2-cube. It is this second definition that motivates giving the dual space basis vectors tλ1 and tλ2 the cosmetic names dt1 and dt2 . Notice that the only locations where the functions fii(x) are "sensed" in this integral are points on the surface x = φ(t) . Combining the two definitions gives ∫φ αx = !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 (10.13.9) with G as in (10.13.6). We redisplay the earlier Fig (10.9.3a) to illustrate the above discussion, where βt = G(t) dt1 ^ dt2 (10.13.10) Here Rm = R3 and the torus is just a sample surface to illustrate the general surface x = φ(t). Comment on orientation Orientation of the domain surface is a tricky business in a k-form integral but can at worst cause confusion about the sign of the result. To make things more "visible", suppose [a,b] = [c,d] = [1,0]. In t-space R2, if [a,b] = (b-a)1 = 1 and [c,d] = (d-c)2= 2, then the 2-cube (unit square) integration domain [0,1]2 can be regarded as being 1 ^ 2 . Recall from the discussion near (4.3.14) that one can regard A ^ B as connected with the signed area of a parallelogram (2-piped), so A ^ B is associated with the area of the front side and B ^ A with the area of the back side, akin to the cross product situation A x B, and these two area have the same magnitude but opposite sign. So we can regard, ∫1 x 2 1 = 1 area of the front side of a unit square is 1 area unit ∫2 x 1 1 = -1 area of the back side of a unit square is -1 area unit (10.13.11) The integral appearing in (10.13.8) is this ∫[0,1]2 G(t) dt1 ^ dt2 = ∫1 x 2 G(t) dt1 ^ dt2 = !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 . (10.13.12) Now consider ∫1 x 2 G(t) dt1 ^ dt2 = ∫1 x 2 [- G(t) dt2 ^ dt1] = – ∫1 x 2 G(t) dt2 ^ dt1 (10.13.13) ∫2 x 1 G(t) dt1 ^ dt2 = [- ∫1 x 2] G(t) dt2 ^ dt1 = – ∫1 x 2 G(t) dt2 ^ dt1 In both these equations a minus sign is generated. In the first the minus sign arises because the 2-form called G(t) dt2 ^ dt1 is the negative of the different 2-form called G(t) dt2 ^ dt1. In the second equation the integration domain 1 ^ 2 refers to the front side of the unit square, while 2 ^ 1 refers to the back side of the unit square, and these domains differ by a minus sign. If both changes are made at once one gets ∫1 x 2 G(t) dt1 ^ dt2 = + ∫2 x 1 G(t) dt2 ^ dt1 (10.13.14) If the integration domain is written simply as [0,1]2, one ends up with ∫[0,1]2 G(t) dt1 ^ dt2 = + ∫[0,1]2 G(t) dt2 ^ dt1 (10.13.15) and this seems to be a contradiction since everyone knows that dt2 ^ dt1 = - dt1 ^ dt2 . The issue here is that the two [0,1]2 domains are not the same, they just look the same. See Sjamaar's page 64 Remark 5 where he treats the domain 2 ^ 1 as a reparametrization of the domain 1 ^ 2 which reverses the orientation of that domain. Further processing: Using the Hodge correspondence suggested in (4.3.17) we define three new function names Fi f12 ≡ F3 f23 ≡ F1 f13 ≡ - F2 (10.13.16) with a minus in the last line since f13 is in anticyclic order compared to (4.3.17). Then (10.13.9) with (10.13.6) for G(t) becomes, ∫φ αx = !Syntax Error, I!Syntax Error, I[ F3(φ(t)) – F2(φ(t)) + F1(φ(t)) ] dt1dt2 = !Syntax Error, I!Syntax Error, I[ F1(φ(t)) – F2(φ(t)) + F3(φ(t)) ] dt1dt2 = !Syntax Error, I!Syntax Error, I[ F1(φ(t)) + F2(φ(t)) + F3(φ(t)) ] dt1dt2 = !Syntax Error, I!Syntax Error, I F(φ(t)) n(t) dt1dt2 (10.13.17) where n below is a vector normal to the surface in R3 at point x = φ(t), n(t) ≡ ( , , ) // Buck p 335, 403 (10.13.18) That n really is a normal vector can be verified by showing that n xuj = 0 for any tangent base vector xuj in the tangent space TxM, where from (E.2) (xuj)i = Rij = ∂jφi , see Buck p 336. But we know that n(t) is a normal vector because the expression for n in (10.13.18) is the same as n' in (10.10.19) (apart from change of notation) and that n' was constructed as a cross product of two vectors on the surface so it was a normal. The vector n(t) is not in general a unit vector, so we define (t) = n(t) / | n(t) | (10.13.19) where | n(t) |2 = [ ]2 + [ ]2 + [ ]2 = det2 + det2 + det2 ≡ K(t)2 (10.13.20) which we recognize as the same object K2 appearing in (10.10.18). Then with n(t) = K(t) (t) (10.13.21) equation (10.13.17) may now be written, ∫φ αx = !Syntax Error, I!Syntax Error, I F(φ(t)) (t) K(t) dt1dt2 . (10.13.22) Going back to the original 2-form αx (10.13.1) one can write, αx = Σ1≤i<i≤3 fii(x) dxi ^ dxi = f12(x) dx1 ^ dx2 + f13(x) dx1 ^ dx3 + f23(x) dx2 ^ dx3 = F1(x) dx2 ^ dx3 + F2(x) dx3 ^ dx1 + F3(x) dx1 ^ dx2 = F1(x) dA1 + F2(x) dA2 + F3(x) dA3 = F(x) dA // dA1 = *dx1 etc, so can say dA = *dx (10.13.23) where in cyclic order we define the following differential area 2-forms dA1 ≡ dx2 ^ dx3 dA2 ≡ dx3 ^ dx1 dA3 ≡ dx1 ^ dx2 . (10.13.24) Then our result (10.13.22) can be concisely written, ∫φ αx = ∫φ F(x) dA = !Syntax Error, I!Syntax Error, I [F(φ(t)) (t) ] K(t) dt1dt2. = !Syntax Error, I!Syntax Error, I [F(φ(t)) (t)] dA dA = K(t) dt1dt2 = !Syntax Error, I!Syntax Error, I F(φ(t)) dA dA = dA = K(t) dt1dt2 (t) . (10.13.25) Normally this is written ∫φ F(x) dA showing the motivation for the cosmetic functional notation dA as defined above. This is the "integral of a vector field F over a surface φ ". Recall now our "no differential forms" surface integration done in (10.10.20) A' <Bn> = ∫S' dA' B(x') ' = ∫S B(F(x)) ' K(x) dx1dx2 . (10.10.17) In the x = φ(t) notation this reads A <Bn> = ∫S dA B(x) = ∫S B(φ(t)) K(t) dt1dt2 (10.13.26) which is the same integral appearing in (10.13.25) with F = B. Therefore, we can interpret (10.13.26) as being the integral of the 2-form, αx = B(x) dA (10.13.27) and one then has ∫φ αx = ∫φ B(x) dA =!Syntax Error, I!Syntax Error, I[B(φ(t)) (t) ] K(t) dt1dt2 = !Syntax Error, I!Syntax Error, I B(φ(t)) dA . (10.13.28) Now return to (10.13.25), ∫φ αx = ∫φ F(x) dA = !Syntax Error, I!Syntax Error, I F(φ(t)) (t) K(t) dt1dt2 . (10.13.25) Suppose the vector field F(φ(t)) happens to be normal to the surface φ for all values of t. In this case, [F(φ(t)) (t) ] = | F(φ(t)) | . (10.13.29) Then setting | F(φ(t)) | = T(φ(t)) we find that ∫φ αx = !Syntax Error, I!Syntax Error, I T(φ(t)) K(t) dt1dt2 (10.13.30) and this shows how the temperature integral of (10.10.20) can be fitted into the 2-form framework. Generalization from φ: R2→ R3 to φ: R2→ Rm If the 2D surface lies in Rm instead of R3, the above results are easily generalized. The 2-form αx is then, αx = Σ1≤i<i≤m fii(x) xλ^I = Σ1≤i<i≤m fii(x) dxi ^ dxi = Σ'I fI(x) dx^I (10.13.31) where there are (m,2) = m(m-1)/2 terms in the ordered sum Σ'I. The pullback also has (m,2) terms, being φ*(αx) = Σ1≤i<i≤m fii(φ(t)) dt1 ^ dt2 = Σ'I fI(φ(t)) det dt1 ^ dt2 = Σ1≤i<i≤m gii(t) dt1 ^ dt2 gii(t) = fii(φ(t)) = G(t) dt1 ^ dt2 G(t) = Σ1≤i<i≤m fii(φ(t)) . (10.13.32) The pullback φ*(αx) = G(t) dt1 ^ dt2 is still a 2-form in dual t-space tΛ2(R2). Then ∫φ αx ≡ ∫[0,1]2 φ*(αx) = ∫[0,1]2 G(t) dt1 ^ dt2 ≡ !Syntax Error, I!Syntax Error, I G(t) dt1dt2 (10.13.33) Here G(t) is a sum of (m,2) terms each of which is a 2x2 Jacobian weighted by a function fii. Example: φ: R2→ R4 αx = f12(x) dx1 ^ dx2 + f13(x) dx1 ^ dx3 + f14(x) dx1 ^ dx4 // 2-form in xΛ2(R4) + f23(x) dx2 ^ dx3 + f24(x) dx2 ^ dx4 + f34(x) dx3 ^ dx4 φ*(αx) = G(t) dt1 ^ dt2 // 2-form in tΛ2(R2) where G(t) = [ f12(x) + f13(x) + f14(x) + f23(x) + f24(x) + f34(x)] and the integrated 2-form is ∫φ αx ≡ ∫[0,1]2 φ*(α) = !Syntax Error, I!Syntax Error, Idt1dt2 G(t) .