Section 10-10 v1a

Section 10.10 v1 PhL 4.29.16 10.10 Integration of functions over surfaces and curves In Section 10.11 we are going to make this claim concerning the integration of an arbitrary differential k-form over a manifold "surface" x' = F(x) embedded in Rm , α' = Σ'I fI(x') λ'^I = Σ'I fI(x') dx'i ^ dx'i .... ^ dx'i // the k-form in x'-space ∫F α' = Σ'I Σ'J∫[0,1]k fI(F(x)) det(RIJ(x)) dxj ^ dxj ^ ...^ dxj R= (DF) = Σ'I Σ'J (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) fI(F(x)) det(RIJ(x)) dxjdxj....dxj . (10.10.1) The general idea is that the k-form α' in dual x'-space is first pulled back to a different k-form in dual x-space, and then the wedge product dxj ^ dxj ^ ...^ dxj of basis functionals appearing in this pulled back k-form is mysteriously replaced by a product of ordinary calculus differentials dxjdxj....dxj . The end result is that ∫F α' is some calculus-computable real number. We assume that the manifold M can be covered by a single mapping x' = F(x), otherwise we create at atlas of mappings as described earlier in Section 10.*. Before delving into this subject, it seems useful to have some discussion of integrals of functions over surfaces and curves in R3 without any mention of differential forms. This discussion takes the form of a set of very simple example problems. Integration over surfaces Example 1: Compute the average temperature on a flat plate in the z = 0 plane in R3. What is the meaning of "average temperature"? We partition the plate into a large array of N x N tiny squares of equal area dAi = ΔxΔy, and we measure the temperatures Ti simultaneously in all N2 locations. The average temperature is then <T> = (1/A) limN→∞ [ΣiTidAi] where A = ab is the area of the plate. So this is an area-weighted average temperature which is cast into a normal 2D Riemann integral, <T> = (1/A) ∫dA T(x) = (1/ab)!Syntax Error, Idx!Syntax Error, Idy T(x,y,0) . dA = dxdy (10.10.2) This same kind of integral would be used to compute the average mass density of a flat plate which has areal mass density ρ(x,y) <ρ> = (1/A) ∫dA ρ(x,y) = (1/ab) ∫dxdy ρ(x,y) . (10.10.3) One could compute the center of mass location of a plate of mass m with similar integrals, <x> = (1/m) ∫dA x ρ(x,y) <y> = (1/m) ∫dA y ρ(x,y) . (10.10.4) Example 2: Compute the average normal component of a magnetic field B on the same flat plate. This question is similar to the temperature question and the result is <Bn> = (1/A) ∫dA Bn(x) = (1/A) ∫dA B(x) = (1/A)∫S dA B(r), dA = dA (10.10.5) where in this problem it happens that = unit normal vector, normal to the surface of the plate at (x,y,0) = = constant . Then <Bn> = <Bz> = (1/ab)!Syntax Error, Idx!Syntax Error, Idy Bz(x,y,0) . (10.10.6) For both these examples, one could consider a round plate instead of a square plate, and then one would use dA = dxdy → (rsinθ)drdθ where the Jacobian J = rsinθ appears. One could show that such differential area patches dA cover the plate surface perfectly with no overlaps and no missed regions. Example 1a: Compute the average temperature on a spherical shell of radius R in R3. Example 2a: Compute the average normal component of a magnetic field B on this shell. We take the general expressions obtained in Examples 1 and 2 above <T> = (1/A) ∫dA T(x) <Bn> = (1/A) ∫dA B(x) . (10.10.7) In spherical coordinates, A = 4πR2, = and dA = R2sinθdθdφ. This area measure can be deduced by looking at a picture of spherical coordinates where dA = (Rdθ)(Rsinθdφ) is a surface patch. Then, <T> = (1/4π) !Syntax Error, Idφ !Syntax Error, Idθ T(Rsinθcosφ, Rsinθsinφ, Rcosθ) <Bn> = <Br> = (1/4π) !Syntax Error, Idφ !Syntax Error, Idθ Br(Rsinθcosφ, Rsinθsinφ, Rcosθ). (10.10.8) In the language of our earlier sections, we can think of this surface being defined by an underlying transformation x = F(θ,φ) : x = Rsinθcosφ y = Rsinθsinφ z = Rcosθ (10.10.9) where we would draw "parameter space" = Rn = R2 on the left (with coordinates θ and φ) and Rm = R3 on the right. So one has (θ,φ)-space on the left, and x-space on the right. In writing dA = R2sinθdθdφ, we are "pulling back" an area patch on the sphere to a rectangular area dθdφ in (θ,φ)-space, and we pick up an area conversion factor R2sinθ. Similarly, the functions T and Br are "pulled back" so they are written in the form T(F(θ,φ) and Br(F(θ,φ). Although nothing has been said about "differential forms", one suspects that this example can somehow be cast into a 2-form scenario. Example 1b: Compute the average temperature on an arbitrary surface S. Example 2b: Compute the average normal component of a magnetic field B on an arbitrary surface S. Again it seems clear that the answers to these questions are as follows, where r = (x,y,z), <T> = (1/A) ∫S dA T(r) <Bn> = (1/A)∫S dA B(r) = (1/A)∫S dA B(r) , dA = dA . (10.10.10) The meaning of these integrals is clear: dA is a local area element at point r on the surface, is a local unit normal at a point r on the surface, and A is the total area of the surface. One just has to figure out what these quantities are for a given surface. Notice that ∫S dA B(r) is the classic "surface integral of a vector field" as one might encounter in an electrostatic flux calculation (B = E) or in a fluid flow situation (B = v). In order to proceed, we shall revert to our earlier notation where the surface is determined by x' = F(x) instead of x = φ(t). In this notation the above solutions are <T> = (1/A') ∫S' dA' T(x') <Bn> = (1/A')∫S' dA' B(x') ' (10.10.11) where the primes are appropriate for x'-space. For example, S' is the arbitrary surface embedded in x'-space over which we wish to integrate, and x'-space is Rm = R3. At point x' on the surface there is a tangent space Tx'M which is spanned by the tangent base vectors u'1 and u'2 which appear in the kinematics package (10.6.a.1). Recall that these vectors are generally not orthogonal. The magnitude of the area of the 2-piped subtended by these vectors is |u'1 x u'2|. But we want a differential 2-piped at point x' with some small extents dξ1 and dξ2 in these two directions, so then dA' = | (dξ1u'1) x (dξ2u'2) | = dξ1dξ2 | u'1 x u'2| . Meanwhile, we know that u'1 = Ru1 and u'2 = Ru2, these being vector transformations under x' = F(x). Therefore R(dx1u1) = dx1u'1 R(dx2u2) = dx2u'2 . (10.10.12) Thus, the small rectangle spanned by (dx1u1,dx2u2) in x-space is mapped into a small 2-piped spanned by (dx1u'1,dx'2u2) in x'-space. We can take this 2-piped to be the 2-piped discussed above by setting dξ1 = dx1 and dξ2 = dx2 and then we have dA' = | u'1 x u'2 | dx1dx2. Recall next that u'3 is constructed "as needed" so the {u'i} form a complete basis for R3 at point x' on the surface S'. We can take u'3 = u'1 x u'2. We then need to know that magnitude of this vector to know dA'. Since Rm = R3 is a Cartesian space, up and down vector component indices are the same, so | u'3|2 = | u'1 x u'2|2 = ( u'1 x u'2) ( u'1 x u'2) = [εiab (u'1)a(u'2)b] [εicd (u'1)c(u'2)d] = εiabεicd(u'1)a(u'2)b(u'1)c(u'2)d = (δacδbd - δadδbc) (u'1)a(u'2)b(u'1)c(u'2)d // see e.g. Tensor (D.10.22) = (u'1)a(u'2)b(u'1)a(u'2)b – (u'1)a(u'2)b(u'1)b(u'2)a = Ra1Ra2 Ra1Rb2 – Ra1Rb2 Rb1Ra2 // kin. package (10.6.a.1) item (e) = Σa (Ra1)2 Σb (Rb2)2 – (ΣaRa1Ra2) (ΣbRb1Rb2) = [Σa (Ra1)2] [Σa (Ra2)2] – [ΣaRa1Ra2]2 ≡ [ K(x) ]2 // since Rij = Rij(x) in general (10.10.13) We could have used the vector identity (A x B) (A x B) = A2B2 – (AB)2 in place of the εε product method, but εε products are good to know about and we give a reasonable source above for the reader interested in their generalizations. On the last line we define K ≡ |u'3|. Finally then we have an expression for differential area dA', dA' = | u'1 x u'2 | dx1dx2 = | u'3| dx1dx2 = K(x) dx1dx2 . (10.10.14) We then also have an expression for the unit normal vector ' at point x', ' = u'3 / | u'3| = (1/K) u'3 = (1/K) u'1 x u'2 (10.10.15) which has components (')i = (1/K) (u'1 x u'2)i = (1/K) εiab (u'1)a(u'2)b = (1/K) εiab Ra1Rb2 . (10.10.16) Finally we are prepared to give solutions to the problems posed above, <T> = (1/A') ∫S' dA' T(x') = (1/A') ∫S T(F(x)) K(x') dx1dx2 <Bn> = (1/A')∫S' dA' B(x') ' = (1/A')∫S B(F(x)) ' K(x) dx1dx2 (10.10.17) where ' and K(x) are as given above. Notice that the resulting integral is over the region S in x-space which maps into the surface S' in x'-space, since the integration variables are those of x-space. The area is of course given by A' = ∫S' dA' = ∫S K(x) dx1dx2 . (10.10.18) The reader will no doubt notice that in writing dA' = K(x) dx1dx2 we are in fact "pulling back" some tilted non-rectangular 2-piped patch dA' on the surface S' in x'-space to a rectangular patch dx1dx2 in x-space and in doing so we pick up a Jacobian-like factor K(x). We are also "pulling back" the integrand functions T(x') and B(x') by writing them as T(F(x)) and B(F(x)). Again we arrive at this "pulling back" concept without ever mentioning "differential forms". The pullback integrals shown above are completely well-defined and it is then just a matter of doing the integrals analytically or numerically. Recall from (10.10.13) that the square of the area transformation factor K is given by, K2 = [ Σa=13 (Ra1)2] [Σa=13 (Ra2)2] – [Σa=13Ra1Ra2]2 . (10.10.19) The reader might be surprised to find that this same K2 can be written in the following manner, K2 = det2 + det2 + det2 . (10.10.20) This is the sum of the squares of the three 2 x 2 minors of the 3 x 2 "tall" R matrix. See for example Buck page 299 where K = k and Rij = aij. The two expressions above for K2 look totally unrelated and it seems rather mysterious that they are equal. Lest one have doubts, we have Maple compute K2 both ways and then subtract the results to demonstrate that the difference is 0: We shall show below the origin of the sum of minors expression for K2. Integration over curves Example 3: Compute the average temperature on a piece of straight wire of length a in R3 . Let be a unit vector which is tangent to the wire at some point x on the wire. Let dx be an arbitrary differential distance vector whose tail is located at position x on the wire. Then ds = dx is a small distance along the wire. In analogy with the flat plate of Example 1, the length-weighted average temperature of a straight wire is <T> = (1/L)∫ds T(x) ds = dx . (10.10.21) In this particular example, the wire is placed on the x axis so L = a, = , ds = dx = dx . Then, <T> = (1/a)!Syntax Error, Idx T(x,0,0) . (10.10.22) Example 4: Compute the average tangential magnetic field on this same straight wire. This question is similar to the temperature question (but → ) and the answer is <Bt> = (1/L) ∫ds Bt(x) = (1/L) ∫ds B(x) = (1/L) ∫dx B(x), ds = dx . (10.10.23) Again setting L = a, = , ds = dx = dx , this becomes <Bt> = (1/a)!Syntax Error, Idx Bx(x) . (10.10.24) Notice that ∫dx B(x) is the classic form of a "line integral of a vector field". Example 3a: Compute the average temperature on a ring of wire of radius R in the x,y plane of R3. Example 4a: Compute the average normal component of a magnetic field B on this ring. The ring is assumed centered at the origin of the x,y plane so we use cylindrical coordinates with z = 0, which then are just polar coordinates. Then = , dx = Rdθ so ds = dx = Rdθ, and L = 2πR. Then <T> = (1/L)∫ds T(x) = (1/2πR)∫Rdθ T(Rcosθ,Rsinθ,0) = (1/2π) !Syntax Error, Idθ T(Rcosθ,Rsinθ,0) <Bt> = (1/L) ∫dx B(x) = 1/2πR)∫ Rdθ Bθ(Rcosθ,Rsinθ,0) = (1/2π)!Syntax Error, I Bθ(Rcosθ,Rsinθ,0) (10.10.25) where the last argument of the integrand functions indicates z = 0 for our placement of the ring in R3. Again, the differential distance element ds = Rdθ is being "pulled back" from x-space = R3 to θ-space = R1, and the integrand functions are pulled back according to T(F(θ)) and Eθ(F(θ)) where x = F(θ) : x = Rcosθ y = Rsinθ z = 0 . (10.10.26) Example 3b: Compute the average temperature on an arbitrary wire in R3. Example 4b: Compute the average normal component of a magnetic field B on this arbitrary wire. As in the treatment of Examples 1b and 2b, we revert to our x' = F(x) notation where the Example 3 and 4 solutions may be written, <T> = (1/L')∫C' ds' T(x') = (1/L)∫C'dx' ' T(x') <Bt> = (1/L') ∫C'ds' B(x') ' = (1/L') ∫C'dx' B(x'), ds' = dx' ' . (10.10.27) We take x'-space to be Rm = R3 and x-space to be Rn = R2. C' is the name of the curve determined by x' = F(x) as we allow x to move along a line segment (0,0) to (0,a) in x-space. In other words, the curve C' in x'-space is being pulled back to a straight line segment C on the x-axis of x-space. To be consistent, we should be calling the x'-space curve F instead of C', and in Example 2b we should call the surface F instead of S', since in both cases the curve and surface are generated by x' = F(x), but we shall sacrifice consistency for clarity. The most pressing issue now is how to compute the unit tangent vector ' . Reaching into our kinematics package (10.6.a.1) and nearby discussion, we realize that ' = u'1 / | u'1| . (10.10.28) This is because the tangent space Tx'M is spanned by the single tangent base vector u'1 , while the other two vectors u'2 and u'3 are selected "as needed" to span the perp space to Tx'M in R3. We invent some differential distance dξ so that dx' = dξ u'1 points along the curve C' at point x'. Meanwhile, we know that u'1 = Ru1, this being a vector transformation under x' = F(x). Therefore R(dx1u1) = dx1u'1 . (10.10.29) Thus, the small differential vector dx = dx1u1 in x-space (tangent to C) is mapped into a small differential vector dx' = dx1u'1 in x'-space, tangent to C' at point x' on C'. Thus we select dξ = dx1 and conclude that dx' = dx1 u'1 ds' = |dx'| = | dx1 u'1| = | u'1| dx1 . (10.10.30) The distance ds' in x'-space is thus being pulled back to distance dx1 in x-space with factor | u'1| . We now compute | u'1| in our Cartesian R3 space ( up and down vector indices the same), | u'1|2 = u'1 u'1 = (u'1)i (u'1)i = Σi=13 (Ri1)2 // kin. package (10.6.a.1) item (e) ≡ K(x)2 // a new and different K from that of Example 2b (10.10.31) Then ' = u'1 / | u'1| = (1/K) u'1 (')i = (1/K) (u'1)i = (1/K) Ri1 ds' = | u'1| dx1 = K(x) dx1 . // x = x u1 = x1u1, dx = dx1 (10.10.32) The solution to our exercise problems is then <T> = (1/L')∫C' ds' T(x') = (1/L') !Syntax Error, Idx1 K(x) T(F(x)) <Bt> = (1/L') ∫C'ds' B(x') ' = (1/L') !Syntax Error, Idx1 K(x) B(x') ' = (1/L') !Syntax Error, Idx1 K(x) Σj=13 Bi(x') (')i = (1/L') !Syntax Error, Idx1 K(x) Σj=13 Bi(x')(1/K(x)) Ri1(x) = (1/L') !Syntax Error, Idx Σj=13 Bi(F(x)) Ri1(x) (10.10.33) where L' = ∫C'ds' = !Syntax Error, Idx1 K(x) = arc length of the curve C' in x'-space K2(x) = Σi=13 (Ri1(x))2 . // = | u'1|2 (10.10.34) The "tall" R-matrix for this problem as we have presented it has m = 3 rows and n = 2 columns and we compute R as follows, x' = F(x) = F(x1,0) // no dependence on x2 in x-space Rij ≡ (∂x'i/∂xj) Ri2 = (∂x'i/∂x2) = 0 (10.10.35) so the entire second column of R is 0 and we have R** = (10.10.36) so in effect, the R matrix is really just R** = . (10.10.37) Notice that K2(x) = Σi=13 (Ri1(x))2 can be written K2 = det2(R11) + det2(R12) + det2(R13) (10.10.38) which is the sum of the minors of the effective R matrix, reminiscent of our K2 result (10.10.20). Comments on the above examples As will be seen formally in the next section, the surface and curve integrations discussed above fall into the realm of 2-form and 1-form integrations. In the above examples, there was no mention of "functionals" or "dual spaces" or "wedge products" or "cosmetic notation" or even of "differential forms". No mention was made of "surface orientation". The calculations were performed on an ad hoc basis as any journeyman might approach these problems. There was, however, some discussion of "pulling back" integrand functions and differential areas and differential lengths from Rm to Rn, but there was no mention of pulling back functionals between the corresponding dual spaces. The method of differential forms provides a systematic method for doing integrations over "surfaces" (manifolds) of any dimension embedded in a space of any same or larger dimension, where the spaces can have arbitrary metric tensors, and where orientation is tracked. \ Compute B := MAMT but don't display messy result ∞