Dr. Allen Back. Dec. 3, 2014

Transcription

1 Dr. Allen Back Dec. 3, 2014

2 forms are sums of wedge products of the basis 1-forms dx, dy, and dz. They are kinds of tensors generalizing ordinary scalar functions and vector fields. They have a skew-symmetry property so that for 1 forms (e.g. dx or dy) ω η = η ω. This implies dx dx = 0 for example. Marsden and Tromba drop the symbol, writing dz dx for dz dx, etc.

3 InR 3 all differential forms have an interpretation as either an ordinary function or a vector field. 0-forms are scalar functions - f (x, y, z) on R 3. 1-forms are linear combinations P dx + Q dy + R dz where P, Q, and R are functions of three variables (x, y, z). Think of this as associated with the vector field (P, Q, R). 2-forms are linear combinations P dy dz + Q dz dx + R dx dy. Think of this as associated with the vector field (P, Q, R). 3-forms are of the form f (x, y, z) dx dy dz. Think of this as associated with the scalar function f (x, y, z).

4 There is an operation exterior differentiation d which takes any k form to a k + 1 form. For ordinary functions f (x, y, z), it looks like the expression for differentials: df = f x dx + f y dy + f z dz.

5 d 2 = 0; i.e. for any (C 2 ) k-form ω d(dω) = 0 So e.g. d(dx) = 0. In the end this is because of the independence of order for mixed partials.

6 Exterior derivative obeys the following anti-commutativity rule for wedge products: d (η ω) = (dη) ω + ( 1) k η (dω). if η is a k-form and ω is an l-form. ω η = ( 1) kl η ω as well.

7 For differential forms, the operations of grad, div, and curl are all special cases of exterior differentiation d! grad is d from 0-forms (functions) to 1-forms as long as we identify the 1-form P dx + Q dy + R dz with the vector field (P, Q, R). curl is d from 1-forms to 2-forms as long as we identify the 1 and 2-forms with their associated vector fields. div is d from 2-forms (vector fields in disguise) to 3-forms (functions in disguise.)

8 Let D = [0, 1] [0, 1] R 2 be a square, Φ : D R 3 a parameterization of a piece of surface S, and ω a 1-form on R 3. There is a pullback operation Φ taking a 2-form (or more generally any k-form) on the image of Φ to a 2-form on R 2. Φ(u, v) = (u 3 + v 3, u v, u + v), Φ (xy dx dy) = (u 3 +v 3 )(u v) (3u 2 du + 3v 2 dv) (du dv).

9 of a differential k-form over a rectangle in R k is the same as an ordinary k-fold multiple integral. We use chains (essentially formal sums of parameterizations, with coefficients like ±1 to express orientation) to reduce the integration of any differential k-form (over e.g. a path, surface, or region in R 3 ) to an ordinary multiple integral.

10 Let D = [0, 1] [0, 1] be a square, Φ : D R 3 a parameterization of a piece of surface S, and ω a 1-form on R 3. Assume Φ( D) = (Φ(D)) S. Then ω = Φ( D) Stokes in R 3 follows from Φ ω = D Φ(D) D d ω d Φ ω on the square in R 2. (A simple version of Green.)

11 Here Φ refers to the pullback operation; e.g. for Φ(u, v) = (u 3 + v 3, u v, u + v), Φ (xy dx dy) = (u 3 +v 3 )(u v) (3u 2 du + 3v 2 dv) (du dv).

12 of differential forms is basically defined in terms of pullback. If c : [a, b] R 3 with c(t) = (x(t), y(t), z(t)), then c dx = x (t)dt, etc. and so the integral of the 1-form η = P dx + Q dy + R dz over C corresponding to c is essentially defined by η = P dx + Q dy + R dz C C = c (P dx + Q dy + R dz) = C b a (Px + Qy + Rz ) dt in line with one of our older classical notations/computation styles for line integrals.

13 Now let s look at the differential form version of the surface integral (x, y, z) ˆn ds S over part of the paraboloid z = 1 x 2 y 2 with z 0.

14 The associated 2-form to the vector field (x, y, z) is µ = x dy dz x dz dx + z dz dy. For the parameterization c(r, θ) = (r cos θ, r sin θ, 1 r 2 ), we have dy = cos θ dr r sin θ dθ, dz = 2r dr etc., so

15 S µ = = = D 1 2π 0 c (x dy dz x dz dx + z dx dy) 0 r cos θ(cos θ dr r sin θ dθ) ( 2r dr) +... which (thinking about ˆn in terms of cross products) can be seen to agree with our usual!

16 of a conservative vector field cartoon.

17 Green s Theorem cartoon.

18 Stokes Theorem cartoon.

19 Both sides of Stokes involve integrals whose signs depend on the orientation, so to have a chance at being true, there needs to be some compatibility between the choices. The rule is that, from the positive side of the surface, (i.e. the side chosen by the orientation), the positive direction of the curve has the inside of the surface to the left. As with all orientations, this can be expressed in terms of the sign of some determinant. (Or in many cases in terms of the sign of some combination of dot and cross products.)

20 Problem: Let S be the portion of the unit sphere x 2 + y 2 + z 2 = 1 with z 0. Orient the hemisphere with an upward unit normal. Let F (x, y, z) = (y, x, e z2 ). Calculate the value of the surface integral F ˆn ds. S

21 Theorem field cartoon.

22 The surface integral side of depends on the orientation, so there needs to be a choice making the theorem true. The rule is that the normal to the surface should point outward from the inside of the region. (For the 2d analogue of (really an application of Green s) F ˆn = (P x + Q y ) dx dy C inside we also use an outward normal, where here C must of course be a closed curve.

23 Problem: Let W be the solid cylinder x 2 + y 2 3 with 1 z 5. Let F (x, y, z) = (x, y, z). Find the value of the surface integral F ˆn ds. W