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Vector Calculus VECTOR CALCULUS TSOGTGEREL GANTUMUR Abstract. Integration and differentiation of vector fields in 2, 3 and 4-dimensions. Duality. Contents 1. Line integrals 1 2. Conservative fields: Scalar potential 4 3. Irrotational fields: Poincar´e'slemma 5 4. Green's theorem 7 5. Curl in 3D 9 6. Surface integrals and fluxes 10 7. The Kelvin-Stokes theorem 12 8. Fluxes, stream functions, and the divergence in 2D 14 9. Vector potentials and the divergence in 3D 16 10. De Rham diagrams 18 11. Duality and the Hodge star 22 12. Electrostatics and Newtonian gravity 26 13. Magnetostatics 29 14. Faraday's law and the Maxwell equations 31 15. The Helmholtz decomposition 33 1. Line integrals Let γ :[a; b] ! Rn be a continuously differentiable function, representing a curve L in Rn. Suppose that u : Rn ! R is a continuously differentiable function. Then f(t) = u(γ(t)) is a function in [a; b], and by the fundamental theorem of calculus, we have ˆ ˆ b b u(γ(b)) − u(γ(a)) = f(b) − f(a) = f 0(t)dt = Du(γ(t))γ0(t)dt: (1) a a This quantity does not depend on the curve L, let alone the parametrization γ, as long as the endpoints γ(a) and γ(b) stay fixed. We are going to interpret the integral in the right hand side as the integral of Du over the curve L, and attempt to generalize it to a class of objects broader than the derivatives of scalar functions. We make the following observations. • The integral must depend not only on the curve L as a subset of Rn, but also on a directionality property of the curve, since switching the endpoints γ(a) and γ(b) would flip the sign of (1). • The derivative Du is a row vector at each point x 2 Rn. Thus it might be possible to generalize (1) from integration of Du to that of F : Rn ! Rn. Date: 2019/11/30 12:54 (EST). Incomplete draft. 1 2 TSOGTGEREL GANTUMUR Let F : Rn ! Rn be smooth, and define ˆ ˆ b F · dγ = F (γ(t)) · γ0(t)dt: (2) γ a The right hand side makes sense, because for each t 2 [a; b], we have F (γ(t)) 2 Rn and γ0(t) 2 Rn, and so F (γ(t)) · γ0(t) 2 R. Objects such as F are called vector fields. Now let us check if the integral (2) depends on the parametrization γ. Suppose that ϕ :[c; d] ! [a; b] is a continuously differentiable function, with ϕ([c; d]) = [a; b], ϕ(c) = a and ϕ(d) = b. Then by applying the change of variables formula ˆ ˆ ϕ(d) d f = (f ◦ ϕ)ϕ0; (3) ϕ(c) c to f(t) = F (γ(t)) · γ0(t), we have ˆ ˆ ˆ b d d F (γ(t)) · γ0(t)dt = F (γ(ϕ(s))) · γ0(ϕ(s))ϕ0(s)ds = F (η(s)) · η0(s)ds; (4) a c c where η(s) = γ(ϕ(s)) is the new parametrization. Hence the integral does not depend on the parametrization, as long as the endpoints of the curve are kept fixed. On the other hand, if ϕ(c) = b and ϕ(d) = a, then we have ˆ ˆ ˆ b a d F (γ(t)) · γ0(t)dt = − F (γ(t)) · γ0(t)dt = − F (η(s)) · η0(s)ds; (5) a b c which tells us that if the endpoints of the curve get switched under reparametrization, then the sign of the integral flips. Therefore, the integral (2) depends only on those aspects of the parametrization γ that specify a certain \directionality" property of the underlying curve. This \directionality" property is called orientation. Remark 1.1. Intuitively, and in practice, an oriented curve is a curve given by some con- crete parametrization γ, with the understanding that one can freely replace it by any other parametrization η = γ ◦ ϕ, as long as ϕ0 > 0. To define it precisely, we need some preparation. Let L ⊂ Rn be a curve, admitting a parametrization γ :[a; b] ! Rn which is a continuously differentiable function with γ0 =6 0 in (a; b). Suppose that P is the set of all such parametriza- n n tions of L. Then for any two parametrizations γ1 :[a1; b1] ! R and γ2 :[a2; b2] ! R from P , 0 there exists a continuously differentiable function ϕ :[a2; b2] ! [a1; b1] with ϕ =6 0 in (a2; b2), such that γ2 = γ1 ◦ ϕ. This gives a way to decompose P into two mutually disjoint classes P1 0 0 and P2: If ϕ > 0, then γ1 and γ2 are in the same class, and if ϕ < 0, then γ1 and γ2 are in different classes. The curve L, together with a choice of P1 or P2, is called an oriented curve. So the classes P1 and P2 are the possible orientations of the curve L. As mentioned before, in practice, we specify an orientated curve simply by giving a concrete parametrization. p 2 2 Example 1.2. Let γ : [0; π] ! R be given by γ(t) = (cos t; sin t), and let η : [0; pπ] ! R 2 2 2 be given by η(s) =p (cos s ; sin s ). Then we have η = γ ◦ ϕ with ϕ(s) = s for s 2 [0; π], and since ϕ0 > 0 in (0; π), these two parametrizations define the same oriented curve. On the other hand, ξ(τ) = (− cos τ; sin τ), τ 2 [0; π], gives the same curve as γ and η, but ξ is in the orientation opposite to that of γ and η, and thus as an oriented curve, ξ is different than γ. Definition 1.3. Let Ω ⊂ Rn be an open set. Then a vector field in Ω is a continuously differentiable function F :Ω ! Rn. For convenience, we restate the definition of integration of vector fields over oriented curves. VECTOR CALCULUS 3 Definition 1.4. Let Ω ⊂ Rn be an open set, and let F be a vector field in Ω. Let γ :[a; b] ! Ω be an oriented curve. Then we define the line integral of F over γ as ˆ ˆ b F · dγ = F (γ(t)) · γ0(t)dt: (6) γ a If γ is a closed curve, that is, if γ(a) = γ(b), then this integral is also called the circulation of F over γ. We have already established that the integral in the right hand side does not depend on the parametrization γ, as long as the orientation is kept fixed. Remark 1.5. The following notations are sometimes used for circulations: ˛ ‰ fi F · dγ; F · dγ; F · dγ: (7) γ γ γ Note that the second and third notations have arrows specifying orientations, and that they make sense only in two dimensions. Example 1.6. (a) Let F : R2 ! R2 be given by F (x; y) = (−y; x), and let γ(t) = (cos t; sin t), t 2 [0; π]. Then we have ˆ ˆ ( ) ( ) ˆ π − sin t T − sin t π F · dγ = dt = dt = π: (8) γ 0 cos t cos t 0 (b) Let F be as before, and let η be the unit circle, oriented counter-clockwise, and let γ be the unit circle, with no orientation specified. Then we have ˆ ˛ ‰ ˆ ( ) ( ) 2π − sin t T − sin t F · dη = F · dη = F · dγ = dt = 2π: (9) η η γ 0 cos t cos t (c) Let ξ be the unit circle, oriented clockwise, and let γ be as before. Then we have ˆ ˛ fi ‰ F · dξ = F · dξ = F · dγ = − F · dγ = −2π: (10) ξ ξ γ γ (d) Let F : R3 ! R3 be given by F (x; y; z) = (−y; x − z; z), and let γ(t) = (cos t; sin t; sin t), t 2 [0; 2π]. Then we have 0 1 0 1 ˆ ˛ ˆ T 2π − sin t − sin t F · dγ = F · dγ = @cos t − sin tA @ cos t A dt = 2π: (11) γ γ 0 sin t cos t Remark 1.7. Another notation for the line integral (6) is ˆ ˆ F1(x)dx1 + ::: + Fn(x)dxn = F · dγ: (12) γ γ For example, with γ(t) = (t; 2t), t 2 [0; 1], we have ˆ ˆ ˆ 1 1 7 xdx + (x + y)dy = (t + (t + 2t) · 2)dt = 7t dt = : (13) γ 0 0 2 4 TSOGTGEREL GANTUMUR 2. Conservative fields: Scalar potential A convenient way to generate vector fields is to differentiate scalar functions. That is, if Ω ⊂ Rn is an open set, and ϕ :Ω ! R is a smooth function, then rϕ(x) = (@1ϕ(x);:::;@nϕ(x)); (14) is a vector field in Ω. This is called the gradient of ϕ. For example, if ϕ(x; y) = x2 + sin y, then rϕ(x; y) = (2x; cos y) 2 R2 for (x; y) 2 R2. The notation gradϕ = rϕ, (15) is also used for the gradient. Definition 2.1. If F = rϕ in Ω ⊂ Rn for some scalar function ϕ, then we say that F is a conservative vector field in Ω, and that ϕ is a scalar potential of F in Ω. Let F = rϕ and let γ :[a; b] ! Ω be an oriented curve. Then we have dϕ(γ(t)) = rϕ(γ(t)) · γ0(t) = F (γ(t)) · γ0(t); (16) dt and so ˆ ˆ b dϕ(γ(t)) F · dγ = = ϕ(γ(b)) − ϕ(γ(a)); (17) γ a dt leading to the following result.
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