ES.182A Topic 46 Notes Jeremy Orloff. 46 Extensions and applications of the divergence theorem

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1 E.182A Topic 46 Notes Jeremy Orloff 46 Extensions and applications of the divergence theorem 46.1 el Notation There is a nice notation that captures the essence of grad, div and curl. The del operator is a boldface triangle pointing down. = x i + y j + k = x, y,. For a function f(x, y, z) and a vector field F = M, N, P, this notation is interpreted as follows: (We use both bracket and i, j, k notation.) grad f = f = x, y, f f = x, f y, f ( grad f = f = x i + y j + ) k f = f x i + f y j + f k divf = F = x, y, M, N, P = M x + N y + P. ( divf = F = x i + y j + ) k (Mi + Nj + P k) = M x + N y + P. curlf = F (more on this later). The symbol is called nabla. According to Wikipedia this is Greek word for the Hebrew or Phoenician harp. There are similar words in Hebrew and Aramaic. Confusingly the notation f is read as del f. Likewise the notations F and F are read as del dot F and del cross F Aside on the Laplacian An important operator on functions is the Laplacian. It is defined in terms of del by Laplacian = 2 f = f = div gradf = 2 f x f y f Extended divergence theorem imilar to Green s theorem, the divergence theorem still holds if the boundary of our solid volume consists of several closed surfaces. That is, for a boundary with two surfaces, divf dv = F dv = A similar statement holds for a boundary with three or more surfaces. 1

2 46 EXTENION AN APPLICATION OF THE IVERGENCE THEOREM 2 We have to be careful that we orient each of the surfaces so their normals point outward from the volume. The figure below indicates a volume which is between the two surfaces 1 and 2. The outward (from ) normals are shown. Example Let F be the gravitational field of a point mass m 1. For a closed surface show that { 4πGm 1 if surrounds m 1 F n d = 0 otherwise. answer: We set our coordinate system so that m 1 is at the origin. We know the graviational field is F(x, y, z) = G m 1 x, y, z ρ 3. We have to be careful because F has a singularity at the origin. We will mimic what we did with Green s theorem and use the extended divergence theorem. First we show that divf = 0. To compute divf, we use the chain rule and the formulas ρ x = x ρ, ρ y = y ρ, etc. o, xρ 3 x = ρ 3 3x 2 ρ 5, yρ 3 x = ρ 3 3y 2 ρ 5, ρ 3 x = ρ 3 3z 2 ρ 5. divf = G m 1 (3ρ 3 3(x 2 + y 2 + z 2 )ρ 5 ) = G m 1 (3ρ 3 3ρ 2 ρ 5 ) = 0. QE In preparation for considering an arbitrary closed surface we compute the flux directly when is a sphere of radius a centered at the origin with outward pointing normals. On, F and n are parallel (in opposite directions). o, F n = F = G m 1 /a 2. Thus, F n d = G m 1 a 2 d = G m 1 a 2 area = G m 1 4π.

3 46 EXTENION AN APPLICATION OF THE IVERGENCE THEOREM 3 In general, for any closed surface, there are two cases, (i) does not surround 0 (ii) surrounds 0. and In case (i) (see picture), F is continuously differentiable on the entire volume inside. o, F n d = divf dv = 0 dv = 0 (as claimed). In case (ii) (see picture), let 1 be a small sphere centered at 0 that is entirely inside. The picture shows 1 with outward normals, so 1 has inward normals. Using the extended divergence theorem, we have (note the signs) F n d = divf dv = 0. 1 o, F n d F n d = 0 1 F n d = F n d = 4πGm 1 1 (again, as claimed). Example Let F be the gravitational field from a mass distribution. how that the flux through a closed surface is 4πGM, where M is the total mass enclosed by. answer: We divide the mass into infinitesimal pieces dm and let df be the field due to dm. By the previous example, if dm is inside then the flux of df through is 4πG dm.

4 46 EXTENION AN APPLICATION OF THE IVERGENCE THEOREM 4 If dm is outside then the flux is 0. That is, each dm contributes 4πG dm if it s inside 0 if it s outside ince the total gravitational field F is the sum of all fields df due to the infinitesimal masses dm, the net flux is just a sum of 4πG dm over all dm inside, that is flux = 4πG M. This example gives what is known in physics as the Gauss-Coulomb Law. An identical law holds for the electric field of a charge distribution Two forms of the Gauss-Coulomb Law Theorem. Let F be a gravitational field of a mass distribution in space and let be a closed surface oriented outwards. We saw in the previous example that the Gauss-Coulomb Law says F n d = 4πGM, (1) where M is the total mass inside. This is called the integral form of Gauss Law. There is an equivalent form, called the differential form of Gauss Law: where ρ is the mass density function. divf (x,y,z) = 4πGρ(x, y, z), (2) Note, neither of these laws is mathematics they are empirical laws of physics. However, their equivalence is a purely mathematical statement that can be proved using the divergence theorem. Proof. To show that the two versions are indeed equivalent we have to argue in both directions. That is, we must show the differential form implies the integral form and vice versa. The differential form implies the integral form: We assume Equation 2 is true. Using the divergence theorem we must show this implies Equation 1 is also true. Let be the solid volume inside. If divf = 4πGρ then F n d = divf dv = 4πGρ dv = 4πGM. The first equality is the divergence theorem. The last equality follows because the integral of density is mass. Thus, Equation 1 is true. QE The integral form implies the differential form: We assume Equation 1 is true. Using the divergence theorem we must show this implies Equation 2 is also true. By the divergence theorem, the integral form can be written as divf dv = 4πGM = 4πGρ dv. (3)

5 46 EXTENION AN APPLICATION OF THE IVERGENCE THEOREM 5 The first equality follows from the divergence theorem and the seconds follows because the integral of density is mass. The basic idea is that if Equation 3 holds for every volume then the integrands must be equal, i.e. divf = rπgρ. Just for completeness we give the formal argument for this basic idea. Basic idea. If f(x, y, z) dx dy dz = g(x, y, z) dx dy dz for every volume, the f and g are the same function. Proof. We proceed to reason by contradiction: that is, we will assume the that f g, i.e. the opposite of what we want to show, and then see that this leads to contradiction. o, assume that at some point (x 0, y 0, z 0 ) we have f(x 0, y 0, z 0 ) g(x 0, y 0, z 0 ). For ease of exposition let s assume that f(x 0, y 0, z 0 ) > g(x 0, y 0, z 0 ). The argument when the inequality is reversed is identical. By continuity, we can choose a sphere 0 centered on P 0 and small enough that f(x, y, z) > g(x, y, z) at all points inside 0. Let 0 be volume inside 0. Using the inequality we have f(x, y, z) dv > g(x, y, z) dv. 0 0 But this contradicts Equation 3, which says the two integrals must be equal. This contradiction proves that the integral form implies the differential form. ince the two forms imply each other, they are equivalent. These are two equivalent statements of the same physical law. The integrated form is perhaps a little easier to understand, since the left hand side is the flux of F through, which is a more intuitive idea than divf. On the other hand, quite a lot of technique is required actually to calculate the flux, whereas very little is needed to calculate the divergence. Note. The Gauss-Coulomb Law also holds for electrostatic fields associated to charge distributions. The differential form of this law is one of Maxwell s equations. You have probably seen it on a t-shirt.