18.02A Topic 46: Applications and interpretations of the divergence theorem.Author: Jeremy OrloffRead: SN: V15 sec. 1 for div in ∇ notation, V15 sec.2 to middle p.3’Del’ Notation: ∇ =∂∂xi +∂∂yj +∂∂zk =∂∂x,∂∂y,∂∂z.grad f = ∇f =∂f∂xi +∂f∂yj +∂f∂zk =∂f∂x,∂f∂y,∂f∂z.divF = ∇ · F =∂∂x,∂∂y,∂∂z· hM, N, P i =∂M∂x+∂N∂y+∂P∂z.curlF = ∇ × F (more on this later).Aside on the Laplacian: Laplacian = ∇2f = ∇ · ∇f = divgradf =∂2f∂x2+∂2f∂y2+∂2f∂z2.Interpretation: divF = “source rate” of F at every point.Reason: Take a small volume around (x0, y0, z0). Call the volume D and its boundary S.The divergence theorem says the flux of F through S =RRRDdivF dV .For small D the triple integral ≈ divF(x0, y0, z0)·volume(D).⇒ divF ≈flux of F through SVol(D)=amount out from (x0, y0, z8)unit volume · unit time= source rate.Extended Divergence Theorem (just like Green’s Theorem)If a solid volume D has a boundary consisting of several closed surfaces thenthe divergence theorem still holds if for each surface we use the normals thatpoint away from the volume.ZZS1+S2=ZZZDdivF dV =ZZZD∇ · F dV.Example: Let F be the gravitational field of a mass m1. For a closed surface S showZZSF · n dS =(−4πGm1if S surrounds m10 otherwiseWe set our coordinate system so m1is at the origin. ⇒ F(x, y, z) = −G m1hx, y, ziρ3.To compute divF use the chain rule and the formula∂ρ∂x=xρ.⇒ divF = −G m11ρ3− 3x2ρ5+ −G m11ρ3− 3y2ρ5+ −G m11ρ3− 3z2ρ5= −G m13ρ3− 3x2+ y2+ z2ρ5= 0.Because F is not defined at the origin we have to be careful in applying the divergencetheorem.118.02A topic 46 2In preparation for considering an arbitrary closed surface we compute the flux directlywhen S is a sphere of radius a centered at the origin with outward pointing normals.On S, F and n are parallel (in opposite directions)⇒ F · n = −|F| = −G m1/a2.⇒ flux =RRSF · n dS =ZZS−G m1a2dS = −G m1a2× area = −G m14π.In general, there are two cases, (i) S does not surround 0 and (ii) S surrounds 0.In case (i) (see picture) F is continuously differentiable on the entire volume inside S.⇒ flux =RRSF · n dS =RRRDdivF dV =RRRD0 dV = 0 (as claimed).In case (ii) (see picture) let S1be a small sphere centered at 0 that isentirely inside S. Using the extended divergence theorem, we have (note the signs)RRS−S1F · n dS =RRRDdivF dV = 0.⇒RRSF · n dS −RRS1F · n dS = 0⇒RRSF · n dS =RRS1F · n dS = −4πGm1. (again, as claimed).Example: Let F be the gravitational field from a mass distribution.Show that the flux through a closed surface S is −4πGM, where M isthe total mass enclosed by S.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 S then the flux of dF through S is −4πG dm.If dm is outside S then the flux is 0.That is, each dm contributes−4πG dm if it’s inside S0 if it’s outside SSince F is the sum of all the dF the net flux is just a sum of −4πG dm over alldm inside S, that is flux = −4πG M.End of topic 46
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