PHZ 3113 Fall 2017 Homework #5, Due Friday, October 13

Transcription

1 PHZ 3113 Fall 2017 Homewok #5, Due Fiday, Octobe Genealize the poduct ule (fg) = f g +f g to wite the divegence Ö (Ù Ú) of the coss poduct of the vecto fields Ù and Ú in tems of the cul of Ù and the cul of Ú. Use this to compute Ö (Öφ Öψ), fo scala fields φ and ψ. This can be done seveal ways. One is to ague as follows: The deivative opeato Ö acts on both Ù and Ú; the poduct ule applies in some fom. The outcome must be a scala, and so cannot contain Ö Ù o Ö Ú, since thee is no way to combine eithe with the emaining vecto to poduce a scala. Thus, the esult must contain Ö Ù and Ö Ú, and to fom a scala must contain Ú (Ö Ù) and Ù (Ö Ú). ompaed to the oiginal, in the fist tem the thee vectos Ö, Ù, and Ú emain in cyclic ode, and so it appeas with a + sign; but in the second tem the cyclic ode is evesed, and so it appeas with a sign: Ö (Ù Ú) = Ú (Ö Ù) Ù (Ö Ú). O, you can wite out all the components (u x,y = u x / y, etc.), in exhaustive detail, Ö (Ù Ú) = u y v z u z v y + u y z v x u x v z + u z x v y u y v x = u y,x v z +u y v z,x u z,x v y u z v y,x +u z,y v x +u z v x,y u x,y v z u x v z,y +u x,z v y +u x v y,z u y,z v x u y v x,z = v x (u z,y u y,z )+v y (u x,z u z,x )+v z (u y,x u x,y ) u x (v z,y v y,z ) u y (v x,z v z,x ) u z (v y,x vx,y) = Ú (Ö Ù) Ù (Ö Ú). O, you can use the ǫ ijk notation: (Ù Ú) i = ǫ ijk u j v k, and Ö (Ù Ú) = Ö i (ǫ ijk u j v k ) = ǫ ijk Ö i (u j v k ) = ǫ ijk [(Ö i u j )v k +u j (Ö i v k )] Fo Ù = Öφ and Ú = Öψ, = v k (ǫ ijk Ö i u j )+u j (ǫ ijk Ö i v k ) = v k (+ǫ kij Ö i u j )+u j ( ǫ jik Ö i v k ) = v k (Ö Ù) k u j (Ö Ú) j = Ú (Ö Ù) Ù (Ö Ú). Ö (Öφ Öψ) = Öψ (Ö Öφ) Öφ (Ö Öψ) = 0, since the cul of a gadient always vanishes fo easons discussed peviously. This also follows fom ǫ ijk Ö k (Ö j φ Ö k ψ).

2 2. Letthevectofield Ú be Ú = y(y2 +z 2 ) 3 ˆÜ+ x(x2 +z 2 ) 3 ˆÝ xyz 3 ˆÞ,whee = Ô x 2 +y 2 +z 2. (a) ompute Ö Ú. ompute Ö Ú. All calculations can make good use of the elation 1 p = and similaly fo y and z. 1 (x 2 +y 2 +z 2 ) p/2 = p 2 2x (x 2 +y 2 +z 2 ) (p+2)/2 = px p+2, The deivatives that appea in the divegence ae = 3xy(y 2 +z 2 ) 5 y = 3xy(x 2 +z 2 ) 5 z = xy(x 2 +y 2 2z 2 ) and the divegence is Ö Ú = + y + z = 4xy 3. The deivatives that appea in the cul ae y = 3x2 y 2 +x 2 z 2 +y 2 z 2 +z 4 = 3x2 y 2 +x 2 z 2 +y 2 z 2 +z 4 = yz(2x2 y 2 z 2 ) z = yz(2x2 y 2 z 2 ) z = xz(2y2 x 2 z 2 ) y = xz(2y2 x 2 z 2 ) and Ö Ú = ˆÜ vz y v y vx + ˆÝ z z v z vy + ˆÞ v x = 0. y

3 (b) ompute Ö(Ö Ú) and Ö 2 Ú. ompute Ö (Ö Ú) two diffeent ways. Again using ( /)(1/ 3 ) = 3x/ etc., it is staightfowad to obtain Ö(Ö Ú) = 4y(3x2 2 ) 5 ˆÜ+ 4x(3y2 2 ) 5 ˆÝ + 12xyz 5 ˆÞ. The Laplacian of the x-component is 2 = 3y(y2 +z 2 )( 2 5x 2 ) 7, y 2 = 3y(2x4 3x 2 y 2 +x 2 z 2 y 2 z 2 z 4 ) 7, z 2 = y(2x4 +x 2 y 2 y 4 11x 2 z 2 +y 2 z 2 +2z 4 ) 7, Ö 2 v x = 4y(2 3x 2 ) 5. The y-component follows fom exchanging x and y, and This gives the familia-looking Ö 2 v z = 12xyz 5. Ö 2 Ú == 4y(3x2 2 ) 5 ˆÜ+ 4x(3y2 2 ) 5 ˆÝ + 12xyz 5 ˆÞ. The cul of the cul diectly is easy, which is the same as Ö(Ö Ú) Ö 2 Ú. Ö (Ö Ú) = Ö (0) = 0. If you think of it, Ö Ú = 0 means that Ú = Öφ fo some φ; and fom v z it is appaent that this potential function is Then, φ = xy. Ö 2 Ú = Ö 2 (Öφ) = Ö(Ö 2 φ) = Ö 4xy 3.

4 (x y) 3. Let Ú(x,y) be the vecto field Ú = ˆÜ+ (x+y) ˆÝ, whee = Ô x 2 +y 2. Let be the cicumfeence of the cicle of adius = 1 in the x-y plane centeed at the oigin. (a) ompute Ú ˆÒds two diffeent ways. To compute integals, we need to wite eveything in tems of some one thing that we can then integate ove. Let a point on the cicle be specified by pola angle θ, x = cosθ, y = sinθ. Then, position and its deivative ae Ö = xˆü+y ˆÝ = cosθ ˆÜ+sinθ ˆÝ, dö = ( sinθ ˆÜ+cosθ ˆÝ)dθ, ds = dö = dθ; and the unit vecto pependicula to the cicle ( unit nomal ) is ˆÒ = cosθ ˆÜ+sinθˆÝ. The diect integal ove θ then gives Ú ˆÒds = [(cosθ sinθ)ˆü+(cosθ+sinθ)ˆý] (cosθ ˆÜ+sinθˆÝ) dθ = cos 2 θ sinθcosθ+cosθsinθ+sin 2 2π θ dθ = dθ = 2π. 0 The othe way is using the two-dimensional divegence theoem, Ú ˆÒds = (Ö Ú)d 2 a. A The divegence is Ö Ú = (x y) + y (x+y) = 1 x(x y) y(x+y) 3 = 1, and the aea integal is also 1 d 2 a(ö Ú) = ddφ 1 A 0 = 2π

5 (b) ompute Ú dö two diffeent ways. The θ-integal is Ú dö = = [(cosθ sinθ)ˆü+(cosθ+sinθ)ˆý] ( sinθ ˆÜ+cosθˆÝ)dθ [ cosθsinθ+sin 2 θ+cos 2 θ+sinθcosθ]dθ = 2π. By Stoke s theoem, this is also Ú dö = (Ö Ú) ˆÒd 2 a. A The cul is leading to the integal Ö Ú = vy v x ˆÞ = 1 y ˆÞ 1 Ú dö = (Ö Ú) ˆÒd 2 a = A 0 ddφ = 2π. (c) What happens when the denominato in Ú is 2 instead of? The values of the integals aound the cicumfeence of the unit cicle ae the same fo any powe p. Without egulaization, the divegence o cul fo denominato 2 would appea to vanish entiely, but with cutoff ǫ and Ö Ú = 2ǫ 2 ( 2 +ǫ 2 ) 2, Ö Ú = 2ǫ 2 ( 2 +ǫ 2 ) 2 ˆÞ, 1 (Ö Ú)d 2 2ǫ 2 2π2 1 a = A 0 ( 2 +ǫ 2 ) 2 2πd = 2 +ǫ 2 = 2π 0 1+ǫ 2 2π. The only contibution to the aea integal is fom the δ-function hiding in the divegence/cul.