Physics 505 Fall 2007 Homework Assignment #5 Solutions. Textbook problems: Ch. 3: 3.13, 3.17, 3.26, 3.27

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1 Physics 55 Fa 7 Homework Assignment #5 Soutions Textook proems: Ch. 3: 3.3, 3.7, 3.6, Sove for the potentia in Proem 3., using the appropriate Green function otained in the text, and verify that the answer otained in this way agrees with the direct soution from the differentia equation. Reca that Proem 3. asks for the potentia etween two concentric spheres of radii a and with > a, where the upper hemisphere of the inner sphere and the ower hemisphere of the outer sphere are maintained at potentia V, and where the other hemispheres are at zero potentia. Since this proem invoves the potentia etween two spheres, we use the Dirichet Green s function G x, x = +,m a + r< a+ r + < r> + r > + Y m ΩY m Ω Because there are no charges etween the spheres, the Green s function soution for the potentia ony invoves the surface integra Φ x = S Φ x G n da Here, we note a sutety in that the oundary surface is actuay disconnected, and incudes oth the inner sphere of radius a and the outer sphere of radius. This means that the potentia may e written as a sum of two contriutions Φ x = r =a Φ x G n a dω r = We now compute the norma derivatives of the Green s function and G r n = G r< =a r = =r =a = a a,m a + r + a Φ x G n dω + r Y m ΩY m Ω G r n = G r> = r = =r = = r a a +,m a + Y m ΩY m Ω r

2 Inserting these expressions into yieds Φ x =,m +,m V a Ω Y m Ω dω V Ω Y m Ω dω a a + a + r + r r a a + + r a Y m Ω Y m Ω This is the genera expression for the soution to the oundary vaue proem where V a Ω is the potentia on the inner sphere and V Ω is the potentia on the outer sphere. For the upper/ower hemispheres proem, we note that azimutha symmetry aows us to restrict the m vaues to m = ony. In this case, the spherica harmonic expansion reduces to a Legendre poynomia expansion Φ x = + V a ζp ζdζ V ζp ζdζ + a + a + r + r a + r a a + r a P cos θ + P cos θ where ζ = cos θ. Since V a = V for ζ > and V = V for ζ <, this aove expression reduces to Φ x = + V N a + a + r a + r + + V N r a a a + r = + V N a + a + r a + r r a + P cos θ + P cos θ a r + P cos θ where N = { = P ζdζ = j+ Γj πj! = j odd

3 If desired, the potentia may e rearranged to read Φ x = + V N a a a + r a + r = V + V j+ 4j Γj a j a j j= 4 πj! a 4j + r P cos θ a j r j + P j cos θ which agrees with the soution to Proem 3. that we have found earier. 3.7 The Dirichet Green function for the unounded space etween the panes at z = and z = L aows discussion of a point charge or a distriution of charge etween parae conducting panes hed at zero potentia. a Using cyindrica coordinates show that one form of the Green function is G x, x = 4 L n= m= e imφ φ sin nπz sin L nπz L nπ nπ I m L ρ < K m L ρ > In cyindrica coordinates, the poar direction φ is periodic with period π. This suggests that the Green s function coud e expanded as a Fourier series in e imφ. Simiary, the oundary conditions G = at z = and z = L motivates the use of a Fourier sine series sinnπz/l in the z coordinate. More precisey, a compete Fourier expansion in φ and z woud give G x, x = gρ, ρ e imφ e im φ nπz n πz sin sin L L m,n,m,n However, it turns out that m and m and n and n do not need to e chosen to e independent. This can e seen from the Green s function equation given here as a differentia equation in x x G x, x = δ 3 x x In cyindrica coordinates, this reads ρ ρ ρ ρ + ρ φ + z Gρ, φ, z; ρ, φ, z = ρ δρ ρ δφ φ δz z 3

4 Using the competeness reations and suggests that we take n= m= e imφ φ = πδφ φ 4 nπz nπz sin sin L L = L δz z G x, x = n= m= gρ, ρ e imφ φ sin nπz nπz sin L L 5 Sustituting this decomposition into 3 gives d ρ dρ ρ d dρ m ρ Making the sustitution nπ L gρ, ρ = 4 Lρ δρ, ρ x = nπρ L converts the homogeneous part of this to a modified Besse equation d dx + x d dx + m x gx, x = 4 Lx δx, x At this stage, the soution ecomes standard. Noting that the modified Besse function I m x ows up as x and the function K m x ows up as x, we are eft with { gx, x AIm x x < x = BK m x x > x where the coefficients A and B are determined y the matching conditions g < = g >, d dx g < = d dx g > + 4 Lx at x = x. This system may e soved to yied A = 4 K m x Lx I mx K m x I m x K mx B = 4 I m x Lx I mx K m x I m x K mx

5 Noting that the modified Besse functions satisfy the Wronskian formua I ν xk νx I νxk ν x = x finay gives { Im xk m x x < x gx, x = 4 L I m x K m x x > x = 4 L I mx < K m x > where x < = minx, x, x > = maxx, x Converting x ack to ρ and sustituting into 5 then gives the desired Dirichet Green s function G x, x = 4 L n= m= e imφ φ sin nπz nπz sin L L I m nπρ< L nπρ> K m L Show that an aternative form of the Green function is G x, x = m= dk e imφ φ J m kρj m kρ sinhkz < sinhkl z > sinhkl This aternative form of the Green s function is derived y expanding in φ and ρ instead of φ and z. For the ρ expansion, we use the integra reation kj ν kρj ν kρ dk = ρ δρ ρ aong with the competeness reation 4 to motivate the decomposition G x, x = m= kdk g k z, z e imφ φ J m kρj m kρ 6 Since the Besse function J m kρ satisfies the Besse equation d dρ + ρ d dρ + k m ρ J m kρ = the sustitution of 6 into the Greens function equation 3 gives d dz k g k z, z = δz z

6 Since g k z, z vanishes at z = and z = L, this is a standard one-dimensiona Green s function proem. Writing { g k z, z A sinhkz z < z = B sinhkl z z > z we find that the matching and jump conditions ecome A sinhkz = B sinhkl z, A coshkz = B coshkl z + k This may e soved to give so that A = k g k z, z = Sustituting this into 6 then yieds G x, x = m= k sinhkl z, B = sinhkz sinhkl k sinhkl k sinhkl sinhkz < sinhkl z > dk e imφ φ J m kρj m kρ sinhkz < sinhkl z > sinhkl 3.6 Consider the Green function appropriate for Neumann oundary conditions for the voume V etween the concentric spherica surfaces defined y r = a and r =, a <. To e ae to use.46 for the potentia, impose the simpe constraint.45. Use an expansion in spherica harmonics of the form G x, x = g r, r P cos γ = where g r, r = r </r + > + f r, r. a Show that for >, the radia Green function has the symmetric form g r, r = r < r> a + rr + + a + r r rr + a+ + + r + r + There are severa approaches to this proem. However, we first consider the Neumann oundary condition.45 G x x n = ndy S

7 For this proem with two oundaries, the surface area S must e the area of oth oundaries ie it is the tota area surrounding the voume. Hence S = a +, and in particuar this is uniform constant in the anges. As a resut, this wi ony contriute to the = term in the expansion of the Green s function. More precisey, we coud write G x x n = ndy g r, r n P cos γ ndy = a + P cos γ Since the Legendre poynomias are orthogona, this impies that g r, r n = ndy a + δ, Noting that the outward norma is either in the ˆr or the ˆr direction for the sphere at a or, respectivey, we end up with two oundary condition equations g r, r r = a a + δ g r, r, r = a + δ, 7 Now that we have written down the oundary conditions for g r, r, we proceed to otain its expicit form. The suggestion of the proem is to write Since g r, r = r < r + > x x = + f r, r r< r> + P cos γ we see that the first term in g r, r is designed to give the singuar source deta function. The remaining term F x, x = f r, r P cos γ then soves the homogeneous equation x F x, x =. But we know how to sove Lapace s equation in spherica coordinates, and the resut is that the radia function must e of the form f r, r = A r + B r + Note that we are taking the Green s function equation to act on the x variae, where x may e thought of as a parameter constant giving the ocation of the deta function source. We thus have g r, r = r < r + > + A r + B r + 8

8 A that remains is to use the oundary conditions 7 to sove for A and B. For the inside sphere at a, we have whie for the outside sphere we have a r + + A a + B a + = δ, a r + + A + B + = δ, a + For we rewrite these equations as a + + A a + = + /r + + B + r which may e soved to give A = B + + a + r = + a + Inserting this into 8 yieds g r, r = r < r + > + r + a + = r < r> a + rr + + = + a + = + + a + r< a + /r + + a + + r a/r / a + + / + a/r + a r + a r < r > a + r + < + r a + r r + a + r rr + a+ + r + + a + r < r > r < r + > r > a + r > + r< + r r + which is vaid for. Note that in the ast few ines we have een ae to rewrite the Green s function in terms of a product of ur < and vr > where u and v satisfies Neumann oundary conditions at r = a and r =, respectivey. r + >

9 This is reated to another possie method of soving this proem. Using the Legendre identity = + P cos γ = δφ φ δcos θ cos θ the Green s function equation may e reduced to the one-dimensiona proem d d dr r + g dr r, r = + δr r Using the genera method for the Sturm-Liouvie proem, the Green s function is given y g r, r = + A u r < v r > where ur and vr sove the homogeneous equation and the constant A is fixed y the Wronskian, W u, v = A /r. For the oundary conditions 7 are homogeneous u r r =a = v r r = = It is easy to see that these are satisfied y ur = r + Computing the Wronskian gives which aows us to identify a + + r + vr = r r + u v u v = + a+ + + r A = a + This gives the resut of the ast ine of. Show that for = g r, r = a r > a + r + fr where fr is aritrary. Show expicity in.46 that answers for the potentia Φ x are independent of fr. The = case invoves a non-homogeneous oundary condition. Hence the resut of wi not work. Of course, we can sti work out the one-dimensiona deta

10 function proem with matching and jump conditions at r = r. However it is more direct to return to 9 and and to simpy sove those conditions for =. Both 9 and resut in a B = a + whie eaving A competey undetermined. Finay, since r is thought of as a parameter, this indicates that A = fr can e an aritrary function of r. The = Green s function is given y 8 g r, r = r > a a + r + fr Incidentay, we note that without the inhomogeneous Neumann oundary condition term /S there wi e no soution to the system 9 and for = uness is taken to. This demonstrates the inconsistency of simpy setting G/ n = for the Neumann Green s function. Note that, y setting fr = a /a + r we otain a symmetrica Green s function g r, r = a r > a + r + r On the other hand, the choice of fr is unphysica. This arises ecause, for the Neumann Green s function, the fr contriution to the potentia is given y Φ x = ρ x fr d 3 x + Φ x ɛ V S n fr da = fr ρ x d 3 x ɛ E x dâ ɛ V S = fr q enc ɛ E x dâ = ɛ y Gauss aw. It is important not to mix up r and r in this derivation. 3.7 Appy the Neumann Green function of Proem 3.6 to the situation in which the norma eectric fied is E r = E cos θ at the outer surface r = and is E r = on the inner surface r = a. a Show that the eectrostatic potentia inside the voume V is r cos θ Φ x = E p 3 + a3 r 3 where p = a/. Find the components of the eectric fied cos θ E r r, θ = E p 3 a3 sin θ r 3, E θ r, θ = E p 3 S + a3 r 3

11 Since there is no charge etween the spheres, the soution to e oundary vaue proem is given y Φ x = Φ x S n G x, x da = E r Ω G x, x dω r = = E G x, x cos θ dω By writing = E r = = P cos γ = r = + g r, r P cos γ cos θ dω m Y m ΩY m Ω and noting that cos θ = /3Y Ω, we end up with the expansion Φ x = E 3,m g r, Y m Ω + = E g r, Y Ω 3 3 = E cos θ g r, 3 Y m Ω Y Ω dω where we have used orthogonaity of the spherica harmonics. Inserting = into then gives Φ x = E cos θ r 3 3 a 3 + a3 3 r = E r cos θ a/ 3 + a3 r 3 3 This is the potentia for a constant eectric fied comined with an eectric dipoe. Defining p = a/, the components of the eectric fied are E r = Φ r = E cos θ p 3 a3 r 3, E θ = Φ r θ = E sin θ p 3 + a3 r 3 Note that the oundary conditions E r r=a = and E r r= = E cos θ are oviousy satisfied. On the other hand, the parae component of the fied, E θ, is non-vanishing on oth surfaces except at the poes. Physicay, this indicates that these surfaces are not conductors. Cacuate the Cartesian or cyindrica components of the fied, E z ad E ρ, and make a sketch or computer pot of the ines of eectric force for a typica case of p =.5.

12 Rewriting 3 as Φ x = E p 3 z + a3 z r 3 = E p 3 z + a 3 z ρ + z 3/ we otain E z = Φ z = E p 3 + a3 3ẑ r 3 E ρ = Φ ρ = E p 3 3a3 ẑ ˆρ r 3 where ˆρ = ρ/r, ẑ = z/r, and r = ρ + z. As indicated aove, this corresponds to a constant eectric fied comined with an eectric dipoe. For E >, a sketch of the eectric fied ines ooks ike This sketch indicates that the radia component of the eectric fied vanishes at the surface of the inner sphere.