As shown in the text, we can write an arbitrary azimuthally-symmetric solution to Laplace s equation in spherical coordinates as:

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1 Remarks: In dealing with spherical coordinates in general and with Legendre polynomials in particular it is convenient to make the sustitution c = cosθ. For example, this allows use of the following simplification of the orthogonality relationship: π P n cosθp m cosθsinθ dθ = 2 2n+ δ nm = P n cp m c dc = 2 2n+ δ nm Since θ = π/2 the equator corresponds to c =, symmetries that correspond to reflection in the equatorial plane correspond to c c symmetry. So the statement P n c = n P n c 2 reports that the n-even P n have even reflection symmetry whereas the n-odd P n have odd reflection symmetry. Finally note that since θ = and π corresponds to c = ±, the statements P n = and P n = n report the ehavior of P n along the positive and negative z axes respectively. As shown in the text, we can write an aritrary azimuthally-symmetric solution to Laplace s equation in spherical coordinates as: φr,θ = A n r n + C n r n+ P n cosθ 3 or equivalently φr,c = A n r n + C n r n+ P n c 4 Example : Consider the prolem of finding φ inside a sphere of radius R where the voltage on the surface of the sphere has een given as a known function Vθ which we will use in the form Vc. First, since nothing singular is happening at the origin, C n = for all n. The A n are determined y the requirement that φ and V agree if r = R: Vc = φr,c = A n R n P n c 5 If we multiply oth sides y P m c and integrate c from to, we can calculate the lhs which of course depends on m and the rhs simplifies ecause of orthogonality: so VcP m c dc = A n R n A m = P n cp m c dc = A m R m 2 2m+ VcP mc dc R m 2 2m+ 6 7 Forexample, iftheappliedvoltage is+v inthenorthernhemisphereand V inthesouthern hemisphere an odd function of c, we can immediately conclude that for n even A n =, and for n odd Mathematica says: A n R n 2 2n+ = 2V P n c dc = V π Γ n/2 Γ3+n/2 8

2 In[]:= A=2 Integrate[LegendreP[n,x],{x,,}] Sqrt[Pi] Out[]= n 3 + n Gamma[ - -] Gamma[-----] 2 2 Mathematica has provided a complex answer for a result that is just a simple rational numer. For your enjoyment, I ll produce a form I can etter understand, ut in the end we ll let Mathematica use its own result. I ll egin y reporting some properties of the Gamma function: Γx+ = xγx 9 Γn+ = n! for n a positive integer π Γ x = sinπx Γx Γ/2 = π 2 x n = xx+x+2 x+n = Γx+n Γx 3 The last formula is for the shifted factorial 2 or Pochhammer Symol defined in class. Note that n is odd which we will write as n = 2m, so m = {,2,3,...} corresponds to n = {,3,5,...}. π Γ n/2 Γ3+n/2 sinπn/2 Γn/2 = = m Γm /2 π Γ3+n/2 Γ/2 Γm+ = m 2 i.e., {, 4, 8, 5 64, , 52,...} Back to Mathematica: m! m 2 = m 2 Γm /2 Γ /2Γm + 4 f[r_,c_]=sum[a 2 n +/2 r^n LegendreP[n,c],{n,,2,2}] ContourPlot[f[Sqrt[x^2+z^2],z/Sqrt[x^2+z^2]],{x,,.9},{z,-.9,.9}, Contours -> {-.9,-.8,-.7,-.6,-.5,-.4,-.3,-.2,-.,,.,.2,.3,.4,.5,.6,.7,.8,.9}, ContourShading->False,RegionFunction->Function[{x, z, q},x^2+z^2< ], AspectRatio->Automatic] Example 2: Consider the prolem of finding φ inside and outside a sphere of radius R where the surface charge density on the surface of the sphere has een given as a known Part of the reason for this complex formula is that Mathematica is showing that n even produces zero result. However it doesn t really matter if you don t recognize the answer as Mathematica can quickly produce the rational numer for any n you want. 2 Note: n = n! more generally x n is n terms multiplied together, starting with x with successive terms one more than the previous.

3 function σθ which we will use in the form σc. First, since nothing singular is happening at the origin, for the inside solution C n = for all n. Since the potential must approach zero as r, for the outside solution A n = for all n. Thus: φr,θ = A n r n P n c for r < R Continuity of φ at r = R produces the requirement: C n r n+p nc for r > R 5 A n R n = C n R n+ 6 The surface charge density can e related to the discontinuity in the radial component of the electric field: σθ = ǫ r φ r=r r φ r=r + 7 = ǫ na n R n +n+c n R n+2 P n c 8 = ǫ 2n+A n R n P n c 9 Theusual FourierTrick multiplyothsidesyp m candintegratefrom tocollapsing the sum to a single term allows A m to e calculated: 2 σcp m c dc = ǫ 2m+A m R m 2 2m+ = ǫ 2A m R m 2 Example 3: Often you can calculate φ along the z axis, ut the off-axis calculation is difficult or impossile. However you can expand φz to produce the full φr,c y a trick. Taylor expand φz to otain a power series expansion: φz = a n z n 22 This formula must agree with the Legendre expansion evaluated on the z axis: φr,c = A n r n P n c = a n z n on the z axis 23 The fact that on axis c = ± and P n ± = ± n allows easy comparison etween these twoseries. Agreement requiresa n usefulforφoff-axis equalsa n determinedonlyknowing φ on-axis. For example, the potential on the z-axis for a ring charge radius R, total charge Q is clearly φz = Q [ z 2 +R 2] /2 Q [ = +z/r 2 ] /2 Q = 2 n z2 /R 2 n 24 4πǫ 4πǫ R 4πǫ n!

4 we can conclude A n = n/2 Q 2 n/2 4πǫ R n n/2! for n even 25 for n odd f[r_,c_]=sum[-^n/2 Pochhammer[/2, n/2] r^n LegendreP[n,c]/n/2!,{n,,2,2}] ContourPlot[f[Sqrt[x^2+z^2],z/Sqrt[x^2+z^2]],{x,,.9},{z,-.9,.9},Contours->6, ContourShading->False,RegionFunction->Function[{x, z, q},x^2+z^2<.8 ], PlotRangePadding->None,AspectRatio->Automatic] Figure : Isopotential contours for Example left and Example 3 right

5 Homework : A physicist aims to suject a sample to a pure quadrupole field n = 2 inside a spherical cavity. The plan is to charge the top and ottom caps of the sphere to V and the remaining and around the equator to a potential of V. Because the applied voltage Vθ is symmetric, terms A n = for n odd. The first important term will then e quadrupole A 2 A corresponds to a constant voltage and so makes no electric field. It would e nice ut not possile to make A 2 the only non-zero term. The est we can do is make A 4 =. Prolem: Find the and angle, θ that makes A 4 =. Find A 6 in this circumstance. Find the values: A, A 2 and A 6. Put the pieces together to express the potential inside the sphere. Have Mathematica produce a contour plot of that voltage. Hint: A 4 { c } VcP 4 c dc = 2 P 4 c dc+ P 4 c dc c Use Mathematica or a root-finding calculator to find the value c to make this quality zero. V z 26 V θ V Homework 2: In a region where there are no currents flowing, we can define a magnetic potential that is exactly analogous to the electric potential: B = φ where: 2 φ = 27 For a pair of Helmholtz coils two identical coaxial coils with centers separated y R; recall Phys 2 las with them, the magnetic potential along the axis is given y: [ ] φz = 5 5R z R/2 6 R 2 +z R/2 + z +R/ R 2 +z +R/2 2 Recall from the 2 la that the spacing of the coils is designed to produce a particularly uniform field etween the coils, and if you reverse the current in one coil you produce a diverging B with B = at the center in short a quadrupole field. Use Mathematica to series expand φ around the origin. Use that series to produce φr,cosθ in a region near the origin. Make a contour plot of φ to confirm that it is nearly uniform. If you have reversed coils a distance aove and elow the origin, φ is given y: [ ] z φz = R 2 +z z + 2 R 2 +z What value of will produce a particularly pure quadrupole field? Make a contour plot of φ to confirm that it is nearly quadrupole.

6 Homework 3: Consider a prolem analogous to Helmholtz coils ut in electrostatics with charged rings. You have a ring radius R, centered on the z axis, in a plane parallel to the xy plane with charge +Q at distance aove the origin, and a similar ring with center at z = with charge Q. Find that will produce the most uniform possile E field in the vicinity of the origin. Explain why the voltage on the z axis is given y: φz = Q 4πǫ [ ] R 2 +z 2 R 2 +z + 2 Expand this result in a power series in z. The term linear in z corresponds to rp c why? and further terms produce a non-uniform E. Determine the value of which makes as many of these further terms zero. Make a contour plot of φr,cosθ for R = to confirm that it is nearly uniform. Example 4: In the case of cylindrical coordinates where φr,θ and not z, we have: φr,θ = A +C lnr+ 3 An r n +C n r n a n cosnθ+c n sinnθ 3 Note that A n,c n,a n,c n are not independent: for example you could multiply oth A n,c n y five and divide oth a n,c n y five and have exactly the same solution. Most commonly one of the terms in parenthesis is reduced to a single term. As usual we have orthogonality as: +π π +π π +π π cosnθ sinmθ dθ = 32 cosnθcosmθ dθ = π δ mn 33 sinnθsinmθ dθ = π δ mn 34 We seek φ outside a cylinder of radius R on which the potential is known to e Vθ = cos 2 θ 35 Since the source extends to infinity, we cannot in general take φ at infinity to e zero; thus the meaning of the potential on the cylinder is amiguous; a convenient solution is to define the constant A = A C lnr a further enefit is the result makes sense dimensionally. φr,θ = A +C lnr/r+ An r n +C n r n a n cosnθ+c n sinnθ 36 Note that in this form the value of C has asolutely no effect on the value of the voltage at r = R; A it of thought should convince you that C is determined y the net chargeper-length on the cylinder. Recall: voltage for a line charge: φ = λ/2πǫ lnr. We will take C =. Since the Vθ is even in θ, c n = ; since the electric field should e regular at infinity A n =. Thus: φr,θ = A + C n r n cosnθ 37

7 Since φr,θ must agree with Vθ we have: cos 2 θ = φr,θ = A + C n R n cosnθ 38 The C n could now e determined using the Fourier Trick, ut a faster way is to use a trig identity to immediately write cos 2 θ in terms of cosnθ: cos 2 θ = cos2θ = A + C n R n cosnθ 39 Simple inspection rather than integration, ut of course integration produces the same result: and C n = for n > 2. So the final result is: 2 = A 4 = C R cosθ 4 2 cos2θ = C 2R 2 cos2θ 42 φr,θ = [ +cos2θ R/r 2 ] 43 2 I hope it is immediately clear that this potential solves Laplace s equation, agrees with Vθ when r = R and represents a cylinder with a simple quadrupole. y V= Vy a x Example 5: Consider an infinite in the z directions rectangular gutter with cross-section etween, and a,. Three of the sides of the gutter are grounded; the fourth, a,y with y, has a specified voltage Vy. Separation of variales yields a general solution: φx,y = We have orthogonality in the form: sin A n sin nπy nπy sinh nπx sinh nπa 44 mπy sin dy = 2 δ mn 45

8 Requiring φx,y to agree with Vy when x = a yields: Vy = φa,y = nπy A n sin 46 With the Fourier Trick yielding: mπy Vy sin dy = 2 A m 47 Consider, for example, the case Vy = constant: So: [ mπy cos mπy sin dy = φx,y = 4 π n odd mπ ] nπy n sin = mπ [ m ] 48 sinh nπx sinh nπa Figure 2: Clockwise from upper left: isopotential contours for Example 5, a slice of Example 5 φ along x,.5, a slice of Example 5 φ along.9,y, isopotential contours for Example 4