Legendre Function

37 Pages·2012·0.49 MB·English

Chapter 11
Legendre Polynomials
and Spherical
Harmonics
11.1 Introduction
Legendre polynomials appear in many different mathematical and physical
situations:
•
They originate as solutions of the Legendre ordinary differential equation
(ODE),whichwehavealreadyencounteredintheseparationofvariables
(Section 8.9) for Laplace’s equation, and similar ODEs in spherical polar
coordinates.
•
They arise as a consequence of demanding a complete, orthogonal set
of functions over the interval [−1,1] (Gram–Schmidt orthogonalization;
Section9.3).
•
Inquantummechanics,they(reallythesphericalharmonics;Section11.5)
representangularmomentumeigenfunctions.Theyalsoappearnaturallyin
problemswithazimuthalsymmetry,whichisthecaseinthenextpoint.
•
Theyaredefinedbyageneratingfunction:WeintroduceLegendrepolyno-
mials here by way of the electrostatic potential of a point charge, which
actsasthegeneratingfunction.
PhysicalBasis:Electrostatics
Legendrepolynomialsappearinanexpansionoftheelectrostaticpotentialin
inverse radial powers. Consider an electric charge q placed on the z-axis at
z=a.AsshowninFig.11.1,theelectrostaticpotentialofchargeq is
1 q
ϕ = · (SIunits). (11.1)
4πε r
0 1
Wewanttoexpresstheelectrostaticpotentialintermsofthesphericalpolar
coordinatesrandθ(thecoordinateϕisabsentbecauseofazimuthalsymmetry,
552
11.1 Introduction 553
Figure11.1
Electrostatic
Potential.Chargeq q
j =
Displacedfrom 4pe r
0 1
Origin
r
r
1
q q
z
z = a
thatis,invarianceunderrotationsaboutthez-axis).Usingthelawofcosines
inFig.11.1,weobtain
q
ϕ = (r2+a2−2arcosθ)−1/2. (11.2)
4πε
0
GeneratingFunction
Consider the case of r > a. The radical in Eq. (11.2) may be expanded in a
binomial series (see Exercise 5.6.9) for r2 > |a2 −2arcosθ| and then rear-
ranged in powers of (a/r). This yields the coefficient 1 of (a/r)0 = 1, cosθ
ascoefficientofa/r, etc.TheLegendre polynomial P (cosθ)(Fig.11.2)is
n
definedasthecoefficientof(a/r)nsothat
(cid:3) (cid:4)
q (cid:2)∞ a n
ϕ = P (cosθ) . (11.3)
4πε r n r
0 n=0
Droppingthefactorq/4πε randusingx=cosθ andt =a/r,respectively,we
0
have
(cid:2)∞
g(t,x)=(1−2xt+t2)−1/2 = P (x)tn, |t|<1, (11.4)
n
n=0
definingg(t,x)asthegeneratingfunctionfortheP (x). Thesepolynomials
n
P (x),showninTable11.1,arethesameasthosegeneratedinExample9.3.1by
n
Gram–Schmidtorthogonalizationofpowersxnovertheinterval−1 ≤ x≤ 1.
Thisisnocoincidencebecausecosθ variesbetweenthelimits±1.Inthenext
section,itisshownthat|P (cosθ)|≤1,whichmeansthattheseriesexpansion
n
[Eq.(11.4)]isconvergentfor|t|<1.1Indeed,theseriesisconvergentfor|t|=1
exceptforx=±1, where|P (±1)|=1.
n
1NotethattheseriesinEq.(11.3)isconvergentforr > aeventhoughthebinomialexpansion
involvedisva√lidonlyforr>(a2+2ar)1/2≥|a2−2arcosθ|1/2sothatr2>a2+2ar,(r−a)2>2a2,
orr>a(1+ 2).
554 Chapter11 LegendrePolynomialsandSphericalHarmonics
Figure11.2
Legendre
P (x)
Polynomials P (x), n
2 1
P (x), P (x),and
3 4
P (x)
5
P
2
0.5
P
3 P
4 P
5
x
−1 0 1
−0.5
Table11.1
P0(x)=1
LegendrePolynomials P1(x)=x
P2(x)= 12(3x2−1)
P3(x)= 12(5x3−3x)
P4(x)= 18(35x4−30x2+3)
P5(x)= 18(63x5−70x3+15x)
P6(x)= 116(231x6−315x4+105x2−5)
P7(x)= 116(429x7−693x5+315x3−35x)
P8(x)= 1128(6435x8−12012x6+6930x4−1260x2+35)
Inphysicalapplications,suchastheCoulomborgravitationalpotentials,
Eq.(11.4)oftenappearsinthevectorform
(cid:7) (cid:3) (cid:4) (cid:3) (cid:4) (cid:8)
1 =(cid:5)r2+r2−2r ·r (cid:6)−1/2 = 1 1+ r2 2−2 r2 cosθ −1/2.
|r −r | 1 2 1 2 r r r
1 2 1 1 1
11.1 Introduction 555
Thelastequalityisobtainedbyfactoringr =|r |fromthedenominator,which
1 1
then,forr >r ,wecanexpandaccordingtoEq.(11.4).Inthisway,weobtain
1 2
(cid:3) (cid:4)
1 = 1 (cid:2)∞ r< nP (cosθ), (11.4a)
|r1−r2| r> n=0 r> n
where
(cid:9)
r> =|r1| for |r |>|r |, (11.4b)
r< =|r2| 1 2
and
(cid:9)
r> =|r2| for |r |>|r |. (11.4c)
r< =|r1| 2 1
EXAMPLE11.1.1 SpecialValues Asimpleandpowerfulapplicationofthegeneratingfunction
gistouseitforspecialvalues(e.g.,x=±1)wheregcanbeevaluatedexplicitly.
Ifwesetx=1,thatis,pointz=aonthepositivez-axis,wherethepotential
hasasimpleform,Eq.(11.4)becomes
(cid:2)∞
1 1
= = tn, (11.5)
(1−2t+t2)1/2 1−t
n=0
usingabinomialexpansionorthegeometricseries(Example5.1.2).However,
Eq.(11.4)forx=1defines
(cid:2)∞
1
= P (1)tn.
(1−2t+t2)1/2 n
n=0
Comparing the two series expansions (uniqueness of power series; Section
5.7),wehave
P (1)=1. (11.6)
n
Ifwelet x = −1inEq.(11.4),thatis,point z= −aonthenegative z-axisin
Fig.11.1,wherethepotentialissimple,thenwesumsimilarly
(cid:2)∞
1 1
= = (−t)n (11.7)
(1+2t+t2)1/2 1+t
n=0
sothat
P (−1)=(−1)n. (11.8)
n
These general results are more difficult to develop from other formulas for
Legendrepolynomials.
Ifwetakex=0inEq.(11.4),usingthebinomialexpansion
1·3···(2n−1)
(1+t2)−1/2 =1− 1t2+ 3t4+···+(−1)n t2n+···, (11.9)
2 8 2nn!
556 Chapter11 LegendrePolynomialsandSphericalHarmonics
wehave2
1·3···(2n−1) (2n−1)!! (−1)n(2n)!
P (0)=(−1)n =(−1)n = (11.10)
2n 2nn! (2n)!! 22n(n!)2
P2n+1(0)=0, n=0,1,2.... (11.11)
TheseresultscanalsobeverifiedbyinspectionofTable11.1. (cid:2)
EXAMPLE11.1.2 Parity If we replace x by −x and t by −t, the generating function is un-
changed.Hence,
g(t,x)= g(−t,−x)=[1−2(−t)(−x)+(−t)2]−1/2
(cid:2)∞ (cid:2)∞
= P (−x)(−t)n= P (x)tn. (11.12)
n n
n=0 n=0
Comparingthesetwoseries,wehave
P (−x)=(−1)nP (x); (11.13)
n n
that is, the polynomial functions are odd or even (with respect to x = 0)
accordingtowhethertheindexnisoddoreven.Thisistheparity3 orreflec-
tionpropertythatplayssuchanimportantroleinquantummechanics.Parity
is conserved when the Hamiltonian is invariant under the reflection of the
coordinates r → −r. For central forces the index n is the orbital angular
momentum[andn(n+1)istheeigenvalueofL2],thuslinkingparityandor-
bitalangularmomentum.Thisparitypropertywillbeconfirmedbytheseries
solutionandforthespecialcasestabulatedinTable11.1. (cid:2)
PowerSeries
Usingthebinomialtheorem(Section5.6)andExercise10.1.15,weexpandthe
generatingfunctionas
(cid:2)∞
(2n)!
(1−2xt+t2)−1/2 = (2xt−t2)n
22n(n!)2
n=0
(cid:2)∞ (2n−1)!!
=1+ (2xt−t2)n. (11.14)
(2n)!!
n=1
Beforeweexpand(2xt−t2)nfurther,letusinspectthelowestpowersoft.
2ThedoublefactorialnotationisdefinedinSection10.1:
(2n)!!=2·4·6···(2n), (2n−1)!!=1·3·5···(2n−1).
3Insphericalpolarcoordinatestheinversionofthepoint(r,θ,ϕ)throughtheoriginisaccom-
plishedbythetransformation[r→r,θ →π−θ,andϕ→ϕ±π].Then,cosθ →cos(π−θ)=
−cosθ,correspondingtox→−x.
11.1 Introduction 557
EXAMPLE11.1.3 Lowest Legendre Polynomials For the first few Legendre polynomials
(e.g., P , P , and P ), we need the coefficients of t0, t1, and t2 in Eq. (11.14).
0 1 2
Thesepowersoftappearonlyinthetermsn=0,1,and2;hence,wemaylimit
ourattentiontothefirstthreetermsoftheinfiniteseries:
0! 2! 4!
(2xt−t2)0+ (2xt−t2)1+ (2xt−t2)2
20(0!)2 22(1!)2 24(2!)2
(cid:3) (cid:4)
3 1
=1t0+xt1+ x2− t2+O(t3).
2 2
Then,fromEq.(11.4)(anduniquenessofpowerseries)weobtain
3 1
P (x)=1, P (x)= x, P (x)= x2− , (11.15)
0 1 2
2 2
confirmingtheentriesofTable11.1.Werepeatthislimiteddevelopmentina
vectorframeworklaterinthissection. (cid:2)
Inemployingageneraltreatment,wefindthatthebinomialexpansionof
the(2xt−t2)nfactoryieldsthedoubleseries
(cid:2)∞ (2n)! (cid:2)n n!
(1−2xt+t2)−1/2 = tn (−1)k (2x)n−ktk
22n(n!)2 k!(n−k)!
n=0 k=0
(cid:2)∞ (cid:2)n (2n)!
= (−1)k ·(2x)n−ktn+k. (11.16)
22nn!k!(n−k)!
n=0k=0
Byrearrangingtheorderofsummation(validbyabsoluteconvergence),Eq.
(11.16)becomes
(cid:2)∞ [(cid:2)n/2] (2n−2k)!
(1−2xt+t2)−1/2 = (−1)k ·(2x)n−2ktn, (11.17)
22n−2kk!(n−k)!(n−2k)!
n=0 k=0
withthetnindependentoftheindexk.4 Now,equatingourtwopowerseries
[Eqs.(11.4)and(11.17)]termbyterm,wehave5
[(cid:2)n/2] (2n−2k)!
P (x)= (−1)k xn−2k. (11.18)
n 2nk!(n−k)!(n−2k)!
k=0
We read off this formula, from k = 0, that the highest power of P (x) is xn,
n
andthelowestpowerisx0 = 1forevennandxforoddn.Thisisconsistent
withExample11.1.3andTable11.1.Also,forneven, P hasonlyevenpowers
n
ofxandthusevenparity[seeEq.(11.13)]andoddpowersandoddparityfor
oddn.
4[n/2]=n/2forneven,(n−1)/2fornodd.
5Equation(11.18)startswithxn.Bychangingtheindex,wecantransformitintoaseriesthat
startswithx0fornevenandx1fornodd.
558 Chapter11 LegendrePolynomialsandSphericalHarmonics
BiographicalData
Legendre, Adrien Marie. Legendre, a French mathematician who was
born in Paris in 1752 and died there in 1833, made major contributions
to number theory, elliptic integrals before Abel and Jacobi, and analysis.
He tried in vain to prove the parallel axiom of Euclidean geometry. His
taste in selecting research problems was remarkably similar to that of his
contemporaryGauss,butnobodycouldmatchGauss’sdepthandperfection.
Hisgreattextbookshadenormousimpact.
LinearElectricMultipoles
Returningtotheelectricchargeonthez-axis,wedemonstratetheusefulness
and power of the generating function by adding a charge −q at z = −a, as
shown in Fig. 11.3, using the superposition principle of electric fields. The
potentialbecomes
(cid:3) (cid:4)
q 1 1
ϕ = − , (11.19)
4πε r r
0 1 2
andbyusingthelawofcosines,wehaveforr >a
(cid:10)(cid:7) (cid:3) (cid:4) (cid:8) (cid:7) (cid:3) (cid:4) (cid:8) (cid:9)
q a a 2 −1/2 a a 2 −1/2
ϕ = 1−2 cosθ + − 1+2 cosθ + ,
4πε r r r r r
0
wherethesecondradicalislikethefirst,exceptthatahasbeenreplacedby
−a.Then,usingEq.(11.4),weobtain
(cid:11) (cid:3) (cid:4) (cid:3) (cid:4) (cid:12)
q (cid:2)∞ a n (cid:2)∞ a n
ϕ = P (cos) − P (cosθ)(−1)n
4πε r n r n r
0 n=0 n=0
(cid:11) (cid:3) (cid:4) (cid:12)
2q a a 3
= P (cosθ) + P (cosθ) +··· . (11.20)
4πε r 1 r 3 r
0
Figure11.3
ElectricDipole j = q 1 - 1
4pe r r
0 1 2
r
2
r
r
1
-q q q
z
z = -a z = a
11.1 Introduction 559
Thefirstterm(anddominanttermforr (cid:14)a)istheelectricdipolepotential
2aq P (cosθ)
ϕ = · 1 , (11.21)
4πε r2
0
with2aqtheelectricdipolemoment(Fig.11.3).IfthepotentialinEq.(11.19)
istakentobethedipolepotential,thenEq.(11.21)givesitsasymptoticbehavior
forlarger.Thisanalysismaybeextendedbyplacingadditionalchargesonthe
z-axis so that the P term, as well as the P (monopole) term, is canceled.
1 0
For instance, charges of q at z = a and z = −a, −2q at z = 0 give rise to a
potentialwhoseseriesexpansionstartswithP (cosθ).Thisisalinearelectric
2
quadrupole. Two linear quadrupoles may be placed so that the quadrupole
term is canceled, but the P , the electric octupole term, survives, etc. These
3
expansionsarespecialcasesofthegeneralmultipoleexpansionoftheelectric
potential.
VectorExpansion
Weconsidertheelectrostaticpotentialproducedbyadistributedchargeρ(r ):
2
(cid:13)
1 ρ(r )
ϕ(r )= 2 dτ . (11.22a)
1 4πε |r −r | 2
0 1 2
Taking the denominator of the integrand, using first the law of cosines and
thenabinomialexpansion,yields
1 =(cid:14)r2−2r ·r +r2(cid:15)−1/2 (11.22b)
|r −r | 1 1 2 2
1 2
(cid:7) (cid:3) (cid:4)(cid:8)
1 2r ·r r2 −1/2
= 1+ − 1 2 + 2 , forr >r
r r2 r2 1 2
1 1 1
(cid:7) (cid:3) (cid:4) (cid:8)
1 r ·r 1r2 3(r ·r )2 r 3
= 1+ 1 2 − 2 + 1 2 +O 2 .
r r2 2r2 2 r4 r
1 1 1 1 1
For r = 1,r = t, and r ·r = xt, Eq. (11.22b) reduces to the generating
1 2 1 2
function,Eq.(11.4).
Thefirstterminthesquarebracket,1,yieldsapotential
(cid:13)
1 1
ϕ (r )= ρ(r )dτ . (11.22c)
0 1 4πε r 2 2
0 1
The integral contains the total charge. This part of the total potential is an
electricmonopole.
Thesecondtermyields
(cid:13)
1 r ·
ϕ (r )= 1 r ρ(r )dτ , (11.22d)
1 1 4πε r3 2 2 2
0 1
wheretheintegralisthedipolemomentwhosechargedensityρ(r )isweighted
2
byamomentarmr .Wehaveanelectricdipolepotential.Foratomicornu-
2
clearstatesofdefiniteparity,ρ(r )isanevenfunctionandthedipoleintegral
2
560 Chapter11 LegendrePolynomialsandSphericalHarmonics
isidenticallyzero.However,inthepresenceofanappliedelectricfieldasuper-
position of odd/even parity states may develop so that the resulting induced
dipole moment is no longer zero. The last two terms, both of order (r /r )2,
2 1
maybehandledbyusingCartesiancoordinates
(cid:2)3 (cid:2)3
(r ·r )2 = x x x x .
1 2 1i 2i 1j 2j
i=1 j=1
Rearrangingvariablestokeepthex insidetheintegralyields
2
(cid:13)
1 1 (cid:2)3 (cid:5) (cid:6)
ϕ (r )= x x 3x x −δ r2 ρ(r )dτ . (11.22e)
2 1 4πε 2r5 1i 1j 2i 2j ij 2 2 2
0 1 i,j=1
This is the electric quadrupole term. Note that the square bracket in the
integrandformsasymmetrictensorofzerotrace.
A general electrostatic multipole expansion can also be developed by
using Eq. (11.22a) for the potential ϕ(r ) and replacing 1/(4π|r −r |) by a
1 1 2
(double)seriesoftheangularsolutionsofthePoissonequation(whicharethe
sameasthoseoftheLaplaceequationofSection8.9).
Beforeleavingmultipolefields,weemphasizethreepoints:
•
First, an electric (or magnetic) multipole has a value independent of the
origin(referencepoint)onlyifalllowerordertermsvanish.Forinstance,
thepotentialofonechargeqatz=awasexpandedinaseriesofLegendre
polynomials.Althoughwerefertothe P (cosθ)terminthisexpansionas
1
adipoleterm,itshouldberememberedthatthistermexistsonlybecause
ofourchoiceofcoordinates.Weactuallyhaveamonopole, P (cosθ),the
0
termofleadingmagnitude.
•
Second, in physical systems we rarely encounter pure multipoles. For
example, the potential of the finite dipole (q at z = a,−q at z = −a)
contained a P (cosθ) term. These higher order terms may be eliminated
3
by shrinking the multipole to a point multipole, in this case keeping the
productqaconstant(a→0,q →∞)tomaintainthesamedipolemoment.
•
Third, the multipole expansion is not restricted to electrical phenomena.
Planetaryconfigurationsaredescribedintermsofmassmultipoles.
Itmightalsobenotedthatamultipoleexpansionisactuallyadecomposition
into the irreducible representations of the rotation group (Section 4.2). The
lthmultipoleinvolvestheeigenfunctionsoforbitalangularmomentum,|lm(cid:15),
one for each component mof the multipole l (see Chapter 4). These 2l +1
componentsofthemultipoleformanirreduciblerepresentationbecausethe
loweringoperatorL−appliedrepeatedlytotheeigenfunction|ll(cid:15)generatesall
othereigenfunctions|lm(cid:15), downtom= −l.Theraisingandloweringopera-
tors L±aregeneratorsoftherotationgroupalongwith Lz, whoseeigenvalue
ism.
11.1 Introduction 561
EXERCISES
11.1.1 Developtheelectrostaticpotentialforthearrayofchargesshown.This
isalinearelectricquadrupole(Fig.11.4).
Figure11.4
LinearElectric
Quadrupole q -2q q
z
z = -a z = a
11.1.2 Calculate the electrostatic potential of the array of charges shown
in Fig. 11.5. This is an example of two equal but oppositely directed
dipoles. The dipole contributions cancel, but the octupole terms do
notcancel.
Figure11.5
LinearElectric -q +2q -2q q
Octupole z
z = -2a -a a 2a
11.1.3 Showthattheelectrostaticpotentialproducedbyachargeq atz= a
forr <ais
(cid:3) (cid:4)
q (cid:2)∞ r n
ϕ(r)= P (cosθ).
4πε a a n
0 n=0
11.1.4 UsingE=−∇ϕ,determinethecomponentsoftheelectricfieldcorre-
spondingtothe(pure)electricdipolepotential
2aqP (cosθ)
ϕ(r)= 1 .
4πε r2
0
Here,itisassumedthatr (cid:14)a.
4aqcosθ 2aqsinθ
ANS. Er =+ 4πε r3 , Eθ =+ 4πε r3 , Eϕ =0.
0 0
11.1.5 Apointelectricdipoleofstrength p(1)isplacedatz=a;asecondpoint
electricdipoleofequalbutoppositestrengthisattheorigin.Keeping
theproduct p(1)aconstant,leta→0.Showthatthisresultsinapoint
electricquadrupole.