The first term (and dominant term forr≫a) is ϕ= 2aq
4π ε0ãP1(cosθ )
r2 , (12.11)
which is the electric dipole potential, and 2aq is the dipole moment (Fig. 12.3). This analysis may be extended by placing additional charges on thez-axis so that theP1term, as well as theP0 (monopole) term, is canceled. For instance, charges ofq atz=a and z= −a,−2qatz=0 give rise to a potential whose series expansion starts withP2(cosθ ).
This is a linear electric quadrupole. Two linear quadrupoles may be placed so that the quadrupole term is canceled but theP3, the octupole term, survives.
Vector Expansion
We consider the electrostatic potential produced by a distributed chargeρ(r2):
ϕ(r1)= 1 4π ε0
ρ(r2)
|r1−r2|d3r2. (12.12a) This expression has already appeared in Sections 1.16 and 9.7. Taking the denominator of the integrand, using first the law of cosines and then a binomial expansion, yields (see Fig. 1.42)
1
|r1−r2| =
r12−2r1ãr2+r22−1/2
(12.12b)
= 1 r1
1+
−2r1ãr2 r12 +r22
r12 −1/2
, forr1> r2
= 1 r1
1+r1ãr2
r12 −1 2
r22 r12+3
2
(r1ãr2)2 r14 +O
r2
r1
3 .
(For r1=1, r2=t, and r1ãr2=xt, Eq. (12.12b) reduces to the generating function, Eq. (12.4).)
The first term in the square bracket, 1, yields a potential ϕ0(r1)= 1
4π ε0 1 r1
ρ(r2) d3r2. (12.12c) The integral is just the total charge. This part of the total potential is an electric monopole.
The second term yields
ϕ1(r1)= 1 4π ε0
r1ã r13
r2ρ(r2) d3r2, (12.12d) where the integral is the dipole moment whose charge densityρ(r2)is weighted by a mo- ment arm r2. We have an electric dipole potential. For atomic or nuclear states of definite parity,ρ(r2)is an even function and the dipole integral is identically zero.
The last two terms, both of order(r2/r1)2, may be handled by using Cartesian coordi- nates:
(r1ãr2)2= 3
i=1
x1ix2i
3 j=1
x1jx2j.
Rearranging variables to take thex1components outside the integral yields ϕ2(r1)= 1
4π ε0 1 2r15
3 i,j=1
x1ix1j 3x2ix2j−δijr22 ρ(r2) d3r2. (12.12e) This is the electric quadrupole term. We note that the square bracket in the integrand forms a symmetric, zero-trace tensor.
A general electrostatic multipole expansion can also be developed by using Eq. (12.12a) for the potentialϕ(r1)and replacing 1/(4π|r1−r2|)by Green’s function, Eq. (9.187). This yields the potentialϕ(r1)as a (double) series of the spherical harmonicsYlm(θ1, ϕ1)and Ylm(θ2, ϕ2).
Before leaving multipole fields, perhaps we should emphasize three points.
• First, an electric (or magnetic) multipole is isolated and well defined only if all lower- order multipoles vanish. For instance, the potential of one charge q atz=a was ex- panded in a series of Legendre polynomials. Although we refer to theP1(cosθ )term in this expansion as a dipole term, it should be remembered that this term exists only because of our choice of coordinates. We also have a monopole,P0(cosθ ).
• Second, in physical systems we do not encounter pure multipoles. As an example, the potential of the finite dipole (q atz=a,−q at z= −a) contained a P3(cosθ ) term. These higher-order terms may be eliminated by shrinking the multipole to a point multipole, in this case keeping the productqaconstant(a→0, q→ ∞)to maintain the same dipole moment.
• Third, the multipole theory is not restricted to electrical phenomena. Planetary configu- rations are described in terms of mass multipoles, Sections 12.3 and 12.6. Gravitational radiation depends on the time behavior of mass quadrupoles. (The gravitational radia- tion field is a tensor field. The radiation quanta, gravitons, carry two units of angular momentum.)
It might also be noted that a multipole expansion is actually a decomposition into the irreducible representations of the rotation group (Section 4.2).
Extension to Ultraspherical Polynomials
The generating function used here,g(t, x), is actually a special case of a more general generating function,
1
(1−2xt+t2)α = ∞
n=0
Cn(α)(x)tn. (12.13)
The coefficients Cn(α)(x)are the ultraspherical polynomials (proportional to the Gegen- bauer polynomials). Forα=1/2 this equation reduces to Eq. (12.4); that is,Cn(1/2)(x)= Pn(x). The casesa=0 and α=1 are considered in Chapter 13 in connection with the Chebyshev polynomials.
Exercises
12.1.1 Develop the electrostatic potential for the array of charges shown. This is a linear elec- tric quadrupole (Fig. 12.4).
12.1.2 Calculate the electrostatic potential of the array of charges shown in Fig. 12.5. Here is an example of two equal but oppositely directed dipoles. The dipole contributions cancel. The octupole terms do not cancel.
12.1.3 Show that the electrostatic potential produced by a chargeqatz=aforr < ais ϕ(r)= q
4π ε0a ∞
n=0
r a
n
Pn(cosθ ).
FIGURE12.4 Linear electric quadrupole.
FIGURE12.5 Linear electric octupole.
FIGURE12.6
12.1.4 Using E= −∇ϕ, determine the components of the electric field corresponding to the (pure) electric dipole potential
ϕ(r)=2aqP1(cosθ ) 4π ε0r2 . Here it is assumed thatr≫a.
ANS.Er= +4aqcosθ
4π ε0r3 , Eθ= +2aqsinθ
4π ε0r3 , Eϕ=0.
12.1.5 A point electric dipole of strengthp(1)is placed atz=a; a second point electric dipole of equal but opposite strength is at the origin. Keeping the productp(1)a constant, let a→0. Show that this results in a point electric quadrupole.
Hint. Exercise 12.2.5 (when proved) will be helpful.
12.1.6 A point chargeq is in the interior of a hollow conducting sphere of radius r0. The charge q is displaced a distance a from the center of the sphere. If the conducting sphere is grounded, show that the potential in the interior produced byq and the dis- tributed induced charge is the same as that produced byqand its image chargeq′. The image charge is at a distancea′=r02/a from the center, collinear withq and the origin (Fig. 12.6).
Hint. Calculate the electrostatic potential fora < r0< a′. Show that the potential van- ishes forr=r0if we takeq′= −qr0/a.
12.1.7 Prove that
Pn(cosθ )=(−1)nrn+1 n!
∂n
∂zn 1
r
.
Hint. Compare the Legendre polynomial expansion of the generating function (a→z, Fig. 12.1) with a Taylor series expansion of 1/r, wherezdependence ofrchanges from ztoz−z(Fig. 12.7).
12.1.8 By differentiation and direct substitution of the series form, Eq. (12.8), show thatPn(x) satisfies the Legendre ODE. Note that there is no restriction uponx. We may have any x,−∞< x <∞, and indeed anyzin the entire finite complex plane.
FIGURE12.7
12.1.9 The Chebyshev polynomials (type II) are generated by (Eq. (13.93), Section 13.3) 1
1−2xt+t2= ∞
n=0
Un(x)tn.
Using the techniques of Section 5.4 for transforming series, develop a series represen- tation ofUn(x).
ANS.Un(x)=
[n/2] k=0
(−1)k (n−k)!
k!(n−2k)!(2x)n−2k.
12.2 R ECURRENCE R ELATIONS AND S PECIAL P ROPERTIES
Recurrence Relations
The Legendre polynomial generating function provides a convenient way of deriving the recurrence relations4and some special properties. If our generating function (Eq. (12.4)) is differentiated with respect tot, we obtain
∂g(t, x)
∂t = x−t
(1−2xt+t2)3/2= ∞
n=0
nPn(x)tn−1. (12.14) By substituting Eq. (12.4) into this and rearranging terms, we have
1−2xt+t2∞
n=0
nPn(x)tn−1+(t−x) ∞
n=0
Pn(x)tn=0. (12.15) The left-hand side is a power series int. Since this power series vanishes for all values of t, the coefficient of each power of t is equal to zero; that is, our power series is unique (Section 5.7). These coefficients are found by separating the individual summations and
4We can also apply the explicit series form Eq. (12.8) directly.
using distinctive summation indices:
∞
m=0
mPm(x)tm−1− ∞
n=0
2nxPn(x)tn+ ∞
s=0
sPs(x)ts+1
+ ∞
s=0
Ps(x)ts+1− ∞
n=0
xPn(x)tn=0. (12.16)
Now, lettingm=n+1, s=n−1, we find
(2n+1)xPn(x)=(n+1)Pn+1(x)+nPn−1(x), n=1,2,3, . . . . (12.17) This is another three-term recurrence relation, similar to (but not identical with) the recur- rence relation for Bessel functions. With this recurrence relation we may easily construct the higher Legendre polynomials. If we taken=1 and insert the easily found values of P0(x)andP1(x)(Exercise 12.1.7 or Eq. (12.8)), we obtain
3xP1(x)=2P2(x)+P0(x), (12.18) or
P2(x)=1 2
3x2−1
. (12.19)
This process may be continued indefinitely, the first few Legendre polynomials are listed in Table 12.1.
As cumbersome as it may appear at first, this technique is actually more efficient for a digital computer than is direct evaluation of the series (Eq. (12.8)). For greater stability (to avoid undue accumulation and magnification of round-off error), Eq. (12.17) is rewritten as
Pn+1(x)=2xPn(x)−Pn−1(x)− 1 n+1
xPn(x)−Pn−1(x) . (12.17a) One starts withP0(x)=1, P1(x)=x, and computes the numerical values of all thePn(x) for a given value of x up to the desired PN(x). The values of Pn(x),0≤n < N, are available as a fringe benefit.
Table 12.1 Legendre Polynomials
P0(x)=1 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)=161(231x6−315x4+105x2−5) P7(x)=161(429x7−693x5+315x3−35x)
P8(x)=1281 (6435x8−12012x6+6930x4−1260x2+35)
Differential Equations
More information about the behavior of the Legendre polynomials can be obtained if we now differentiate Eq. (12.4) with respect tox. This gives
∂g(t, x)
∂x = t
(1−2xt+t2)3/2= ∞
n=0
Pn′(x)tn, (12.20) or
1−2xt+t2∞
n=0
Pn′(x)tn−t ∞
n=0
Pn(x)tn=0. (12.21) As before, the coefficient of each power oft is set equal to zero and we obtain
Pn′+1(x)+Pn′−1(x)=2xPn′(x)+Pn(x). (12.22) A more useful relation may be found by differentiating Eq. (12.17) with respect toxand multiplying by 2. To this we add(2n+1)times Eq. (12.22), canceling thePn′ term. The result is
Pn+1′ (x)−Pn−1′ (x)=(2n+1)Pn(x). (12.23) From Eqs. (12.22) and (12.23) numerous additional equations may be developed,5in- cluding
Pn′+1(x)=(n+1)Pn(x)+xPn′(x), (12.24) Pn′−1(x)= −nPn(x)+xPn′(x), (12.25) 1−x2
Pn′(x)=nPn−1(x)−nxPn(x), (12.26) 1−x2
Pn′(x)=(n+1)xPn(x)−(n+1)Pn+1(x). (12.27) By differentiating Eq. (12.26) and using Eq. (12.25) to eliminatePn′−1(x), we find that Pn(x)satisfies the linear second-order ODE
1−x2
Pn′′(x)−2xPn′(x)+n(n+1)Pn(x)=0. (12.28) The previous equations, Eqs. (12.22) to (12.27), are all first-order ODEs, but with poly- nomials of two different indices. The price for having all indices alike is a second-order
5Using the equation number in parentheses to denote the left-hand side of the equation, we may write the derivatives as 2ãdxd (12.17)+(2n+1)ã(12.22)⇒(12.23),
1 2
'(12.22)+(12.23)(
⇒(12.24), 1
2
'(12.22)−(12.23)(
⇒(12.25), (12.24)n→n−1+xã(12.25)⇒(12.26),
d
dx(12.26)+nã(12.25)⇒(12.28).
differential equation. Equation (12.28) is Legendre’s ODE. We now see that the polynomi- alsPn(x)generated by the power series for(1−2xt+t2)−1/2satisfy Legendre’s equation, which, of course, is why they are called Legendre polynomials.
In Eq. (12.28) differentiation is with respect tox (x=cosθ ). Frequently, we encounter Legendre’s equation expressed in terms of differentiation with respect toθ:
1 sinθ
d dθ
sinθdPn(cosθ ) dθ
+n(n+1)Pn(cosθ )=0. (12.29)
Special Values
Our generating function provides still more information about the Legendre polynomials.
If we setx=1, Eq. (12.4) becomes 1
(1−2t+t2)1/2 = 1 1−t =
∞
n=0
tn, (12.30)
using a binomial expansion or the geometric series, Example 5.1.1. But Eq. (12.4) forx=1 defines
1
(1−2t+t2)1/2 = ∞
n=0
Pn(1)tn.
Comparing the two series expansions (uniqueness of power series, Section 5.7), we have
Pn(1)=1. (12.31)
If we letx= −1 in Eq. (12.4) and use 1
(1+2t+t2)1/2= 1 1+t, this shows that
Pn(−1)=(−1)n. (12.32)
For obtaining these results, we find that the generating function is more convenient than the explicit series form, Eq. (12.8).
If we takex=0 in Eq. (12.4), using the binomial expansion 1+t2−1/2
=1−1 2t2+3
8t4+ ã ã ã +(−1)n1ã3ã ã ã(2n−1)
2nn! t2n+ ã ã ã, (12.33) we have6
P2n(0)=(−1)n1ã3ã ã ã(2n−1)
2nn! =(−1)n(2n−1)!!
(2n)!! =(−1)n(2n)!
22n(n!)2 (12.34)
P2n+1(0)=0, n=0,1,2. . . . (12.35)
These results also follow from Eq. (12.8) by inspection.
6The double factorial notation is defined in Section 8.1:
(2n)!! =2ã4ã6ã ã ã(2n), (2n−1)!! =1ã3ã5ã ã ã(2n−1), (−1)!! =1.
Parity
Some of these results are special cases of the parity property of the Legendre polynomials.
We refer once more to Eqs. (12.4) and (12.8). If we replacex by −x andt by−t, the generating function is unchanged. Hence
g(t, x)=g(−t,−x)=
1−2(−t )(−x)+(−t )2 −1/2
= ∞
n=0
Pn(−x)(−t )n= ∞
n=0
Pn(x)tn. (12.36)
Comparing these two series, we have
Pn(−x)=(−1)nPn(x); (12.37) that is, the polynomial functions are odd or even (with respect tox=0, θ=π/2) according to whether the indexnis odd or even. This is the parity,7or reflection, property that plays such an important role in quantum mechanics. For central forces the indexnis a measure of the orbital angular momentum, thus linking parity and orbital angular momentum.
This parity property is confirmed by the series solution and for the special values tabu- lated in Table 12.1. It might also be noted that Eq. (12.37) may be predicted by inspection of Eq. (12.17), the recurrence relation. Specifically, ifPn−1(x)andxPn(x)are even, then Pn+1(x)must be even.
Upper and Lower Bounds for Pn(cos θ )
Finally, in addition to these results, our generating function enables us to set an upper limit on|Pn(cosθ )|. We have
1−2tcosθ+t2−1/2
=
1−t eiθ−1/2
1−t e−iθ−1/2
=
1+12t eiθ+38t2e2iθ+ ã ã ã
ã
1+12t e−iθ+38t2e−2iθ+ ã ã ã
, (12.38)
with all coefficients positive. Our Legendre polynomial,Pn(cosθ ), still the coefficient of tn, may now be written as a sum of terms of the form
1 2am
eimθ+e−imθ
=amcosmθ (12.39a)
with all theampositive andmandnboth even or odd so that Pn(cosθ )=
n m=0 or 1
amcosmθ. (12.39b)
7In spherical polar coordinates the inversion of the point(r, θ, ϕ)through the origin is accomplished by the transformation [r→r, θ→π−θ, andϕ→ϕ±π]. Then, cosθ→cos(π−θ )= −cosθ, corresponding tox→ −x(compare Exercise 2.5.8).
This series, Eq. (12.39b), is clearly a maximum whenθ=0 and cosmθ =1. But forx= cosθ=1, Eq. (12.31) shows thatPn(1)=1. Therefore
Pn(cosθ )≤Pn(1)=1. (12.39c) A fringe benefit of Eq. (12.39b) is that it shows that our Legendre polynomial is a linear combination of cosmθ. This means that the Legendre polynomials form a complete set for any functions that may be expanded by a Fourier cosine series (Section 14.1) over the interval[0, π].
• In this section various useful properties of the Legendre polynomials are derived from the generating function, Eq. (12.4).
• The explicit series representation, Eq. (12.8), offers an alternate and sometimes supe- rior approach.
Exercises
12.2.1 Given the series
α0+α2cos2θ+α4cos4θ+α6cos6θ=a0P0+a2P2+a4P4+a6P6, express the coefficientsαi as a column vector αand the coefficients ai as a column vector a and determine the matricesAandBsuch that
Aα=a and Ba=α.
Check your computation by showing thatAB=1 (unit matrix). Repeat for the odd case α1cosθ+α3cos3θ+α5cos5θ+α7cos7θ=a1P1+a3P3+a5P5+a7P7. Note.Pn(cosθ )and cosnθ are tabulated in terms of each other in AMS-55 (see Addi- tional Readings of Chapter 8 for the complete reference).
12.2.2 By differentiating the generating functiong(t, x)with respect tot, multiplying by 2t, and then addingg(t, x), show that
1−t2
(1−2t x+t2)3/2= ∞
n=0
(2n+1)Pn(x)tn.
This result is useful in calculating the charge induced on a grounded metal sphere by a point chargeq.
12.2.3 (a) Derive Eq. (12.27), 1−x2
Pn′(x)=(n+1)xPn(x)−(n+1)Pn+1(x).
(b) Write out the relation of Eq. (12.27) to preceding equations in symbolic form analogous to the symbolic forms for Eqs. (12.23) to (12.26).
12.2.4 A point electric octupole may be constructed by placing a point electric quadrupole (pole strengthp(2) in the z-direction) atz=a and an equal but opposite point elec- tric quadrupole atz=0 and then lettinga→0, subject top(2)a=constant. Find the electrostatic potential corresponding to a point electric octupole. Show from the con- struction of the point electric octupole that the corresponding potential may be obtained by differentiating the point quadrupole potential.
12.2.5 Operating in spherical polar coordinates, show that
∂
∂z
Pn(cosθ ) rn+1
= −(n+1)Pn+1(cosθ ) rn+2 .
This is the key step in the mathematical argument that the derivative of one multipole leads to the next higher multipole.
Hint. Compare Exercise 2.5.12.
12.2.6 From
PL(cosθ )= 1 L!
∂L
∂tL
1−2tcosθ+t2−1/2
t=0
show that
PL(1)=1, PL(−1)=(−1)L. 12.2.7 Prove that
Pn′(1)= d dxPn(x)
x=1=1
2n(n+1).
12.2.8 Show that Pn(cosθ )=(−1)nPn(−cosθ ) by use of the recurrence relation relating Pn, Pn+1, andPn−1and your knowledge ofP0andP1.
12.2.9 From Eq. (12.38) write out the coefficient oft2in terms of cosnθ,n≤2. This coeffi- cient isP2(cosθ ).
12.2.10 Write a program that will generate the coefficientsas in the polynomial form of the Legendre polynomial
Pn(x)= n s=0
asxs.
12.2.11 (a) CalculateP10(x)over the range[0,1]and plot your results.
(b) Calculate precise (at least to five decimal places) values of the five positive roots of P10(x). Compare your values with the values listed in AMS-55, Table 25.4. (For the complete reference, see Additional Readings of Chapter 8.)
12.2.12 (a) Calculate the largest root ofPn(x)forn=2(1)50.
(b) Develop an approximation for the largest root from the hypergeometric represen- tation of Pn(x)(Section 13.4) and compare your values from part (a) with your hypergeometric approximation. Compare also with the values listed in AMS-55, Table 25.4. (For the complete reference, see Additional Readings of Chapter 8.)
12.2.13 (a) From Exercise 12.2.1 and AMS-55, Table 22.9, develop the 6×6 matrixBthat will transform a series of even-order Legendre polynomials throughP10(x)into a power series5
n=0α2nx2n.
(b) CalculateAasB−1. Check the elements ofAagainst the values listed in AMS-55, Table 22.9. (For the complete reference, see Additional Readings of Chapter 8.) (c) By using matrix multiplication, transform some even power series 5
n=0α2nx2n into a Legendre series.
12.2.14 Write a subroutine that will transform a finite power seriesN
n=0anxninto a Legendre seriesN
n=0bnPn(x). Use the recurrence relation, Eq. (12.17), and follow the technique outlined in Section 13.3 for a Chebyshev series.
12.3 O RTHOGONALITY
Legendre’s ODE (12.28) may be written in the form d
dx
1−x2
Pn′(x) +n(n+1)Pn(x)=0, (12.40) showing clearly that it is self-adjoint. Subject to satisfying certain boundary condi- tions, then, it is known that the solutions Pn(x) will be orthogonal. Upon comparing Eq. (12.40) with Eqs. (10.6) and (10.8) we see that the weight functionw(x)=1,L= (d/dx)(1−x2)(d/dx),p(x)=1−x2, and the eigenvalueλ=n(n+1). The integration limits onxare±1, wherep(±1)=0. Then form=n, Eq. (10.34) becomes
1
−1
Pn(x)Pm(x) dx=0,8 (12.41) π
0
Pn(cosθ )Pm(cosθ )sinθ dθ =0, (12.42) showing thatPn(x)andPm(x)are orthogonal for the interval[−1,1]. This orthogonality may also be demonstrated by using Rodrigues’ definition ofPn(x)(compare Section 12.4, Exercise 12.4.2).
We shall need to evaluate the integral (Eq. (12.41)) whenn=m. Certainly it is no longer zero. From our generating function,
1−2t x+t2−1
= ∞
n=0
Pn(x)tn 2
. (12.43)
Integrating fromx= −1 tox= +1, we have 1
−1
dx 1−2t x+t2=
∞
n=0
t2n 1
−1
Pn(x)2dx; (12.44)
8In Section 10.4 such integrals are interpreted as inner products in a linear vector (function) space. Alternate notations are 1
−1
Pn(x)∗Pm(x) dx≡ Pn|Pm ≡(Pn, Pm).
The form, popularized by Dirac, is common in the physics literature. The ( ) form is more common in the mathematics literature.
the cross terms in the series vanish by means of Eq. (12.42). Using y=1−2t x+t2, dy= −2t dx, we obtain
1
−1
dx
1−2t x+t2= 1 2t
(1+t )2 (1−t )2
dy y =1
t ln 1+t
1−t
. (12.45)
Expanding this in a power series (Exercise 5.4.1) gives us 1
t ln 1+t
1−t
=2 ∞
n=0
t2n
2n+1. (12.46)
Comparing power-series coefficients of Eqs. (12.44) and (12.46), we must have 1
−1
Pn(x)2dx= 2
2n+1. (12.47)
Combining Eq. (12.42) with Eq. (12.47) we have the orthonormality condition 1
−1
Pm(x)Pn(x) dx= 2δmn
2n+1. (12.48)
We shall return to this result in Section 12.6 when we construct the orthonormal spherical harmonics.
Expansion of Functions, Legendre Series
In addition to orthogonality, the Sturm–Liouville theory implies that the Legendre polyno- mials form a complete set. Let us assume, then, that the series
∞
n=0
anPn(x)=f (x) (12.49)
converges in the mean (Section 10.4) in the interval[−1,1]. This demands thatf (x)and f′(x)be at least sectionally continuous in this interval. The coefficientsanare found by multiplying the series by Pm(x)and integrating term by term. Using the orthogonality property expressed in Eqs. (12.42) and (12.48), we obtain
2
2m+1am= 1
−1
f (x)Pm(x) dx. (12.50)
We replace the variable of integrationxbytand the indexmbyn. Then, substituting into Eq. (12.49), we have
f (x)= ∞
n=0
2n+1 2
1
−1
f (t )Pn(t ) dt
Pn(x). (12.51)
This expansion in a series of Legendre polynomials is usually referred to as a Legendre series.9Its properties are quite similar to the more familiar Fourier series (Chapter 14). In
9Note that Eq. (12.50) givesamas a definite integral, that is, a number for a givenf (x).
particular, we can use the orthogonality property (Eq. (12.48)) to show that the series is unique.
On a more abstract (and more powerful) level, Eq. (12.51) gives the representation of f (x)in the vector space of Legendre polynomials (a Hilbert space, Section 10.4).
From the viewpoint of integral transforms (Chapter 15), Eq. (12.50) may be considered a finite Legendre transform off (x). Equation (12.51) is then the inverse transform. It may also be interpreted in terms of the projection operators of quantum theory. We may take Pmin
[Pmf](x)≡Pm(x)2m+1 2
1
−1
Pm(t ) f (t ) dt
as an (integral) operator, ready to operate onf (t ). (Thef (t )would go in the square bracket as a factor in the integrand.) Then, from Eq. (12.50),
[Pmf](x)=amPm(x).10
The operatorPmprojects out themth component of the functionf.
Equation (12.3), which leads directly to the generating function definition of Legendre polynomials, is a Legendre expansion of 1/r1. This Legendre expansion of 1/r1or 1/r12 appears in several exercises of Section 12.8. Going beyond a Coulomb field, the 1/r12 is often replaced by a potentialV (|r1−r2|), and the solution of the problem is again effected by a Legendre expansion.
The Legendre series, Eq. (12.49), has been treated as a known functionf (x)that we arbitrarily chose to expand in a series of Legendre polynomials. Sometimes the origin and nature of the Legendre series are different. In the next examples we consider unknown functions we know can be represented by a Legendre series because of the differential equation the unknown functions satisfy. As before, the problem is to determine the un- known coefficients in the series expansion. Here, however, the coefficients are not found by Eq. (12.50). Rather, they are determined by demanding that the Legendre series match a known solution at a boundary. These are boundary value problems.
Example 12.3.1 EARTH’SGRAVITATIONALFIELD
An example of a Legendre series is provided by the description of the Earth’s gravitational potentialU(for exterior points), neglecting azimuthal effects. With
R=equatorial radius=6378.1±0.1 km GM
R =62.494±0.001 km2/s2, we write
U (r, θ )=GM R
R r −
∞
n=2
an R
r n+1
Pn(cosθ )
, (12.52)
10The dependent variables are arbitrary. Herexcame from thexinPm.
a Legendre series. Artificial satellite motions have shown that a2=(1,082,635±11)×10−9, a3=(−2,531±7)×10−9, a4=(−1,600±12)×10−9.
This is the famous pear-shaped deformation of the Earth. Other coefficients have been computed throughn=20. Note that P1 is omitted because the origin from which r is measured is the Earth’s center of mass (P1would represent a displacement).
More recent satellite data permit a determination of the longitudinal dependence of the Earth’s gravitational field. Such dependence may be described by a Laplace series (Sec-
tion 12.6).
Example 12.3.2 SPHERE IN AUNIFORMFIELD
Another illustration of the use of Legendre polynomials is provided by the problem of a neutral conducting sphere (radius r0) placed in a (previously) uniform electric field (Fig. 12.8). The problem is to find the new, perturbed, electrostatic potential. If we call the electrostatic potential11V, it satisfies
∇2V =0, (12.53)
Laplace’s equation. We select spherical polar coordinates because of the spherical shape of the conductor. (This will simplify the application of the boundary condition at the surface of the conductor.) Separating variables and glancing at Table 9.2, we can write the unknown potentialV (r, θ )in the region outside the sphere as a linear combination of solutions:
V (r, θ )= ∞
n=0
anrnPn(cosθ )+ ∞
n=0
bnPn(cosθ )
rn+1 . (12.54)
FIGURE12.8 Conducting sphere in a uniform field.
11It should be emphasized that this is not a presentation of a Legendre-series expansion of a knownV (cosθ ). Here we are back to boundary value problems of PDEs.