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**Example 12.3.1 EARTH’S GRAVITATIONAL FIELD **

An example of a Legendre series is provided by the description of the Earth’s gravitational potential U (for exterior points), neglecting azimuthal effects. With

R = equatorial radius = 6378.1 ± 0.1km

we write

(12.52)

a Legendre series. Artiﬁcial satellite motions have shown that

This is the famous pear-shaped deformation of the Earth. Other coefﬁcients have been computed through n = 20. Note that P_{1} is omitted because the origin from which r is measured is the Earth’s center of mass ( P_{1} would 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.

**Example 12.3.2 SPHERE IN A UNIFORM FIELD **

Another illustration of the use of Legendre polynomials is provided by the problem of a neutral conducting sphere (radius r_{0} ) 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 potential V , it satisﬁes

∇^{2} V = 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 potential V(r, θ ) in the region outside the sphere as a linear combination of solutions:

**FIGURE 12.8 Conducting sphere in a uniform ﬁeld.**

No ϕ -dependence appears because of the axial symmetry of our problem. (The center of the conducting sphere is taken as the origin and the z-axis is oriented parallel to the original uniform ﬁeld.) It might be noted here that n is an integer, because only for integral n is the θ dependence well behaved at cos θ =±1. For nonintegral n the solutions of Legendre’s equation diverge at the ends of the interval [−1, 1], the poles θ = 0,π of the sphere. It is for this same reason that the second solution of Legendre’s equation, Q_{n} , is also excluded.

Now we turn to our (Dirichlet) boundary conditions to determine the unknown an and bn of our series solution, Eq. (12.54). If the original, unperturbed electrostatic ﬁeld is E_{0} , we require, as one boundary condition,

V(r →∞) =−E_{0z} =−E_{0r} cos θ =−E_{0r}P_{1}(cos θ). (12.55)

Since our Legendre series is unique, we may equate coefﬁcients of P_{n} (cos θ) in Eq. (12.54) (r →∞) and Eq. (12.55) to obtain

a_{n} = 0,n > 1 and n = 0,a1 =−E0. (12.56)

If a_{n} ≠ 0 for n> 1, these terms would dominate at large r and the boundary condition (Eq. (12.55)) could not be satisﬁed.

As a second boundary condition, we may choose the conducting sphere and the plane θ = π/2 to be at zero potential, which means that Eq. (12.54) now becomes

In order that this may hold for all values of θ , each coefﬁcient of P_{n} (cos θ) must vanish.^{12}

Hence

b_{0} = 0,^{13} b_{n} = 0, n ≥ 2, (12.58)

whereas

(12.59)

The electrostatic potential (outside the sphere) is then

(12.60)

it was shown that a solution of Laplace’s equation that satisﬁed the boundary conditions over the entire boundary was unique. The electrostatic potential V ,asgiven by Eq. (12.60), is a solution of Laplace’s equation. It satisﬁes our boundary conditions and therefore is the solution of Laplace’s equation for this problem.

It may further be shown that there is an induced surface charge density

(12.61)

**ELECTROSTATIC POTENTIAL OF A RING OF CHARGE **

As a further example, consider the electrostatic potential produced by a conducting ring carrying a total electric charge q (Fig. 12.9). From electrostatics the potential ψ satisﬁes Laplace’s equation. Separating variables in spherical polar coordinates (compare Table 9.2), we obtain

(12.63a)

Here a is the radius of the ring that is assumed to be in the θ = π/2 plane. There is no ϕ (azimuthal) dependence because of the cylindrical symmetry of the system. The terms with positive exponent in the radial dependence have been rejected because the potential must have an asymptotic behavior,

(12.63b)

The problem is to determine the coefﬁcients c_{n} in Eq. (12.63a). This may be done by evaluating ψ(r, θ ) at θ = 0,r = z, and comparing with an independent calculation of the

**FIGURE 12.9 : Charged, conducting ring.**

potential from Coulomb’s law. In effect, we are using a boundary condition along the z - axis. From Coulomb’s law (with all charge equidistant),

The last step uses the result of Exercise 8.1.15. Now, Eq. (12.63a) evaluated at θ = 0,r = z (with P_{n} (1) = 1), yields

(12.63d)

Comparing Eqs. (12.63c) and (12.63d), we get c_{n }= 0 for n odd. Setting n = 2s ,wehave

(12.63e)

and our electrostatic potential ψ(r, θ ) is given by

The magnetic analog of this problem appears in Example 12.5.3.

**ALTERNATE DEFINITIONS OF LEGENDRE POLYNOMIALS**

**Rodrigues’ Formula**

The series form of the Legendre polynomials (Eq. (12.8)) of Section 12.1 may be transformed as follows. From Eq. (12.8),

(12.64)

For n an integer,

Note the extension of the upper limit. The reader is asked to show in Exercise 12.4.1 that the additional terms [n/2]+ 1to n in the summation contribute nothing. However, the effect of these extra terms is to permit the replacement of the new summation by (x^{ 2} − 1)^{n} (binomial theorem once again) to obtain

(12.65)

This is Rodrigues’ formula. It is useful in proving many of the properties of the Legendre polynomials, such as orthogonality. The Rodrigues deﬁnition is deﬁne the associated Legendre functions. it is used to identify the orbital angular momentum eigenfunctions.

**Schlaeﬂi Integral**

Rodrigues’ formula provides a means of developing an integral representation of P_{n} (z). Using Cauchy’s integral formula.

(12.66)

with

(12.67)

we have

(12.68)

with the contour enclosing the point t = z

This is the Schlaeﬂi integral. Margenau and Murphy^{ 14} use this to derive the recurrence relations we obtained from the generating function.

The Schlaeﬂi integral may readily be shown to satisfy Legendre’s equation by differentiation and direct substitution (Fig. 12.11). We obtain

nothing new. A contour about t =+1 and t =−1 will lead to a second solution, Qν (z) .

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