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!

1 d sin dPlm (cos ) + l (l + 1) m2 P m (cos ) = 0 (16.27) again

l

sin2

sin d d ;

Let us assume we have a source at the north pole, where = 0. Far away from the source,

the term m2 = sin2 in the last term is much smaller than the constant l (l + 1).

Problem a: Show that the words \far away from the source" stand for the requirement

m

pl (l + 1)

sin (16.55)

and show that this implies that the approximation that we will derive will break

down near the north pole as well as near the south pole of the employed system of

spherical coordinates. In addition, the asymptotic expression that we will derive will

be most accurate for large values of the angular order l.

Problem b: Just as in the previous section we will transform the rst derivative in the

di erential equation (16.27) away, here this can be achieved by writing Plm (cos ) =

(sin ) glm ( ). Insert this substitution in the di erential equation (16.27), show

that the rst derivative dglm =d disappears when = and that the resulting

;1=2,

m ( ) is given by:

di erential equation for gl

( )

d2 glm + l + 1 2 m2 1 cos2 gm ( ) = 0 : (16.56)

l

sin2 4 sin2

d2 2 ; ;

CHAPTER 16. NORMAL MODES

224

Problem c: If the terms m2= sin2 and cos2 =4 sin2 would be absent this equation

would be simple to solve. Show that these terms are small compared to the constant

2

l + 1 when the requirement (16.55) is satis ed.

2

Problem d: Show that under this condition the associated Legendre functions satisfy the

following approximation

1

cos l + 2 +

Plm (cos ) A (16.57)

p

sin

where A and are constants.

Just as in the previous section the constants A and cannot be obtained from this analysis

because (16.57) satis es the approximate di erential equation for any values of these

constant. As shown in expression (2.5.58) of Ref. 20] the asymptotic relation of the

associated Legendre functions is given by:

r2 1 (2m + 1) 4 + O(l;3=2 )

(;l)m

Plm (cos ) cos l + 2 (16.58)

l sin ;

This means that the spherical harmonics also have the same approximate dependence on

the angle . Just like the Bessel functions the spherical harmonics behave like stand-

p

ing wave given by a cosine that is multiplied by a factor 1= sin that modulates the

amplitude.

i(l+1/2) Î¸

A(Î¸)e

Figure 16.3: An expanding wavefront on a spherical surface at a distance from the

source.

Problem e: Use a reasoning as you used in problem d of section 16.5 to explain that

this amplitude decrease follows from the requirement of energy conservation. In

doing so you may nd gure 16.3 helpful.

Problem f: Deduce from (16.58) that the wavelength of the associated Legendre functions

1

measured in radians is given by 2 = l + 2 .

16.6. LEGENDRE FUNCTIONS ARE DECAYING COSINES 225

This last result can be used to nd the number of oscillations in the spherical harmonics

when one moves around the globe once. For simplicity we consider here the case of a

spherical harmonic Yl0 ( ') for degree m = 0. When one goes from the north pole to

the south pole, the angle increases from 0 to . The number of oscillations that t in

this interval is given by =wavelength, according to problem f this number is equal to

= 2 = l + 2 = l + 1 =2. This is the number of wavelenths that t on half the globe.

1

2

1

When one returns from the south pole to the north pole one encounters another l + 2

oscillations. This means that the total number of waves that t around the globe is given

by l + 1 . It may surprise you that the number of oscillations that one encounters making

2

one loop around the globe is not an integer. One would expect that the requirement of

constructive interference dictates that an integer number of wavelengths should \ t" in

this interval. The reason that the total number of oscillations is l + 1 rather than the

2

integer l is that near the north pole and near the south pole the asymptotic approximation

(16.58) breaks down, this follows from the requirement (16.55).

1

The fact that l + 2 rather than l oscillations t on the globe has a profound e ect

on quantum mechanics. In the rst attempts to explain the line spectra of light emitted

by atoms, Bohr postulated that an integer number of waves has to t on a sphere, this

H

can be expressed as kds = 2 n, where k is the local wave-number. This condition could

not explain the observed spectra of light emitted by atoms. However, the arguments of

this section imply that the number of wavelengths that t on a sphere should be given by

the requirement I 1

kds = 2 n + 2 : (16.59)

This is the Bohr-Sommerfeld quantization rule, which was the earliest result in quantum

mechanics that provided an explanation of the line-spectra of light emitted by atoms.1

1

More details on this issue and the cause of the factor 2 in the quantization rule can be

found in the Refs. 59] and 12].

The asymptotic expression (16.58) can give a useful insight in the relation between

modes and travelling waves on a sphere. Let us rst return to the modes on the string,

which according to (16.5) are given by sin kn x. For simplicity, we will leave out normal-

ization constants in the arguments. The wave motion associated with this mode is given

by the real part of sin kn x exp (;i!nt), with !n = kn =c. These modes therefore denote

a standing wave. However, using the decomposition sin kn x = eikn x e;ikn x =2i, the ;

mode can in the time domain also be seen as a superposition of two waves ei(kn x;!n t) and

The fact that (l+1=2) oscillations of the spherical harmonics t on the sphere appears to be in contrast

1

with the statement made in section 16.3 that the spherical harmonic Ylm has exactly l ; m oscillations.

The reason for this discrepancy is that the sperical harmonics are only oscillatory for an angle that

satis es the inequality (16.55). This means for angular degree m that is nonzero the spherical harmonics

only oscillate with wavelength 2 = (l + 1=2) on only part of the sphere. This leads to a reduction of the

number of oscillations of the spherical harmonics between the poles with increasing degree m. However,

the quantization condition (16.59) holds for every degree m because a proper treatment of this quatization

condition stipulates that the integral is taken over a region where the modes are oscillatory. This means

that one should not integrate from pole-to-pole, but that the integration must be taken over a closed path

that is not aligned with the north-south direction on the sphere. One can show that one encounters exactly

(l + 1=2) oscillations while making a closed loop along such a path. This issue is explained in a clear and

pictorial way by Dahlen and Henson 18].

CHAPTER 16. NORMAL MODES

226

e;i(knx+!nt) . These are two travelling waves that move in opposite directions.

Problem g: Suppose we excite a string at the left side at x = 0. We know we can

account for the motion of the string as a superposition of standing waves sin kn x.

However, we can consider these modes to exist also as a superposition of waves e ikn x

that move in opposite directions. The wave eikn x moves away from the source at

x = 0. However the wave e;ikn x moves towards the source at x = 0. Give a physical

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