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Problem d: The result is now expressed in the rst term as an integral over the boundary

of the body. Let us assume that the modes satisfy on this boundary one of the

three boundary conditions: (i) u = 0, (ii) n ru = 0 (where n is the unit vector

^ ^

perpendicular to the surface) or (iii) n ru = u (where is a constant). Show

^

that for all of these boundary conditions the surface integral in (16.40) vanishes.

16.4. NORMAL MODES AND ORTHOGONALITY RELATIONS 219

The last result implies that when the modes satisfy one of these boundary conditions that

Z

kp ; kq 2

2 uq updN x = 0 : (16.41)

Let us rst consider the case that the modes are equal, i.e. that p = q. In that case the

R

integral reduces to p 2 dN x which is guaranteed to be positive. Equation (16.41) then

ju j

implies that kp = kp2 , so that the wave-numbers kp must be real: kp = kp . For this reason

2

the complex conjugate of the wave-numbers can be dropped and (16.41) can be written

Z

as:

2 2

kp kq uq up dN x = 0 : (16.42)

;

Now consider the case of two di erent modes for which the wave-numbers kp and kq are

2 2

di erent. In that case the term kp kq is nonzero, hence in order to satisfy (16.42) the

;

modes must satisfy Z

uq updN x = 0 for kp = kq : (16.43)

6

This nally gives the orthogonality relation of the modes in the sense that it states that

R

gi f g dN x. Note that

the modes are orthogonal for the following inner product: hf

the inner product for which the modes are orthogonal follows from the Helmholtz equation

(16.1) that de nes the modes.

Let us now consider this orthogonality relation for the modes of the string, the drum

and the spherical surface of the previous sections. For the string the orthogonality relation

was derived in problem d of section 16.1 and you can see that equation (16.9) is identical

to the general orthogonality relation (16.43). For the circular drum the modes are given

by equation (16.19).

Problem e: Use expression (16.19) for the modes of the circular drum to show that the

orthogonality relation (16.43) for this case can be written as:

Z RZ 2

Jm1 (kn1 1 ) r)Jm2 (kn2 2 ) r)ei(m1 ;m2 )' d' rdr = 0

(m (m for kn1 1 ) = kn2 2 )

(m (m

6

0 0

(16.44)

Explain where the factor r comes from in the integration.

Problem f: This integral can be 2separated in an integral over ' and an integral over r.

R

The '-integral is given by 0 ei(m1 ;m2 )' d'. Show that this integral vanishes when

m1 = m2: Z2

6

ei(m1 ;m2 )' d' = 0 for m1 = m2 : (16.45)

6

0

Note that you have derived this relation earlier in expression (13.9) of section 13.2

in the derivation of the residue theorem.

Expression (16.45) implies that the modes un1 m1 (r ') and un2 m2 (r ') are orthogonal

when m1 = m2 because the '-integral in (16.44) vanishes when m1 = m2 . Let us now

6 6

consider why the di erent modes of the drum are orthogonal when m1 and m2 are equal

to the same integer m. In that case (16.44) implies that

ZR

(m) (m)

Jm (kn1 r)Jm (kn2 r) r dr = 0 for n1 = n2 : (16.46)

6

0

CHAPTER 16. NORMAL MODES

220

(m) (m)

Note that we have used here that kn1 = kn2 when n1 = n2 . This integral de nes

6 6

an orthogonality relation for Bessel functions. Note that both Bessel functions in this

relation are of the same degree m but that the wave-numbers in the argument of the Bessel

functions di er. Note the resemblance between this expression and the orthogonality

relation of the modes of the string that can be written as

Z 2R

sin kn x sin km x dx = 0 for n = m : (16.47)

6

0

The presence of the term r in the integral (16.46) comes from the fact that the modes of

the drum are orthogonal for the integration over the total area of the drum. In cylinder

coordinates this leads to a factor r in the integration.

Problem g: Take you favorite book on mathematical physics and nd an alternative

derivation of the orthogonality relation (16.46) of the Bessel functions of the same

degree m.

Note nally that the modes un1 m1 (r ') and un2 m2 (r ') are orthogonal when m1 = m2 6

because the '-integral satis es (16.45) whereas the modes are orthogonal when n1 = n2 6

but the same order m because the r-integral (16.46) vanishes in that case. This implies

that the eigen-functions of the drum de ned in (16.19) satisfy the following orthogonality

relation: Z RZ 2

u (r ')un2 m2 (r ') d' rdr = C n1n2 m1 m2 (16.48)

0 0 n1 m1

where ij is the Kronecker delta and C is a constant that depends on n1 and m1 .

A similar analysis can be applied to the spherical harmonics Ylm ( ') that are the

eigen-functions of the Helmholtz equation on a spherical surface. You may wonder in

that case what the boundary conditions of these eigen-functions are because in the step

from equation (16.40) to (16.41) the boundary conditions of the modes have been used. A

closed surface has, however, no boundary. This means that the surface integral in (16.40)

vanishes. This means that the orthogonality relation (16.43) holds despite the fact that

the spherical harmonics do not satisfy one of the boundary conditions that has been used

in problem d. Let us now consider the inner product of two spherical harmonics on the

RR

sphere: Yl1 m1 ( ')Yl2 m2 ( ')d .

Problem h: Show that the '-integral in the integration over the sphere is of the form

R 2 exp i (m m ) d' and that this integral is equal to 2 .

2; 1 m1 m2

0

This implies that the spherical harmonics are orthogonal when m1 = m2 because of the

6

'-integration. We will now continue with the case that m1 = m2 , and denote this common

value with the single index m.

Problem i: Use the general orthogonality relation (16.43) to derive that the associated

Legendre functions satisfy the following orthogonality relation:

Z

Plm(cos )Plm (cos ) sin d = 0 when l1 = l2 : (16.49)

6

0 1 2

Note the common value of the degree m in the two associated Legendre functions.

Show also explicitly that the condition kl1 = kl2 is equivalent to the condition l1 = l2 .

6 6

16.5. BESSEL FUNCTIONS ARE DECAYING COSINES 221

Problem j: Use a substitution of variables to show that this orthogonality relation can

also be written as

Z1

Plm (x)Plm (x)dx = 0 when l1 = l2 : (16.50)

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