More on Bessel functions. Infinite domain, δ-function normalization

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1 More on Bessel functions Infinite domain, δ-function normalization Consider Bessel s equation on the domain < ρ < as R ½ Bessel s equation, (375) or (393), says ρ d ρ dj ν(kρ) + k 2 ν 2 ρ 2 J ν (kρ) = As in class, multiply this equation by ρj ν (k ρ) and integrate from ρ = to R: R ρj ν (k d ρ) ρ dj ν(kρ) + k 2 ν 2 ρ ρ 2 J ν (kρ) = Integrate the first term by parts, R djν (k ρ) dj ν (kρ) ρ + k 2 ν 2 ρ 2 J ν (k ρ)j ν (kρ) = Exchanging the roles of k and k, and subtracting leads to R (k 2 k 2 ) ρj ν (k ρ)j ν (kρ) = ρj ν (kρ) dj ν(k ρ) ρj ν (k ρ) dj ν(kρ) ρj ν (k ρ) dj R ν(kρ) R As ρ, both terms on the right-hand side have the same leading behavior, (kk ρ 2 ) ν /Γ(ν), and so cancel for any value of ν The next-leading terms go as (k 2 k 2 )ρ 2 (kk ρ 2 ) ν, and so do not cancel but vanish as ρ for ν > The surface term on the right-hand side then contains only the contribution from ρ = R, R ρj ν (k ρ)j ν (kρ) = k RJ ν (kr)j ν(k R) krj ν (k R)J ν(kr) k 2 k 2 (a) Using recursion relations, this can also be written R ρj ν (k ρ)j ν (kρ) = k R J ν (kr) J ν (k R) kr J ν (k R) J ν (kr) k 2 k 2 (b) Although J ν (kr) vanishes as R /2 for large R, the presence of the factor of R means we must investigate in detail the behavior of the Bessel functions at large argument The Bessel functions for large argument are given in (39), J ν (kρ) Ö 2 πkr cos kr νπ 2 π 4

2 2 Making use of the trig identity cosαcosβ = 2 [cos(α β)+cos(α+β)], either (a) or (b) then leads to R ρj ν (k ρ)j ν (kρ) sin[(k k )R] π kk (k k ) cos[(k +k )R νπ] π kk (k +k (2) ) The first term oscillates as R ½ and so averages to zero, unless k = k, when it becomes large, behavior we expect for a δ-function; the second term averages to zero for all k, k In more detail, the function δ ǫ (x) = sin(x/ǫ) πx has value δ ǫ = /ǫπ at x =, oscillates rapidly outside the interval x = ǫπ, andhas integral dxδ ǫ (x) = Thus, the limit ǫ is a representation of the Dirac δ-function (cf entry [34] in the Math- World page on δ-functions, This functional form also appears in time-dependent perturbation theory in quantum mechanics, where it gives the energy-conserving δ-function The limit in (2) then becomes ρj ν (k ρ)j ν (kρ) = lim δ kk /R (k k ) R δ /R k +k R (ν 2 )π = δ(k k ) kk = δ(k k ) k The second term does not contribute, because the argument never vanishes I was assigned this problem when I was a graduate student, and I have written solutions to it at various times in the past, but this version is more correct than any of those earlier attempts The figure shows both the exact J ν result of (a/b) (red) and the cosine approximation (2) for large argument (dotted, blue) plotted as a function of k for k =, R =, and ν =

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4 4 Green s function Green s function constructions always follow a similar pattern The Green s function is a solution to Ö 2 G = Ö 2 G = 4πδ(Ü Ü ), with the boundary condition that G as r ½ or r ½ Since the δ-function vanishes almost everywhere, we can expand G(Ü) in the modes which are solutions to Ö 2 G =, G(Ü) = dka m (k)j m (kρ)e imφ e ±kz, where the coefficients A m (k) will depend on ρ, φ, z We could guess more about them, but the result will follow systematically from this starting point The Bessel function satisfies the boundarycondition G as ρ ½, but for thez-dependence we need different expressions for the two regions z > z and z < z : G < (Ü) = dka < m(k)j m (kρ)e imφ e +kz (z < z ), G > (Ü) = dka > m(k)j m (kρ)e imφ e kz (z > z ) Except at ρ = ρ, φ = φ, the Green s function must be continuous at z = z, and so we must have A < m(k)e +kz = A > m(k)e kz = C m (k) This leads to the single expression G(Ü) = dkc m (k)j m (kρ)e imφ e k(z > z < ), where z < and z > are the smaller and larger of z, z The remainder of the construction uses the δ-function information, Ö 2 G = 4πδ(Ü Ü ) Integrate this equation over z from z = z ǫ to z = z +ǫ for ǫ to obtain z +ǫ dz Ö 2 G z=z +ǫ G = z ǫ z z=z ǫ = 4π δ(ρ ρ ) ρ δ(φ φ ) ( ) (the ρ and φ derivatives produce factors of k and m that are finite term by term, and so those contributions vanish as ǫ ) Exchanging the roles of k and ρ in part (a), we have an expansion of δ(ρ ρ ) in Bessel functions, and the representation of δ(φ φ ) in e imφ is elementary Writing both sides of ( ) as series expansions gives dkc m (k)j m (kρ)e imφ e k(z > z < ) z=z +ǫ z z=z ǫ = 4π kdkj m (kρ)j m (kρ ) ) 2π eim(φ φ

5 5 Writing the derivative explicitly and identifying coefficients of J m (kρ)e imφ on left and right hand sides we obtain ) C m (k) z e k(z z z) z e k(z = 2kC m (k) z=z = 4πkJ m (kρ ) 2π e imφ = 2kJ m (kρ )e imφ Thus, C m (k) = J m (kρ )e imφ, and we have completed the Bessel function representation of the Green s function for free space, Ü Ü = dkj m (kρ)j m (kρ )e im(φ φ ) e k(z > z < ) In class we constructed Green s functions for the square (see also JDJ Problem 25) and for the volume between two spherical surfaces (see also JDJ Section 39); and Jackson does a different Bessel function construction in Section 3 Some Bessel function relations Take the Green s function and evaluate for Ü On the left-hand side Ü Ü = ρ 2 +ρ 2 2ρρ cos(φ φ )+(z z ) 2 ρ 2 +z 2; while, since J m (x) (x/2) m, only m = contributes to the series on the right-hand side as ρ Thus, dkj (kρ)e k z = ρ 2 +z 2 Now, evaluate this first result at ρ = ρ 2 +ρ 2 2ρρ cos(φ φ ), ρ 2 +ρ 2 2ρρ cos(φ φ )+z 2 = dkj [k ρ 2 +ρ 2 2ρρ cos(φ φ )]e k z But, this is also just the Green s function evaluated at z =, but arbitrary ρ, ρ 2 +ρ 2 2ρρ cos(φ φ )+z 2 = dkj m (kρ)j m (kρ )e im(φ φ ) e k z The integral over k amounts to a Laplace transform, and the theory of Laplace transforms assures us that the transformation is unique and invertible, and so the integrands on the right-hand side must be equal: J [k ρ 2 +ρ 2 2ρρ cos(φ φ )] = J m (kρ)j m (kρ )e im(φ φ )

6 6 Evaluate the relation at the bottom of the previous page at kρ = kρ = x, φ φ = ; recall that J m (x) = ( ) m J m (x), and Jm(x) 2 = [J (x)] 2 +2 [J k (x)] 2 = J () = k= Take the result and evaluate for φ = in the limit ρ becomes large In this limit the square Õ ρ 2 +ρ 2 2ρρ cosφ = ρ ρ cosφ+ç( ρ2 ρ ), and the large argument limit of the Bessel functions then gives root becomes Ö 2 πkρ cos kρ kρcosφ π 4 Ö 2 = J m (kρ) πkρ cos kρ mπ 2 π e imφ, 4 correct up to terms of order /ρ Use that cos(α β) = cosαcosβ+sinαsinβ, and equate separately the coefficients of cos(kρ π 4 ) and sin(kρ π 4 ) on each side (equality must hold for all values of ρ ), cos(kρcosφ) = J m (kρ) cos( mπ 2 )eimφ, sin(kρcosφ) = J m (kρ) sin( mπ 2 )eimφ Finally, add i times the second to the first, e ikρcosφ = J m (kρ)e imπ 2 e imφ Take the real and imaginary parts of the complex exponential evaluated at kρ = x, φ = : Re i m J m (x) = k= Im i m J m (x) ( ) k J 2k (x) = J (x)+2 = 2 k= k= ( ) k+ J 2k+ (x) = sinx ( ) k J 2k (x) = cosx, Integral representation Take the complex exponential, evaluate at kρ = x, multiply by e imφ, and integrate over φ: 2π dφ 2π 2π e imφ i m J m (x)e im φ = i m dφ J m (x) = m = 2π eixcosφ imφ