I. CONSTRUCTION OF THE GREEN S FUNCTION

Transcription

1 I. CONSTRUCTION OF THE GREEN S FUNCTION The Helmohltz equation in 4 dimensions is 4 + k G 4 x, x = δ 4 x x. In this equation, G is the Geen s function and 4 efes to the dimensionality. In the vey end, we will set the fequency to zeo to obtain the Geen s function of the Laplacian in 4d. In spheical coodinates, x =, θ, ψ, φ whee is the adius and the est ae the angula coodinates. A constant value of the pola angle θ defines a cone in 4d. The cone constains θ but is spheically symmetic with espect to ψ and φ just like the azimuthal symmety of 3d cone. So we can expand the Geen s function as G 4 x, x = lm g4 lm, θ,, θ Y lm ψ, φylmψ, φ The enomalization is intoduced to use the analogy with the Lapalacian in 3d. Then, Eq. takes the fom l + / + k g 4 lm = δ δcos θ cos θ 3 Note that the Laplacian in the last equation is defined in d: = + θ θ. We should beak up the Geen s function in the coodinate θ, not because the bounday condition is defined on a suface of constant θ ı.e., cone. So we need a completeness elation in the function space of the othe vaiable. The latte is known by the Kontovich-Lebedev tansfom and is given by dλλ sinh λ k iλ / k iλ / = δ, whee k ν is the spheical Bessel function of the ode ν. Using this elation, it is not had to show that the following solves Eq. 3 lm = κ l dλλ k iλ / κk iλ / κ Γiλ + l + iγiλ l iλ / cos θ < iλ / cos θ. 4 g 4 Hee, κ is the imaginay fequency k = iκ, θ < = minθ, θ and similaly defined θ. We have not intoduced the boundaies yet, hence the subscipt in g. The completeness elation togethe with a useful fom of the Wonskian of the Legende functions can be used to justify the last equation. Note that we cannot take the limit κ at this point. Howeve, by a constuction simila to Ref. [], we can analytically continue the complex ode of the Bessel and Legende functions back to the eal axis. A caeful implementation of this method gives g 4 lm = κ n=l+ i κ < k κ n Γn + l + Γn l l / cos θp cos θ. 5 Now the asymmety aises in the adial vaiable while the angula coodinates ae teated on equal footing. I used the symmety x = l+n+ x fo intege l, n with l < n, to estoe the symmety in the angles. Only now I can take the limit κ. Putting all the pieces in Eq. while taking this limit, we find G 4 x, x = n= l= m= l Γn + l + < Γn l n+ cos θ cos θ Y lm ψ, φy lmψ, φ. 6 The subscipt indicates that this is the fee Geen s function. In ode to impose the bounday condition though, we have to go back to Eq. 4 in which epesentation, it is not had to satisfy the bounday condition lm = κ l dλλ k iλ / κk iλ / κ g 4 iλ / cos θ < iλ / cos θ Γiλ + l + iγiλ l iλ / cos θ iλ / cos θ iλ / l / cos θpiλ / cos θ. 7

2 It is immediately clea that this is the Geen s function. The fist tem in the last line of this equation gives the delta function once applied to the Helmholtz equation while the second tem does not contibute to the delta function it has no discontinuity in θ = θ but guaantees that the Geen s function vanishes on the cone as θ θ. We should implement the same pocedue to otate to the eal axis. Taking the limit κ then gives G 4 x, x = k l= m= l ρk < l ρ k+ ρ k / cos θ ρk ρ k / cos θ Γρ k + l + sinρ k Γρ k l ρ k / cos θ The intege k epesents the k-th oot of the tanscendental equation ρ k / cos θ Y lm ψ, φy lmψ, φ. 8 ρ k / cos θ =. 9 In geneal, ρ k depends on θ and l. This Geen s function is the basis fo the following computations. II. THE LOOP COMPUTATION The fist coection to the Geen s function comes fom a one-loop diagam. The extenal legs in this diagam coespond to the Geen s function G defined in the pevious section which knows about the bounday condition but computed in the fee theoy. Naively we might think that the Geen s function uns in the loop. Howeve this leads to a divegence in this diagam and must be egulaized. It tuns out that we should eplace the Geen s function in the loop by G G whee G is the Geen s function in the fee space. While the Geen s function in the loop must be summed ove all quantum numbes k, l, m, those of the extenal legs must be evaluated only fo the lowest numbe k =, l, m = fo a eason simila to Cady s analysis [] G k=,l,m= = < ρ 8 ρ+ ρ sin ρ P / ρ / cos θ ρ P / ρ / cos θ P / ρ / The Legende function of the degee / can be expessed in tems of elementay functions P / ρ / cos θ = cos θ P / ρ cos / θ. sin ρ θ ρ. Then, ρ is detemined by the fist oot of P / ρ cos θ / which tanslates to sin ρ θ = ρ = Putting all these togethe, Eq. becomes G k=,l,m= = < ρ 4 ρ+ I define the exponent η accoding to Cady [] as the powe of <. So, θ. sin ρ θ sin ρ θ, η = ρ = θ θ. 3 The subscipt on η means RW. Now we can compute the -loop diagam G loop = α d 4 x Gx, x k= G G x, x Gx, x k=. 4 Hee α is the coupling constant. Fo the N-component φ 4 field theoy, this is α = N +u = N+8 N+8 ɛ N = ɛ [? ]. I choose the point x at the lesse adius close to the tip and x to be at the geate adius. I follow and justify Cady in choosing the domain of integation as < <. The expected logaithm in comes out tivially fom the latte domain d 3 ρ ρ ρ + ρ = ρ + ρ + log/.

3 The coefficient of this tem is detemined by the integation of the angula coodinates in the loop and the sum ove all quantum numbes. It is instuctive to do this fo the special case of an infinite plate but in the epesentation that we developed hee. Fo an infinite plate, we can diectly constuct the Geen s function fom the fee Geen s function by method of images G = G x, x G x, x T, whee T denotes the mio image of a cetain point. The explicit Geen s function can be obtained fom Eq. 6. Again the Geen s function in the extenal legs is obtained by substituting the lowest numbe in its expansion G k=,l,m= = < 3 cos θ cos θ, 5 and the Geen s function in the loop must be subtacted by G. By noting that Gx, x G x, x = Gx, x T, we find G G 4 x, x = n= l= m= l < n+ Γn + l + Γn l l / cos θp cos θ Y lm ψ, φylmψ, φ. 6 The agument of the second Legende function in the last equation is cos θ not cos θ due to the mio imaging. The -loop computation gives G loop = α 3 log/ cos θ cos θ l + Γn + l + 8 Γn l n= l= / dθ cos θ l / P cos θ cos θ. 7 The integal ove ψ and φ is tivial as nothing else depends on them. Futhemoe, the sum ove m can be explicitly done to find the facto l+ in the last equation. The full Geen function including the -loop coection then becomes G + G loop = 3 cos θ cos θ + α log/ l + Γn + l + 8 Γn l n= l= / dθ cos θ l / P cos θ cos θ 3. 8 Supisingly, we find that only the fist tem in the sum gives the total contibution. Othe tems ae not zeo but vanish once summed ove all values of l fo any n. I do not undestand this completely but it can be due to the huge symmety of an infinite plate. The last equation then becomes G + G loop = 3 cos θ cos θ + α 6 log/ = +ɛ/8 3+ɛ/8 cos θ cos θ. 9 In the last line, we substituted α = ɛ and exponentiated the logaithm. The exponent η up to -loop coection is I have not checked this against the liteatue. η = η + η loop = + ɛ/8. A. Loop Comp. fo a Single Needle The coefficient of the ɛ-expansion can be fomally witten fo a geneal half-opening angle. But I will do this fo small angles in which case I can find analytical esults. The main challenge lies in computing G G which uns in the loop. At small angles, thee ae some analytic expessions fo the oots of the Legende functions. We do not need these fomulae but only use the fact that at small

4 angles and at the leading ode in θ, we can set l to zeo: we only need the l = channel of G G. Note that the Legende function of the degee / ae given by Eq.. By doing some algeba, we get G l= G l= = 4 k= Now we should compute GG G G. A caeful algeba gives sin k θ θ sin kθ k sin. θ G + G loop ρ 4 ρ + + α 6 3 log/.99 θ + θ log θ. The coefficient of the logaithm is analytical while that of the linea tem is found numeically. We can then ead off the -loop coection to the Geen s function keeping in mind that α = ɛ η = η θ, ɛ +.6 θ + 4 θ log θ ɛ. 3 The function η is given in Eq. 37 of Ref. [3] assuming that η in my notes is thei β η θ, ɛ = Γ ɛ/ Γ/ Γ/ ɛ/ θ ɛ. 4 Expanding the latte in ɛ and eplacing in the equation fo η, we find η θ. + θ θ log θ ɛ +.6 θ + 4 θ log θ 4 ɛ. 5 The fist tem in this equation is η as computed in the fee theoy in 4d, then comes the ɛ expansion in dimension but in the fee theoy, finally the last paentheses is the ɛ expansion in the inteaction but in fixed fou dimensions. The appeaance of logaithms is suggestive of anothe exponentiation; this time in θ η.6ɛ θ 3 4 ɛ ɛ=.6 θ /4. 6 B. Loop Comp. fo a Single Needle Attached to a Plate We can cay out the same computation fo the case that the cone is attached to a plate. Ou fist task is to obtain the fee Geen s function which solves the bounday conditions. We cannot use the method of images because it does not lead to a symmetic Geen s function. Let s setup the bounday poblem as follows. I conside an inne cone defined by θ < θ attached to an oute cone defined by θ θ. The egion θ < θ < θ defines the fee space. Spaing the technical steps, we should change the second line of Eq. 7 by iλ / cos θ iλ / cos θ iλ / cos θ < iλ / cos θ iλ / cos θ iλ / cos θ iλ / cos θ iλ / cos θ < iλ / cos θ iλ / cos θ iλ / cos θ iλ / cos θ. We ae inteested in a cone attached to a plate. So we choose θ θ and θ = /. A little bit of algeba gives lm =κ l dλλ k iλ / κk iλ / κ Γiλ + l + iγiλ l iλ / cos θ iλ / cos θ iλ / cos θ iλ / cos θ < iλ / cos θ iλ / cos θ < iλ / cos θ iλ / cos θ. g 4 7 8

5 5 By otating back to the eal axis, the full Geen s function becomes G 4 x, x = k l= m= l ρk < l ρ k+ ρ k / Γρ k + l + sinρ k Γρ k l l / cos θ Pρ k / cos θ ρ k / cos θ ρk ρ k / cos θ ρ k / cos θ ρ k / cos θ ρ k / cos θ Y lm ψ, φy lmψ, φ. 9 The ode ρ k is the k-th oot of the tanscendental equation ρ k / cos θ ρ k / cos θ =. Again the main challenge is constucting G G. By the same line of easoning, we find an analog of Eq. fo a small angle cone attached to a plate Simila computations give the exponent η To find η θ, ɛ, we have to solve G l= G l= = 4 η = η θ, ɛ + k= sin k / θ / θ sin kθ k sin. 3 θ.4 θ + θ log θ ɛ. 3 P δ ρ +δ cos θ P δ ρ +δ cos θ = δ = d 3. 3 Using the expansion of the Legende function fo aguments close to and - P ν µ cos θ = Γ µ θ / µ + Oθ, P ν µ Γ µ cos θ = Γ µ νγ µ + ν θ / µ + Oθ sinνγµ θ / µ + Oθ, the condition 3 gives Γ + δ θ / δ Γδ Γ ργρ + δ θ / δ =. Note that in the limit θ, this gives ρ = so η = β = as it should. At small angles, I find Then, Exponentiating θ, we get ρ + Γ + δ ΓδΓ + δ θ / δ. 33 η + 4θ.5 + θ 4 θ log θ ɛ +.76 θ + θ log θ ɛ η +.76ɛ θ 3 4 ɛ ɛ= +.5 θ /4. 35 The exponent γ is elated to the othe exponents via γ = ν η. Equations 6 and 35 then define γ γ ν = η c η cp = +.5 θ /4. 36 [] M. F. Maghebi et al.,. [] J. L. Cady, Jounal of Physics A Mathematical Geneal 6, [3] E. Ben-Naim and P. L. Kapivsky, Jounal of Physics A Mathematical Geneal 43, 57.