1. Introduction
The Goos-Hänchen phenomenon refers to a deviation of the classical trajectory of a light beam predicted by the geometrical optics theory, when the light beam is incident upon a dielectric interface at an angle larger than the critical angle (see Fig. 1). The first experimental observation dates from 1947 [1], and since then, many authors have tried to give a theoretical solution to the problem. Lotsch [2] gave an extended discussion on the whole subject, saying that no rigorous solution was known back then, but that an analitical solution definitely exists. Renard [3] assumed that there is an energy flow associated to the evanescent wave that penetrates the dielectric interface, then, using conservation of energy, arrives to an analytical expression that, however, is not valid in a close neighborhood of θc. Horowitz and Tamir [4], using direct integration methods, arrived to the first theoretical important result, since they were the first to give an explicit formula valid for angles close to θc. Cowan and Anic̆in [5], using a bounded microwave beam, investigated the lateral displacement, and compared their results with the theoretical results of Horowitz and Tamir. They found that their results disagree with the theoretical curve near the critical angle, they argue that this disagreement may be due to the fact that they were using a short wide beam. Later on, Lai, Cheng and Tang [6] slighty improved the result presented by Horowitz and Tamir, when they used an expansion of the reflectance that retains terms of second order. Their results give a lateral shift that varies continuously and smoothly around the critical angle. Also, Chan and Tamir [7], using the same expansion, investigated the lateral displacement and other effects that were not studied before: focal shift, angular shift and the beam waist modification.
This effect has also been studied from the view point of physical applications. For example, Chiu and J. J. Quinn [8] considered a wave packet and interpret the Goos-Hänchen effect as a time delay scattering process. Kogelnik and Weber [9], investigated the light propagation through a dielectric waveguide. Using ray methods and considering the Goos-Hänchen effect, arrived to a predicted phase and group velocity that agree with the usual energy conservation approach. Since then, various aplications of this physical proccess were given. Some include, aplications in acoustics [10], quantum mechanics [11-13] and nonlinear optics [14]. Using photonic crystals, Soboleva, Moskalenko and Fedyanin [15] enhanced the Goos-Hänchen effect and observed that the displacement is, at least, one order of magnitude larger than in a dielectric surface. Chremmos and Efremidis [16] give the lateral displacement for an Airy beam. Finally, Prajapati and Ranganathan [17], using numerical methods, give the total Goos-Hänchen effect for a three dimensional Hermite-Beam for two orthogonal components.
Despite all the results above, an analytical expression for the lateral shift for a Hermite-Gaussian beam does not exist. Gaussian beam is just one of many solutions (or modes) of the Huygens-Fresnel integral. It is the purpose of the present paper to show the theoretical behavior of the lateral displacement for a higher mode solution. To this end, we use a higher unidimensional mode solution of the Huygens-Fresnel integral and use the direct integration method used by Horowitz and Tamir to derive the theoretical total displacement of the field incident upon a dielectric interface. The final result is valid for angles that are in a close neighborhood of the critical angle.
In Sec. 2 we give a derivation of the reflected field for a Hermite-Gaussian beam in one transverse dimension. We write the incident field as a normalized one dimensional higher order Hermite-Gaussian field, this field being a solution to the Hygens-Fresnel equation (strictly speaking, it is a eigenfunction of the Hyghens-Frensnel integral equation). Then, we write this field in a simplified form, assuming that the distance of propagation is so small that we can ignore the Guoy phase shift. This approximation allows us to express the field only in terms of the Hermite polynomials and a exponential. Thereafter, we study separately the case for even and odd Hermite polynomials. Then, the field is expressed as a superposition of plane waves using the Fourer Transform. With this expression we write the reflected field in terms of the reflectance and a correction factor. The reason to express the field in such a way is to simplify the integration of the reflected field. In Sec. 3 we determine the general expression for the lateral displacement. This formula shows explicity the dependence with the incidence angle and the mode of the field. In Sec. 4 we present numerical results that show the behavior of the lateral displacement for different modes of the field and for two different beam widths. An interesting result is that, in general, the lateral displacement increases as the mode of the field increases and then decreases again. All important calculations are derived in the Appendix A and B.
2. Reflected Field
We will work under the paraxial approximation. This means that the field is a solution to the Huygens-Fresnel integral. Then, assume a monocromatic one transverse dimensional Hermite-Gaussian beam incident upon a dielectric interface in the plane y=0. In addition, we will assume a linearly polarized field and troughout the discussion a time dependence is implied and supressed. The index of refraction of the dielectric is taken to be n=k/k0>1, where k and k0 are the wavenumbers associated to the incident and dielectric media, respectively. We will make the assumption that the beam is well defined, that is kw≫1, where w is the beam waist. The geometry of the problem is presented in Fig. 2. We have three set of coordinates: the interface coordinates (x,y), the incident coordinates (xi,yi) and the reflected coordinates (xr,yr). We are going to work out the problem in the rotated plane that is paralell to the interface plane. By this construction, these set of coordinates are related by
xi=xcosθ-y+sinθ,yi=xsinθ+cosθ,
(1)
and
xr=xcosθ+y-sinθ,yr=xsinθ-y-sinθ.
(2)
where y+=y+h, y-=y-h and h is the position of the plane where the beam is and parallel to the dielectric interface.
The incident beam is located at yi=0. According to Siegman and Sziklas [18] (see also [19] Chapter 16, Eq. 54), a higher-order Hermite Gaussian beam in one transverse dimension, along the yi=0 axis, can be represented as
ψn,inc(xi,y)=(2π)1/4(12nn!w)1/2q0q(y)[q0q0*q*(y)q(y)]n/2×Hn[2xiw(y)]exp[-ikxi22q(y)],
where
1q(y)=1R(y)-iλπw2(y),
(3)
where R(y) is the radius of curvature and w(y) the beam waist as a function of propagation distance. This field is generated by a point source (see [20]), so if we assume that the wave is at long distance from the source we will have 1/R(y)→0. We can suppose, also, that the distance of propagation is so small that we can take q(z)≃q0 and w(y)≃w. In the same manner, the coordinate xi, written in the rotated axis paralell to the dielectric interface, is xi=xcosθ. In that way, we can write the equation for a Hermite-Gaussian beam, of order n, as
ψn,inc(xi,y)=αnHn2c2xexp[-c2x2]πw,
(4)
where
αn=(2π3)1/4(w2nn!)1/2,
(5)
c=cosθw.
(6)
The final field propagating along the y axis will be approximated by adding a plane-wave variation exp(ikyi) along the same axis. Using Eq. (1), the coordinates along the rotated plane y=-h will give for the field
ψn,inc(xi,y)=αnHn2c2xexp[-c2x2+ik¯x]πw,
(7)
where
k¯=ksinθ.
(8)
We need now an explicit representation for the Hermite polynomials. Hermite polynomialas can be of even or odd order, for each case we must derive an expression for the lateral displacement. We consider, as a first case, an even order Hermite-Gaussian beam, which can be written as:
ψ2n,inc(xi,y)=α2n∑j=0nfjx2jexp[-c2x2+ik¯x]πw,
(9)
where
fj=(-1)n+j(2n)!(2j)!(n-j)!(22c2)2j.
(10)
The reflected wave can be written as a continuous superposition of plane waves, each of one are affected by the Fresnel reflectance rkx, i.e.,
ψ2n,r(x,y)=12π∫-∞∞r(kx)ψ~2n(kx)×expdkx,
(11)
where
r(kx)=(k2-kx2)1/2-m(k02-kx2)1/2(k2-kx2)1/2+m(k02-kx2)1/2,
(12)
and kx, ky are the components of the wave number k and are related by kx2+ky2=(2π/λ)2. The term k0 is k0=ksinθc. The constant m is the coefficient that depends on the polarization of the incident wave, i.e.,
m=1orm=n2,
(13)
for normal and parallel polarization to the plane of incidence, respectively, and
ψ~2n(kx)=∫-∞∞ψ2n,inc(x,y)exp(-ikxx) dx,
(14)
is the inverse Fourier Transform. Substituting Eq. (9) into Eq. (14) yields
ψ~2n(kx)=α2nπw∑j=0nfj∫-∞∞x2jexp[-c2x2-i(kx-k¯)x]dx,
(15)
after completig the square and changing variable we will obtain
ψ~2n(kx)=α2ncosθ∑j=0n∑k=0jfjgk(kx-kc)2j-2kexp[-14(kx-kc)2],
(16)
where
gk=(2j)!k!(2j-2k)!(-i)2j-2k12c2j.
(17)
It is important to note that the lateral displacement will depend upon the polarization of the field through the function r(kx) (which, in turn, depends on polarization through the constant m). We will make all calculations assuming a field in which the polarization is normal to the plane of incidence. The effect of parallel polarization will only re-scale the graphics and will not affect the shape of the final curves.
Proceeding now like Horowitz and Tamir, we can write the reflectance in such a way that we may extract the part responsible for the geometric-optics result,
r(kx)=r(k¯)[1+rc(kx)],
(18)
where
rc(kx)=-1-r(kx)r(k¯).
(19)
The term rc(kx) represents a correction term and takes into account the ondulatory behavior of the incident field; that is, the function rc(kx) is responsable for the deviation of the Hermite-Gaussian beam from the geometric-optics trajectory. This way, we can write the reflected field like
ψ2n,r(x,y)=ψ2n,rg(x,y)[1+ψ2n,ro(x,y)],
(20)
where
ψ2n,rg(x,y)=r(k¯)2πcosθ∫-∞∞ψ~(kx)×expdkx,
(21)
and
ψ2n,ro(x,y)=r(k¯)2πψrg(x,y)∫-∞∞rc(kx)ψ~(kx)×expdky.
(22)
Now, if we substitute Eq. (16) into Eq. (21) we arrive to
ψ2n,rg(x,y)=r(k¯)2πcosθ∑j=0n∑k=0jfjgk∫-∞∞(kx-kc)2j-2kexp[-14(kx-kc)2]exp[i(kxx-kyy-)]dkx.
(23)
The reason to express the reflected field ψ2n,r(x,y) in terms of ψ2n,rg(x,y) and ψ2n,ro(x,y) is that in such way we can simplify the integration because it is esier to handle the function rc(kx) (see Appendix A.2 where we made the integration using series expansion for rc(kx)) than the function r(kx). Of course, we can still make a calculation using r(kx), but in the present work we only focus in the much simplified version of the problem. In future papers we will handle the problem using r(kx) an we will compare those results with the ones presented here.
Now, making the integration (see Appendix A.1) we get the following expression for the reflected field (geometric-optical field):
ψ2n,rg(x,y)=-α2nΠπwr×∑j=0n∑k=0jfjhkH2(j-k)(xr/wr),
(24)
where H2(j-k) is the Hermite polynomial of order 2(j-k) and
Π=exp[-(xrwr)2]exp(ikyr),
(25)
wr2=w2-i2zksecθ,
(26)
hk=(-2wwr)2j-2kgk.
(27)
For the correction field we subsitute Eq. (16) into Eq. (22)
ψ2n,ro(x,y)=r(k¯)2πψrg(x,y)∑j=0n∑k=0j∫-∞∞rc(kx)(kx-kc)2j-2kexp[-14(kx-kc)2]exp[i(kxx-kyy-)]dkx.
(28)
After performing and substituting Eq. (25) into Eq. (28), (see Appendix B) we will obtain,
ψ2n,ro(x,y)=C1(θ)P2n(xr/wr)∑j=0n∑k=0j∑l=02j-2kfjgk×
(29)
(kwδ)2j-2k-l[(-δ)1/2slHl(γ/2)-
(30)
βplF1/2+l(γ)],
(31)
where
C1(θ)=4mcos2θcsinθcos1/2θ(sinθ+sinθc)1/2×1[cosθ2+m2(sin2θ-sin2θc)],
(32)
δ=(sinθ-sinθc)secθ,
(33)
P2n(x/w)=∑j=0n∑k=0jfjhkH2(j-k)(x/w),
(34)
sl=(2j-2k)!l!(2j-2k-l)!(i)l,
(35)
pl=(2j-2k)!l!(2j-2k-l)!(l)l/2exp[-iπ2l],
(36)
β=23/4(kw)1/2exp(iπ/4),
(37)
γ=2(ikwrδ2-xrwr),
(38)
F1/2+l(γ)=exp[γ2/4]D1/2+l(γ),
(39)
and Hl and D1/2+l are the Hermite polynomial and the Parabolic cylinder function of order l and 1/2+l, respectevily. The term P2n(xr/wr) is just the geometric-optical field without the constant part. Notice that the real part of γ is a measure of the distance between the observation point and the reflected axis, while the imaginary part is a measure of the deviation of the incidence angle from the critical angle.
Now, if we consider the odd Hermite polynomials, the incident field will be
ψ2n+1,inc(x,y)=α2n+1∑j=0nf^jx2j+1×expπw,
(40)
where
f^j=(2n+1)!(2j+1)!(n-j)!(-1)j(22c2)2j+1.
(41)
Making the same procedure we did before, we arrive to the following expression for the correction term
ψ2n+1,ro(x,y)=A(θ)P2n(xr/wr)∑j=0n∑k=0j∑l=02j-2kf^jg^k(kwδ)2j-2k-l[(-δ)1/2slHl(γ/2)-βplF1/2+l(γ)],
(42)
where
g^k=(2j)!k!(2j-2k)!(-1)ki2c2j1c.
(43)
Equations (29) and (42) are expressions for the correction term; it is not, strictly speaking, the total reflected field. Nevertheless, the reflected field is expressed in terms of the classical reflected field (geometric-optical field) and the correction factor. For our purpouses, the correction term can be used to arrive to an expression for the lateral displacement, as it will be discussed in the following section. Notice that the difference between the correction factor for the even and odd cases are just the summation terms fj and gk, the rest of the terms are exactly the same.
3. Lateral Displacement
The objective of the present section is to obtain the lateral shift. To this end we need to rewrite the reflcected field in such way that the equation shows explicitly the trajectory of the beam. In order to obtain this expression (for the even Hermite-Gaussian beam), we first recognize that if the distance of propagation is sufficiently small compared to the beam width, then the beam will remain well collimated, that means wr≃w. Then, we rewrite Eq. (20) in the following form
ψ2n,r(x,y)=-απwP2n(xr/w)exp[-(xr/w)2+ln(1+ψ2n,ro)]exp[iky],
(44)
if we expand the logarithm into a Taylor series around (xr/w)=0 (γ=ikwδ/2) we get
ln[1+ψ2n,ro(x,y)]≃a0+a1(xr/w)+a2(xr/w)+...,
(45)
where
an=1n!dnd(xr/w)nln(1+ψ2n,ro)|(xr/w)=0.
(46)
Making the substitution and after algebraic manipulations, we arrive to the following expression:
exp[-(xr/w)2+ln[1+ψro]]≃exp(-1-a2w2)xexp[xr-L2n],
(47)
where L2n has dimensions of length. Notice now the argument of the exponential, we can interpret L2n as a distance that is displaced with respect to xr and is, therefore, a quantity that can be interpreted as a lateral displacement for the even Hermite-Gaussian beam. Notice that the important step in obtaining such result is present in Eq. (44), where we rewrite the correction factor as a natural logarithm, this shows the importance of rewriting the reflected field in terms of the correction factor. This allow us to make a series expansion and, finally, re-arrange the terms so that we get the lateral shift L2n. This lateral shift is, then, given by
L2n≃w2cosθRea11-a2,
(48)
where
a1=A(θ)P2n(0)f2(δ)f1(δ),
(49)
a2=A(θ)P2n(0){12f3(δ)f1(δ)-12[f2(δ)f1(δ)]2},
(50)
and
f1(δ)=1+A(θ)P2n(0)x∑j=0n∑k=0j∑l=02j-2kfjgk(kwδ)2j-2k-lx[(-δ)1/2slHl(γ0/2)-βplF1/2+l(γ0)],
(51)
f2(δ)=∑j=0n∑k=0j∑l=02j-2kfjgk(kwδ)2j-2k-l[(-δ)1/2sl(2l)Hl-1(γ0/2)-βpl(1/2+l)F-1/2+l(γ0)],
(52)
f3(δ)=∑j=0n∑k=0j∑l=02j-2kfjgk(kwδ)2j-2k-l[(-δ)1/2sl(4l)(l-1)Hl-2(γ0/2)-βpl(-1/4+l2)F-3/2+l(γ0)].
(53)
For the lateral displacement for the odd case, we use
ln[1+ψ2n+1,ro(x,y)]≃b0+b1(xr/w)+b2(xr/w)+…,
(54)
where
bn=1n!dnd(xr/w)nln(1+ψ2n+1,ro)|(xr/w)=0.
(55)
For this case, L2n+1 will be
L2n+1≃w2cosθReb11-b2,
(56)
where
b1=A(θ)P2n(0)g2(δ)g1(δ),
(57)
b2=A(θ)P2n(0){12g3(δ)g1(δ)-12[g2(δ)g1(δ)]2},
(58)
and
g1(δ)=1+A(θ)P2n(0)∑j=0n∑k=0j∑l=02j-2kf^jg^k(kwδ)2j-2k-l×[(-δ)1/2slHl(γ0/2)-βplF1/2+l(γ0)],
(59)
g2(δ)=∑j=0n∑k=0j∑l=02j-2kf^jg^k(kwδ)2j-2k-l×[(-δ)1/2sl(2l)Hl-1(γ0/2)-βpl(1/2+l)F-1/2+l(γ0)],
(60)
g3(δ)=∑j=0n∑k=0j∑l=02j-2kf^jg^k(kwδ)2j-2k-l×[(-δ)1/2sl(4l)(l-1)Hl-2(γ0/2)-βpl(-1/4+l2)F-3/2+l(γ0)].
(61)
Equations (48) and (56) are our final result. It should be noted that we are taking the positive value, since the displacement is always a positive magnitude. Notice, also, that we take terms of second order in Eqs. (48) and (56) since in our case, the terms a2 and b2 cannot be neglected. In the next section, we use Eqs. (48) and (56) with Eqs. (46-50) and (54-58), respectevily, to obtain the lateral displacement for beams of different order.
4. Numerical Results
We did calculations (programming was done using Wolfram Mathematica 7.0) for the lateral displacement for a Hermite-Gaussian beam, of even order and of width kw=10000, reflected from a dielectric interface of index of refraction n=1.52, for the following values: N=0, 2, 4, 6, 8, and 10.
The results are shown in the Fig. 3 in a semilog scale. We see that the lateral displacement increases for N=2 until N=4, and then starts to decrease. This peculiar behavior may be due to the fact that we are cosidering just one dimensional transverse beam and the fact that we are not considering the Guoy phase shift. We also did the calculations for the odd case; that is, for N=1, 3, 5, 7, 9 and 11 (see Fig. 4). We see the same behavior as the even case; that is, an increase in the magnitude of the lateral displacement, and then a decrease in the value of the lateral displacement.
Finally, we did the same calculation, but for a very low beam width kw=10. In Fig. 5 and 6 it is shown the results for N=0,2,6,8,10 and N=1,3,5,7,9,11 in a semilog scale. In this case, we see that the behavior is, except for the case N=2 where the displacement is larger than N=0, a steadily decrease of the lateral displacement as the order of the mode of the beam increases. We see, also, that the maximum displacement occurrs at different angles for the even and odd case. For the even case, the displacement attains its maximum for angles above δ=0.001∘, and for the odd case, for angles below δ=0.001∘.
One point of consideration, as was mentioned before, is that all the present calculations were made assuming a wave which is normal to the plane of incidence, i.e., for the case of normal polarization. This dependence is given by the constant m in the reflectance (see Eq. (13)). For the case of a wave of parallel polarization to the plane of incidence we must include the index of refraction n, this only will re-scale all the curves and wil not affect the shape of these curves.
Take into account that the present simulations are new and, to our knowledge, haven’t been made before, so what we do here is only compare the results with the case for a Gaussian beam, result that is known since the results presented by Horowitz and Tamir. In future papers we will make an equivalent procedure to obtain the lateral shift and compare the results with the ones presented here.
5. Conclusions
Previously, Horowitz and Tamir arrived to an analitic equation for the lateral displacement for a Gaussian beam of zero order. Some years after, Lai, Cheng and Chang , using an alternative method, arrived to a similar formula. However, until now, no analitic formula has been derived for a more general Hermite-Gaussian beam. By the same methods described above, we could arrive to an expression for the Goos-Hänchen shift for the more general case of a Hermite-Gaussian beam. Using such formula, we gave numerical results of the lateral displacement in fuction of the order of the Hermite-Gaussian beam and the beam width. We found the peculiar behavior that the lateral displacement first increases and then decreases steadily as the order of the beam increases. Assumptions, such as not taking in account the Gouy phase shift as the beam propagates, has been made. If such phase is taking into account, the mathematical problem becomes, analitically, inmanageable and numerical methods should be used. We will consider, for a future paper, the behavior of the lateral displacement when the beam incides on a finite dielectric interface. Another investigation that will be considered in another paper is the behavior of the lateral displacement when we use the reflectance r(kx) and not the correction factor rc(kx); then, using these results we will make a comparision with the results presented in the present work. Also, we will consider the efect of the Guoy phase shift in the problem; such consideration will requiere the use of numerical methods to solve the integrals.
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Appendix
A. Reflected Field - Geometric Optical field
The first integral we need to solve is
I1=∫-∞∞(kx-kc)2j-2kexp[-14(kx-kc)2]×exp[i(kxx-kyy-)]dkx.
(A.1)
To accomplish this, we first need to express ky as a function of kx. Since ky=(k02-kx2)1/2 is not managable the way it is, we expand ky in a Taylor series, to second order, around the critical value k0=ksinθc. The result is
ky=kcosθ+kμsinθ-k(μ2/2)sec2θ,
(A.2)
where
μ=(sinθ-kx/k)/cosθ.
(A.3)
Substituting Eq. (A.2) into Eq. (A.1), we get
I1=Π(kw)2j-2k∫-∞∞μ2j-2k×exp[-(kwr2)2(μ+i2xwr2k)2]dμ,
(A.4)
where Π and wr are given by Eqs. 25 and 27. The integral can be solved easily if we make u=(kwr/2)[σ+i(2x/kwr2)] and using Newton’s binomial,
I1=Π(2/kwr)(2w/wr)2j-2k∑l=02j-2k(2j-2k)!l!(2j-2k-l)!×[-ixrwr]2j-2k-l∫-∞∞ulexp(-u2)du.
(A.5)
Using now the fact that
∫-∞∞ulexp(-u2)du={(2l)!22ll!πforleven,0forlodd,
(A.6)
we get as a result
I1=Π(2/kwr)(2w/wr)2j-2k×∑l=0j-k(2j-2k)!l!(2j-2k-2l)![ixrwr]2j-2k-2lπ22l.
(A.7)
We can simplify the previous result by observing that
∑l=0j-k(2j-2k)!l!(2j-2k-2l)![ixrwr]2j-2k-2l×π22l=π(-12)2j-2k×∑l=0[2(j-k)/2](2j-2k)!l!(2j-2k-2l)!(-1)l×(2xrwr)2j-2k-2l.
(A.8)
We recognize that the term in the summatory are the Hermite polynomials of order 2(j-k). For this reason, our final result is
I1=Π(2/kwr)(2w/wr)2j-2kπ(-12)2j-2k×H2(j-k)(xr/wr).
(A.9)
B. Reflected Field - Correction Factor
The other integral we need to solve is
I2=∫-∞∞rc(kx)(kx-kc)2j-2kexp[-14(kx-kc)2]×exp[i(kxx-kyy-)]dkx.
(B.1)
For this case, it is convenient to use the following variable
ν=μ-δ=(sincθ-kx/k)/cosθ.
(B.2)
Since we are interested in a small region of δ, the main contribution to the integral arise for a small values of ν. Also, the principal contribution arise from values of ky in a neighborhood centered around the value kx=ksinθ. In terms of this variable, ky can be expanded in Taylor series around ν=-δ (the axis of the geometric-optics reflected beam) giving
ky=kcosθ+k(ν+δ)tanθ-k[(ν+δ)2/2]secθ2.
(B.3)
We now focus our attention on the functions r(kx) and rc(kx). First, the second term of the numerator and the denomitaror in the reflectance is (k02-kx2)1/2. Second, we observe that this function varies rapidly around kx≃k0 due to the singularity at kx=k0. For this reason it is possible to make the change of variable to ν1/2. We then rewrite the function r(kx) (see Eq. (20)) as
rc(ν1/2)=1r(k¯)p(ν1/2)q(ν1/2)-1
(B.4)
where
r(k¯)=r(ksinθ)=cosθ-m(sin2θc-sin2θ)1/2cosθ+m(sin2θc-sin2θ)1/2
(B.5)
pν12= cosθ+(ν+δ)sinθ-(ν+δ)22secθ+mcosθ×[ξ+2(ν+δ)2tanθ+(ν+δ)2]12,
(B.6)
q(ν1/2)=cosθ+(ν+δ)sinθ-(ν+δ)22secθ+mcosθ×[ξ+2(ν+δ)tanθ+(ν+δ)2]1/2,
(B.7)
and
ξ=(sinθc2-sinθ2)secθ2.
(B.8)
We now expand rc(ν1/2) in Tylor series in powers of ν1/2 around ν=-δ, obtaining (keeping only terms of first order)
rc(ν)=C0-C1ν1/2-(-δ)1/2,
(B.9)
where
C0=rcν| nu=-δ,C1=-drc(ν)dν|ν=-δ.
(B.10)
The evaluation of C0 gives zero. The evaluation of C1 is easy but cumbersome. First, we obserbe that
C1=1r(k¯)p(ν)q'(ν)-p'(ν)q(ν)q(ν)2|ν=-δ,
(B.11)
where
p(ν)|ν=-δ=cosθ[1-mξ1/2],
(B.12)
q(ν)|ν=-δ=2(-δ)1/2[1-mξ-1/2],
(B.13)
p'(ν)|ν=-δ=2(-δ)1/2[1+mξ-1/2].
(B.14)
and
q'(ν)|ν=-δ=2(-δ)1/2[1+mξ-1/2].
(B.15)
Substituting Eqs. (B.12)-(B.15) into Eq. (B.11) gives (after algebraic simplification)
C1(θ)≡C1=4mcos2θcsinθcos1/2θ(sinθc+sinθ)1/2×1[cosθ2+m2(sin2θ-sin2θc)]
(B.16)
So, the function rc(ν1/2) is, finally:
rc(ν1/2)=-C1(θ)[ν1/2-(-δ)1/2].
(B.17)
The integral I2 is then
I2=(kcosθ)(kw)2j-2kΠexp[γ2/2]C1(θ)×∫-∞∞[ν1/2-(-δ)1/2](ν+δ)2j-2k×exp[-(kwr2)2ν2+i(kγwr2)ν]dν.
(B.18)
If now we use Newton’s binomial we get
I2=(kcosθ)(kw)2j-2kΠexp[γ2/2]C1(θ)×∑l=02j-2k(2j-2k)!l!(2j-2k-l)!δ2j-2k-l×[J1-(-δ)1/2J2],
(B.19)
where
J1=∫-∞∞ν1/2+l×exp[-(kwr2)2ν2+i(kγwr2)ν]dν,
(B.20)
and
J2=∫-∞∞νl×exp[-(kwr2)2ν2+i(kγwr2)ν]dν.
(B.21)
The integral J1 can be split into two parts:
J1=∫-∞0ν1/2+l×exp[-(kwr2)2ν2+i(kγwr2)ν]dν+∫0∞ν1/2+l×exp[-(kwr2)2ν2+i(kγwr2)ν]dν,
(B.22)
after changing variable in the first integral
J1=(-1)li∫0∞ν1/2+lexp[-(kwr2)2ν2-i(kγwr2)ν]dν+∫0∞ν1/2+l×exp[-(kwr2)2ν2+i(kγwr2)ν]dν.
(B.23)
Each integral can be found in tables (see 3.462-1 in [21]). We obtain
J1=(kwr2)-(3/2+l)Γ(32+l)exp(-γ24)×[(-1)liD-(3/2+l)(iγ)+D-(3/2+l)(-iγ)],
(B.24)
after using the identity 9.248-1 in [21] we get
J1=(kwr2)-(3/2+l)2πexp[-iπ4(1-2l)]×exp(-γ24)D1/2+l(γ).
(B.25)
The integral J2 can be solved directly after completing the square an changing the variable of integration. We obtain
J2=(2kwr)πexp(-γ22)×(-ikwr)lHl(γ/2).
(B.26)
Substitution of Eq. (B.25) and (B.26) into Eq. (B.19) gives the final expression for the integral I2.