13.4. The Green's function in different coordinate systems

Rectangular coordinates

Rectangular coordinates—near-field

The Green's function in rectangular coordinates was given by Eq. (13.4) as
g(x,y,z|x0,y0,z0)=ejkR/(4πR),
image (13.31)
where R is the Euclidean distance, which is given by
R=(xx0)2+(yy0)2+(zz0)2.
image (13.32)
However, problems are often encountered when using this expression because the space variables are all enclosed in a square-root sign and therefore cannot be separated. This makes finding analytical solutions very difficult, and one often has to resort to using moveable-origin coordinate systems. This limits its use to numerical integration in the Kirchhoff–Helmholtz surface integral. Unfortunately, the Green's function of Eq. (13.31) is not particularly amenable to numerical integration because it is singular at the origin and leads to oscillatory integrands at high frequencies. The dipole integral is even more problematic because the Green's function normal gradient has a 1/R 2 term, which leads to diverging numerical and analytical integrals. Furthermore, where the surface of integration encloses one or more sources, we can only calculate the field on the side of the surface where there are no sources and the waves are diverging. In other words we cannot solve the reverse problem and calculate the field in which there are sources and the waves converge toward them. A more powerful formula is given by
g(x,y,z|x0,y0,z0)=j8π2ej(kx(xx0)+ky(yy0)+kz|zz0|)kzdkxdky
image (13.33)
where
kz={k2k2xk2y,k2x+k2yk2jk2x+k2yk2,k2x+k2y>k2
image (13.34)
where k x , k y , and k z represent the spatial frequency components in the x, y, and z directions respectively of a plane wave of spatial frequency k traveling in an arbitrary direction. For example, if the direction of travel subtends an angle θ with the z-axis, then the trace velocity seen along the z-axis is c/cos θ and the wave number is k z   =   k cos θ. Hence the wavelength will appear to be longer along the z-axis. To gain a better understanding of Eq. (13.33) we may compare it with Eq. (7.113) for the pressure field inside an enclosure by letting k x   =   /l x , k y   =   /l y , and k z   =   k mn . We also replace the infinite integrals with summations. In other words, Eq. (13.33) may be thought of as the spatial distribution of an infinite enclosure in which traveling plane waves of any wavelength may exist as opposed to standing ones of particular wavelengths that correspond to the dimensions of the finite enclosure. The fact that a point source can be represented as integral over all spatial frequencies is not so surprising when we consider that an infinite impulse contains all frequencies.
It may seem counterintuitive to introduce two extra integrals, but the troublesome 1/R term has vanished along with the square-root sign in the exponent. When used in the Kirchhoff–Helmholtz integrals, we will show in Section 13.19 that this integral form of the Green's function is an inverse Fourier transform. This leads to an important theorem that forms the basis of near-field acoustical holography in which the dipole Kirchhoff–Helmholtz integral evaluated over one plane is the Fourier transform of the pressure distribution in that plane. The sound field spectra is then propagated in k-space to another parallel plane in which the Green's function is the inverse Fourier transform that gives the pressure in that plane. Furthermore, we can solve the so-called reverse problem where there are one or more sources in the field of interest. This method of calculation is particularly amenable to the digital processing of sound fields captured by planar microphone arrays to calculate the entire sound field of interest. In other words, if there are sources on one side of the array, we can plot the pressure field on both sides of the array. This is not possible using the Euclidean form of the Green's function of Eq. (13.31).

Proof of the Fourier Green's function in rectangular coordinates

To derive the Fourier Green's function, we shall apply a triple Fourier transform, one for each Cartesian ordinate, to the Green's function in the spatial domain to convert it to the spatial frequency domain or k-space.
G(kx,ky,kz)=g(x,y,z|x0,y0,z0)ej(kxx+kyy+kzz)dxdydz,
image (13.35)
where G(k x , k y , k z ) is the Fourier transform of g(x, y, z|x 0, y 0, z 0). The inverse transform is
g(x,y,z|x0,y0,z0)=18π3G(kx,ky,kz)ej(kxx+kyy+kzz)dkxdkydkz.
image (13.36)
To solve for G(k x , k y , k z ), we take the Fourier transform of Eq. (13.17)
(2+k2)g(x,y,z|x0,y0,z0)ej(kxx+kyy+kzz)dxdydz=δ(xx0)δ(yy0)δ(zz0)ej(kxx+kyy+kzz)dxdydz,
image (13.37)
where
2=2x2+2y2+2z2.
image (13.38)
Using the general property of the Dirac delta function from Eq. (154) of Appendix II and noting that
2ej(kxx+kyy+kzz)=(k2xk2yk2z)ej(kxx+kyy+kzz)
image
yields
(k2k2xk2yk2z)g(x,y,z|x0,y0,z0)ej(kxx+kyy+kzz)dxdydz=ejk(kxx0+kyy0+kzz0),
image (13.39)
which after substituting in Eq. (13.35) gives us the Green's function in k-space:
G(kx,ky,kz)=ej(kxx0+kyy0+kzz0)k2x+k2y+k2zk2.
image (13.40)
Applying the inverse Fourier transform of Eq. (13.36) then gives us a Fourier Green's function in terms of k-parameters:
g(x,y,z|x0,y0,z0)=18π3ej(kx(xx0)+ky(yy0)+kz(zz0))k2x+k2y+k2zk2dkxdkydkz=18π3ej(kx(xx0)+ky(yy0)+kz(zz0))(kz+k2k2xk2y)(kzk2k2xk2y)dkxdkydkz,
image (13.41)
which has two poles: one at k z   =   +σ and the other at k z   =   σ, where
σ=k2k2xk2y.
image
We now convert Eq. (13.41) from a volume integral to a surface one by integrating over k z using the residue theorem, which states that
f(x)ejxtdx={2πj(sumofresiduesoff(x)ejxtatallitspolesαiabovetherealaxis),t02πj(sumofresiduesoff(x)ejxtatallitspolesαionorbelowtherealaxis),t<0,
image (13.42)
where each residue is defined by
(xαi)f(x)ejxt|xαi.
image (13.43)
Applying this to Eq. (13.41) to solve the integral over k z gives
g(x,y,z|x0,y0,z0)=j8π2ej(kx(xx0)+ky(yy0)+σ|zz0|)σdkxdky.
image (13.44)
If we let k z   =   σ, this then gives us Eq. (13.33).

Rectangular coordinates—far-field

At a large distance R, Eq. (13.31) simplifies to
g(x,y,z|x0,y0,z0)|R=ejk(x(xx0)+y(yy0)+z(zz0)/r4πr=ejkr4πrejk(xx0+yy0+zz0)/r,
image (13.45)
where
r=x2+y2+z2.
image (13.46)

Cylindrical coordinates

If we substitute
x=wcosϕ,y=wsinϕ,x0=w0cosϕ0,y0=w0sinϕ0
image (13.47)
in Eq. (13.31) and use
sinϕsinϕ0+cosϕcosϕ0=cos(ϕϕ0),
image
we obtain
g(w,ϕ,z|w0,ϕ0,z0)=ejkR/(4πR),
image (13.48)
where
R=w2+w202ww0cos(ϕϕ0)+(zz0)2.
image (13.49)
However, this expression is of limited use and suffers from all the same drawbacks as were described in reference to the Euclidean Green's function in rectangular coordinates given by Eq. (13.31). A more powerful formula [4] is given by
g(w,ϕ,z|w0,ϕ0,z0)=j4πn=0(2δn0)cosn(ϕϕ0)0Jn(kww)Jn(kww0)ejkz|zz0|kzkwdkw,
image (13.50)
where
kz={k2k2w,0kwkjk2wk2,kw>k,
image (13.51)
which is known as the Lamb–Sommerfeld integral [5,6]. This equation can be considered as the integral over all spatial frequencies of radial standing waves in an infinite cylinder, which are also summed over all azimuthal harmonics of order n. The component in the z direction is planar as represented by the exponent term. The reason why we have radial standing waves is that incoming waves pass through the z axis (or w   =   0) before traveling back out again. In doing so, the imaginary part of the Hankel function, or Y n function, changes sign. Thus the Y n function is canceled leaving just the J n function. This can be considered as the same phenomena as the incoming waves being reflected back from a rigid termination at w   =   0. Hence the standing waves. In the case of axial symmetry, we exclude all azimuthal harmonics but the n   =   0 term:
g(w,z|w0,z0)=j4π0J0(kww)J0(kww0)ejkz|zz0|kzkwdkw.
image (13.52)
We will apply this formula to problems with cylindrical symmetry such as circular sources.

Proof of the Fourier Green's function in cylindrical coordinates

If we substitute x 0   =   w 0 cos ϕ 0, y 0   =   w 0 sin ϕ 0, x   =   w cos ϕ, y   =   w sin ϕ, k x   =   k w cos φ, and k y   =   k w sin φ in Eq. (13.33) and use the identity of Eq. (46) in Appendix II, we obtain
g(w,ϕ,z|w0,ϕ0,z0)=j8π22π00ejkw(wcos(φϕ)w0cos(φϕ0))ejk2k2w|zz0|k2k2wkwdkwdφ.
image (13.53)
We then expand the first exponent term using Eq. (110) of Appendix II to give
g(w,ϕ,z|w0,ϕ0,z0)=j8π2m=0n=0(2δm0)(2δn0)jmn×2π00cosm(φϕ0)cosn(φϕ)Jm(kww0)Jn(kww)ejk2k2w|zz0|k2k2wkwdkwdφ,
image (13.54)
where the angular integral over φ is solved using
2π0cosm(φϕ0)cosn(φϕ)dφ={2πcosn(ϕϕ0)2δn0m=n0,mn,
image (13.55)
so that the double expansion of Eq. (13.54) reduces to the single one of Eq. (13.50).

Spherical coordinates

The Green's function in rectangular coordinates was given by Eq. (13.4):
g(x,y,z|x0,y0,z0)=ejkR/(4πR),
image (13.56)
where
R=(xx0)2+(yy0)2+(zz0)2.
image (13.57)
If we substitute
x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθx0=r0sinθ0cosϕ0,y0=r0sinθ0sinϕ0,z0=r0cosθ0
image (13.58)
and use
sinϕsinϕ0+cosϕcosϕ0=cos(ϕϕ0),
image
we obtain
g(r,θ,ϕ|r0,θ0,ϕ0)=ejkR/(4πR),
image (13.59)
where
R=r2+r202rr0(sinθsinθ0cos(ϕϕ0)+cosθcosθ0).
image (13.60)
However, as in the cylindrical and rectangular cases, this expression is of limited use and suffers from all the same drawbacks as were described in reference to the Euclidean Green's function in rectangular coordinates given by Eq. (13.31). A more powerful formula [7] is given by
g(r,θ,ϕ|r0,θ0,ϕ0)=jk4πn=0(2n+1)nm=0(2δm0)(nm)!(n+m)!cosm(ϕϕ0)×Pmn(cosθ0)Pmn(cosθ){jn(kr0)h(2)n(kr),r>r0jn(kr)h(2)n(kr0),r<r0.
image (13.61)
In the case of axial symmetry, we exclude all terms from the summation in m except for the m   =   0 term:
g(r,θ|r0,θ0)=jk4πn=0(2n+1)Pn(cosθ0)Pn(cosθ){jn(kr0)h(2)n(kr),r>r0jn(kr)h(2)n(kr0),r<r0.
image (13.62)
By relocating the source to θ 0   =   π, r 0     ∞, we obtain the expansion for a plane wave:
ejkrcosθ=n=0(j)n(2n+1)jn(kr)Pn(cosθ).
image (13.63)
Like the Fourier Green's functions in rectangular and cylindrical coordinates, this expansion form in spherical coordinates can be applied to reverse problems (see Ref. [8], pp. 210–211, Eqs. (6.107)–(6.110) for compact Kirchhoff–Helmholtz integrals. Note that G N and G D are not equivalent to the Green's function given by Eq. (13.61) above or its normal derivative but can be derived from it using the method we shall apply in Section 13.5).

Spherical–cylindrical coordinates

Spherical–cylindrical coordinates—near-field

If we substitute
x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθ,x0=w0cosϕ0,y0=w0sinϕ0
image (13.64)
in Eq. (13.31) and use
sinϕsinϕ0+cosϕcosϕ0=cos(ϕϕ0),
image
we obtain
g(r,θ|w0,ϕ0)=ejkR/(4πR),
image (13.65)
where
R2=r2+w20+z202r(w0sinθcos(ϕϕ0)+z0cosθ).
image (13.66)
If we set z 0   =   0 and ϕ   =   0, this simplifies to
R2+r2+w202rw0sinθcosϕ0.
image (13.67)
Again, this expression is of limited use, and a more powerful formula [9] is given by
g(r,θ,ϕ|w0,ϕ0)={jk4πn=0(2n+1)h(2)n(kr)jn(kw0)Pn(sinθcos(ϕϕ0)),w0rjk4πn=0(2n+1)jn(kr)h(2)n(kw0)Pn(sinθcos(ϕϕ0)),w0r,
image (13.68)
where
Pn(sinθcos(ϕϕ0))=m=0(2δm0)(1)mPmn(0)Pmn(cosθ)cos(m(ϕϕ0)),
image
which is a modified form of the Gegenbauer addition theorem or multipole expansion. We shall use it to derive near-field expressions for axisymmetric planar sources.

Spherical–cylindrical coordinates—far-field

At a large distance r, the terms containing r in Eq. (13.66) dominate. Hence the remaining terms can be replaced with ones that enable R to be factorized as follows:
R2=r22r(w0sinθcos(ϕϕ0)+z0cosθ)+w20+z20r22r(w0sinθcos(ϕϕ0)+z0cosθ)+(w0sinθcos(ϕϕ0)+z0cosθ)2=(rw0sinθcos(ϕϕ0)z0cosθ)2.
image (13.69)
Thus we can write the far-field Green's function as
g(r,θ,ϕ|w0,ϕ0,z0)|r=ejk(rw0sinθcos(ϕϕ0)z0cosθ)4πr.
image (13.70)
We will use this formula to derive far-field expressions for axisymmetric planar sources.
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