An excellent route for learning about radiation in general, with applications to scattering and resonators.
This paper assumes prior knowledge of the properties of sound waves and the use of the vector potential. For an introduction to acoustics through plane waves, see Sound Waves. For an explanation of the wave equation, and the properties of its solutions, including amplitudes, interference and the use of phasors, see Wave Functions. For an exposition of the vector calculus, the velocity potential, and spherical wave solutions, refer to Waves in Three Dimensions. All this forms a comprehensive course in sound waves, but not in the vibration of material bodies, a subject usually closely associated with acoustics.
The wave equation implies wave solutions: that is, disturbances that independently carry energy forward without a continuous motion of the medium supporting the waves. This is the very essence of wave motion, well shown by the plane wave solutions. But the wave energy must have had some source, and we must extract energy from the wave in order that it have some effect, so propagation by itself is not the whole story. When there is a localized source in a large volume, the energy moves outward along radial lines, at least at distances large compared to the size of the source, in the well-known rays, so this is phenomenon is called radiation.
Radiation involves not only the waves, but also their physical source and the interaction of the two, so its study is much more complicated than that of wave motion by itself. Electromagnetic radiation is one example of very great utility and interest, but it is rather complicated, not least because the vector nature of the waves must be explicitly treated. Also, it travels at the speed of light, which is a fundamental constant of relativity, making light a very special case of wave motion. Acoustic radiation is much simpler, because the waves may be described by a scalar, the velocity potential, they travel at lower and more comfortable speeds where relativity is not an issue, and their interpretation is concrete, the mechanical motion of a material medium. Many of the general properties of radiation will be common to light and sound, however, so that the study of acoustic radiation offers an understanding of electromagnetic radiation as well. A direct, plodding approach to radiation is impossibly difficult and unenlightening. It is necessary to apply intelligence and ingenuity to this problem, as in so many other endeavours, in addition to mathematical skill.
Our method of approach is as follows. We postulate that the wave equation is satisfied, so that we can make free use of superpositon. This allows us to make a Fourier transformation from time to frequency space or, what is the same thing, to assume a time variation eiωt. This agrees very nicely with the harmonic vibrations of solids that will be our sources. Then we look for solutions that represent outgoing (and perhaps incoming) waves, which seem appropriate to the geometry. If the constants in a solution can be adjusted so that the normal velocity is continuous at a bounday, then we assume that the result is the unique solution to our problem. If we had another solution that satisfied the same boundary conditions, then the difference of the two solutions would have a zero normal velocity on all boundaries, and would, therefore, be identically zero, showing that the two assumed solutions must actually be the same. This uniqueness proof deserves careful mathematical attention, which it has received. We simply use the results of these investigations in gratitude that the hard work has been done for us.
The simplest problem is a point source in infinite space, where we assume that all motion vanishes at infinity. The origin of coordinates is taken at the source. The simplest solution for this case is φ = C e-ikr/r, where C is a constant, k is the wavenumber, 2π/λ, and r the distance from the source. The radial velocity is then v = C e-ikr/r2 + ik C e-ikr/r. The flux through a sphere of radius r is 4πr2v. As r is made very small compared to 1/k, the second term vanishes, while the first term gives 4πC, a constant, showing that matter is created at the point at the rate 4πC = q cm3/s. Of course, all these quantities are amplitudes multiplying the common time factor eiωt. We have found that the monopole source q gives rise to a velocity potential qe-ikr/4πr.
It is instructive to note that the velocity consists of two terms with different radial dependences. When kr is small, the first term predominates, and is approximately q/4πr2, precisely what you would expect if the medium were incompressible and the speed of sound infinite. The velocity is in phase with the source. When kr is large, the second term predominates. The velocity falls off only as 1/r, and is 90° advanced in phase. The first region is called the near field, differing little from potential flow in an incompressible medium, and the second the far field, where the 1/r dependence shows that energy in the wave is propagated to unlimited distance. The far field velocity results from differentiation of the retardation factor e-ikr. Radiation is, it may be concluded, a consequence of the finite speed of propagation of the wave.
The connection with incompressible flow is helpful in our study, since incompressible flow is easier to work with. In this case, the velocity potential satisfies Laplace's equation, which gives us many analogies with electrostatics and potential theory. To pass to the general case, we simply multiply by the retardation factor. Thus, φ = q/4πr becomes φ = qe-ikr/4πr. Note that we cannot apply retardation directly to the velocity, since the velocity does not satisfy the simple wave equation. By using frequency space and potentials, we easily glide by some very difficult mathematics with no loss of understanding.
The monopole source is not quite general enough for practical use. In electromagnetism, it does not exist at all, because of the conservation of charge. Two monopole sources of strengths +q and -q, separated by a distance a, form a new type of source when they are conceived as strengthening while approaching one another so that the product p = qa remains constant. This product is called the dipole moment, and its direction is considered as from the -q to the +q. Of course, dipoles are very familiar from electromagnetism. To find the potential of a dipole, we simply differentiate the monopole potential of a source q with respect to a distance x, obtaining φ = qxdx/4πr3, and pass to the limit qdx = p. The result is most conveniently expressed as φ = p cosθ/4πr2, where θ is the angle between r and the direction of the dipole. The sign is easily determined by considering the velocity along θ = 0, which must be away from this positive end of the dipole. The potential in the general case is then (p cosθ)e-ikr/4πr2.
The velocity for a dipole radiator can now be found by differentiation. There is a near field term with a radial component p(1 + ikr)e-ikrcosθ/4πr3 and a transverse component half as large, varying as sin θ instead of cos θ. The far field term is purely radial, -k2p cos θ e-ikr/4πr. Once p has been determined for the near field by applying the boundary conditions, it is also known for the far field. This is the connection between the boundary conditions and radiation.
Now we can solve a very interesting case. Consider a sphere of radius a, such that ka<<1, that oscillates along the x-axis with velocity u = Ueiωt. If the amplitude of the velocity is small, the radial velocity at the surface of the sphere is U cos θ. This exactly matches the radial velocity for a dipole source, p cos θ / 2πa3, at the distance a. Therefore, our solution is p = 2πa3U. Hence, the radial velocity in the far field is -k2Ua3 cos θ e-ikr/2r. We have solved a radiation problem, and this is a rare treat. The oscillating sphere can be replaced by a dipole at its centre, so far as motion beyond the sphere is concerned.
The rate at which energy is carried away from the vibrating sphere can now be found. The intensity of a harmonic plane wave of amplitude A is ρcω2A2/2, or ρcV2/2 if V is the velocity amplitude. Far away from the dipole, we have essentially a plane wave, and its intensity is ρck4U2a6 cos2 θ/8r2. To find the total rate of emission, this must be integrated over the whole sphere. A surface element 2πr2sin θ d&theta is appropriate, integrated from 0 to π. Then W = 2πρck4U2a6/3 erg/s. In terms of the dipole moment p, this is W = ρck4p2/24π, which is easier to compare with the electromagnetic result.
The rate of radiation from a monopole can be found the same way, and the result is W = ρck2q2/8π. The frequency dependence is not as extreme, merely varying as the square of the frequency. Going the other way, if we differentiate the velocity potential for a monopole twice, we get a quadrupole source, and three times, an octupole. From dimensional considerations alone, we expect a frequency dependence of k6 for the quadrupole, and k8 for the octupole. In all these cases, the rate of radiation is multiplied by factors of (ka)2, and as this is assumed small, the rate of radiation is decreasing proportionally for the higher multipoles.
Near the oscillating sphere, air moves from the leading edge to the trailing edge, along streamlines that are identical to the lines of force of an electric dipole. The electric field and the velocity field are both the gradients of potentials of exactly the same form. When a surface vibrates, air moves from the parts moving outwards to the parts moving inwards, over distances that are small compared with a wavelength. When a bell vibrates in its lowest mode, there are four meridional nodal lines, and velocities are alternately outward and inward, suggesting a quadrupole. In the higher modes, the decrease in wavelength may dominate the decrease in size, favouring radiation in spite of the higher multipole order. A bar vibrating laterally would resemble a dipole source, perhaps a stronger one than that formed by the ends of a bar vibrating longitudinally. The radiation, in most cases, is rather weak, affecting the vibrations of solid bodies very slightly.
It was noticed by J. Leslie in 1837 that a bell which sounded loudly in air was scarcely perceptible when it rang in hydrogen. The effect was much greater than could be expected from the differences in density or acoustic impedance. The puzzle was resolved by Stokes in 1868, who pointed out that the higher speed of sound in hydrogen allowed the quicker equalization of pressures over the bell as it vibrated, and correspondingly less radiation. As we have seen, an infinite phase velocity means no radiation at all. If the rate of radiation is expressed in terms of the amplitude of vibration and the frequency, we find that it is proportonal to ρc as before, but inversely proportional to the sixth power of the phase velocity for the same acoustic impedance. This ratio of these factors for air (M=29) with respect to hydrogen (M=2) is about 800, which is certainly a significant difference.
To estimate the effect of radiation on the movement of a body, let us think of a sphere of mass M vibrating under the restoring force of a spring of spring constant K. If A is the instantaneous amplitude of vibration, then its total energy E = KA2/2, which decreases at the rate W, evaluated above. Setting dE/dt = -W, we find dA/dt + A/τ = 0, where the relaxation time τ is given by τ/T = M/π(ka)3M', where M' is the mass of air displaced by the sphere, and T is the period of vibration. It is easy to see that τ is very much greater than T in any practical case, which means that the vibration is very slightly damped. We conclude that radiation from a radiator much smaller than a wavelength in size is very inefficient.
If the energy of the sphere decreases, this means that it is acted upon by a force associated with the radiation, the radiation reaction. Let's try to find the force on the sphere directly, by integrating the pressure over the surface of the sphere. The static pressure does not contribute, of course, and the varying pressure, or overpressure, is equal to the density times the time derivative of the velocity potential. We have everything we need for this, and the integration is left as an exercise for the reader. The additional force is F' = 2πρa3dU/dt, or one-half of the mass of the fluid displaced time the acceleration of the sphere. This means that the dynamics of the sphere would be the same if the medium were removed, and the mass of the sphere increased by half the mass of the fluid it displaced. The force is due to the near-field motion of the medium, not the radiation reaction, which should be proportional to the velocity. Finding the radiation reaction directly requires a yet finer analysis.
In these cases, as well as in potential (irrotational) flow in general, there are tangential velocities at the vibrating surface. The viscosity, however small, leads to the creation of the well-known viscous boundary layer in which the tangential velocity becomes zero at the surface. So long as the depth of this boundary layer is small compared with the other dimensions, it should have little effect, and dissipate a negligible amount of energy, even less than is radiated. Nevertheless, these are only fond hopes, and should be checked in a deeper investigation, or at least recognized as a possible source of departures from our theory.
Many techniques familar from electromagnetism can also be used here, but care must be taken. As an example, suppose we have a monopole source on a rigid infinite plane at x = 0. This is the limiting case of a monopole source a short distance in front of the plane. The condition of zero normal velocity is satisfied if we superimpose the field of an identical source an equal distance behind the plane. In the electric case, this image charge would be of opposite sign, but not here, since the boundary conditions are different. In the limit, we have a source of twice the strength, and the rate of radiation is quadrupled. It is difficult to achieve the ideal case in practice, but the strengthening of radiation by providing a baffle is undoubted. A loudspeaker in a baffle is approximately a monopole source. If we have a sphere oscillating towards and away from the plane, the image dipole is in the opposite direction, and in the limit there is no radiation at all, the two dipoles cancelling (which might be expected from the geometry). If the sphere oscillates parallel to the plane, then its radiation is quadrupled, as in the case of the monopole.
Suppose a plane wave fall on fixed, rigid sphere. The obstacle disturbs the flow in two ways. First, the rigidity of the sphere means that it does not compress and expand like the medium it displaces. Second, as it surges back and forth the medium must flow around the fixed obstruction. We appeal to superposition to create an equivalent flow field whose properties we better know. The first disturbance is equivalent to a simple source at the centre of the sphere that would source and sink sufficient volume to cancel the changes that would occur in its absence. The second disturbance is the same as for a sphere oscillating in a medium at rest with a velocity opposite to that in the wave. Therefore, we can replace the sphere by a single and a double source located at its centre, and this will have the same effect on the flow at large distances as the sphere did. We assume that ka << 1, so that the properties of the plane wave are approximately constant across the sphere. The monopole and dipole will radiate energy in all directions, so the energy in the plane wave will be partially scattered.
The next step is to evaluate the strength of the sources. The condensation s in the incident wave is the change in volume per unit volume, so the total volume compressed into the volume of the sphere is Qs, where Q is the volume of the sphere, 4πa3/3. This would be roughly the same for a body of any shape. The strength of the monopole is the time derivative of this, which gives a monopole strength of q = -k2Qφ0, since the condensation is the time derivative of the velocity potential, divided by c2 from the equation of motion, and then the wave equation expresses the time derivative of this in terms of the space derivatives, from which the factor k arises. φ has been subscripted to show that it refers to the incident wave. The dipole strength will be p = 2πa3U with U = ikφ0, the negative of the particle velocity. The resultant velocity potential from these two sources is: φ = -(ka)2(2πaφ0/3)(2 + 3 cosθ)e-ikr/r + (ka)(2πaφ0)cosθ e-ikr/r2. The first term is the one giving the radiation, the second is purely a near-field flow.
From the expression for the velocity potential, we can see that the scattered intensity is proportional to: (a) the intensity of the incident wave, (b) (ka)4, (c) the cross-sectional area of the sphere, and (d) 4 + 12 cosθ + 9 cos2θ. These deductions are the same as for the scattering of an electromagnetic wave, except for the angular dependence, which is very different, because in the electromagnetic case there is only the dipole, and it is oriented quite differently. On integrating over all angles to find the total scattered intensity, the interference term in cos θ disappears, and the result is the same as if the monopole and dipole radiated independently. The amplitude of the velocity potential in a plane wave is the phase velocity c times the amplitude of particle displacement A. For the total scattered intensity, we find W = (πa2)(ρcω2A2/2)(ka)4(4/9 + 1/3), the 4/9 coming from the monopole, and the 1/3 from the dipole. If we divide by the first factor, the cross-sectional area of the sphere, we find the cross-section for scattering. The intensity in the incident wave falling on this area is the amount scattered.
The scattering by a sphere is predominantly into the forward hemisphere; very little is scattered backwards. This results mainly from the interference of the monopole and dipole components. (If I have the phases wrong, then the scattering will be backwards, but I have tried to be careful.) Scattering by a flat disc will have a very different angular distribution, since the monopole contribution will be absent (the volume of the disc is zero). In this case, forward and backward scattering can be expected to be roughly equal. This is different from the electromagnetic case, where the angular distribution is roughly independent of the shape of the scattering centre.
Interference between the scattered amplitude and the amplitude of the incident wave will account for the loss of energy from the incident wave and its appearance in the scattered wave. To investigate this, one starts by expanding the e-ikx for the incident wave in spherical partial waves, and then adds it to the scattered amplitude. The result is a wave slighly distorted close to the scatterer, and this distortion spreads widely until the wave is again indistinguishable from a plane wave, but with the subtle reduction in amplitude due to the scattering.
If a pulse of acoustic radiation is sent into a collection of scatterers, such as a wood, we expect a faint echo of scattered radiation, in which the higher frequencies are more prominent than in the original pulse. This phenomenon was remarked by Rayleigh, who was famous for his studies of scattering. Arnold Sommerfeld first rigorously worked out the diffraction of sound from a straight edge as a preliminary to his famous work on electromagnetic diffraction. Acoustic radiation has not been neglected by the great investigators.
Consider a Florence flask (for concreteness) of 500 ml volume (Q), with a neck 10 cm long (L), and 1 cm2 in area (A). Air moving down the neck will cause the pressure in the flask to rise, and this pressure will oppose the motion. If it's only the momentum of the air in the neck that sustains the motion, it will soon cease, and then reverse, as the pressure drives the air out, producing momentum. The motion will overshoot, and the decreasing pressure in the flask will soon bring it to a halt. The external pressure will now drive air into the flask, and we are back to our starting point. This is an oscillation, with energy moving back and forth between potential (pressure) and kinetic (movement). Let us now find the natural period of oscillation.
We know that if the dimensions are small compared to a wavelength, the motion of the air is essentially incompressible, and the problem is much easier. Let us then assume that the dimensions of the flask are small compared to a wavelength, and check that this is the case later. We also assume that the velocity is important only in the neck, and negligible both within the flask and outside. Then, the pressure difference across the neck is sρc2, and the net force acting on the plug of air in the neck is this times A, which must equal the product of the mass and acceleration of the plug. Its mass is ρLA, and the displacement x is related to s by s = -xA/Q, or x = -Qs/A. The negative sign means only that an increase in x in the direction of the pressure decreases s. From all this, the equation of motion is found to be sρc2A + (ρLA)(Q/A)d2s/dt2. The natural angular frequency of oscillation is, therefore, ω = c sqrt(A/LQ). For our Florence flask, the natural wavelength is about 444 cm, a comfortable amount larger than its diameter of about 10 cm. The frequency is 75 Hz.
A resonator may well have some other form of aperture than a tube of length L and area A, as shown in the Figure. The result can be generalized by means of an electrical analogy. The velocity potential is related to the velocity just as the electric field intensity is related to the electrostatic potential. Hence the velocity flow through the aperture is analogous to a problem in electrical conduction. The velocity potential will be practically constant inside the resonator (since the velocities are small) at some value φ, and also constant outside, where the velocities are again small, and this value can be chosen to be zero. The velocity is then φ/L for the tube aperture, and the total flux dq/dt = (A/L)φ. The equivalent electrical result is I = (σA/L)φ, so for the flow we have simply σ = 1. For other shapes, there will be some different value K as the analogue of the conductivity, so in general we can write dq/dt = Kφ. As a generalized coordinate, the volume q of fluid, assumed incompressible as it moves in and out, is chosen.
Now, dφ/dt = c2s, and q = -sQ, so d2φ/dt2 = -Kc2q/Q. This equation of motion gives ω2 = c2(K/Q), or k = (K/Q)1/2. The wavelength depends only on the shape and dimensions of the resonator. Restricting the aperture, which decreases K, makes the wavlength larger, as does increasing the volume Q of the resonator. The kinetic energy T = mv2/2 = (1/2)(ρLA)(dq/dt/A)2 = (ρL/2A)(dq/dt)2 = (ρ/2K)(dq/dt)2, where K has replaced A/L. The potential energy V = pq/2 = ρc2s2Q/2 = (ρc2/2Q)q2. For a circular aperture in a thin wall, K equals the diameter of the circle. In this case, the frequency varies as A1/4Q-1/2. When the aperture is a neck, its length L is supposed small compared to a wavelength, but large compared to the diameter of the neck.
Taking another tack, we have already found solutions of the wave equation appropriate to spherical symmetry, e-ikr/r and eikr/r, corresponding to outward and inward waves, respectively. Inside a sphere, we expect that the wave will be reflected from the surface, so let us superimpose the outgoing and incoming waves in equal amounts. The only combination that will give a finite val ue at the origin, r = 0, is φ = sin kr / r. The condition that the velocity vanish at the surface of the sphere r = a, is that the radial derivative of φ vanish there, giving the condition tan ka = ka; This is a famous transcendental equation best solved by successive approximation. The smallest solutions are ka = 1.4303π, 2.4590π, 3.4709π,..., approaching (n + 1/2)π for large n. For our flask of 4.9 cm radius, the lowest wavelength is 6.85 cm, or about 4.88 kHz. Now the wavelengths are comparable to the dimensions of the resonator, and the approximation of incompressible flow cannot be used. The locations of kinetic and potential energy are no longer separate, and the neck plays no role, except perhaps as an outlet for the sound. These radial normal modes have spherical nodal surfaces sin kr = 0. The nth mode has n nodal surfaces.
Just as we found the dipole velocity potential by differentiating the monopole potential, we can do the same thing here. The derivative with respect to x yields: φ = (kr cos kr - sin kr)cos θ/r2. Again applying the condition that the velocity vanish at the surface of the sphere, we find tan ka = 2ka / 2 - (ka)2, of which the solutions are ka = 0.6625π, 1.891π, 2.930π, 3.948π, 4.959π, .... For our Florence flask, the lowest of these modes has λ = 14.8 cm, and ν = 2.26 kHz. These modes all have nodal planes at θ = π. In the lowest mode, which is the lowest of any mode of vibration, the air sloshes from side to side of the flask.
A rectangular room is another soluble case. If φ = (Ceiωt) cos Kx cos Ly cos Mz, the normal derivatives vanish on the planes x = 0, y = 0, z = 0. If the velocity potential is to vanish at x = a, then sin Ka = 0, or Ka = kπ, where k = 0, 1, 2, .... Similar requirements at y = b and z = c give integers l and m. Hence, φ = C cos(kπx/a)cos(lπy/b)cos(mπz/c). There is one normal mode for each triplet of numbers k,l,m except 0,0,0, which gives an uninteresting constant value for φ. The wave equation is satisfied if K2 + L2 + M2 = ω 2, from which the frequencies can be obtained once k,l,m are specified.
An important property of these normal modes of vibration is that they are periodic. If the system is subjected to a periodic disturbance at the frequency of a normal mode, that mode will be gradually reinforced, like pushing a swing, until it dissipates as much energy, by radiation or friction, as it receives from the exciting source.
The first kind of resonators that we considered, those small compared to a wavelength, have some unexpected properties. If a small source is brought near one of these resonators, the resonator will induce a greater emission of radiation, through making the source 'work harder' against an increased pressure. When the source is distant, the resonator will scatter radiation of frequencies near its natural frequency quite efficiently, perhaps redirecting it conveniently. Brass vessels serving as resonators were used in ancient theatres to take advantage of these properties, as Vitruvius explains in de Architectura.
The calculation of the greater emission of a source in the presence of a resonator is, as Lamb says, a matter of delicacy. The resonator is excited to a large amplitude of vibration, and is an efficient simple source of radiation. It is merely a passive instrument as far as energy is concerned, all of which must come from the source. The increased radiation from the mouth of the resonator acts on the source to increase the rate at which it radiates, by providing an increased pressure in phase with the velocity. The distance b from the source to the aperture of the resonator must be small compared to a wavelength. The intensification for a simple source is (kb)-2, and for a double source (kb)-4. The details are in Lamb.
When a plane wave falls on a resonator, the scattering cross section of the resonator is λ2/π. This resonance cross section is huge compared to the usual scattering cross section of a small object. The explanation of it is the interference of its radiation with the incident wave. All these effects are stronger if the aperture of the resonator is in an infinite baffle surrounding it.
Composed by J. B. Calvert
Created 3 June 2000
Last revised 4 June 2000