Scattering in Acoustic Oceanography

1. Scattering (Part 1 of 2) • Topics to be covered today: • Sound received by a sonar, scattered from targets submerged in the water (as opposed to, e.g., the seafloor) • Reflection of sound from simple targets (e.g., spheres) • Reflections from multiple closely-spaced targets • Schools of fish, zooplankton • Clouds of bubbles • Suspended silts / sediment • Next week, Part 2: • Special types of targets like bubbles, fish, sediment • Reading: ambient noise of bubbles generated by typhoon breaking waves Oc679 Acoustical Oceanography

2. Pulsed Sonar • Consider the following basic setup for a monostatic ocean sonar • Sonar transmits a short duration ping • E.g., for a ping of duration 100 μs, what is its length in the water? • Ping travels through the water, can approximate as a plane wave • Ping hits an object (“target”), which reflects sound. We are considering targets in the water, surface/bottom not considered for now. • Some of the reflected sound is directed back towards the sonar • Q: what is the time elapsed between transmit and receive, for a target at range R? • Received echo will have same frequency as the transmitted ping, and usually a comparable duration, but reduced in amplitude (why?). • Some sound will travel around/past the target, and may then be reflected by other targets further in range. By listening over time, will hear multiple echos from different targets. Oc679 Acoustical Oceanography

3. Sound scattered by a body: Terminology(reference: Medwin & Clay Ch 7) scattering is the combined processes of reflection, refraction and diffraction at surfaces marked by inhomogeneities in , c - these may be external or internal to a scattering volume ( internal inhomogeneities important when considering scattering from fish, for example ) net result of scattering is a redistribution of sound pressure in space – changes in both direction and amplitude for a monostatic system (most sonars), we are interested in the sound reflected back to the source/receiver – this is termed backscatter scattering is wavelength- (frequency, sort of) dependent we will refer to the objects doing the scattering as scatterers or targets (same thing) the sum total of scattering contributions from all scatterers is termed reverberation • this is heard as a long, slowly decaying quivering tonal blast following the ping of an active sonar system to start, we consider simple, hard, individual scatterers Oc679 Acoustical Oceanography

4. transmission is a gated sinusoid or ping of duration tp • Figure shows crests of sinusoidal waves, indicated as a sequence of wave fronts • in this sketch ,  dependencies of incident wave are suppressed (this would be due to source beam pattern) • for simplicity wavefronts drawn as if coming from center of object – for a complex object, as shown, there would be many interfering wave fronts spreading from the object • within shadow, interference of incident and scattered waves is destructive as incident and scattered waves arrive at the same time with same amplitude, but out-of-phase • outside of shadow, interference of incident and scattered waves forms a penumbra (partial shadow) Oc679 Acoustical Oceanography

5. , = 0, otherwise , = 0, otherwise As viewed on an oscilloscope: Traces of transmitted and received sound pressures on a sonar • ping has duration tp • two-way travel time for sound to go from sonar to target and back is 2R/c, where R is range between the two 2 incident sound pressure: scattered sound pressure: pscat(t) = Pscat ei2πf(t-R/c) Complex Acoustical Scattering Length, consider amplitude and ignore phase: (This is a definition) dimension is length, but is not related to any length scale of the body or Oc679 Acoustical Oceanography

6. Differential Scattering Cross-section, dimension area [ m2 ] Again, this is just a definition/notation, we have not introduced any physics yet In practice, s, L, and Pscat all depend on the geometry of the measurement and the carrier frequency, f, of the ping (really they depend on wavelength) The dependence on  and  are relevant for cases where the source and receiver are at different locations (bi-static) When source and receiver are at the same position (monostatic), it is called backscatter: in that case,  =  = 0, and differential backscattering cross-section Oc679 Acoustical Oceanography

7. So far we have only considered “differential” (and have used ) – that is, the portion of the power that is radiated from the scattering body in one particular direction The total scattering cross-section is defined as the integral over all possible angles: In other words it is the ratio of total power scattered vs. incident intensity scat is the total power scattered by the body Iincident is the intensity (power per unit area) of the incident wave Oc679 Acoustical Oceanography

8. target strength No unified language / notation exists for acoustic scattering, unfortunately… Yet another notation used is target strength (TS), a logarithmic measure of differential cross-section. Might consider several variants depending on the context: note, referenced to 1 m2 for backscatter in terms of scattering length in terms of backscattering length Oc679 Acoustical Oceanography

9. Putting it all together, now return to the sonar problem. What is the sound received from a target with backscatter length Lbs? The incident wave is: Pincei2πft = P0 R0 /Rei2πf(t-R/c) 10-αR/20 The scattered wave is: Pscatei2πft = Pinc |Lbs(f)| ei2πf(t-R/c) 10-αR/20 = P0 R0 /R2 ei2πf(t-2R/c) |Lbs(f)| 10-2αR/20 Monostatic sonar where P0 is incident pressure as measured at an arbitrary reference distance R0 (e.g., 1 m) Note, travel time source to scatterer (one-way) is R/c sound spreads spherically outwards from source, and then again from the return path. Q: How is this accounted for in the equations? Absorption losses are included (α). Q: what are the units of α ? Note the use of backscattering length, Lbs. Verify for yourself that the units work out. Oc679 Acoustical Oceanography

10. Scattering by spheres • simple geometry, can solve wave equation exactly, well studied • an acoustically small non-spherical body scatters in about the same way as a sphere of same volume and same average physical characteristics (, c) • Or, at the very least, “on average” as the object rotates around and is viewed at different angles • Scattering depends on the ratio of wavelength to sphere size: quantified by ka, where k is the acoustic wavenumber (k = 2π f / c) and a is the radius of the sphere • We will discuss three specific cases: • rigid sphere ka >> 1 reflection dominates, geometrical or specular scattering • rigid sphere ka << 1 diffraction dominates, Rayleigh scattering • fluid sphere – includes transmission through medium Oc679 Acoustical Oceanography

11. Warm-up Problem: The “Ideal” Sphere Simplest possible model: A sphere that radiates sound equally in all directions • Neglects sound absorption into the sphere • Neglects diffraction of waves around the sphere • Sound field = sum of incident wave and radiated wave Oc679 Acoustical Oceanography

12. Warm-up Problem: The “Ideal” Sphere (check: integral over unit sphere gives σs=1, as defined above) Oc679 Acoustical Oceanography

13. small target compared to λ ka << 1 Rayleigh scattering large target compared to λ ka << 1 geometrical scattering

14. short wavelengths ka >> 1 i.e. high frequency or big objects geometrical scatter from a rigid sphere ( ka >> 1 ) specular (mirrorlike) reflection in the Kirchoff approximation (see Medwin & Clay) plane waves reflect from an area as if the local, curved surface is a plane scatter consists of a spray of reflected waves each obeying simple reflection law – that is, angle of reflection = angle of incidence  i.e., we are thinking in terms of a ray solution Assume (for now): • no diffraction effects - these would be important at edge of shadow and behind sphere • incident sound is a plane wave of intensity Iinc (Recall: units W/m2) • no energy absorption in medium • no energy penetrates into the sphere Oc679 Acoustical Oceanography

15. short wavelengths ka >> 1 short wavelengths ka >> 1 i.e. high frequency or big objects adi dSicosi 2asini a i dSi Consider di , a single “ring” increment on the sphere 1st we need to know the incoming power at angle i surface area increment (corresponding to grey shaded area) component in direction of incident wave input power to ring (remember, incident intensity Iinc is power per unit area) Oc679 Acoustical Oceanography

16. short wavelengths ka >> 1 short wavelengths ka >> 1 i.e. high frequency or big objects incident rays within angular increment di at angle i are scattered within increment ds=2di at angle s=2i now, calculate scattered power… the geometrically-scattered power measured at a range R is  using the geometric argument in above figure assume this scattering accounts for all the incident power [ no power loss ] Then , where r.h.s. is what we got on previous slide Plug in the 2 eqn’s for dπgs and dπinc (from above, and prev. slide) Now compare this to the defn. of |L|, result: scattered intensity is independent of angle of incidence – as in our idealized case, but is not the case in general in the case of the sphere, this means that all differential geometrical scattering cross-sections, including backscattering cross-sections, are equal , so using Pgs / Pinc from previous line above, Oc679 Acoustical Oceanography

17. short wavelengths ka >> 1 short wavelengths ka >> 1 i.e. high frequency or big objects From this geometric ray theory derivation, the differential scattering cross-section is: and the total geometrical scattering cross-section is Note: These results should look familiar… >> >> , where a2 is the cross-sectional area However: gs does not include the effects of diffraction, so it is not the true total scattering cross-section the geometric (ray) solution is deceptively simple: • is accurate in the backscatter direction 0-90 • but completely ignores everything beyond 90 (the sphere casts a perfect “shadow”) Complete calculations (using full wave theory, not rays) show that the total scattering cross-section approaches twice the geometrical cross-section (i.e., 2a2) for large ka In other words, diffraction effects increase the total scattering x-section. What is the physical meaning of this increase? How might it be explained intuitively? Q: What would be involved in “using wave theory” to calculate the true solution? Oc679 Acoustical Oceanography

18. long wavelengths ka << 1 i.e. low frequency or small objects Rayleigh scatter from a sphere ka << 1 when the wavelength is large compared to the sphere radius, scatter is due solely to diffraction. Scattering is no longer geometric, but comes from two contributions: • monopole radiation, AKA isotropic scattering, “breathing mode” – the part of scattering that radiates in all directions from the sphere center. Can be due to pure rigid reflection, or due to the body’s response as it is compressed/expanded by the sound pressure wave • dipole radiation – if the sphere’s density (1) is much greater than that of the medium (0), the body’s inertia will cause it to lag behind as the plane wave oscillates (sloshes back and forth). This motion is equivalent to the water being at rest and the body being in oscillation. This motion generates a dipole reradiation. When 1< 0, the effect is the same but the phase is reversed (when might this happen in the ocean?). In general, when 1 0, the scattered pressure is proportional to cos, where  is the angle between scattered and incident directions. Oc679 Acoustical Oceanography

19. long wavelengths ka << 1 long wavelengths ka << 1 i.e. low frequency or small objects peaks, troughs at ka>1 due to interference between diffracted wave around periphery and wave reflected at front surface of sphere geometrical scattering Rayleigh scattering M&C develop solutions for the monopole and dipole radiation independently and then sum them. First, here are the results for the Rigid Sphere: finite density, but zero compressibility Monopole scattering Dipole backscatter is determined by setting  = 0 Note: As ka is reduced, scattering x-section in the Rayleigh regime becomes exceedingly small compared to geometric regime. Sound waves bend around and are almost unaffected by acoustically small, non-resonant bodies Oc679 Acoustical Oceanography

20. Comparison to Light Scattering scattering of light follows essentially the same scattering laws as sound but light wavelengths are much smaller than sound - O(100s of nm) almost all scattering bodies in seawater are large compared to optical wavelengths and have optical cross-sections equal to their geometrical cross-sections the sea is turbid to light on the other hand, scattering bodies found in seawater are typically small compared to acoustic wavelengths (at 300 kHz,  5 mm, 4 orders of magnitude larger)  acoustic scattering is dominated by Rayleigh scattering by comparison the sea is transparent to sound - what limits the propagation of 300 kHz sound is usually not scattering, but absorption What about Rayleigh scattering from extremely small particles in water (e.g. molecules)? Answer (thanks Pat!) is that in completely pure water Rayleigh scattering causes attenuation that limits visibility beyond about 30m. Note that pure water is very rare, requires lab conditions to avoid microbubbles, dust, etc. Oc679 Acoustical Oceanography

21. why is the sky blue? For Rayleigh scattering, intensity of backscatter is proportional to (ka)4 Air is mostly N2, which has diameter ~3 Angstrom (3x10-10 m) Blue light has wavelength ~400 nm Hence, light in the atmosphere obeys Rayleigh scattering. Blue light is scattered more than red light (lower wavelength, hence smaller k). Why, then, is the sky not red? ka ~ 0.002 << 1 Oc679 Acoustical Oceanography

22. why is the sky blue? For Rayleigh scattering, intensity of backscatter is proportional to (ka)4 Air is mostly N2, which has diameter ~3 Angstrom (3x10-10 m) Blue light has wavelength ~400 nm Hence, light in the atmosphere obeys Rayleigh scattering. Blue light is scattered more than red light (lower wavelength, hence smaller k). Why, then, is the sky not red? (A: Multiple scattering brings the sunlight to our eyes, except when looking directly at the sun) What about sunrise / sunset? ka ~ 0.002 << 1 Oc679 Acoustical Oceanography

23. why is the sky blue? For Rayleigh scattering, intensity of backscatter is proportional to (ka)4 Air is mostly N2, which has diameter ~3 Angstrom (3x10-10 m) Blue light has wavelength ~400 nm Hence, light in the atmosphere obeys Rayleigh scattering. Blue light is scattered more than red light (lower wavelength, hence smaller k). Why, then, is the sky not red? (A: Multiple scattering brings the sunlight to our eyes, except when looking directly at the sun) What about sunrise / sunset? at low azimuth, light passes through extended range of atmosphere blue gets completely scattered out, while red is not scattered as strongly What about the “Green Flash”? occurs at the very end of the sunset (beginning of sunrise, when sun has passed below horizon) • Shorter wavelengths are refracted more effectively than longer wavelengths • Blue is scattered away, as in sunset • Red is not effectively refracted (wavelength is too long) so runs into Earth • Green is “just right” to bend over the horizon (sometimes, for < 1s) ka ~ 0.002 << 1 Oc679 Acoustical Oceanography

24. Non-Rigid (fluid) sphere: Rayleigh’s Result ( ka << 1) more general case, when sphere is an elastic fluid • ’/ = ratio of sphere’s density to medium density • κ’/ κ = compressibility (inverse of bulk modulus) ratio •  = angle between incident and scatter directions monopole component (aka “breathing mode”) dipole component [ backscatter case is when  = 0 ] Most bodies in the sea have values of ’/ and κ’/ κ close to unity, hence both terms are of similar importance. Also, in most cases the backscatter cross-section is comparable to the rigid body case E.g. Sand:’/ ~ 2.6, and κ’/ κ ~ 0.1. Plugging in, backscatter x-section is about ½ that of rigid case An exception is bubbles (next lecture): Highly compressible, κ’/ κ ~ 19,000 (!!) - scattering cross-sections can be several orders of magnitude greater than geometrical Oc679 Acoustical Oceanography

25. Non-Rigid (Elastic) Sphere When the scatterer is not rigid, it can support additional waves due to compression and even shear. Solution requires more math. Some references: • Faran (1951): Original theory and solutions (beware of typos) • Hickling (1962, 1964): Refined version of theory • Morse & Ingard (1968) “Theoretical Acoustics”: Textbook • Guanard & Uberall (1983): Good reference for numerical calculations Example: Calculations of backscatter length for a Tungsten Carbide sphere. Note this is a pretty “rigid” material! Compressibility effects Shear wave effects Oc679 Acoustical Oceanography

26. Non-Rigid (Elastic) Sphere The general case for scattering from a non-rigid sphere involves many considerations: • Compression/expansion of the sphere (breathing mode) • Movement (dipole radiation) • Diffraction waves in water that are trapped on sphere surface (aka creeping waves, circumferential waves) • Waves traveling inside the sphere (compressional & shear) • Direct path • “Whispering Gallery” • …and note these have different sound speeds than in water These are all accounted for in the full theory (references in previous slide), though their contributions are not obvious just from looking at a plot of |L| Pinc Oc679 Acoustical Oceanography

27. Experimental Data: Scattering from an Aluminum Cylinder • Neubauer et al. (1969) “Theory of Creeping Waves in Acoustics and their Experimental Demonstration” • Schlieren visualization of 3 types of scattering from an aluminum cylinder, ka~50 • (B) Specular reflection: bounces straight off of the cylinder • (C) Franz waves: wrap around the cylinder, travel in water • (D) Rayleigh-type waves: enter the aluminum at a critical angle and travel on circumference before exiting

28. Example of Experimental Data Simple experiment: Scattering by a small Tungsten Carbide sphere, for range of ka ~ 20-80 (actually was for a calibration experiment, but we will discuss calibration in a later lecture)

29. Example of Experimental Data • Arrival of different wave types can cause a train of pulses to be scattered from the sphere • At least for metal spheres, this can be very accurately predicted by the theoretical |L(ka)| (as shown) • Interference of these waves result in huge peaks and valleys in |L|, which are very sensitive to the material of the target, the length and shape of the pulse, etc. • Note, is less important for small targets (roughly ka>5) Echo from a single elastic sphere Circumferential waves Specular reflection Same data converted to scattering length