Definition of the speed of sound

Question

I am reading about dispersion relations and got stuck with a connection I cannot follow. I have studied the mono-atomic chain which yields a dispersion relation

$w(k) \propto |\sin(\frac{ka}{2})|$.

Now, looking at small $k$, we can approximate that $\sin(ka/2) = ka/2$ so that $w(k)\propto ka$. The textbook states that this limit describes sound waves and this is exactly what I do not quite understand. Which formula is pointing out that we are dealing with sound waves? I have read Why can the dispersion relation for a linear chain of atoms (connected by springs) be written as $\omega(k)=c_s \lvert k\rvert$? and the corresponding comments where it says:

is just the speed of sound, which connects (in your linear dispersion relation) the frequency with the momentum. So, this is just the speed of sound by definition.

and

To elaborate upon what qmd, QuantumMechanics said, the speed of sound is defined either the "phase velocity" ω/k, or the "group velocity", dω/dk. Whenever ω=Ak for some A, one may see that the group velocity and the phase velocity coincide and both are equal to A.

where I understand how the group velocity coincides with the phase velocity but now which of the two corresponds to the speed of sound? or is it necessary for them to coincide in order to have sound waves? I feel like, I just lack a definition of what sound really is.

Answer 1

I will try to explain the answer in a really symple way. Proabably as you know, the dispersion relation that you have written above is obtained doing some approximations (i.e. the first neighbour interaction, and the approximation that the interaction between atoms of the chain is harmonic). At this point you can imagine your 1-D mono-atomic crystal as a set of harmonic oscillators, so, if you perturbe your system, you will get an "harmonic response". The reason why in this case you can talk about speed of sound, is that you can map the nuclear part of the Born-Oppeneimer equation into a 1-D D'Almbert equation in which this speed of sound $v_{s}$ appears (i will not demonstrate it, but trust me on the fact that you can do this). As you know this equation reads $$ (\frac{1}{v_{s}^{2}} \frac{\partial^{2}}{\partial t^{2}}- \frac{\partial^{2}}{\partial x^{2}}) f(x,t) = 0 $$ Now, let's consider the classical elastic theory for continuos medium, in which the atomic nature is negligible and let's construct a parallelism with the atomic chain theory. If you perturbe a continuos medium you will obtain a wave travelling into the medium with a dispersion relation: $$\omega= v_{s} k$$ where the speed of sound (i.e. the speed of propagation of the wave due to the perturbation induced) is classicaly defined as: $$v_{s} = \sqrt{\frac{C}{\rho}}$$ where $C$ is the Young-modulus of the medium and $\rho$ is the density of the medium. At this point, let's link this classical elastic theory with the one associated to the mono-atomic chain. In other words, let's consider that your medium is made by mono-atomic chains in such a way that you have a set of atomic layers of mass $M$, section $S$ and each layer is distant $a$ from another one ($a$ is the reticular constant). For the sake of simplicity imagine that in every layers there are $N$ atoms and, finally, imagine that this medium has got a total lenght at rest $l= Na$, where N is the number of atoms that compose the linear chain. If you ideally compress it by a quantity $dl$ the response of the system is harmonic (it is, beacuase of the approximation that you have done in your linear-chain theory) and so, you will have that the response-force can be written as: $$dF = (C S)\frac{dl}{l} $$ Now setting $dl \approx a$ and remembering that the force is harmonic $dF = k(dl) \approx k a$, one has: $$ka = (C S)\frac{a}{N a}$$ notice that $S a$ is clearly the volume $V$ occupied by a layer of atoms, then multiply both members of the equation by $a$. In this way: $$k a^2 = \frac{CV}{N}$$ and using $\rho =\frac{M}{V} \rightarrow V= \frac{M}{\rho}$ one gets: $$\frac{k a^2}{M} = \frac{C}{\rho N}$$ Let's identify $m= \frac{M}{N}$ as the mass of a single atom, and so, in conclusion we get to: $$\boxed{ \frac{K a^2}{m} = \frac{C}{\rho} = v_{s}^{2} } (1)$$ In this way you have identified the speed of sound by the parameters of your mono-atomic chain theory. Now, if you see at the dispersion relation obtained in the thoery of the linear chain, you can simple derive the same exact expression that is at the left side of the equation (1) for the group velocity $\frac{d \omega(k)}{d k}$ computed for small $k$, exactly as you did in your question (apart from a dimensional factor that is in front of the dispersion relation that you have omitted). This means that you can identify the oscillation modes of the linear chain at big wave-lenght ( or $k \approx 0$) like a sound wave that is propagating into a continuos medium!

Answer 2

The complex analysis of sound velocity must satisfy the ratio of distance to the arrival time for the pressure wave in the medium.

The complexity arises due to differences in propagation spe3ed in the 3 dimensional complex medium between 2 points. There are a variety of factors that affect sensitivity to distance, frequency, temperature, air pressure, medium density, temperature, pressure, and molecular spectrum of different gasses, solids and liquids.

The propagation modes of longitudinal waves, shear waves and dispersion with a sensitivity to the previous factors eluded many 17th century scientists including Netwton, Laplace, Mersene, and Boyle. They understood the fundamentals better than most others with variables of compressibility, elasticity, density, shear waves , gas dispersion, with different speeds even at the same frequency due to these effects.

The Fourier Spectral analysis of thunder would show a wide range of results with variations in spectral velocity due to the slower 3D effects of dispersion vs. the linear longitudinal compression waves vs the faster effects of shear waves. Also the thermal gradient effects of reduced attenuation from curvature over water improves clarity as the spectral bandwidth improves with a more constant speed thus much greater clarity than the delayed attenuated dispersed sounds.

Group delay is a phenomena that characterizes a narrow channel's phase response. I believe it does not have to coincide with velocity of the impulse wave except for the dominant frequencies band of the channel that make up the impulse response.

In order to reduce the acoustical distortion of sound velocity in air for music, I expect the best results in near field where the loss factors are reduced and the emitter is linear phase plane wave rather than a polar radiator.

This reminds me that when I attended the 1st US Festival in San Bernadino, California, in the early 80's that it was best sound system I had heard from a stage from a mountain top to see/hear Police, Pat Benatar and 30 or so other bands projected on two large drive-in sized screens, all the way down to 50 yards from the stage. The source was a array 1meter square speakers stacked to the size of a drive-in theater screen, in stereo and dual projector screens of similar size. With something like 500 kilowatts of stage sound and 100kW of delayed sound on scaffolding half way up the mountain, there was none of the echo problems common to arenas because there was no to bounce off up the horizon.