Consider a lattice of massive points connected by harmonic springs, with zero or periodic boundary conditions. If we make a repeating pattern of $N$ varying masses, the system will have $N$ bands of eigenfrequencies with real wave vectors and band gaps between these bands, where wave vectors are imaginary. If we then put these lattices around another lattice with some eigenfrequencies in band gap of outer lattices, we'll get something like "quantum well" for mechanical waves (similar to phonon confinement in heterostructures). This way we'll get evanescent waves in outer lattice.

But this model has a problem: its evanescent waves are envelope waves, i.e. in the limit of lattice constant $a\to0$ the waves won't have smooth waveform. To have a smooth waveform for evanescent waves one'd need a band gap between zero frequency and bottom-most band. But as I add smaller and smaller eigenfrequencies in the first band for inner lattice (so that they were smaller then smallest eigenfrequency of outer one), I only get linear waveforms in outer lattice – they never curve enough to become exponential, i.e. the wavevector never becomes imaginary.

I begin to think that at least in purely mechanical case evanescent waves must be envelope waves, and there's no way to make a true (non-envelope) wave with imaginary wave vector.

Is this true? How can this be (dis)proved?

**EDIT**: as noted by @WetSavannaAnimalAkaRodVance, setting up a medium with different wave velocities allows one to get true evanescent waves in a faster medium via total internal reflection. This works fine, I've checked it. But total internal reflection requires at least two dimensions, so I still wonder whether true evanescent mechanical waves can be created in one dimension.

Rephrasing my question with the remark above in mind: can a one-dimensional purely mechanical system be created, in which evanescent waves *without oscillating structure* would exist? If not, how to prove this impossibility?

Here's what I mean by wave with oscillating structure – it is **not** what I'm looking for:

## Best Answer

I believe there is an acoustic analogy to evanescent waves, therefore a spring and mass lattice model must approximate evanescent waves in the same way that a finite difference model of the continuous medium example below must simulate evanescent waves.

In homogeneous mediums scalar acoustic fields fulfil the Helmholtz equation:

$$\left(\nabla^2 + \frac{\omega^2}{c_a^2}\right) \psi = 0$$

where $\omega$ is the waves angular frequency, $c_a = \sqrt{\frac{K}{\rho}}$ the speed of sound in the medium, $\rho$ the medium's mass density and $K$ its bulk modulus.

Therefore, if you have a plane wave travelling in a medium of relatively low speed and it is incident at a glancing angle on a plane interface with a medium of relatively high speed and if the angle of incidence $\theta$ is greater than the critical angle $\theta_c$ where:

$$\sin\theta_c = \frac{c_{a,1}}{c_{a,2}} = \sqrt{\frac{K_1\,\rho_2}{K_2\,\rho_1}}$$

where $c_{a,1}$ is the speed of sound in the first (slower) medium and $c_{a,2}$ that of the second (faster) medium, then there will be

total internal reflexion. Therefore, there will be an evanescent wave tunnelling into the second (faster) medium, with a Goos-Hänchen phase shift. Scalar wave theory (i.e. waves fulfilling the Helmholtz equation) is all that is needed to account for this phenomenon in optics, so there is a precise mathematical analogy between this acoustic example and my answer here where I derive the evanescent wave and the Goos-Hänchen phase shift.So now, if you have a discretised model of such a system:

i.e.two "mediums" made of a spring and mass lattices with a plane interface between them, you can have total internal reflexion and the continuum, Helmholtz-equation-fulfilling waves as long as $\lambda = 2\pi c_a/\omega$ is much longer than the lattice spacing in both mediums. Therefore, you must get approximately evanescent waves on the high velocity medium side.It is likely even possible for discretised exactly exponentially decaying waves to be derived mathematically in the lattice without thinking of them as approximation to the continuum Helmholtz equation modes: the lattice will support discretised sine waves. You could probably derive the theory with $z$-Transformed discretised propagation equations instead of Laplace / Fourier transformed Helmholtz equations.

I have added the acoustics tag to your question because there is definitely an acoustic element to this question. Hopefully an acoustics expert can confirm the above and describe mechanical evanescent waves more directly.

## Further answer after new comments

This is a very good point. I agree with you that you need two or more dimensions for total internal reflexion and this is a fundamental limitation because TIR is a glancing phenomenon - the result of a component of the wavevector projected onto an interface such that the wavevector component $k_x^\prime$ is higher than the wavevector magnitude $k^\prime$ in the new medium so that the new axial component $k_z^\prime$, which fulfils the Fourier transformed Helmholtz equation ${k_z^\prime}^2 + {k_x^\prime}^2 = k^2$ so that ${k_z^\prime}^2 < 0$.

If you are interested in one dimension, then I believe that truly there are no truly one dimensional evanescent plane waves.

For general background, you may wish to take a look at the following document:

S.W. Rienstra & A. Hirschberg (Eindhoven University of Technology), "An Introduction to Acoustics", 14 December 2013

They deal with evanescent waves in §3.3 and one dimensional systems in §4. An interesting piece of trivia is that the acoustic coupling between the loudspeaker and the listener's ear in a Sony Walkman (the authors are perhaps letting slip their age here - presumably this comment includes things like modern earpieces for iPods and the like) is through near-field, evanescent wave transmission. This of course has the advantage of being not bothersome to other nearby people because the radiative component of the acoustic field is minimised. Evanescent waves come up again in §7.2, where they look at cut-off modes in ducts. However, all these examples are two dimensional: a circular duct supports cylindrical Bessel modes whose pressure varies like $J_n(U_{n,m} r/r_0) \cos(n\,\phi+\delta)\,e{i\,k_z\,z}$ where $r$ is the radial co-ordinate, $\phi$ the azimuthal co-ordinate, $z$ the axial co-ordinate, $r_0$ the duct radius and $U_{n,m}$ the $m^{th}$ zero of the first derivative of $n^{th}$ order first kind Bessel function $J_n$. The transverse component of the wavevector has magnitude $U_{n,m}/r_0$ and the substitution of the solution into Helmholtz's equation yields:

$$k_z^2 + \frac{U_{n,m}^2}{r_0^2} = k^2$$

thus leading to "cut-off" modes (i.e. evanescent ones with imaginary axial wavenumber) when $U_{n,m} > k\,r_0$.

In one dimension, something that might be construed as an evanescent wave is the transitional acoustic field in a tuned

acoustic grating. In §4.4.1 of the Rienstra &Hirschberg reference above, they derive the transmission $T$ and reflection $R$ coefficients across a plane interface and they are precisely analogous to these electromagnetic case for transverse electromagnetic waves on two-conductor, translationally invariant transmission-lines, namely:$$R = \frac{Z_1-Z_2}{Z_1+Z_2};\quad T = \frac{2\,Z_2}{Z_1+Z_2}$$

where $Z_1, Z_2$ are the acoustic impedances of the two mediums in question. Thus if we assemble a stack of alternating, uniform media, we can build lossless acoustic reflectors, and the wave penetrating into the reflector dwindles exponentially with depth as the grating losslessly converts the forward running into the reflected field. If:

$$\psi_j=\left(\begin{array}{c}a_j\\b_j\end{array}\right)$$

represents the field at the $j^{th}$ interface in the stack, then the transfer matrix across the interface from medium 1 into medium 2 is:

$$K_j=\frac{1}{1+R}\left(\begin{array}{cc}-R&1\\1&-R\end{array}\right)$$

so that the transfer matrix through a layer of thickness $t_1$ of medium 1 followed by a layer of thickness $t_2$ of medium 2 is:

$$P=\frac{1}{1-R^2}\left(\begin{array}{cc}e^{i\,k_1\,t_1}&0\\0&e^{-i\,k_1\,t_1}\end{array}\right)\left(\begin{array}{cc}-R&1\\1&-R\end{array}\right)\left(\begin{array}{cc}e^{i\,k_2\,t_2}&0\\0&e^{-i\,k_2\,t_2}\end{array}\right)\left(\begin{array}{cc}R&1\\1&R\end{array}\right)$$

which has eigenvalues:

$$\lambda = a\pm\sqrt{a^2-b^2}$$

where:

$$a = \cos (k_1 t_1-k_2 t_2)-R^2\,\cos (k_1 t_1+k_2 t_2));\qquad b = 1-R^2$$

By checking stationary points of this expression we can show that the eigenvalues both lie inside the closed unit circle and that there are choices for $k_1\,t_1\,k_2 \,t_2$ for any $R$ such that both eigenvalues lie

strictlywithin the unit circle. Therefore, because $\psi_{2\,j} = P^j \psi_0$, we see that this means $|\psi_j|^2$ decreases exponentially with grating cycle number $j$ and is of the order of $|\lambda_{max}|^{2\,j} |\psi_0|^2$, where $\lambda_{max}$ is the maximum magnitude eigenvalue, which we can choose to have either exactly unit magnitude, or strictly less than unit magnitude. So the grating steadily converts forwards running waves into backwards running waves as the wave penetrates deeper into the reflector. This, I believe, is a pretty good analogue for the penetration and gradual conversion of wave direction that happens beyond a totally internally reflecting optical interface.We can even chirp the grating to get essentially any frequency response of the reflexion that we like. This is what is done in one-dimensional gratings in one-moded fibre optic waveguides. Leon Paladian at the University of Sydney came up with a general method to design the band structure of the photonic crystal (grating) to achieve a given frequency response in the Mid 1990s. Check his personal page (and here) for publications in this field. I should declare a friendship with Leon, but was not involved in this particular work.