The properties of Cylindrical Internal-Surface Acoustic Waves (CISAW) propagating on the inside surface of a high purity fused quartz tubular fiber and its gravity effect are derived in this study from basic principles. The CISAW have a small group velocity and the effect of gravity on CISAW propagation is large. Therefore, one can build an array of CISAW based gravity wave detectors that are able to observe gravity wave images.

I was unable to find descriptions in scientific literature of the modes of CISAW or its gravity effect. The wavelength of the CISAW is of the same order of magnitude as the fiber inside circumference. The CISAW are similar to Rayleigh waves [1] [2] . The effect of gravity on CISAW is analyzed in this study. These fibers are designed for Cylinder Internal-Surface Acoustic Wave propagation, and not for light wave propagation.

This analysis uses the Earth’s surface as an approximate Inertial Reference Frame. A point on the surface of the Earth moves with an approximately constant linear velocity during the time it takes an Acoustic Surface Wave to propagate through the tubular fiber.

Here a Classical Continuum Mechanics model is used. An infinitesimal mass in a solid consists of an infinitesimal volume with a sufficient number of atoms to be considered as a continuum. This infinitesimal mass does not move more than an infinitesimal distance. But each infinitesimal mass bumps against the next infinitesimal mass etc. This slightly compresses the infinitesimal mass. In each collision it transfers Kinetic Energy and Momentum and when it re-expands, it also transfers Elastic Potential Energy to the next infinitesimal mass. These Energy Momentum Packets (EMP) can travel any distance. The EMP’s have mass. Therefore, in this study the displacement vector u describes the position of the EMP’s, not the position of the mass increments. The displacement is a wave. The Quantum Mechanical version of the CISAW are a type of Phonons. The displacement of the EMP’s have longitudinal and transverse components.

If one neglects the small effect of the air, there is no force on the inside surface of the fiber tube on which the CISAW propagates. Therefore, the stress in the solid produced by the longitudinal and transverse displacement components has to be zero on the inside surface of the tubular fiber. This couples the longitudinal and transverse displacement components to form Elastic Surface Wave modes.

The longitudinal displacement component wave is shifted by 90 degrees with respect to the transverse component wave. When the transverse displacement component wave is zero and the longitudinal displacement component wave is at its maximum, the total displacement is longitudinal. When the longitudinal displacement component wave is zero and the transverse displacement component wave is at its maximum, the total displacement is transverse to the inside surface. But at the next zero value of the transverse displacement component wave the longitudinal component wave is at its maximum negative value, therefore, the total displacement is moving in the reverse direction. This makes the total displacement to move in a loop along the wave. Thus, the EMP’s in SAW’s, move in loops along the wave.

Over the distance of a few meters on Earth the gravitational force is constant in the “down” direction. But only a component of the gravity force acts on an object in the direction of its motion. If the object periodically changes direction, the component of the gravity force in the direction of the object motion will also change periodically. This is how the gravity force acts on a pendulum bob. The component of the gravity force acts in the direction of the pendulum bob motion. This gravity force component has a non-zero value during the downwards swing of the pendulum bob, and it is equal to zero when the pendulum bob moves horizontally at the bottom of its swing.

As described above the EMP’s have mass and are affected by gravity. When an EMP moves longitudinally on top and bottom of the loop, it does not interact with gravity. When an EMP moves vertically in the loop, its interaction with gravity is a maximum. Therefore, the part of the gravitational force acting on the EMP’s have a similar looping wave motion as the EMP’s. Thus, SAW’s interact with gravity. The interaction of the energy momentum packets in the SAW’s is similar to the interaction of the pendulum bob with gravity. The effect of gravity on the EMP’s of CISAW’s is also similar to the effect of gravity on Ocean Surface Waves (OSW) [3] on a deep ocean.

There are interactions with the curvature of space-time. The small effect of the curvature of space time on CISAW’s is neglected here.

I think this is the first published analysis of the modes of CISAW’s. I was unable to find any analysis in literature of Surface Acoustic Waves propagating on the inside surface of a cylinder. But, described below, is research somewhat related to this study:

I remember Dr. Stephen Tehan and Stephen Wanuga at the Electronics Laboratory of the General Electric Company in Syracuse, NY about 1975 propagated Surface Acoustic Waves on the inside surface of simple tubular fused quartz fibers. They wanted to use CISAW pulses propagating in the tubular fibers as a digital memory. But I was unable to find any calculations or other documentation of this work. The CISAW’s propagated with relatively little amplitude loss.

Wen Wang et al. [4] of the Acoustic Institute of the Chinese Academy of Science developed a Surface Acoustic Wave (SAW) based gyroscope on a LiTaO5 wafer. The gyroscope consists of two SAW transducers and two resonators in a loop on a piezoelectric LiTaO₅ wafer. It uses the carioles force. The infinitesimal energy momentum packets under the transducers have inertia and tend to move in a straight line while the gyroscope moves with the Earth’s rotation. This carioles force effect on the SAW is similar to the effect of gravity acting on the SAW in the tubular fiber described in this study. There are many papers published describing SAW Devices on piezoelectric wafers such as [5] .

R. E. Bunney et al. [6] propagated Surface Acoustic Waves on the outside of stainless steel and various aluminum alloy cylinders submerged in water. These cylinders had diameters of 10 to 400 times of the acoustic wavelengths.

Denos C. Gazis [7] [8] describes Elastic Waves in the material of hollow cylinders. He calculates the modes of Longitudinal waves, Equivoluminal Lame Type elastic waves, and Torsional elastic waves. All these elastic modes are bulk waves propagating in a solid material. In longitudinal and shear elastic waves where the displacements are horizontal or vertical and therefore, the EMP’s don’t interact with gravity in the Newtonian model. None of these bulk waves exhibit looping motion of the EMP’s. Only Surface Acoustic Waves, Rayleigh waves, exhibit looping motion of the EMP’s that couple to gravity. The EMP’s circular motion in Torsional waves might interact with gravity. However, since the Torsional waves are also on an outside surface, they are not suitable for many applications.

The gravity wave detectors based on CISAW interferometers can have a large effect on astronomical observations. The effect of gravity on CISAW’s is much larger than on light waves used in the current (year 2024) gravity wave detectors of the Light Interferometer Gravity-wave Observatory (LIGO) experiment [9] - [16] . Therefore, interferometer gravity wave detectors based on CISAW’s would be much smaller than gravity wave detector interferometers based on light waves. This would facilitate the building of CISAW gravity wave detector arrays. These Gravity wave interferometer arrays would for the first time be able to observe space-time curvature mode images. Gravity waves detected to date, have frequencies of a few hundred cycles or wavelengths of about 43 km to 1500 km [9] - [16] . The detectors in these arrays would be spaced at intervals that are much less than the wavelengths of a gravity wave. These arrays would perform sub-wavelength imaging [17] with gravity waves. Sub-wavelength imaging is currently used in photolithography in nanometer-scale semiconductor device manufacturing. It is interesting what future discoveries can be made using gravity-wave imaging.

This analysis uses the Earth’s surface as an approximate Inertial Reference Frame. The propagation time of the elastic wave in the 1257 meter long fiber was approximately 0.430321 seconds. A point on the surface of the Earth moved approximately 141.132 m during this time at a latitude of 45 degrees. This represents only 31.295 micro-radians of arc or 1.79305 milli-degrees of the Earth’s rotation. Therefore, during this period it can be assumed that a point on the surface of the Earth moves with an approximate constant linear velocity. The Earth also moves approximately linearly in its orbit during this time period.

The analysis starts with the symmetrized second rank strain tensor components ${\Sigma }_{\alpha \beta }$ which have the following form:

${\Sigma }_{\alpha \beta }=\frac{1}{2}\left({\partial }_{\beta }{u}_{\alpha }+{\partial }_{\alpha }{u}_{\beta }\right)$(1)

where $u_{\alpha }\ni ℝ\left(3\right)$ is a displacement component of an infinitesimal Energy Momentum Packet

(EMP), $x_{\alpha }\ni ℝ\left(3\right)$ is a coordinate component, where $\alpha =1,2,3$ and ${\partial }_{\alpha }=\frac{\partial }{\partial x_{\alpha }}$. As described in the introduction it is the EMP’s that travel in the CISAW.

Next, to obtain the elastic properties of the fused silicon material, the components of the second rank stress tensor elements ${\sigma }_{\alpha \beta }$ are expanded to first order in the strain tensor elements ${\Sigma }_{\gamma \delta }$:

$\begin{array}{l}\text{a}){\sigma }_{\alpha \beta }=\frac{\partial {\sigma }_{\alpha \beta }}{\partial {\Sigma }_{\gamma \delta }}{\Sigma }_{\gamma \delta }+\cdots \\ \text{b}){\sigma }_{\alpha \beta }=c_{\alpha \beta \delta \gamma }{\Sigma }_{\gamma \delta }+\cdots \end{array}$(2)

where the summation over the repeated indices is implied. Here, we retain only the linear expansion terms where $c_{\alpha \beta \gamma \delta }$ is a component of the covariant fourth rank elastic constant tensor of the fused silicon material. There exists much smaller higher order non-linear elastic constant terms that contribute to the losses mentioned in the introduction. The effect of gravity is represented by the gravitational potential density φ. In the Newtonian model used here, the gravitational potential density is represented by the fused quartz material density ρ times the acceleration of gravity g times the elevation h(u) of the EMP.

$\varphi \Rightarrow \rho gh\left(u\right)$(3)

The properties of the CISAW are described by a wave equation. In order to derive this wave equation, a Newtonian Mechanics Lagrangian density $ℒ$ is hypothesized. This is the first hypothesis. The Lagrangian density $ℒ$ is the difference between the Kinetic Energy density, the Elastic and the gravitational Potential Energy densities.

$ℒ=\frac{1}{2}\rho {\stackrel{\dot{}}{u}}_{\alpha }{\stackrel{\dot{}}{u}}_{\alpha }-\frac{1}{2}{\Sigma }_{\alpha \beta }{c}_{\alpha \beta \gamma \delta }{\Sigma }_{\gamma \delta }-\rho gh\left(u_{\alpha }\right)$(4)

The gravitational potential energy density could have the form $\varphi \approx -\frac{\rho MG}{r}$ where ρ is the mass density, M is a mass, G is Newton’s gravitational constant, and r is the distance between the centers of mass of objects.

Substituting Equation (1) into Equation (4) in order to formulize the Lagrangian density in terms of the displacement components:

$\begin{array}{c}ℒ=\frac{1}{2}\rho {\stackrel{\dot{}}{u}}_{\alpha }{\stackrel{\dot{}}{u}}_{\alpha }-\frac{1}{2}{\partial }_{\beta }{u}_{\alpha }{c}_{\alpha \beta \gamma \delta }{\partial }_{\delta }{u}_{\gamma }-\frac{1}{2}{\partial }_{\alpha }{u}_{\beta }{c}_{\alpha \beta \gamma \delta }{\partial }_{\delta }{u}_{\gamma }\\ -\frac{1}{2}{\partial }_{\beta }{u}_{\alpha }{c}_{\alpha \beta \gamma \delta }{\partial }_{\gamma }{u}_{\delta }-\frac{1}{2}{\partial }_{\alpha }{u}_{\beta }{c}_{\alpha \beta \gamma \delta }{\partial }_{\gamma }{u}_{\delta }-\rho gh\left(u_{\alpha }\right)\end{array}$ (5)

At this point the Euler Lagrange [18] equation could have been used to obtain a wave equation. But, I prefer the derivation from basic principles using a variational process. This requires the forming of a constant action integral I which states that the integral of the Lagrangian density over the time period and volume considered is a constant. In this study there is nothing outside the volume and time period considered. Therefore, an integral over this volume and time period is a constant, and the variation of this integral is equal to zero.

We form an action integral $ℐ$ over both the time period and volume considered.

$\begin{array}{c}ℐ=\frac{1}{2}{\displaystyle {\int }_{{\tau }_a}^{{\tau }_b}\rho {\stackrel{\dot{}}{u}}_{\alpha }\left(x_{\mu },\xi ,\tau \right){\stackrel{\dot{}}{u}}_{\alpha }\left(x_{\mu },\xi ,\tau \right)\text{d}\tau }\\ -\frac{1}{2}{\displaystyle {\int }_{s_a}^{s_b}\{c_{\alpha \beta \gamma \delta }[{\partial }_{\beta }{u}_{\alpha }\left(x_{\mu },\xi ,\tau \right){\partial }_{\delta }{u}_{\gamma }\left(x_{\mu },\xi ,\tau \right)}\\ +{\partial }_{\alpha }{u}_{\beta }\left(x_{\mu },\xi ,\tau \right){\partial }_{\delta }{u}_{\gamma }\left(x_{\mu },\xi ,\tau \right)+{\partial }_{\beta }{u}_{\alpha }\left(x_{\mu },\xi ,\tau \right){\partial }_{\gamma }{u}_{\delta }\left(x_{\mu },\xi ,\tau \right)\\ +{\partial }_{\alpha }{u}_{\beta }\left(x_{\mu },\xi ,\tau \right){\partial }_{\gamma }{u}_{\delta }\left(x_{\mu },\xi ,\tau \right)]+2\rho gh\left(u_{\alpha }\right)\}\text{d}v\end{array}$(6)

Because the analysis here uses three-dimensional space and time separately, the action integral over time and the three spatial dimensions were performed separately. Here, dv is a volume increment in three-dimensions and $s_a$ and $s_b$ are the two surfaces at the start and end of the volume of integration respectively. Dummy variables $\tau \in ℝ\left(1\right)$ and $\xi \in ℝ\left(1\right)$ were introduced to facilitate the variation. The variation of the constant action integral is equal to zero.

$\delta ℐ=0$(7)

The variation of the action integral was performed in Appendix A. We obtained a wave equation from Equation (A16) in Appendix A:

$\begin{array}{l}\rho {\ddot{u}}_{\alpha }=\frac{1}{2}[c_{\alpha \beta \gamma \delta }{\partial }_{\beta }{\partial }_{\delta }{u}_{\gamma }+c_{\gamma \beta \alpha \delta }{\partial }_{\delta }{\partial }_{\beta }{u}_{\gamma }+c_{\beta \alpha \gamma \delta }{\partial }_{\delta }{\partial }_{\beta }{u}_{\gamma }+c_{\gamma \beta \alpha \delta }{\partial }_{\delta }{\partial }_{\gamma }{u}_{\beta }\\ +c_{\alpha \beta \gamma \delta }{\partial }_{\beta }{\partial }_{\gamma }{u}_{\delta }+c_{\delta \beta \gamma \alpha }{\partial }_{\gamma }{\partial }_{\beta }{u}_{\delta }+c_{\beta \alpha \gamma \delta }{\partial }_{\beta }{\partial }_{\gamma }{u}_{\delta }+c_{\delta \beta \gamma \alpha }{\partial }_{\gamma }{\partial }_{\delta }{u}_{\beta }]\\ +\rho g\frac{\partial h}{\partial u_{\alpha }}+\frac{1}{2}\frac{{\partial }^2\varphi }{\partial u_{\alpha }\partial u_{\beta }}{u}_{\beta }\end{array}$(8)

The change $\rho g\frac{\partial h}{\partial u_{\alpha }}$ of the gravity potential density ρgh with the displacement $u_{\alpha }$ of the EMP is the component of the gravity force in the direction of the EMP’s motion. $\frac{\partial h}{\partial u_{\alpha }}$ is a dimensionless direction cosine that will be discussed later. There is a second-order smaller Newtonian gravity effect $\frac{1}{2}\frac{{\partial }^2\varphi }{\partial u_{\alpha }\partial u_{\beta }}{u}_{\beta }$ that was not part of this derivation. However, this second-order term is much smaller than the first order term, and is here neglected. This term is the effect of gravity on longitudinal and shear elastic waves.

In an isotropic material the Wave Equation is given in terms of the Lame constants [17] . In the next step we show that the wave equation, Equation (8) derived here is equal to this reported wave equation. The elastic constant tensors are symmetric in the first two indices $c_{\alpha \beta \gamma \delta }=c_{\beta \alpha \gamma \delta }$ and in the last two indices $c_{\alpha \beta \gamma \delta }=c_{\alpha \beta \delta \gamma }$.

$\begin{array}{l}\rho {\ddot{u}}_{\alpha }=\frac{1}{2}[4c_{\alpha \beta \gamma \delta }{\partial }_{\beta }{\partial }_{\delta }{u}_{\gamma }+c_{\gamma \beta \alpha \delta }{\partial }_{\delta }{\partial }_{\beta }{u}_{\gamma }+c_{\gamma \beta \alpha \delta }{\partial }_{\delta }{\partial }_{\gamma }{u}_{\beta }+c_{\delta \beta \gamma \alpha }{\partial }_{\gamma }{\partial }_{\beta }{u}_{\delta }\\ +c_{\delta \beta \gamma \alpha }{\partial }_{\gamma }{\partial }_{\delta }{u}_{\beta }]+\rho g\frac{\partial h}{\partial u_{\alpha }}\end{array}$ (9)

Isotropic symmetry requires that we contract Equation (9) with respect to the indices, γ and δ to obtain:

$\begin{array}{l}\text{a})\rho {\ddot{u}}_{\alpha }=\frac{1}{2}\left(4c_{\alpha \beta \delta \delta }+c_{\delta \beta \alpha \delta }+c_{\delta \beta \delta \alpha }\right){\partial }_{\beta }{\partial }_{\gamma }{u}_{\gamma }+\frac{1}{2}\left(c_{\delta \beta \alpha \delta }+c_{\delta \beta \delta \alpha }\right){\partial }_{\gamma }{\partial }_{\gamma }{u}_{\beta }\\ +\rho g\frac{\partial h}{\partial u_{\alpha }}\\ \text{b})\rho {\ddot{u}}_{\alpha }=\left(\lambda +\mu \right){\delta }_{\alpha \beta }{\partial }_{\beta }{\partial }_{\gamma }{u}_{\gamma }+\mu {\delta }_{\alpha \beta }{\partial }_{\gamma }{\partial }_{\gamma }{u}_{\beta }+\rho g\frac{\partial h}{\partial u_{\alpha }}\end{array}$(10)

where the Lame constants in terms of the elastic constant tensor elements are:

$\text{a})\lambda {\delta }_{\alpha \beta }=2c_{\alpha \beta \delta \delta }\text{b})\mu {\delta }_{\alpha \beta }=\frac{1}{2}\left(c_{\delta \beta \alpha \delta }+c_{\delta \beta \delta \alpha }\right)$(11)

Equation (10b) can be written in vector form using the Lame constants λ and µ. The Lame constants for fused quartz are λ = 15.87 GPa and µ = 31.26 GPa. The density of fused quartz is ρ = 2200 kg/m³. The vector form of Equation (10b) is the wave equation for the EMP’s¹ [19] .

$\rho \ddot{u}=\left(\lambda +\mu \right)\nabla \left(\nabla ·u\right)+\mu {\nabla }^{2}u+{\hat{a}}_{\alpha }\rho g\frac{\partial h}{\partial u_{\alpha }}$(12)

where bold fonts denote vectors. Here ${\hat{a}}_{\alpha }$ is a unit vector in the direction of the EMP motion. Equation (12) can be rewritten using the vector calculus identity [19] $\nabla \times \nabla \times u=\nabla \left(\nabla ·u\right)-{\nabla }^{2}u$ for ${\nabla }^{2}u$.

$\rho \ddot{u}=\left(\lambda +\text{2}\mu \right)\nabla \left(\nabla ·u\right)-\mu \nabla \times \nabla \times u+\hat{a}\rho g\frac{\partial h}{\partial u_{\alpha }}$(13)

At this point of the calculation, an attempt can be made to separate the displacement vector u into two parts:

$u=u_{\parallel }+u_{\perp }$(14)

where

$\begin{array}{l}\text{a})\text{Longitudinal component with}\nabla \times u_{\parallel }=\text{0}\\ \text{b})\text{Transverse component with}\nabla ·u_{\perp }=\text{0}\end{array}$ (15)

Substituting Equation (14) into Equation (13), using Equation (15), and dividing the resulting equation by ρ.

${\ddot{u}}_{\parallel }+{\ddot{u}}_{\perp }=\frac{\lambda +2\mu }{\rho }\nabla \left(\nabla ·u_{\parallel }\right)-\frac{\mu }{\rho }\nabla \times \nabla \times u_{\perp }+{\hat{a}}_{\alpha }g\frac{\partial h}{\partial u_{\alpha }}$(16)

The longitudinal and shear velocities squared are:

$\begin{array}{l}\text{a})\frac{\lambda +2\mu }{\rho }:=v_{\ell }^2\text{Longitudinal velocity squared}\\ \text{b})\frac{\mu }{\rho }:=v_s^2\text{Shear velocity squared}\end{array}$(17)

The longitudinal velocity v_λ = 5969.23933 m per second and the shear velocity v_s = 3769.494782 m per second for fused quartz. Substituting Equation (17) into Equation (16).

${\ddot{u}}_{\parallel }+{\ddot{u}}_{\perp }=v_{\ell }^2\nabla \left(\nabla ·u_{\parallel }\right)-v_s^2\nabla \times \nabla \times u_{\perp }+\hat{a}g\frac{\partial h}{\partial u_{\alpha }}$(18)

Making a transformation of variables to normal mode displacement vectors:

$\text{a})u=u_{\parallel }+u_{\perp }\text{b})s=\frac{v_{\ell }^2}{v_{\ell }^2+v_s^2}{u}_{\parallel }-\frac{v_s^2}{v_{\ell }^2+v_s^2}{u}_{\perp }$(19)

Equation (19a) is a restatement of Equation (14). Here u is a wave displacement mode vector of an EMP and s is a stationary mode vector also of an EMP. Inverting Equation (19)

$\begin{array}{l}\text{a})u_{\parallel }\left(t,r\right)=s\left(r\right)+\frac{v_s^2}{v_{\ell }^2+v_s^2}u\left(t,r\right)\\ \text{b})u_{\perp }\left(t,r\right)=-s\left(r\right)+\frac{v_{\ell }^2}{v_{\ell }^2+v_s^2}u\left(t,r\right)\end{array}$(20)

By substituting Equation (20) into Equation (18) and collecting terms, one obtains equations for the wave-like mode and the stationary mode of the EMP.

$\begin{array}{l}\text{a})\ddot{u}\left(r,t\right)=\frac{v_s^2{v}_{\ell }^2}{v_{\ell }^2+v_s^2}\left\{\nabla \left[\nabla ·u\left(r,t\right)\right]-\nabla \times \nabla \times u\left(r,t\right)\right\}+{\hat{a}}_{\alpha }g\frac{\partial h}{\partial u_{\alpha }}\text{wave mode}\\ \text{b})0=v_{\ell }^2\nabla \left[\nabla ·s\left(r\right)\right]-v_s^2\nabla \times \nabla \times s\left(r\right)\text{stationary mode}\end{array}$(21)

The wave-like mode of the EMPs is similar to the orbital mode, and the stationary mode of the EMPs is similar to the center of mass mode of an orbiting system. Defining the velocity v_c:

$v_c^2\equiv \frac{v_s^2{v}_{\ell }^2}{v_s^2+v_{\ell }^2}$ (22)

where v_c = 3187.198854 m per second.

Proceeding with the calculation of the wave-like mode of Equation (21a). In Cylindrical coordinates, only the wave equation for the z direction coordinate is independent. The wave equations for the two perpendicular displacement components are coupled. The gravity force in the horizontal fiber is in the down direction. Therefore, the gravity force has no z direction component. Substituting Equations (22) and (B4) of Appendix B for the z displacement component into Equation (21a).

${\ddot{u}}_z=v_c^2\left[\frac{1}{r}\frac{\partial u_z}{\partial r}+\frac{{\partial }^2{u}_z}{\partial r^2}+\frac{1}{r^2}\frac{{\partial }^2{u}_z}{\partial {\theta }^2}+\frac{{\partial }^2{u}_z}{\partial z^2}\right]$(23)

We hypothesize that the displacement components were in the form of waves:

$u_z=U_z\left(r\right)\cos \theta \exp \left(j\omega t-jkz\right)$(24)

In the θ direction this mode has a standing wave. Substituting Equation (24) into Equation (23)

$\begin{array}{l}\text{a})-{\omega }^2{U}_z\left(r\right)=v_c^2\left[\frac{1}{r}\frac{\partial U_z\left(r\right)}{\partial r}+\frac{{\partial }^2{U}_z\left(r\right)}{\partial r^2}-\frac{1}{r^2}{U}_z\left(r\right)-k^2{U}_z\left(r\right)\right]\\ \text{b})\frac{1}{r}\frac{\partial U_z\left(r\right)}{\partial r}+\frac{{\partial }^2{U}_z\left(r\right)}{\partial r^2}-\frac{1}{r^2}{U}_z\left(r\right)+\left(\frac{{\omega }^2}{v_c^2}-k^2\right)U_z\left(r\right)=0\end{array}$(25)

Defining the radial wave vector w_z for the z displacement component.

$w_z^2\equiv k^2-\frac{{\omega }^2}{v_c^2}$ (26)

Substituting Equation (26) into Equation (25b) and dividing the resulting Equation by $w_z^2$.

$\frac{1}{w_{z}r}\frac{\partial U_z\left(w_{z}r\right)}{\partial \left(w_{z}r\right)}+\frac{{\partial }^2{U}_z\left(w_{z}r\right)}{\partial {\left(w_{z}r\right)}^2}-\left[\frac{1}{{\left(w_{z}r\right)}^2}+1\right]U_z\left(w_{z}r\right)=0$ (27)

Equation (27) is a Bessel equation with solutions of a Modified Bessel Function of the second kind of order 1, $K_1\left(w_{z}r\right)$ in this cylindrically symmetric system.

$U_z\left(w_{z}r\right)=B_z{K}_1\left(w_{z}r\right)$(28)

where B_z is an Amplitude constant.

Returning to the wave equation of the r and θ displacement components. Substituting Equations (22) and (B4) of Appendix B, into Equation (21a) to obtain a wave equation for the r displacement component.

${\ddot{u}}_r=v_c^2\left[\frac{1}{r}\frac{\partial u_r}{\partial r}+\frac{{\partial }^2{u}_r}{\partial r^2}-\frac{u_r}{r^2}-\frac{2}{r^2}\frac{\partial u_{\theta }}{\partial \theta }+\frac{1}{r^2}\frac{{\partial }^2{u}_r}{\partial {\theta }^2}+\frac{{\partial }^2{u}_r}{\partial z^2}\right]+g\frac{\partial h}{\partial u_r}$(29)

where $g\frac{\partial h}{\partial u_r}$ is the gravitational force component per unit mass in the r displacement direction. Substituting Equations (22) and (B4) of Appendix B into Equation (21a) for the θ displacement component. Similar to Equation (29) but with the gravity force component per unit mass $g\frac{\partial h}{\partial u_{\theta }}$ in the θ direction.

${\ddot{u}}_{\theta }=v_c^2\left[\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}+\frac{{\partial }^2{u}_{\theta }}{\partial r^2}-\frac{u_{\theta }}{r^2}+\frac{1}{r^2}\frac{{\partial }^2{u}_{\theta }}{\partial {\theta }^2}+\frac{{\partial }^2{u}_{\theta }}{\partial z^2}+\frac{2}{r^2}\frac{\partial u_r}{\partial \theta }\right]+g\frac{\partial h}{\partial u_{\theta }}$(30)

Similar to Equation (29) $g\frac{\partial h}{\partial u_{\theta }}$ is the gravity force component per unit mass

in the direction of the θ displacement. We hypothesize that the displacement components were in the form of waves, but the transverse displacement components have the same angular dependence as the gravitational effect on the horizontal fiber:

$\begin{array}{l}\text{a})u_r=U_r\cos \theta \exp \left(j\omega t-jkz\right)\\ \text{b})u_{\theta }=U_{\theta }\sin \theta \exp \left(j\omega t-jkz\right)\end{array}$(31)

As described in the introduction, the EMP move in loops in the CISAW and the gravity components in the direction of the motion of the EMP’s have the same wave form as the displacement components. The gravity components act on the EMP’s.

$\begin{array}{l}\text{a})g\frac{\partial h}{\partial u_r}=g{G}_r\left(r\right)\cos \theta \exp \left(j\omega t-jkz\right)\\ \text{b})g\frac{\partial h}{\partial u_{\theta }}=g{G}_{\theta }\left(r\right)\sin \theta \exp \left(j\omega t-jkz\right)\end{array}$ (32)

Therefore, the gravity functions G_r and G_θ are dimensionless parameters that have the same r functional form as the r and θ displacement components. Here U_r and U_θ. are radial coordinate dependent r and θ displacement amplitudes. Substituting Equations (31) and (32) into Equations (29) and (30).

$\begin{array}{l}\text{a})-{\omega }^2{U}_r\cos \theta =v_c^2\left[\frac{1}{r}\frac{\partial U_r}{\partial r}+\frac{{\partial }^2{U}_r}{\partial r^2}-\frac{2U_r}{r^2}-k^2{U}_r\right]\cos \theta \\ +g{G}_r\cos \theta -j{v}_c^2\frac{2}{r^2}{U}_{\theta }\cos \theta \\ \text{b})-{\omega }^2{U}_{\theta }\sin \theta =v_c^2\left[\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}+\frac{{\partial }^2{u}_{\theta }}{\partial r^2}-\frac{2U_{\theta }}{r^2}-k^2{U}_{\theta }\right]\sin \theta \\ +g{G}_{\theta }\sin \theta -j{v}_c^2\frac{2}{r^2}{U}_r\sin \theta \end{array}$ (33)

where in Equation (33a) cosθ was cancelled and in Equation (33b) sinθ was cancelled. Dividing Equations (33) by $v_c^2$ and collecting terms.

$\begin{array}{l}\text{a})0=\frac{1}{r}\frac{\partial U_r}{\partial r}+\frac{{\partial }^2{U}_r}{\partial r^2}-\frac{2U_r}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2\right)U_r+\frac{g{G}_r}{v_c^2}-j\frac{2}{r^2}{U}_{\theta }\\ \text{b})0=\frac{1}{r}\frac{\partial U_{\theta }}{\partial r}+\frac{{\partial }^2{U}_{\theta }}{\partial r^2}-\frac{2U_{\theta }}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2\right)U_{\theta }+\frac{g{G}_{\theta }}{v_c^2}-j\frac{2}{r^2}{U}_r\end{array}$(34)

Assuming one can express the functions $U_r\left(r\right)$, $U_{\theta }\left(r\right)$, $G_r\left(r\right)$, and $G_{\theta }\left(r\right)$, in as yet not defined functions $P_r\left(r\right)$ and $P_{\theta }\left(r\right)$ of the r coordinate multiplied by constants.

$\begin{array}{l}\text{a})U_r=B_r{P}_r\left(r\right)\text{b})U_{\theta }=B_{\theta }{P}_{\theta }\left(r\right)\\ \text{c})G_r\equiv \frac{B_r}{{\gamma }_r}{P}_r\left(r\right)\text{d})G_{\theta }\equiv \frac{B_{\theta }}{{\gamma }_{\theta }}{P}_{\theta }\left(r\right)\end{array}$(35)

where $P_r\left(r\right)$ and $P_{\theta }\left(r\right)$ are as of yet not specified functions of r and $B_r$ and $B_{\theta }$ are constants. Here ${\gamma }_r$ and ${\gamma }_{\theta }$ are gravity effect length associated with the displacement Amplitudes as described in the Introduction. Substituting Equation (35) into Equation (34).

$\begin{array}{l}\text{a})0=\left[\frac{1}{r}\frac{\partial P_r}{\partial r}+\frac{{\partial }^2{P}_r}{\partial r^2}-\frac{2P_r}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2+\frac{g}{{\gamma }_r{v}_c^2}\right)P_r\right]B_r-j\frac{2}{r^2}{P}_{\theta }{B}_{\theta }\\ \text{b})0=\left[\frac{1}{r}\frac{\partial P_{\theta }}{\partial r}+\frac{{\partial }^2{P}_{\theta }}{\partial r^2}-\frac{2P_{\theta }}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2+\frac{g}{{\gamma }_{\theta }{v}_c^2}\right)P_{\theta }\right]B_{\theta }-j\frac{2}{r^2}{P}_r{B}_r\end{array}$(36)

The determinate of Equation (36) must be equal to zero.

$\begin{array}{l}-j\frac{2}{r^2}{P}_r\left[\frac{1}{r}\frac{\partial P_r}{\partial r}+\frac{{\partial }^2{P}_r}{\partial r^2}-\frac{2P_r}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2+\frac{g}{{\gamma }_r{v}_c^2}\right)P_r\right]\\ +j\frac{2}{r^2}{P}_{\theta }\left[\frac{1}{r}\frac{\partial P_{\theta }}{\partial r}+\frac{{\partial }^2{P}_{\theta }}{\partial r^2}-\frac{2P_{\theta }}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2+\frac{g}{{\gamma }_{\theta }{v}_c^2}\right)P_{\theta }\right]=0\end{array}$(37)

Defining the radial wave vectors w_r and w_θ for the r and θ displacement components.

$\text{a})w_r^2=k^2-\frac{{\omega }^2}{v_c^2}-\frac{g}{{\gamma }_r{v}_c^2}\text{b})w_{\theta }^2=k^2-\frac{{\omega }^2}{v_c^2}-\frac{g}{{\gamma }_{\theta }{v}_c^2}$(38)

If we neglect the small gravitational effects, the two terms of Equation (37) become equal:

$P\left[\frac{1}{r}\frac{\partial P}{\partial r}+\frac{{\partial }^{2}P}{\partial r^2}-\frac{2P}{r^2}+\left(\frac{{\omega }^2}{v_c^2}-k^2\right)P\right]\approx 0$(39)

Neglecting the gravity effects, substituting Equation (38) into Equation (39) and dividing the resulting Equation by $w^2$.

$P\left\{\frac{1}{wr}\frac{\partial P}{\partial \left(wr\right)}\frac{{\partial }^{2}P}{\partial {\left(wr\right)}^2}-\left[\frac{2}{{\left(wr\right)}^2}+1\right]P\right\}\approx 0$ (40)

Equation (40) has two solutions for P, P = 0 or P is equal to a Modified Bessel function of the second kind of order 2.

$\text{a})P_r\approx K_2\left(w_{r}r\right)\text{b})P_{\theta }\approx K_2\left(w_{\theta }r\right)$(41)

where we took the liberty to reintroduce the small gravitational effects. Substituting Equation (41) into Equation (35) and the resulting Equations into Equation (31) we obtain the form of the three displacement components:

$\begin{array}{l}\text{a})u_r=B_r{K}_2\left(w_{r}r\right)\cos \theta \exp \left(j\omega t-jkz\right)\\ \text{b})u_{\theta }=B_{\theta }{K}_2\left(w_{\theta }r\right)\sin \theta \exp \left(j\omega t-jkz\right)\\ \text{c})u_z=B_z{K}_1\left(w_{z}r\right)\cos \theta \exp \left(j\omega t-jkz\right)\end{array}$(42)

where $w_r$ and $w_{\theta }$ are given by Equation (38) and Equation (42c) is derived from Equations (24) and (26).

We hypnotize that we neglected forces due to the air on the inner surface of the tubular fiber at r = a, where a is the core radius. The strain components in cylindrical coordinates are given in Appendix C. The stress ${\sigma }_{rr}$ on the inside surface at r = a, must be equal to zero.

$\begin{array}{l}\text{a}){\sigma }_{rr}=0\\ \text{b})\left(\lambda +\mu \right){\frac{\partial u_r}{\partial r}]}_{r=a}-\lambda \left\{\frac{u_r}{a}+\frac{1}{r}{\frac{\partial u_{\theta }}{\partial \theta }]}_{r=a}\right\}-\lambda {\frac{\partial u_z}{\partial z}]}_{r=a}=0\end{array}$(43)

Substituting Equations (42) into (43) and collecting terms with the same angular dependence.

$\begin{array}{l}{B}_r\left\{\left(\lambda +\mu \right)w_r{\frac{\text{d}{K}_2\left(w_{r}r\right)}{\text{d}\left(w_{r}r\right)}]}_{r=a}-\frac{\lambda }{a}{K}_2\left(w_{r}a\right)\right\}\cos \theta \\ -\frac{\lambda }{a}{B}_{\theta }{K}_2\left(w_{\theta }a\right)\cos \theta +j\lambda k{B}_z{K}_1\left(w_{z}a\right)\cos \theta =0\end{array}$ (44)

Substituting Equation (D1) of Appendix D into Equation (44) for the derivative of the Modified Bessel Function of the second kind.

$B_r\left[-\frac{1}{2}\left(\lambda +\mu \right)w_r\left(K_1+K_3\right)-\frac{\lambda }{a}{K}_2\right]-B_{\theta }\frac{\lambda }{a}{K}_2+j{B}_z\lambda k{K}_1=0$(45)

The shear stresses from Equations (C1d) and (C1e) of Appendix C on the inside surface must also be equal to zero at r = a.

The symmetrized shear stress component ${\sigma }_{r\theta }$ represents a force acting in the r direction on a surface that is perpendicular to the θ direction. This is the rz surface. It also represents a force in the θ direction acting on a surface perpendicular to the r direction. This is the zθ surface.

The symmetrized shear stress component ${\sigma }_{rz}$ represents a force acting in the r direction on a surface that is perpendicular to the z direction. This is the rθ surface. It also represents a force in the z direction on a surface that is perpendicular to the r direction. This is the zθ surface.

$\begin{array}{l}\text{a}){\sigma }_{r\theta }=0\text{b}){\sigma }_{rz}=0\\ \text{c}){\sigma }_{r\theta }=\frac{\mu }{2}\left({\frac{\partial u_{\theta }}{\partial r}]}_{r=a}-\frac{u_{\theta }}{a}+\frac{1}{a}\frac{\partial u_r}{\partial \theta }\right)\text{d}){\sigma }_{rz}=\frac{\mu }{2}\left(\frac{\partial u_r}{\partial z}+{\frac{\partial u_z}{\partial r}]}_{r=a}\right)\end{array}$(46)

Substituting Equations (42) into (46) and taking the angular dependences into consideration we obtain:

$\begin{array}{l}\text{a})\frac{\mu }{2}\left\{B_{\theta }{w}_{\theta }{\frac{\text{d}{K}_2\left(w_{\theta }r\right)}{\text{d}\left(w_{\theta }r\right)}]}_{r=a}-B_{\theta }\frac{1}{a}{K}_2\left(w_{\theta }a\right)-B_r\frac{1}{a}{K}_2\left(w_{r}a\right)\right\}\sin \theta =0\\ \text{b})\frac{\mu }{2}\left\{-jk{B}_r{K}_2\left(w_{r}a\right)+B_z{w}_z{\frac{\text{d}{K}_1}{\text{d}\left(w_{z}r\right)}]}_{r=a}\right\}\cos \theta =0\end{array}$(47)

Observe from Equation (47b) that the phase of the r displacement component u_r is shifted by 90 degrees with respect to the z displacement component u_z. Observe from Equation (47a) that the amplitudes of the radial displacement u_r and the angular displacement u_θ components are in phase. Therefore, the Energy Momentum packets move in loops that perform a rolling motion in the rz plane and the zθ plane. Gravity couples to these rolling motions similar to the rolling motion of a looping roller coaster car described in the introduction. The effect of gravity is to slightly distort this rolling motion.

Substituting Equation (D1) for the derivatives of the Modified Bessel Functions of the second kind in Equation (47).

$\begin{array}{l}\text{a})\frac{\mu }{2}\left[-B_{\theta }\frac{1}{2}{w}_{\theta }\left(K_1+K_3\right)-B_{\theta }\frac{1}{a}{K}_2-B_r\frac{1}{a}{K}_2\right]=0\\ \text{b})\frac{\mu }{2}\left[-jk{B}_r{K}_2-B_z\frac{1}{2}{w}_z\left(K_0+K_2\right)\right]=0\end{array}$(48)

Equations (45) and (48) are written in matrix form as follows:

$\left|\begin{array}{ccc}-\frac{1}{2}\left(\lambda +\mu \right)w_r\left(K_1+K_3\right)-\frac{\lambda }{a}{K}_2& -\frac{\lambda }{a}{K}_2& j\lambda k{K}_1\\ \frac{1}{a}{K}_2& \frac{1}{2}{w}_{\theta }\left(K_1+K_3\right)+\frac{1}{a}{K}_2& 0\\ jk{K}_2& 0& \frac{1}{2}{w}_z\left(K_0+K_2\right)\end{array}\right|=0$(49)

The determinate of the matrix of Equation (49) must be equal to zero.

$\begin{array}{l}\frac{1}{8}\left(\lambda +\mu \right)w_r{w}_{\theta }{w}_z{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)+\frac{\lambda }{4a}{w}_{\theta }{w}_z\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ +\frac{1}{4a}\left(\lambda +\mu \right)w_r{w}_z\left(K_1+K_3\right)\left(K_0+K_2\right)K_2+\frac{\lambda }{2a^2}{w}_z\left(K_0+K_2\right)K_2^2\\ -\frac{\lambda }{2a^2}{w}_z\left(K_0+K_2\right)K_2^2-\frac{1}{2}\lambda k^2{w}_{\theta }\left(K_1+K_3\right)K_1{K}_2-\lambda \frac{1}{a}{k}^2{K}_1{K}_2^2=0\end{array}$ (50)

In order to formalize Equation (50) in dimensionless form we multiply Equation (50) by the constants a³ and divide by λ.

$\begin{array}{l}\frac{1}{8}\frac{\lambda +\mu }{\lambda }{w}_r{w}_{\theta }{w}_z{a}^3{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}{w}_{\theta }{w}_z{a}^2\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ +\frac{1}{4}\frac{\lambda +\mu }{\lambda }{a}^2{w}_r{w}_z\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2}{k}^2{a}^3{w}_{\theta }\left(K_1+K_3\right)K_1{K}_2-k^2{a}^2{K}_1{K}_2^2=0\end{array}$(51)

Substituting Equations (26) and (38) for the radial wave vectors w_r, w_θ and w_z into Equation (51) and factoring out k.

$\begin{array}{l}\frac{\lambda +\mu }{8\lambda }{k}^3{a}^3\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}}\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}-\frac{g}{k^2{\gamma }_r{v}_c^2}}\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}-\frac{g}{k^2{\gamma }_{\theta }{v}_c^2}}{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}{k}^2{a}^2\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}}\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}-\frac{g}{k^2{\gamma }_{\theta }{v}_c^2}}\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ +\frac{1}{4}\frac{\lambda +\mu }{\lambda }{k}^2{a}^2\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}}\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}-\frac{g}{k^2{\gamma }_r{v}_c^2}}\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2}{k}^3{a}^3\sqrt{1-\frac{{\omega }^2}{k^2{v}_c^2}-\frac{g}{k^2{\gamma }_{\theta }{v}_c^2}}\left(K_1+K_3\right)K_1{K}_2-k^2{a}^2{K}_1{K}_2^2=0\end{array}$ (52)

Defining a dimensionless parameter

$q^2\equiv \frac{{\omega }^2}{k^2{v}_c^2}$ (53)

Substituting the parameter q² of Equation (53) into Equation (52) and factoring out 1 – q².

$\begin{array}{l}\frac{\lambda +\mu }{8\lambda }{k}^3{a}^3{\left(1-q^2\right)}^{\frac{3}{2}}\sqrt{1-\frac{g}{{\gamma }_r{k}^2{v}_c^2\left(1-q^2\right)}}\sqrt{1-\frac{g}{\left(1-q^2\right)k^2{\gamma }_{\theta }{v}_c^2}}{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}{k}^2{a}^2\left(1-q^2\right)\sqrt{1-\frac{g}{k^2{\gamma }_{\theta }{v}_c^2\left(1-q^2\right)}}\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ +\frac{1}{4}\frac{\lambda +\mu }{\lambda }{k}^2{a}^2\left(1-q^2\right)\sqrt{1-\frac{g}{k^2{\gamma }_r{v}_c^2\left(1-q^2\right)}}\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2}{k}^3{a}^3\sqrt{1-q^2}\sqrt{1-\frac{g}{k^2{\gamma }_{\theta }{v}_c^2\left(1-q^2\right)}}\left(K_1+K_3\right)K_1{K}_2-k^2{a}^2{K}_1{K}_2^2=0\end{array}$ (54)

Expanding Equation (54) to lowest order in $\frac{g}{k^2{\gamma }_r{v}_c^2\left(1-q^2\right)}$ and $\frac{g}{k^2{\gamma }_{\theta }{v}_c^2\left(1-q^2\right)}$.

$\begin{array}{l}\frac{\lambda +\mu }{8\lambda }{k}^3{a}^3{\left(1-q^2\right)}^{\frac{3}{2}}\left[1-\frac{g}{2k^2{v}_c^2\left(1-q^2\right)}\left(\frac{1}{{\gamma }_r}+\frac{1}{{\gamma }_r}\right)\right]{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}{k}^2{a}^2\left(1-q^2\right)\left[1-\frac{g}{2k^2{\gamma }_{\theta }{v}_c^2\left(1-q^2\right)}\right]\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ +\frac{1}{4}\frac{\lambda +\mu }{\lambda }{k}^2{a}^2\left(1-q^2\right)\left[1-\frac{g}{2k^2{\gamma }_r{v}_c^2\left(1-q^2\right)}\right]\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2}{k}^3{a}^3\sqrt{1-q^2}\left[1-\frac{g}{2k^2{\gamma }_{\theta }{v}_c^2\left(1-q^2\right)}\right]\left(K_1+K_3\right)K_1{K}_2-k^2{a}^2{K}_1{K}_2^2=0\end{array}$(55)

The CISAW mode can be calculated by neglecting the small gravity term in Equation (55). For this case all transverse wave vectors w_r, w_θ, and w_z are equal and the argument x of the Modified Bessel Functions of the second kind are:

$\text{a})K_{\eta }=K_{\eta }\left(x\right)\text{b})x\equiv aw\text{c})x=ka\sqrt{1-q^2}$(56)

Neglecting the small effect of gravity and substituting Equation (56) into Equation (55) to calculate the first order angular standing wave CISAW mode.

$\begin{array}{l}\text{a})\xi =0\\ \text{b})\xi =\frac{1}{8}\frac{\lambda +\mu }{\lambda }{x}^3{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}\left(\frac{\lambda +\mu }{\lambda }+1\right)x^2\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{x^3}{2\left(1-q^2\right)}\left(K_1+K_3\right)K_1{K}_2-\frac{x^2}{1-q^2}{K}_1{K}_2^2\end{array}$(57)

Equation (57) is nonlinear in x because the Modified Bessel Functions of the second kind $K_{\eta }\left(x\right)$ are functions of x. The asymptotic approximation [20] for the Modified Bessel Functions of the second kind from Equations (D3a) and (D3b) of Appendix D are used for the numeric solution of Equation (57). Chipmunk Basic software is used for the numeric calculations.

Equation (57), has one root x for each value of q. These roots x are obtained by calculating $\xi$ from Equation (57) for successive values of x at a fixed value of q until a sufficient small value of $\left|\xi \right|<{10}^{-6}$ is obtained. The value of x giving the smallest value of $\xi$ is the root x of Equation (57).

At this point in the calculation, Lord Rayleigh [1] obtained a cubic equation. His calculation of SAW’s on a flat surface involve sin’s, cos’s, and exponentials, while SAW’s propagating on a cylindrical surface involved Modified Bessel Functions of the second kind.

The highly non-linear Equation (57) does not seem to have roots x for values of q less than q = 0.88 and more than q = 0.94. There are also two ranges of roots x for q between q = 0.88 and q = 0.900016 and between q = 0.900016 and q = 0.94. At the boundary of the two ranges at q = 0.89992, x has a value of x = 1.01283 and at the value of q = 0.900016, x has a value of x = 0.025405.

The dispersion relation can be calculated from the roots x and the values q of Equation (57). Figure 1 shows a plot of the longitudinal wave vector k as a function of frequency for values of q between q = 0.900016 and q = 0.94. This is the dispersion relation of the CISAW.

The group velocity v_g is the change of the radial velocity ω with the longitudinal wave vector k.

$v_g=\frac{\text{d}\omega }{\text{d}k}$(58)

The group velocity v_g as a function of the frequency is given in Figure 2 .

Substituting Equation (56c) for x into Equation (55)

$\begin{array}{l}\frac{\lambda +\mu }{8\lambda }{x}^3\left(1-\frac{g{a}^2}{\gamma v_c^2{x}^2}\right){\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}\left(\frac{\lambda +\mu }{\lambda }+1\right)x^2\left(1-\frac{g{a}^2}{2\gamma v_c^2{x}^2}\right)\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2\left(1-q^2\right)}{x}^3\left(1-\frac{g{a}^2}{2\gamma v_c^2{x}^2}\right)\left(K_1+K_3\right)K_1{K}_2-k^2{a}^2{K}_1{K}_2^2=0\end{array}$(59)

where for simplicity we assumed that ${\gamma }_{\theta }={\gamma }_r$. In order to calculate the change ∆k of the wave vector k we hypothesize that the parameter x consists of x and a small change ∆x. That is, $x\to x+\Delta x$ in Equation (59). By multiplying out Equation (59) to lowest order.

$\begin{array}{l}\frac{\lambda +\mu }{8\lambda }\left(x^3+3x^2\Delta x-\frac{g{a}^2}{\gamma v_c^2}x\right){\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{\lambda +\mu }{4\lambda }w\left(K_1+K_3\right)\left(K_0+K_2\right){\left[\frac{\text{d}{K}_1}{\text{d}\left(wr\right)}+\frac{\text{d}{K}_3}{\text{d}\left(wr\right)}\right]}_{r=a}\Delta x+\cdots \\ +\frac{1}{4}\left(\frac{\lambda +\mu }{\lambda }+1\right)\left(x^2+2x\Delta x-\frac{g{a}^2}{2\gamma v_c^2}\right)\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2\left(1-q^2\right)}\left(x^3+3x^2\Delta x-\frac{g{a}^2}{2\gamma v_c^2}x\right)\left(K_1+K_3\right)K_1{K}_2-k^2{a}^2{K}_1{K}_2^2=0\end{array}$(60)

Substituting Equation (57) into Equation (60) and for simplicity neglecting the expansions of the Modified Bessel Functions of the second kind in ∆x one obtains:

$\begin{array}{l}\frac{\lambda +\mu }{8\lambda }\left(3x^2\Delta x-\frac{g{a}^2}{\gamma v_c^2}x\right){\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}\left(\frac{\lambda +\mu }{\lambda }+1\right)\left(2x\Delta x-\frac{g{a}^2}{2\gamma v_c^2}\right)\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{1}{2\left(1-q^2\right)}\left(3x^2\Delta x-\frac{g{a}^2}{2\gamma v_c^2}x\right)\left(K_1+K_3\right)K_1{K}_2=0\end{array}$(61)

One can solve Equation (61) for ∆x

$\begin{array}{l}\Delta x=\frac{g{a}^2}{2\gamma v_c^2}[\frac{\lambda +\mu }{4\lambda }x{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}\left(\frac{\lambda +\mu }{\lambda }+1\right)\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{x}{2\left(1-q^2\right)}\left(K_1+K_3\right)K_1{K}_2][\frac{3\left(\lambda +\mu \right)}{8\lambda }{x}^2{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ {+\frac{1}{2}\left(\frac{\lambda +\mu }{\lambda }+1\right)x\left(K_1+K_3\right)\left(K_0+K_2\right)K_2-\frac{3x^2}{2\left(1-q^2\right)}\left(K_1+K_3\right)K_1{K}_2]}^{-1}\end{array}$ (62)

Similar to Equation (56c)

$\Delta x\approx \Delta ka\sqrt{1-q^2}$(63)

Substituting Equation (63) into Equation (62) and solving for ∆k.

$\begin{array}{l}\Delta k=\frac{ga}{2\gamma v_c^2\sqrt{1-q^2}}[\frac{\lambda +\mu }{4\lambda }x{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ +\frac{1}{4}\left(\frac{\lambda +\mu }{\lambda }+1\right)\left(K_1+K_3\right)\left(K_0+K_2\right)K_2\\ -\frac{x}{2\left(1-q^2\right)}\left(K_1+K_3\right)K_1{K}_2][\frac{3\left(\lambda +\mu \right)}{8\lambda }{x}^2{\left(K_1+K_3\right)}^2\left(K_0+K_2\right)\\ {+\frac{1}{2}\left(\frac{\lambda +\mu }{\lambda }+1\right)x\left(K_1+K_3\right)\left(K_0+K_2\right)K_2-\frac{3x^2}{2\left(1-q^2\right)}\left(K_1+K_3\right)K_1{K}_2]}^{-1}\end{array}$ (64)