Waves – EthnoPhysics

Outline

To summarize, so far EthnoPhysics has considered a generic particle P as a repetitive chain of events written as $\Psi^{\mathsf{P}} = \left( \mathsf{\Omega}_{1}, \, \mathsf{\Omega}_{2}, \, \mathsf{\Omega}_{3} \, \ldots \, \right) \! .$ Each repeated cycle $\mathsf{\Omega}$ is a bundle of sensations. And these cycles must be similar to each other so that P can be recognized.

The replication of sensation may be presented in a spatial display, like a recurring pattern in a carpet or textile. But often $\mathsf{\Omega}$ is a bundle of sensation that is repeated as time goes by, like the chorus in a song. Or perhaps $\mathsf{\Omega}$ is more like telephone poles along a road. The pattern is fixed if viewed while standing, but seems to fluctuate when driving by.

So whether a sensory pattern is local or extended, or whether it is steady or changing, can depend on a point-of-view. An experience may be objectified as point-like or wave-like according to the perspective of an observer. This ambivalence is called wave-particle duality and it has been confusing and contentious during the development of physics.

However for EthnoPhysics there is no quandary. Scientific facts and theories are founded on sensation, and whether we call these perceptions particles or waves is just a question of convenience. If feelings are more-or-less steady and local, then we discuss particles. But if experiences are fluctuating and extended, then we use words like wave, wavenumber and wavelength. In between, we might speak of excited states. Anyway, on this page we talk about waves.

The first thing to say about waves is that they move, and there is usually some sort of rhythmic regularity in the motion. So waves undulate, oscillate, fluctuate and vibrate. Basically a wave is anything that repetitively shifts to and fro. But this is imprecise. So next we are going to develop a mathematical quark-based description of waves.

The Quark Flux

Consider a particle P described by some repetitive chain of events written as $\Psi^{\mathsf{P}} \! = \! ( \mathsf{\Omega}_{1}, \, \mathsf{\Omega}_{2}, \, \mathsf{\Omega}_{3} \ldots )$ where each repeated cycle is a bundle of quarks like

$\mathsf{\Omega}^{\mathsf{P}} = \left( \mathsf{q}_{1}, \, \mathsf{q}_{2}, \, \mathsf{q}_{3} \; \ldots \; \mathsf{q}_{i} \; \ldots \; \mathsf{q}_{N} \right)$

Let each quark be described by its phase $\delta_{\theta}$ and its radius vector $\overline{\rho} \, .$ Use the phase to sort quarks into a pair of sets, $\mathsf{P}_{\! \mdsmwhtcircle}$ and $\mathsf{P}_{\! \mdsmblkcircle} \hspace{0.5pt} ,$ so that all quarks of the same phase are in the same set. Then $\mathsf{P}_{\! \mdsmwhtcircle}$ and $\mathsf{P}_{\! \mdsmblkcircle}$ are called phase components of P, and they are out of phase with each other. We write

$\mathsf{\Omega}^{\mathsf{P}} = \left\{ \mathsf{P}_{\! \mdsmblkcircle} \, , \, \mathsf{P}_{\! \mdsmwhtcircle} \rule{0pt}{9px} \right\}$

and

$\delta_{\theta} ( \mathsf{P}_{\! \mdsmblkcircle} ) = - \delta_{\theta} ( \mathsf{P}_{\! \mdsmwhtcircle} ) = \pm 1$

A difference in the radius-vectors of these phase-components can characterize any particle. But it is especially relevant for waves because traditionally, the phase distinguishes between daytime and nighttime conditions. So a customary interpretation of any phase-related difference describes daily changes. And there is one absolutely outstanding example of a regular variation in radii that has been known globally over all human history; the daily rising and falling of sea levels. Little waves may splash about chaotically at every shoreline. But gross tidal motion is incessant and world-wide. Waves are historically exemplified by the flux and flow of oceanic tides.

For modern physics, phase is understood much more generally than a daily variation. But oceanic oscillations are still the semantic undercurrent for identifying a flux with a phase-difference. Thus we define the quark flux vector as

$\Delta \overline{\rho}^{\, \mathsf{P}} \equiv \delta_{\theta} ( \mathsf{P}_{\! \mdsmblkcircle} ) \left( \; \overline{\rho} ( \mathsf{P}_{\! \mdsmblkcircle} ) - \overline{\rho} ( \mathsf{P}_{\! \mdsmwhtcircle} ) \, \rule{0px}{12px} \right)$

This flux is specified relative to $\mathsf{P}_{\! \mdsmblkcircle} \hspace{0.5pt} .$ Recall that the radius-vector for a composite quark is just the sum of the radius-vectors of its components. So

$\overline{\rho} ( \mathsf{P}_{\! \mdsmblkcircle} ) = \hspace{-8px} \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmblkcircle}} \overline{\rho}^{\, \mathsf{q}} }$

Thus the quark-flux vector can also be stated in terms of individual quarks as

$\begin{align*} \Delta \overline{\rho}^{\, \mathsf{P}} \, &= \delta_{\theta} ( \mathsf{P}_{\! \mdsmblkcircle} ) \left( \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmblkcircle}} \overline{\rho}^{\, \mathsf{q}} } \right) - \delta_{\theta} ( \mathsf{P}_{\! \mdsmblkcircle} ) \left( \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmwhtcircle}} \overline{\rho}^{\, \mathsf{q}} } \right) \\ &= \delta_{\theta} ( \mathsf{P}_{\! \mdsmblkcircle} )\left( \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmblkcircle}} \overline{\rho}^{\, \mathsf{q}} } \right) + \delta_{\theta} ( \mathsf{P}_{\! \mdsmwhtcircle} ) \left( \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmwhtcircle}} \overline{\rho}^{\, \mathsf{q}} } \right) \rule{0px}{30px} \\ &= \left( \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmblkcircle}} \delta_{\theta}^{\hspace{0.5pt} \mathsf{q}} \, \overline{\rho}^{\, \mathsf{q}} } \right) + \left( \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P}_{\mdsmwhtcircle}} \delta_{\theta}^{\hspace{0.5pt} \mathsf{q}} \, \overline{\rho}^{\, \mathsf{q}} } \right) \rule{0px}{30px} \\ &= \displaystyle{\sum_{\; \; \mathsf{q} \, \in \, \mathsf{P} } \delta_{\theta}^{\hspace{0.5pt} \mathsf{q}} \, \overline{\rho}^{\, \mathsf{q}} } \rule{0px}{30px} \end{align*}$

Now recall that any particle $\mathsf{P}$ and its conjugate twin $\overline{\mathsf{P}}$ have symmetrically opposed radius-vectors such that $\overline{\rho} ( \mathsf{P} ) \! = \! - \overline{\rho} ( \mathsf{\overline{P}} ) .$ Then by substitution their quark-flux vectors are related as

$\Delta \overline{\rho} \hspace{1px} ( \mathsf{P} ) = - \Delta \overline{\rho} \hspace{1px} ( \mathsf{\overline{P}} )$

We also define an areal flux density through some surface area $A$ from the ratio

$\overline{\varkappa} \equiv \Delta \overline{\rho} / A$

Then the flux-densities of conjugate-twins are related by $\overline{\varkappa} ( \mathsf{P} ) \! = \! - \overline{\varkappa} ( \mathsf{\overline{P}} )$ as well. If P is described by the simple particle model where a pair of concentric spheres designate its inside and outside surfaces, then the areal flux density through these surfaces is noted by $\overline{\varkappa}_{in}$ or $\overline{\varkappa}_{out} \hspace{0.5pt} .$ In the following articles we invariably use this model.

Waves In Vacuo

Let P be any sort of particle, perhaps a photon, or maybe something more material. We define the wavevector in vacuo of P as

$\overline{\kappa}_{\mathsf{o}} \equiv \, \overline{\varkappa}_{in} - \overline{\varkappa}_{out}$

Recall that any particle $\mathsf{P}$ and its conjugate twin $\overline{\mathsf{P}}$ have symmetrically opposed flux-densities such that $\overline{\varkappa} ( \mathsf{P} ) \! = \! - \overline{\varkappa} ( \mathsf{\overline{P}} ) \, .$ So if quarks are swapped with anti-quarks, without altering their phases, then by substitution

$\overline{\kappa}_{\mathsf{o}} ( \mathsf{P} ) = - \overline{\kappa}_{\mathsf{o}} ( \overline{\mathsf{P}} )$

Now consider the norm of this vector. It is called the wavenumber in vacuo and has units of (m^-1). We write

$\kappa_{\mathsf{o}} \equiv \left\| \, \overline{\varkappa}_{in} - \overline{\varkappa}_{out} \right\|$

The wavenumber in vacuo may be zero if P is in its ground state. But norms are never negative so $\kappa_{\mathsf{o}} \! \geq \! 0 .$ Thus conjugate-twins are related as

$\kappa_{\mathsf{o}} ( \mathsf{P} ) = \kappa_{\mathsf{o}} ( \mathsf{\overline{P}} )$

Waves ebb and flow. Their movement is described by a number called the quantity-of-motion in vacuo defined by

$p_{\mathsf{o}} \equiv \dfrac{h}{2\pi} \kappa_{\mathsf{o}}$

This number does not depend on the frame of reference. So it is intrinsic. But it does rely on assumptions that are built into the simple particle model. If P is in its ground state then the quantity-of-motion is nil. But the wavenumber is never negative so $p_{\mathsf{o}} \! \geq \! 0 \hspace{1px} .$ Accordingly conjugate-twins are related as

$p_{\mathsf{o}} ( \mathsf{P} ) = p_{\mathsf{o}} ( \mathsf{\overline{P}} )$

The wavenumber in vacuo can also used to obtain another quantity $\lambda_{\mathsf{o}}$ called the wavelength in vacuo. This length is essential for the analysis and reporting of experimental observations. It is defined by

$\lambda_{\mathsf{o}} \equiv \begin{cases} \hspace{15 px} 0 \; & \mathsf{\text{if}} \; \kappa_{\mathsf{o}} =0 \\ \; 2\pi / \kappa_{\mathsf{o}} \; & \sf{\text{if}} \; \kappa_{\mathsf{o}} \ne 0 \end{cases}$

We note that if $\kappa_{\mathsf{o}} \! \! = 0$ then $\lambda_{\mathsf{o}}$ takes a discontinuous jump. And it leaps to zero, so perhaps we could say it collapses. But keep in mind that this fracture is just one of many. The wavelength is discontinuous for all values because it is defined from quark coefficients and quark coefficients are always integers.

Wavenumbers are never negative, so the wavelengths of conjugate-twins are related as

$\lambda_{\mathsf{o}} ( \mathsf{P} ) = \lambda_{\mathsf{o}} ( \mathsf{\overline{P}} )$

Use of the term in-vacuo is historical. It suggests that P is situated in a vacuum. But for EthnoPhysics, empty space is not defined. So here we take it to mean that P is undisturbed by any interaction with other particles. We use the suffix in-vacuo to indicate that P is presumed to be perfectly isolated.

P may be isolated, but we still consider that boundary conditions might constrain its environment. Indeed we employ the surface areas $A_{in}$ and $A_{out}$ from the simple particle model to write the wavevector-in-vacuo as

$\overline{\kappa}_{\mathsf{o}} = \dfrac{\Delta \overline{\rho}}{A_{in}} - \dfrac{\Delta \overline{\rho}}{A_{out}}$

Then expressing surface-areas in terms of their spherical radii $\mathcal{R}_{in}$ and $\mathcal{R}_{out}$ gives the constrained wavevector in vacuo as

$\displaystyle \overline{\kappa}_{\mathsf{o}} = \dfrac{1}{4\pi} \left( \dfrac{1}{\mathcal{R}_{in}^{2}} - \dfrac{1}{\mathcal{R}_{out}^{2}} \right) \Delta \overline{\rho}$

Next the definitions for $\mathcal{R}_{in}$ and $\mathcal{R}_{out}$ are substituted into this expression to state the wavevector in terms of the coefficients of down quarks as

$\displaystyle \overline{\kappa}_{\mathsf{o}} = \frac{2\pi k_{\mathsf{F}}}{hc} \left[ \left( \frac{8}{ \Delta n^{\mathsf{D}} } \right)^{\! 2} - \left( \frac{8}{ N^{\mathsf{D}} } \right)^{\! 2} \right] \Delta \overline{\rho}$

But for a perfectly free particle, $\left| \Delta n^{\mathsf{D}} \right| = 8$ and $N^{\mathsf{D}} \! \to \infty$ . So a free wavevector in vacuo can be put plainly as

$\displaystyle \overline{\kappa}_{\mathsf{o}} = \dfrac{2\pi k_{\mathsf{F}}}{hc} \, \Delta \overline{\rho}$

Now recall that quarks are conserved. And the radius vector $\overline{\rho}$ is defined from sums of quarks. So if some free particles interact like $\mathbb{X} + \mathbb{Y} \leftrightarrow \mathbb{Z} \, ,$ then their radii combine as $\overline{\rho}^{\mathbb{X}} \! + \overline{\rho}^{\mathbb{Y}} \! = \overline{\rho}^{\mathbb{Z}} .$ Also let the frame of reference be fixed and firm. A steady frame assumption implies that phases do not change during the description of some interaction, so we can treat the phase of each quark as a constant. Then, since the foregoing expression for $\overline{\kappa}_{\mathsf{o}}$ is a sum over quarks, we obtain

$\overline{\kappa}^{\mathbb{X}}_{\mathsf{o}} + \overline{\kappa}^{\mathbb{Y}}_{\mathsf{o}} = \overline{\kappa}^{\mathbb{Z}}_{\mathsf{o}}$

Photons in Vacuo

The foregoing equations assess the wave-like character of P, a generic particle that might be material. But next we specifically look at photons with their anti-symmetric phase arrangements. So let P be a photon, noted by $\boldsymbol{\gamma} ,$ that is described by a repetitive chain of events written as $\Psi\! \left( \boldsymbol{\gamma} \right) \! = \! ( \mathsf{\Omega}_{1}, \, \mathsf{\Omega}_{2}, \, \mathsf{\Omega}_{3} \ldots )$ where each repeated cycle is a finite bundle of quarks noted as

$\mathsf{\Omega}\! \left( \boldsymbol{\gamma} \right) = \left( \mathsf{q}_{1}, \, \mathsf{q}_{2}, \, \mathsf{q}_{3} \; \ldots \; \mathsf{q}_{i} \; \ldots \; \mathsf{q}_{N} \right)$

For $\boldsymbol{\gamma}$ , the constrained wavevector in vacuo which was discussed above, is written as

$\displaystyle \overline{\kappa}_{\mathsf{o}} \! \left( \boldsymbol{\gamma} \right) = \dfrac{1}{4\pi} \left( \dfrac{1}{\mathcal{R}_{in}^{2}} - \dfrac{1}{\mathcal{R}_{out}^{2}} \right) \Delta \overline{\rho}$

where $\Delta \overline{\rho}$ is the quark flux vector. Expanding this vector in terms of individual quarks obtains

Here radius vectors are marked by $\overline{\rho} .$ And $\delta_{\theta}$ notes the phase of each quark $\mathsf{q}$ in $\boldsymbol{\gamma} .$ We use this phase to sort quarks into a pair of anti-symmetric phase components $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ such that

$\mathsf{\Omega} \left( \boldsymbol{\gamma} \right) = \left\{ \mathcal{A}_{\mdsmwhtcircle} \, , \, \mathcal{A}_{\mdsmblkcircle} \rule{0px}{10px} \right\}$

and

$\delta_{\theta} \! \left( \mathcal{A}_{\mdsmblkcircle} \right) = - \, \delta_{\theta} \! \left( \mathcal{A}_{\mdsmwhtcircle} \right)$

Then the sum over quarks is expanded as

$\begin{align*} \displaystyle \sum_{i=1} ^{N} \delta_{\theta}^{\, i} \; \overline{\rho}^{ \, i} &= \delta_{\theta}^{\, \mathcal{A}_{\mdsmwhtcircle} } \hspace{-6px} \sum_{\mathsf{q} \, \in \, \mathcal{A}_{\mdsmwhtcircle} } \hspace{-4px} \overline{\rho}^{\, \mathsf{q}} \, + \, \delta_{\theta}^{\, \mathcal{A}_{\mdsmblkcircle} } \hspace{-6px} \sum_{\mathsf{q} \, \in \, \mathcal{A}_{\mdsmblkcircle} } \overline{\rho}^{\, \mathsf{q}} \\ &= \rule{0px}{18px} \delta_{\theta}^{\, \mathcal{A}_{\mdsmwhtcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle} } + \; \delta_{\theta}^{\, \mathcal{A}_{\mdsmblkcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmblkcircle} } \end{align*}$

But $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ are out of phase. So $\delta_{\theta} \! \left( \mathcal{A}_{\mdsmblkcircle} \right) = - \, \delta_{\theta} \! \left( \mathcal{A}_{\mdsmwhtcircle} \right)$ and therefore

$\displaystyle \sum_{i=1} ^{N} \delta_{\theta}^{\, i} \; \overline{\rho}^{ \, i} = \; \delta_{\theta} (\mathcal{A}_{\mdsmwhtcircle}) \left( \, \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle}} - \, \overline{\rho}^{\, \mathcal{A}_{\mdsmblkcircle}} \rule{0px}{17px} \right)$

Also the radius-vectors of any particle and its conjugate twin are symmetrically opposed. So $\overline{\rho} \! \left( \mathcal{A}_{\mdsmblkcircle} \right) \! =- \, \overline{\rho} \hspace{0.5pt} ( \overline{\mathcal{A}_{\mdsmblkcircle}} )$ and therefore

$\displaystyle \sum_{i=1} ^{N} \delta_{\theta}^{\, i} \; \overline{\rho}^{ \, i} = \; \delta_{\theta} (\mathcal{A}_{\mdsmwhtcircle}) \left( \, \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle}} + \, \overline{\rho}^{\, \overline{\mathcal{A}_{\mdsmblkcircle}} } \rule{0px}{17px} \right)$

But the phase-components of a photon are anti-symmetric, so $\mathcal{A}_{\mdsmwhtcircle} \! = \overline{\mathcal{A}_{\mdsmblkcircle}}$ and $\overline{\mathcal{A}_{\mdsmblkcircle}}$ can be eliminated to obtain

$\displaystyle \sum_{i=1} ^{N} \delta_{\theta}^{\, i} \; \overline{\rho}^{ \, i} = \delta_{\theta} \! \left(\mathcal{A}_{\mdsmwhtcircle} \right) \; 2 \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle}}$

Next this expression is put back into the equation for the constrained wavevector yielding

$\overline{\kappa}_{\mathsf{o}} \! \left( \boldsymbol{\gamma} \right) = \delta_{\theta} (\mathcal{A}_{\mdsmwhtcircle}) \left( \dfrac{1}{\mathcal{R}_{in}^{2}} - \dfrac{1}{\mathcal{R}_{out}^{2}} \right) \dfrac{\; \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle}}}{2\pi}$

A wavenumber is the norm of a wavevector. So $\kappa_{\mathsf{o}} \! \equiv \left\| \, \overline{\kappa}_{\mathsf{o}} \, \right\|$ gives the constrained wavenumber in vacuo as

$\kappa_{\mathsf{o}} \! \left( \boldsymbol{\gamma} \right) = \dfrac{1}{2\pi} \left( \dfrac{1}{\mathcal{R}_{in}^{2}} - \dfrac{1}{\mathcal{R}_{out}^{2}} \right) \left\| \, \overline{\rho}^{\, \mathcal{A}} \right\|$

The subscript on $\mathcal{A}$ can be dropped because both phase-components have the same norm.

Waves in Situ

The foregoing talk has all been about waves in-vacuo. So far we have just assumed that waves are vibrating to and fro in perfect isolation from other particles. This is a big approximation.

Scientific descriptions always require a frame of reference. And the frame itself may absorb and emit particles that affect wave motion. Perhaps by gravitation? Possibly via other fields? Or maybe just by direct contact? These departures from the idealized in-vacuo case can be very significant. So next we expand the discussion of reference-frames so that we can theoretically assess waves as they are found in-situ.

Let the reference frame F, for some particle P, be composed from two parts. Consider a large component noted by $\mathtt{F}$ that includes the bulk of F. And a smaller part $\mathtt{f} ,$ formed from just the surroundings of P. We use $N$ to note the total number of quarks in a particle. This arrangement is expressed as

$\mathsf{F} = \left\{ \mathtt{F} , \, \mathtt{f} \rule{0px}{10px} \right\}$

$N \hspace{-0.5pt} (\mathsf{F} ) \simeq N \hspace{-0.5pt} ( \mathtt{F} )$

and

$N \hspace{-0.5pt} ( \mathtt{f} ) < N \hspace{-0.5pt} ( \hspace{0.5pt} \mathtt{F} \hspace{0.5pt} )$

Let the large part $\mathtt{F}$ account for any interactions with P involving gravitons. So the size of $\mathtt{F}$ is like the Earth. Or perhaps much larger if the Sun is explicitly considered. We call $\mathtt{F}$ the central body in a description of P. And we say that the quarks in $\mathtt{F}$ establish an external environment for P.

Let the small part $\mathtt{f}$ account for any interactions with P that do not involve gravitons. So the size of $\mathtt{f}$ is like our classrooms and laboratories. Or perhaps much smaller.¹Perhaps very much smaller because experimental physicists often arrange for $\mathtt{f}$ to be extremely unusual as a way of separating variables and untangling causal relationships. Such extraordinary conditions often need to be tightly enclosed. We call $\mathtt{f}$ the surroundings in a description of P. And we say that the quarks in $\mathtt{f}$ establish a local environment for P.

Gravity-Free Environments

Next we develop some specific vocabulary for different environments. If any interactions between P and a graviton are negligible, then we say that $\mathtt{F} ,$ the external environment of P, is a gravity-free environment and write $\overline{\kappa}_{\mathsf{o}} \! \left( \hspace{0.5pt} \mathtt{F} \hspace{0.5pt} \right) = \left( 0, \, 0, \, 0 \right) .$

Particles in a gravity-free environment are not necessarily completely secluded, just isolated from gravitons. They might still interact with photons and other particles, as in space stations for example. And almost anywhere could be considered gravity-free if observations are quick enough.

Inertial Environments

If P never has any interactions with anything then we say that F, the reference frame for P, provides an inertial environment. We write $\overline{\kappa}_{\mathsf{o}} \! \left( \hspace{0.5pt} \mathsf{F} \hspace{0.5pt} \right) = \left( 0, \, 0, \, 0 \right)$ and we call F an inertial frame of reference.

Particles in an inertial environment are completely isolated. Their external environment is gravity-free. And their local environment does not permit any interaction with other particles. So inertial reference frames may provide the idealized environment associated with in-vacuo waves.

If we assume that a frame is inertial, then in addition, we implicitly presume that gravitational effects are negligible. In an inertial environment $\overline{\kappa}_{\mathsf{o}} \! \left( \hspace{0.5pt} \mathtt{F} \hspace{0.5pt} \right) = \left( 0, \, 0, \, 0 \right) .$

Dispersive Environments

If any interaction between P and its surroundings is negligible, then we say that $\mathtt{f} ,$ the local environment of P, is a non-dispersive environment and write $\overline{\kappa}_{\mathsf{o}} \! \left( \hspace{0.5pt} \mathtt{f} \hspace{0.5pt} \right) = \left( 0, \, 0, \, 0 \right) .$ However, if interactions are appreciable, then we say that $\mathtt{f}$ is a dispersive environment. Particles in a dispersive environment are situated among other particles, so we say that they are located in situ. And they are described using characteristics that depend on $\mathtt{f} .$ So next we consider some of these characteristics in detail.

Refraction describes the shimmering iridescence evoked by this lustrous silken textile from Indonesia. — Tampan, Paminggir people. Sumatra 19th century, 58 x 61 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop.

The Wavevector

In a dispersive environment particles are nominally described by the wavevector-in-situ. But usually this is just called the wavevector because dispersion is so commonplace. It is defined for some generic particle P, relative to a frame of reference F, by

$\overline{\kappa}^{\, \mathsf{P}} \equiv \dfrac{ \, N^{\mathsf{F}} \, \overline{\kappa}_{\mathsf{o}}^{\, \mathsf{P}} - N^{\mathsf{P}} \, \overline{\kappa}_{\mathsf{o}}^{\, \mathsf{F}} \, }{N^{\mathsf{F}}}$

where $\overline{\kappa}_{\mathsf{o}}$ is the wavevector in vacuo and $N$ notes the total number of quarks in a particle. This relative wavevector represents the contrast between P and a scaled-down version of F. It is related to the wavenumber, by $\kappa \equiv \left\| \, \overline{\kappa} \, \right\| \! .$ So

$\kappa^{\mathsf{P}} = \dfrac{ \, \left\| \, N^{\mathsf{F}} \, \overline{\kappa}_{\mathsf{o}}^{\, \mathsf{P}} - N^{\mathsf{P}} \, \overline{\kappa}_{\mathsf{o}}^{\, \mathsf{F}} \, \right\| \, }{N^{\mathsf{F}}}$

Now recall that if some free particles interact like $\mathbb{X} \! + \! \mathbb{Y} \! \leftrightarrow \! \mathbb{Z} \, ,$ then their wavevectors-in-vacuo combine like $\overline{\kappa}_{\mathsf{o}}^{\mathbb{X}} \! + \overline{\kappa}_{\mathsf{o}}^{\mathbb{Y}} \! = \overline{\kappa}_{\mathsf{o}}^{\mathbb{Z}} .$ And since $\mathsf{F} = \left\{ \mathtt{F} , \, \mathtt{f} \rule{0px}{10px} \right\}$ we have $\overline{\kappa}_{\mathsf{o}}^{\, \mathsf{F}} = \overline{\kappa}_{\mathsf{o}}^{\, \mathtt{F}} + \overline{\kappa}_{\mathsf{o}}^{\, \mathtt{f}} .$ Let the external environment $\mathtt{F}$ be gravity-free. Then $\overline{\kappa}_{\mathsf{o}}^{\, \mathtt{F}} \! \! = \! \left( 0, \, 0, \, 0 \right)$ and $\overline{\kappa}_{\mathsf{o}}^{\, \mathsf{F}} \! \! = \overline{\kappa}_{\mathsf{o}}^{\, \mathtt{f}} .$ All dispersion is due to local interactions, and the wavenumber can be written as

Refraction

Let us consider some particle P, that interacts with the quarks in some local environment $\mathtt{f} ,$ which is itself situated within some gravity-free external environment $\mathtt{F} .$ This scenario leads to the fascinating study of optics . And here we start by defining a dimensionless number called the index of refraction as

$\eta \equiv \dfrac{ \, \left\| \, N^{\mathtt{ F}} \, \overline{\kappa}^{\, \mathsf{P}}_{\mathsf{o}} - N^{\mathsf{P}} \, \overline{\kappa}_{\mathsf{o}}^{\, \mathtt{f}} \, \right\| \, }{ N^{\mathtt{F}} \, \kappa^{\mathsf{P}}_{\mathsf{o}} }$

By this definition, in-vacuo and in-situ wavenumbers are related as

$\kappa = \eta \kappa_{\mathsf{o}}$

The index $\eta$ is useful because it is often very close to the number one. Then in-situ and in-vacuo characteristics are the same as each other. The index is a function of both the particle and frame. So for example, if $N^{\mathsf{P}} \! \! \ll \! N^{\mathtt{F}}$ then $\eta \! \simeq \! 1 \hspace{0.5pt} .$ And as a different example, if $\mathtt{f}$ is a classroom or laboratory, then the local environment is filled with common atmospheric gases. And the refractive index for these gases is equal to one within a few parts in $10^{4}.$ So for cases like these we can often write $\kappa \! = \! \kappa_{\mathsf{o}} \hspace{0.5pt} .$

The Wavelength

Recall that we have previously discussed $\lambda_{\mathsf{o}}$ as the wavelength in vacuo for use in a non-dispersive environment. So conversely, in a dispersive environment we nominally employ the wavelength-in-situ. But usually, this quantity is just called the wavelength. And it is defined by the ratio

$\lambda \equiv \dfrac{ \, \lambda_{\mathsf{o}} \, }{ \eta}$

Here $\eta$ is the index of refraction in a dispersive local environment $\mathtt{f} ,$ which is presumably situated within some gravity-free external environment $\mathtt{F} .$ Substituting-in definitions for $\eta$ and $\lambda_{\mathsf{o}}$ gives

$\lambda = \dfrac{ 2\pi N^{\mathtt{F}} }{ \, \left\| \, N^{\mathtt{F}} \, \overline{\kappa}^{\, \mathsf{P}}_{\mathsf{o}} - N^{\mathsf{P}} \, \overline{\kappa}_{\mathsf{o}}^{\, \mathtt{f}} \, \right\| \, }$

So if $N^{\mathsf{P}} \! \! \ll \! N^{\mathtt{F}} \! ,$ or perhaps if $\mathtt{f}$ is a classroom or laboratory, then we can often write $\eta \! = \! 1$ and $\lambda \! = \! \lambda_{\mathsf{o}} \hspace{0.5pt} .$

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References
1	Perhaps very much smaller because experimental physicists often arrange for $\mathtt{f}$ to be extremely unusual as a way of separating variables and untangling causal relationships. Such extraordinary conditions often need to be tightly enclosed.