Spectacular Photons – EthnoPhysics

Outline

Waves

This article is about photons. But before we get to that, we have to consider some related concepts like waves and fields. So far EthnoPhysics has considered a generic particle P by objectifying some repetitive chain of events written as

$\Psi^{\mathsf{P}} = \left( \mathsf{\Omega}_{1}, \, \mathsf{\Omega}_{2}, \, \mathsf{\Omega}_{3} \, \ldots \, \right)$

Each cycle $\mathsf{\Omega}$ is defined by sensation. And they must be repetitive so that P can be recognized, so $\mathsf{\Omega}_{1} = \mathsf{\Omega}_{2} = \mathsf{\Omega}_{3} \, \ldots$ and so on. Perhaps like a pattern in a carpet or textile. But we might just as well understand such a recurring sequence of sensations to be a wave train. That is, $\Psi$ could represent some sort of of periodically undulating or fluctuating perception. This interpretive ambivalence is called wave-particle duality and it has been 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 localized, then we talk about particles. Or if sensory phenomena seem to have some extended quality, then we often use words like wave, wavenumber, wavelength, etc. In between, we might speak of particles that are in excited states. These terms are developed from a discussion of quarks as follows. Let each cycle of P be a bundle of $N_{ \! \mathsf{q}}$ quarks written as

$\mathsf{\Omega} = \left( \mathsf{q}_{1}, \, \mathsf{q}_{2}, \, \mathsf{q}_{3} \, \ldots \, \mathsf{q}_{i} \, \ldots \, \mathsf{q}_{ N_{\! \mathsf{q}} } \rule{0px}{13px} \right)$

And let each quark be described by its phase $\delta_{\theta}$ along with its radius vector $\overline{\rho}$ . Then the wavevector of P is defined as

$\displaystyle \overline{\kappa} \equiv \left( \dfrac{1}{\rho_{in}^{2}} - \dfrac{1}{\rho_{out}^{2}} \right) \sum_{i=1} ^{N_{\! \mathsf{q}}} \delta_{\theta}^{\, i} \; \overline{\rho}^{i}$

where $\rho_{in}$ is the inner radius and $\rho_{out}$ is the outer radius of P. Substituting-in the definitions for these radii gives the wavevector in terms of the coefficients of down quarks as

$\displaystyle \overline{\kappa} = \frac{ k_{\mathsf{F}}}{hc} \left[ \left( \frac{8}{ \Delta n^{\mathsf{D}} } \right)^{\! 2} - \left( \frac{8}{ N^{\mathsf{D}} } \right)^{\! 2} \right] \cdot \sum_{i=1}^{N_{\! \mathsf{q}}}\delta_{\theta}^{\, i} \; \overline{\rho}^{i}$

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

$\displaystyle \overline{\kappa} = \frac{ k_{\mathsf{F}} }{hc} \sum_{i=1}^{N_{\! \mathsf{q}}} \delta_{\theta}^{\, i} \; \overline{\rho}^{i}$

Recall that quarks are conserved and $\overline{\rho}$ is defined from sums of quarks. So if some free particles interact like $\mathbb{X} + \mathbb{Y} \leftrightarrow \mathbb{Z} ,$ their radii will combine as $\overline{\rho}^{\mathbb{X}} + \overline{\rho}^{\mathbb{Y}} = \overline{\rho}^{\mathbb{Z}} .$ Also remember that $\overline{\kappa}$ is a relative characteristic because the phase depends on the juxtaposition of P with some frame of reference. Let this frame be steady so that quarks do not change phase. Then wavevectors are related as

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

The radius vector also relates particles and anti-particles by $\overline{\rho} \mathsf{(P)} = - \overline{\rho} \mathsf{(\overline{P})}$ . So if quarks are swapped with anti-quarks, without altering the phase, then

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

The average wavevector describes some hypothetical typical quark in P using the ratios $\widetilde{\kappa} \equiv \overline{\kappa} / N_{\! \mathsf{q}}$ where $N_{ \! \mathsf{q}}$ is the total number of quarks in P.

Inertial Frames of Reference

Let some well-known particle F be employed as a frame of reference. This just means that we use F to describe changing phenomena. Otherwise a reference frame is a compound quark like any other particle, so it can be characterized by its quark coefficients and wavevector $\overline{\kappa}$ . For example, a rigid frame of reference always has the same radius vector.

Another important special case is when each component of the average wavevector of F is zero

$\widetilde{\kappa} \equiv \dfrac{ \overline{\kappa} }{N_{\! \mathsf{q}}} = \left( 0, 0, 0 \right)$

then we say that F provides a perfectly inertial frame of reference. This condition is approximated when the total number of quarks $N_{ \! \mathsf{q}}$ is enormous. Because if the total number of quarks is huge, there will likely be some mix of quarks and anti-quarks making the $\Delta n$ terms in the radii of the numerator tend toward zero, even as the denominator gets larger.

Wavelengths

A wavenumber can be specified by the norm of a wavevector. It is written without an overline as $\kappa \equiv \left\| \, \overline{\kappa} \, \right\|$ . Then the wavelength of P is defined as

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

Thus $\lambda$ takes a logically required, discontinuous jump when $\kappa =0$ . 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.

Sensory Interpretation: Radius vectors are defined from dynamic quarks, not baryonic quarks. So the wavevector, wavenumber and wavelength can only represent somatic and visual sensations, not thermal perceptions. And, if a frame of reference is inertial, it is big and greyish.

Waves of repetitive sensation are graphically displayed in this rattan weaving from Indonesia. — Ajat basket, Penan people. Borneo 20th century, 20 (cm) diameter by 35 (cm) height. Hornbill motif. Photograph by D Dunlop.

Excited Particles

Consider a particle P described by some repetitive chain of events written as

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

where each repeated cycle is a bundle of quarks

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

and each quark is described by its phase $\delta_{\theta}$ . Use this phase to sort quarks into a pair of sets, $\mathsf{P}_{\mdsmwhtcircle}$ and $\mathsf{P}_{\mdsmblkcircle}$ , 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

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

and

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

Some sub-set of the quarks in $\mathsf{P}_{\mdsmwhtcircle}$ may be matched with quarks of the same type in $\mathsf{P}_{\mdsmblkcircle}$ . These quarks have phase symmetry with each other, so we use $\mathcal{S}_{\mdsmwhtcircle}$ and $\mathcal{S}_{\mdsmblkcircle}$ to symbolize these sub-sets.

A different sub-set of quarks might be matched with anti-quarks of the same type. These quarks have phase anti-symmetry with each other, so we note them as $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ .

Some quarks in $\mathsf{P}_{\mdsmwhtcircle}$ might not correspond with any quarks in $\mathsf{P}_{\mdsmblkcircle}$ and vice versa. But such lopsided possibilities seem to be superfluous so we do not consider them further.

Thus P is represented by the union of two entirely symmetric, and two purely anti-symmetric components. Quarks in the symmetric sets may vary independently of the quarks in the anti-symmetric sets. This is expressed mathematically as

$\mathsf{\Omega}^{\mathsf{P}} = \left\{ \left\{ \mathcal{S}_{\mdsmblkcircle}, \, \mathcal{A}_{\mdsmblkcircle} \right\}, \ \left\{ \mathcal{S}_{\mdsmwhtcircle}, \, \mathcal{A}_{\mdsmwhtcircle} \right\} \rule{0px}{11px} \right\}$

where

$\mathcal{S}_{\mdsmblkcircle} = \mathcal{S}_{\mdsmwhtcircle}$

$\mathcal{A}_{\mdsmblkcircle} = \overline{\mathcal{A} _{\mdsmwhtcircle} }$

$\delta_{\theta}^{ \mathcal{S}_{\mdsmblkcircle} } =- \, \delta_{\theta}^{ \mathcal{S}_{\mdsmwhtcircle} }$

and

$\delta_{\theta}^{ \mathcal{A}_{\mdsmblkcircle} } =- \, \delta_{\theta}^{ \mathcal{A}_{\mdsmwhtcircle} }$

This arrangement provides a general way of describing particles that are moved or excited by the absorption of additional quarks. P is defined as an excited particle, or said to be in an excited-state, if it contains at least one anti-symmetric pair of quarks

$\mathcal{A}_{\mdsmblkcircle} = \overline{\mathcal{A}_{\mdsmwhtcircle}} \ne \left\{ \varnothing \right\}$

These anti-symmetric quark-pairs may be due to the absorption of a photon. Or more generally, to interactions with any field quanta.

Ground State Particles

We say that P is in its ground state if it has perfect phase symmetry. Then P has no anti-symmetric quark-pairs

$\mathcal{A}_{\mdsmblkcircle} = \overline{ \mathcal{A}_{\mdsmwhtcircle} } = \left\{ \varnothing \right\}$

This definition constrains particle-models so that quark-coefficients must all be integer multiples of two when in the ground-state. It is why many nuclear particle models show patterns like 2-4-6 instead of 1-2-3.

If P is a free particle, then we can also evaluate its wavevector which is given by the sum

$\displaystyle \overline{\kappa}^{\, \mathsf{P}} = \dfrac{k_{\mathsf{F}}}{hc} \sum_{i=1}^{N_{\! \mathsf{q}}} \delta_{\theta}^{\,i} \, \overline{\rho}^{i}$

Since the asymmetric sets are empty, the sum is taken over just the symmetric sets, so

$\begin{align*} \overline{\kappa }^{\, \mathsf{P}} &= \dfrac{k_{\mathsf{F}}}{hc} \left( \delta_{\theta}^{\, \mathcal{S}_{\mdsmwhtcircle} } \hspace{-5px} \sum_{\mathsf{q} \, \in \, \mathcal{S}_{\mdsmwhtcircle} } \! \overline{\rho}^{\, \mathsf{q}} \; + \; \delta_{\theta}^{\, \mathcal{S}_{\mdsmblkcircle} } \hspace{-5px} \sum_{\mathsf{q} \, \in \, \mathcal{S}_{\mdsmblkcircle} } \! \overline{\rho}^{\, \mathsf{q}} \right) \\ &= \dfrac{k_{\mathsf{F}}}{hc} \left( \delta_{\theta}^{ \mathcal{S}_{\mdsmwhtcircle} } \, \overline{\rho}^{ \mathcal{S}_{\mdsmwhtcircle} } \; + \; \delta_{\theta}^{ \mathcal{S}_{\mdsmblkcircle} } \, \overline{\rho}^{ \mathcal{S}_{\mdsmblkcircle} } \right) \\ &= \dfrac{k_{\mathsf{F}}}{hc} \left( \delta_{\theta}^{ \mathcal{S}_{\mdsmwhtcircle} } + \delta_{\theta}^{ \mathcal{S}_{\mdsmblkcircle} } \right) \, \overline{\rho}^{ \mathcal{S}_{\mdsmwhtcircle} } \ \ \mathsf{\text{because}} \ \ \overline{\rho}^{ \mathcal{S}_{\mdsmwhtcircle} } = \overline{\rho}^{ \mathcal{S}_{\mdsmblkcircle} } \\ &= (0, 0, 0) \ \ \ \ \mathsf{\text{because}} \ \ \ \ \delta_{\theta}^{\mathcal{S}_{\mdsmwhtcircle}} = - \, \delta_{\theta}^{\mathcal{S}_{\mdsmblkcircle}} \rule{0px}{16px} \end{align*}$

Thus the wavenumber of a free particle in its ground state is always zero

$\kappa^{\mathsf{P}} \equiv \left\| \, \overline{\kappa}^{\mathsf{P}} \right\| = \left\| \, (0, 0, 0) \, \vphantom{<meta charset="utf-8">\overline{\kappa}^{\mathsf{P}} } \right\| = 0$

The Principal Quantum Number

Whenever P interacts by absorbing or emitting a photon, its degree or level of excitation changes too because photons contain anti-symmetric quark-pairs. This relationship is articulated by defining the principal quantum number as

$\mathrm{n} \equiv \dfrac{ \; n^{\mathsf{d}} \; }{4}$

where $n^{\mathsf{d}}$ is the number of ordinary down quarks in P. Note that the italic letter $n$ is employed for quark coefficients, whereas the upright font $\mathrm{n}$ is reserved for the new quantum number. We use $\mathrm{n}$ to describe the level of excitation, and so down-quarks retain an important role despite having almost no internal energy.

Remember that $n^{\mathsf{d}} \ge 0$ so the principal quantum number is never negative. And recall that quarks are conserved. So if P changes from some initial state $i$ to some final state $f$ by emitting a photon $\boldsymbol{\gamma}$ , then we can write $\mathsf{P}_{i} \to \mathsf{P}_{f} + \boldsymbol{\gamma}$ and ${\mathrm{n}} \! \left( \mathsf{P}_{i} \right) = {\mathrm{n}} \! \left( \mathsf{P}_{f} \right) + {\mathrm{n}} \! \left( \boldsymbol{\gamma} \right).$ Thus conservation of down-quarks implies that the photon’s principle quantum number is

$\mathrm{n} \! \left( \boldsymbol{\gamma} \right) = \mathrm{n} \! \left( \mathsf{P}_{i} \right) - \mathrm{n} \! \left( \mathsf{P}_{f} \right)$

But also, by definition, the photon’s principal quantum number can be written as

$\begin{align*} {\mathrm{n}} \left( \boldsymbol{\gamma} \right) &\equiv \frac{ n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{4} \\ &= \left[ \frac{n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{8} + \frac{n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{8} \right] + \; 0 \\ &= \left[ \frac{n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{8} + \frac{n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{8} \right] + \left[ \frac{n^{\sf{\overline{d}}}\left( \boldsymbol{\gamma} \right) }{8} - \frac{n^{\sf{\overline{d}}}\left( \boldsymbol{\gamma} \right) }{8} \right] \\ &= \left( \frac{n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{8} + \frac{n^{\sf{\overline{d}}}\left( \boldsymbol{\gamma} \right) }{8} \right) - \left( \frac{n^{\sf{\overline{d}}}\left( \boldsymbol{\gamma} \right) }{8} - \frac{n^{\sf{d}}\left( \boldsymbol{\gamma} \right) }{8} \right) \\ &= \frac{N^{\sf{D}}\left( \boldsymbol{\gamma} \right) }{8} - \frac{\Delta n^{\sf{D}}\left( \boldsymbol{\gamma} \right) }{8} \end{align*}$

Then comparing expressions for $\mathrm{n} \! \left( \boldsymbol{\gamma} \right)$ shows that the conservation law is always satisfied if

${\rm{n}} \! \left( {\mathsf{P}}_{i} \right) = \dfrac{N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \right) }{8}$

and

${\rm{n}} \! \left( {\mathsf{P}}_{f} \right) = \dfrac{\Delta n^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \right) }{8}$

Excited states are a mix of symmetric and asymmetric patterns, like this bead panel from Kalimantan. — Baby Carrier panel, Kayan or Kenyah people. Borneo 20th century, 27 x 27 cm. Photograph by D Dunlop.

Fields

A particle that is formed entirely from ethereal, imaginary and neutral components can be difficult to characterize and distinguish from other more salient phenomena. So if there are a lot of these elusive particles in a description then it may be more convenient to group them together and refer to them collectively as a field. Different sorts of fields are defined by different quark distributions.

The fields that we discuss most are electromagnetic fields. When considered as individual particles they are called photons. Photons are defined from quarks that are paired with their matching anti-quarks. Different types of photons are modeled by different combinations of these $\mathsf{q \overline{q}}$ pairs. For example the gross spectrum of hydrogen is accurately described by rotating and electrochemical quark pairs.

Other sorts of fields like gravity can also be modeled using these $\mathsf{q \overline{q}}$ particles. Different combinations are associated with different forces. So we use them like building blocks and call them field quanta.

Field Quanta

A simple field quantum is formed by a pair of quarks that are out of phase anti-particles to each other. Then the net number of quarks is zero, except for down-quarks. So these simple quanta have no mass, charge, baryon-number, lepton-number or strangeness. But they do have distinct temperatures and momenta that vary by quark-type. Here is a list of simple field quanta.

We use the symbol $\mathscr{F}$ to denote a simple field quantum. For any pair of quarks there are two possible arrangements that depend on their phases. That is, arrangements depend on the helicity of the reference frame as noted by $\mathsf{F_{\mdsmwhtcircle}}$ or $\mathsf{F_{\mdsmblkcircle}}$ . For example consider the following pairs of negative quarks that have their phases illustrated by background shading.

$\mathscr{F} \! ( \mathsf{e} ) \equiv \left\{ \left\{ \mathsf{e}, \mathsf{F_{\mdsmblkcircle}} \right\}, \left\{ \overline{\mathsf{e}}, \mathsf{F_{\mdsmwhtcircle}} \right\} \rule{0px}{14px} \right\}$

$\overline{\mathscr{F}} \! ( \mathsf{e} ) \equiv \left\{ \left\{ \overline{\mathsf{e}}, \mathsf{F_{\mdsmblkcircle}} \right\}, \left\{ \mathsf{e}, \mathsf{F_{\mdsmwhtcircle}} \right\} \rule{0px}{14px} \right\}$

Roughly speaking, dynamic field quanta are like little bits of momentum. They can be used as components of photons. Field quanta can move around between Newtonian particles by catching a ride on any passing photon. When baryonic field quanta move they do not carry momentum, but they produce thermal currents.

Momentum is conserved. So if another particle interacts with a field quantum, then it experiences a force that is proportional to the momentum of $\mathscr{F}$ .

A field quantum $\mathscr{F}$ and its anti-particle $\overline{\mathscr{F}}$ have their momenta pointed in opposite directions. So if one increases, then the other decreases the total momentum of any absorbing particle. Hence, the impressed force has two possibilities; like an attraction or a repulsion, or perhaps a push vs a pull. The direction depends on phase relationships.

Dark Quanta

In addition to the foregoing $\mathsf{q \overline{q}}$ pairs, we also include $\mathsf{dd}$ and $\mathsf{\overline{d} \overline{d}}$ pairs as simple field quanta. This is because their internal energy is so small that any distinction between down quarks and down anti-quarks is imperceptible. Pauli’s exclusion principle is not violated because they are still distinguished by their phase.

Thus there are four quanta composed from out of phase pairs of down quarks. They all have the same temperature of -760 (K), which means that they are very stable. And they all have an internal energy that is very close to zero, about -54 micro electronvolts. This is utterly negligible in the realm of nuclear reactions where particle energies are typically trillions of times larger and measured in (MeV). It is also imperceptible in most atomic and chemical reactions where energies are about a million times larger. Indeed, almost the only experimental access we have to these elusive quanta comes from extremely precise observations of fine structure in atomic spectra. So by convention we almost always ignore them.

These field components are known as dark quanta. They are illustrated in the following images which use background shading to indicate the phase.

$\overline{\mathscr{F}} \! ( \mathsf{d} )$

$\overline{\mathscr{F}} \! ( \mathsf{2d} )$

Individually, these quanta rarely make any noticeable contribution to energetics. But since their internal energy is not exactly zero, they may be collectively relevant if there are enough of them. And enormous quantities are possible when considering astronomical distances. Then dark quanta may give rise to dark energy .

Subatomic Quanta

Atomic and molecular interactions are mediated by fields and forces too. Long-distance forces may be carried by photons. But brief, short-range forces from collisions, explosions and chemical reactions can also be accurately modeled using $\mathsf{q \overline{q}}$ pairs. When these quanta interact within atoms, they are called subatomic quanta. For example see the following list of particles used to provoke transitions between different excited states of hydrogen.

The combination of quarks noted as $\text{\L}$ in the foregoing list is called the Lamb quantum. It is used a lot in spectroscopic models because interacting with $\text{\L}$ changes the orbital angular momentum without altering the total angular momentum. For more about these quanta, please see the discussion of fine structure in the spectrum of hydrogen.

Strange Quanta

Simple field quanta are the building blocks of photons, but they may also be assembled into other force-carrying particles as well. For example, consider nuclear particles like the kaons. These strange particles have their rotating quarks arranged in unusual ways. So their interactions involve some different quanta too. The following strange quanta carry weak forces.

The relationship between field quanta and their forces can be developed in much more detail, but next we go back and focus on exactly what we mean by a photon.

Definitions of Photons

Let particle P be characterized by some repetitive chain of events written as

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

where each repeated cycle is a bundle of quarks

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

and each quark is described by its phase $\delta_{\theta}$ . Use this phase to sort quarks into a pair of sets, $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ , so that all quarks of the same phase are in the same set. Then $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ are called phase components of P, and they are out of phase with each other. We write

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

and

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

Now let P be an almost perfectly phase anti-symmetric particle so that $\mathcal{A}_{\mdsmwhtcircle} = \overline{\mathcal{A}_{\mdsmblkcircle}}$ for all types of quarks except down quarks. Then we define a photon $\boldsymbol{\gamma}$ as a particle like P, that also satisfies the conditions

$N^{\mathsf{D}} = N^{\mathsf{U}} \! \pm 8$

and

$\left| {\Delta}n^{\mathsf{Z}} \rule{0px}{10px} \right| = \begin{cases} \ \ 0 \ &\mathsf{\text{if}} \ {\mathsf{Z \ne D}} \\ \; \ge 8 \ &\mathsf{\text{if}} \ {\mathsf{Z = D}} \end{cases}$

These constrain the angular momentum $\textsl{\textsf{J}}$ and the inner radius $\rho_{in}$ so that for all photons

$\textsl{\textsf{J}} \left( \boldsymbol{\gamma} \right) = 1$

and

$\rho_{in} \! \left( \boldsymbol{\gamma} \right) \ge \sqrt{ \dfrac{hc}{2\pi k_{\mathsf{F}}} \rule{0px}{17px} }$

Notice that solitary photons are always located away from the spatial origin. Their inner radius is never zero. So under some conditions, we may be able to say that $\boldsymbol{\gamma}$ is a free particle.

Photons have almost perfect phase anti-symmetry, perhaps somewhat like this bead panel from Indonesia. — Baby Carrier panel, Ngaju people. Borneo 20th century, 33 x 26 cm. Photograph by D Dunlop.

Types of Photons

Consider a photon $\boldsymbol{\gamma}$ described by an out of phase pair of components

$\mathsf{\Omega} \left( \boldsymbol{\gamma} \right) = \left\{ \mathcal{A}_{\mdsmwhtcircle}, \, \mathcal{A}_{\mdsmblkcircle} \rule{0px}{9px} \right\}$

that are almost perfectly anti-symmetric so that

$\mathcal{A}_{\mdsmwhtcircle} \simeq \overline{\mathcal{A}_{\mdsmblkcircle}}$

for all sorts of quarks except down quarks. Many photon attributes are nil because phase anti-symmetry implies that almost every quark is matched with a corresponding anti-quark somewhere in the photon. So for most types of quark Z, the net number of quarks is zero

$\Delta n^{\mathsf{Z}} \equiv n^{\mathsf{\overline{z}}} - n^{\mathsf{z}} = 0 \; \; \; \; \; \; \forall \; \mathsf{Z} \ne \mathsf{D}$

As for the down-quarks, $\Delta n^{\mathsf{D}}$ is not zero. But recall that by convention, the internal energy of a down-quark is so small that it is usually taken to be zero. Then any imbalance between ordinary down quarks and down anti-quarks can be ignored. Substituting these conditions into the definitions for charge, strangeness, lepton number, baryon number and enthalpy gives

$q(\boldsymbol{\gamma})=0$

$S(\boldsymbol{\gamma})=0$

$L(\boldsymbol{\gamma})=0$

$B(\boldsymbol{\gamma})=0$

and

$H(\boldsymbol{\gamma}) \simeq 0$

Recall that the lepton-number, baryon-number and charge are conserved, so a particle may freely absorb or emit countless photons without altering its own values for these quantum numbers. And almost no work is required to assemble a photon because down quarks are so small. Thus $W(\boldsymbol{\gamma}) \simeq 0$ too. But not all photon characteristics are null; the outer radius $\rho_{out}$ and the inner radius $\rho_{in}$ may be greater than zero. Also consider the wavevector $\overline{ \kappa }$ which is found from the sum

$\begin{align*} \displaystyle \overline{\kappa} &\equiv \left( \frac{1}{\rho_{in}^{\, 2}} - \frac{1}{\rho_{out}^{\, 2}} \rule{0px}{18px} \right) \sum_{\mathsf{q} \, \in \, {\boldsymbol{\gamma}} } \delta_{\theta}^{\, \mathsf{q}} \; \overline{\rho}^{\, \mathsf{q}} \\ &= \left( \frac{1}{\rho_{in}^{\, 2}} - \frac{1}{\rho_{out}^{\, 2}} \rule{0px}{18px} \right) \left[ \; \rule{0px}{22px} \delta_{\theta}^{\, \mathcal{A}_{\mdsmwhtcircle} } \hspace{-6px} \sum_{\mathsf{q} \, \in \, \mathcal{A}_{\mdsmwhtcircle} } \hspace{-4px} \overline{\rho}^{\, \mathsf{q}} \ + \ \delta_{\theta}^{\, \mathcal{A}_{\mdsmblkcircle} } \hspace{-6px} \sum_{\mathsf{q} \, \in \, \mathcal{A}_{\mdsmblkcircle} } \overline{\rho}^{\, \mathsf{q}} \; \right] \\ &= \left( \frac{1}{\rho_{in}^{\, 2}} - \frac{1}{\rho_{out}^{\, 2}} \rule{0px}{18px} \right) \left[ \, \rule{0px}{16px} \delta_{\theta}^{\, \mathcal{A}_{\mdsmwhtcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle} } \; + \; \delta_{\theta}^{\, \mathcal{A}_{\mdsmblkcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmblkcircle} } \, \right] \end{align*}$

But $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ are out of phase, so $\delta_{\theta}^{\mathcal{A}_{\mdsmwhtcircle}} =- \, \delta_{\theta}^{\mathcal{A}_{\mdsmblkcircle}} = \pm 1.$ And the radius vectors of particles and anti-particles are symmetrically opposed, so $\overline{\rho}^{\mathcal{A}_{\mdsmwhtcircle}} \! =- \, \overline{\rho}^{\mathcal{A}_{\mdsmblkcircle}} .$ These two negative factors cancel each other such that $\delta_{\theta}^{\, \mathcal{A}_{\mdsmwhtcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle} } \! = \delta_{\theta}^{\, \mathcal{A}_{\mdsmblkcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmblkcircle} } .$ Then we can express the wavevector as

$\overline{\kappa} = 2\left( \dfrac{1}{\rho_{in}^{\, 2}} - \dfrac{1}{\rho_{out}^{\, 2}} \rule{0px}{18px} \right) \delta_{\theta}^{\, \mathcal{A}_{\mdsmwhtcircle} } \, \overline{\rho}^{\, \mathcal{A}_{\mdsmwhtcircle} }$

The wavenumber is given by the norm of the wavevector, so

$\kappa = 2\left( \dfrac{1}{\rho_{in}^{\, 2}} - \dfrac{1}{\rho_{out}^{\, 2}} \rule{0px}{18px} \right) \left\| \, \overline{\rho}^{\, \mathcal{A}} \, \rule{0px}{11px} \right\|$

We can drop the subscript on $\mathcal{A}$ because both phase components have the same norm. And recall that $W^{\mathcal{A}} = k_{\mathsf{F}} \left\| \, \overline{\rho}^{\mathcal{A}} \rule{0px}{8px} \right\|$ is the work required to build one of these phase-components. So in energetic terms, the photon’s wavenumber can be written as

$\kappa = \dfrac{2W^{\mathcal{A}}}{k_{\mathsf{F}}} \left( \dfrac{1}{\rho_{in}^{\, 2}} - \dfrac{1}{\rho_{out}^{\, 2}} \rule{0px}{18px} \right)$

Then the wavelength of a photon is

$\lambda = \dfrac{2\pi}{\kappa} = \dfrac{\pi k_{\mathsf{F}}}{\rule{0px}{11px} \, W^{\mathcal{A}} \, } \left( \dfrac{1}{\rho_{in}^{\,2}} - \dfrac{1}{\rho_{out}^{\,2}} \rule{0px}{18px} \right)^{\! -1}$

Finally recall that by definition, the inner radius $\rho_{in}$ is constrained such that for all photons

Photon Type	𝜆 (m)
a gamma-ray	$\lesssim 10^{-12}$
an X-ray	$10^{-11} \sim 10^{-8}$
an ultraviolet photon	$\sim 10^{-8}$
a visible photon	$\sim 10^{-7}$
an infrared photon	$10^{-6} \sim 10^{-3}$
a microwave	$10^{-3} \sim 1$
a radio-wave	$1 \sim 10^{8}$

$\rho_{in} \! \left( \boldsymbol{\gamma} \right) \ge \sqrt{ \dfrac{hc}{2\pi k_{\mathsf{F}}} \rule{0px}{17px} }$

Then if $\boldsymbol{\gamma}$ is a free particle where $\rho_{out} \to \infty$ and $\rho_{in}$ is as small as possible, the wavelength will be

$\lambda = \dfrac{hc}{\rule{0px}{11px} \,2 W^{\mathcal{A}} }$

Photons are classified by wavelength as noted in the accompanying table. A more general treatment considers that a photon’s wavelength may also depend on its surroundings. Then the symbol $\lambda _{\mathsf{o}}$ is used to indicate a wavelength where any such environmental effects are negligible.

Anti-Photons

For EthnoPhysics anti-photons are just like other anti-particles. So $\overline{\boldsymbol{\gamma}}$ is defined from $\boldsymbol{\gamma}$ by exchanging ordinary-quarks with anti-quarks of the same type, while leaving the phase and other relationships unchanged. In a photon, $\Delta n = 0$ for all types of quarks except down-quarks. So photons and anti-photons have just about the same characteristics as each other

$\textsl{\textsf{J}} \left( \boldsymbol{\gamma} \right) = \textsl{\textsf{J}} \left( \overline{\boldsymbol{\gamma}} \right)$

and

$\rho_{in} \! \left( \boldsymbol{\gamma} \right) = \rho_{in} \! \left( \overline{\boldsymbol{\gamma}} \right)$

But $\Delta n^{\mathsf{D}} \left( \boldsymbol{\gamma} \right) = - \Delta n^{\mathsf{D}} \left( \overline{ \boldsymbol{\gamma}} \right)$ . And photons also have relative characteristics which may differ between $\overline{\boldsymbol{\gamma}}$ and $\boldsymbol{\gamma}$ depending on their juxtaposition with a frame of reference. For example, the wavevector $\overline{\kappa}$ depends on the phase so that

$\overline{\kappa} \left( \boldsymbol{\gamma} \right) = - \, \overline{\kappa} \left( \overline{ \boldsymbol{\gamma}} \right)$

and the two photons have symmetrically opposed wavevectors. So photons and anti-photons are mostly the same as each other, but moving in opposite directions.

Photon types are distinguished by somatic and phase relationships, perhaps somewhat like the patterns in this beaded pattern from Indonesia. — Baby Carrier panel, Kayan people. Borneo 20th century, 31 x 27 cm. Photograph by D Dunlop.

Gross Hydrogen Spectrum

Photons that are absorbed or emitted by atomic hydrogen $\mathbf{H}$ , are collectively known as the spectrum of hydrogen. They are mostly involved in the atomic and molecular interactions of everyday experience, not nuclear reactions. Energies are typically measured in (eV) rather than (MeV).

All hydrogen-spectrum photons are linked to changes in the excited states of atomic hydrogen. When $\mathbf{H}$ goes from some initial state $i ,$ to some final state $f ,$ by emitting a photon $\boldsymbol{\gamma}$ , we write

$\mathbf{H}_{i} \to \mathbf{H}_{f} + \boldsymbol{\gamma}$

Particles are described by their principal quantum number $\rm{n} .$ This quantity is always conserved because quarks are conserved, and $\rm{n}$ is directly proportional to the number of down quarks. So for any interaction, the conservation of down quarks guarantees that

${\rm{n}} \! \left( {\mathbf{H}}_{i} \right) = {\rm{n}} \! \left( {\mathbf{H}}_{f} \right) + {\rm{n}} \! \left( \boldsymbol{\gamma} \right)$

Let us write ${\rm{n}}_{i} \equiv {\rm{n}} \! \left( {\mathbf{H}}_{i} \right)$ and ${\rm{n}}_{f} \equiv {\rm{n}} \! \left( {\mathbf{H}}_{f} \right)$ . Then, as shown earlier, quark conservation will automatically be obtained if

${\rm{n}}_{i} = \dfrac{N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \right) }{8}$

and

${\rm{n}}_{f} = \dfrac{\Delta n^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \right) }{8}$

Here $\Delta n^{\mathsf{D}}$ and $N^{\mathsf{D}}$ note the photon’s coefficients for down quarks (i.e. $n$ is different from $\rm{n}$ ). These relationships can be used to understand the hydrogen spectrum. Atomic interactions involving hydrogen emit many ultraviolet, visible and infrared photons. There are also some microwaves. But usually, there are no gamma-rays. We can describe photons that are not gamma-rays by first assessing their momentum as follows.

The gross hydrogen spectrum is shown in this photograph of four lines; red, green, blue and violet. — The first quantized account of hydrogen was achieved by Johann Balmer, a Swiss high-school teacher. Here is a photo of what he described mathematically.

Consider finding the mechanical energy $E$ , of a photon $\boldsymbol{\gamma}$ , that is specified by a pair of phase components $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle}$ written as

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

As discussed earlier, the wavenumber of this photon can be written as

$\kappa \! \left( \boldsymbol{\gamma} \right) = 2 \left\| \, \overline{\rho}^{\mathcal{A}} \rule{0px}{10px} \right\| \! \left( \dfrac{1}{ \rho_{in}^{2} } - \dfrac{1}{ \rho_{out}^{2} } \rule{0px}{18px} \right)$

where $\rho_{in}$ is the photon’s inner radius, $\rho_{out}$ is its outer radius and $\overline{\rho}$ is a radius vector. The subscript on $\mathcal{A}$ is dropped because both phase-components have the same norm. We can use this wavenumber to express the momentum of the photon, in a perfectly inertial reference frame, as

$p \! \left( {\boldsymbol{\gamma}} \right) = \dfrac{h}{2\pi} \kappa = \dfrac{h}{\pi} \left\| \, \overline{\rho}^{\mathcal{A}} \rule{0px}{10px} \right\| \! \left( \dfrac{1}{\rho_{in}^{2}} - \dfrac{1}{\rho_{out}^{2}} \rule{0px}{18px} \right)$

Writing-out the norm in terms of the radial components of $\mathcal{A}$ gives

$\begin{equation*} \left\| \, \overline{\rho}^{\mathcal{A}} \, \right\| = \left| \begin{split} & \; k_{mm} \rho_{m}^{2} + k_{ee} \rho_{e}^{2} + k_{zz} \rho_{z}^{2} \\ & + 2 k_{em} \rho_{m} \rho_{e} + 2k_{mz}\rho_{m} \rho_{z} \\ & \hspace{30px} + 2 k_{ez}\rho_{e} \rho_{z} \; \end{split} \right|^{\frac{1}{2}} \end{equation*}$

This expression can be simplified if the photon is not a gamma-ray. For long-wavelength photons the coefficients of leptonic quarks must all be zero because even one of these high energy quarks is enough to yield a gamma-ray. So the electric and magnetic radii of $\mathcal{A}$ are null. That is, $\rho_{m} = \rho_{e} = 0 .$ And recall that $k_{zz} = 1 .$ Then $\left\| \, \overline{\rho}^{\mathcal{A}} \rule{0px}{10px} \right\| = \left| \rho_{z} \right|$ and so

$p \! \left( {\boldsymbol{\gamma}} \right) = \dfrac{h}{\pi} \left| \rho_{z} \right| \! \left( \dfrac{1}{\rho_{in}^{2}} - \dfrac{1}{\rho_{out}^{2}} \rule{0px}{18px} \right)$

The photon’s momentum is proportional to the absolute-value of its polar radius which is defined as

$\rho_{z} \equiv \, \rho_{core} + \dfrac{ \Delta n^{\mathsf{U}} U^{\mathsf{U}} - \Delta n^{\mathsf{D}} U^{\mathsf{D}} }{ k_{\mathsf{F}} }$

where

$\rho_{core} \equiv \dfrac{\, \left| H_{chem} \rule{0px}{9px} \right| \, }{k_{\mathsf{F}}}$

and $H_{chem}$ is the enthalpy due to any chemical quarks. This expression can be simplified too because by convention $U^{\mathsf{D}} \! =0$ . Moreover, for the high-energy up-quarks $\Delta n^{\mathsf{U}}$ must be zero or else the photon would be a gamma-ray. Thus the polar radius is just

$\rho_{z} = \dfrac{\, \left| H_{chem}^{\mathcal{A}} \rule{0px}{9px} \right| \, }{k_{\mathsf{F}}}$

And the photon’s momentum can be stated as

$p \! \left( {\boldsymbol{\gamma}} \right) = \dfrac{h}{\pi k_{\mathsf{F}} } \left| H_{chem}^{\mathcal{A}} \rule{0px}{13px} \right| \! \left( \dfrac{1}{\rho_{in}^{2}} - \dfrac{1}{\rho_{out}^{2}} \rule{0px}{18px} \right)$

But photons are ethereal. Their mass is zero, so

$E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \rule{0px}{10px} \; } = cp$

The mechanical energy of the photon is directly proportional to its momentum and can be written as

$E \! \left( \boldsymbol{\gamma} \right) = \dfrac{hc}{\pi k_{\mathsf{F}} } \left| \, H_{\sf{chem}}^{\mathcal{A}} \rule{0px}{13px} \right| \left( \dfrac{1}{\rho_{in}^{2}} - \dfrac{1}{\rho_{out}^{2}} \right)$

Then substituting-in definitions for the radii gives the photon energy in terms of quark coefficients as

$E \! \left( \boldsymbol{\gamma} \right) = 2 \left| \, H_{chem}^{\mathcal{A}} \rule{0px}{13px} \right| \cdot \left[ \dfrac{64}{\left( \Delta n^{\mathsf{D}} \right)^{2} } - \dfrac{64}{ \left( N^{\mathsf{D}} \right)^{2} } \right]$

This formula shows the strong influence of any down-quarks on these low-energy photons. Substituting-in the relationships with $\mathrm{n}$ discussed above gives

$E \! \left( \boldsymbol{\gamma} \right) = 2 \left| \, H_{chem}^{\mathcal{A}} \rule{0px}{13px} \right| \cdot \left( \dfrac{1}{ {\rm{n}}_{f}^{2} } - \dfrac{1}{ {\rm{n}}_{i}^{2} } \right)$

This expression is used to make quark-models for hydrogen by adjusting the distribution¹Other electrochemical quark distributions are used to model other atoms and molecular bonds. of electrochemical quarks so that

$\left| \, H_{chem}^{\mathcal{A}} \rule{0px}{13px} \right| = \dfrac{hc}{2} \mathcal{R}_{\mathbf{H}}$

where $\mathcal{R}_{\mathbf{H}}$ is the Rydberg constant of hydrogen. Using this constraint to eliminate $H_{chem}$ then defines the energy of a photon for the so-called gross structure of hydrogen spectroscopy

$E_{\mathsf{gross}} \equiv hc \mathcal{R}_{\mathbf{H}} \left( \dfrac{1}{{\rm{n}}_{f}^{2}} - \dfrac{1}{{\rm{n}}_{i}^{2}} \right)$

Measurements of photons report their wavelength which is related to the momentum by $\lambda = h / p .$ So for ethereal particles $\lambda = hc/E .$ Wavelengths may also depend on a photon’s surroundings. Then the symbol $\lambda _{\mathsf{o}}$ is used to indicate a wavelength where any such environmental effects are negligible. We assume this is the case to write

$\dfrac{1}{\lambda_{\mathsf{o}}} = \mathcal{R}_{\mathbf{H}} \left( \dfrac{1}{{\rm{n}}_{f}^{2}} - \dfrac{1}{{\rm{n}}_{i}^{2}} \right)$

The description of hydrogen was first put in this form by Johannes Rydberg . Models for these photons are built exclusively from electrochemical and rotating quarks.²Baryonic and leptonic quarks are not used because we are not considering gamma-rays and nuclear processes. The rotating quarks all provide an angular momentum of $\textsl{\textsf{J}} = \! 1$ in various ways. But the electrochemical quarks are the same for every photon because all these photons are associated with hydrogen.

Comparisons are made with experimental observations.³Kramida, A., Ralchenko, Yu., Reader, J., and the NIST ASD Team (2018). NIST Atomic Spectra Database (version. 5.6.1). National Institute of Standards and Technology, Gaithersburg, MD, USA. For this reference, the lines from transitions between levels designated only by principal quantum numbers correspond to observations in which any finer structure is completely unresolved. Reported energy levels are the centre of groups of all fine-structure levels having the same $\rm{n}$ . Their values are based on observations of the Sun. Some models provide very accurate descriptions that are nonetheless outside of experimental uncertainty. This is indicated with an X in the following tables.

Lyman Series

Balmer Series

Paschen Series

Brackett Series

Other Series

In the models above, agreement with experiment is good to a few parts in a million. This would be outstanding in almost any other scientific discipline. But for atomic spectroscopy, it is mediocre. However, it is good enough to distinguish between competing quark-models. And so these specific quark combinations are taken to define each photon. Then later, after a discussion of atomic hydrogen, we use these models to make a more accurate description of fine structure in the hydrogen spectrum.

Calculated results depend on how quarks are distributed between phase components. For that level of detail please see the spreadsheet titled Atoms & Photons.

X-Rays

The following quark-models of X-rays use only rotating and electrochemical quarks. These sorts of arrangements are called atomic X-ray models. It is also possible to model X-rays using leptonic quarks, but we reserve that design for the X-rays coming from nuclear-decay. Atomic X-rays are different. They are typically produced by bombarding some metallic atom $\mathbf{A}$ , with electrons that have been accelerated to high speeds by absorbing vast quantities of photons containing electrochemical quarks. So the total number of quarks used for these models is not constrained, and $N_{ \! \mathsf{q}}$ rises into the thousands. Pauli’s principle is not violated because all absorbed photons presumably have different phase angles.

Atomic X-rays have distinct peaks in their observed energies. This phenomenon is linked to a few specific combinations of rotating quarks. Like other photons, the distribution of rotating quarks is described by the principal quantum number $\mathrm{n}$ , and the total angular momentum quantum number $\textsl{\textsf{J}}$ , as expressed using the spectroscopic notation for X-rays. These quantities are reviewed in the table below. The wavelength of an X-ray has been established above as

$\lambda_{\mathsf{o}} = \dfrac{hc}{\rule{0px}{11px} \,2 W^{\mathcal{A}} }$

where $W$ notes the work required to build $\mathcal{A}$ , a phase component of the X-ray. The wavelength is related to the X-ray’s energy by

$E \! \left( \boldsymbol{\gamma} \right) = \dfrac{hc}{\lambda_{\mathsf{o}}} = 2W^{\mathcal{A}}$

This relationship can be used to calculate the photon energies directly from quark-coefficients. And so here are some quark-models for the X-rays obtained by bombarding zinc, copper, iron and calcium. They are demonstrative, rather than definitive. The very close agreement with observed⁴R.D. Deslattes, E.G. Kessler Jr., P. Indelicato, L. de Billy, E. Lindroth, J. Anton, J.S. Coursey, D.J. Schwab, C. Chang, R. Sukumar, K. Olsen, and R.A. Dragoset (2005), X-ray Transition Energies Database (version 1.2). National Institute of Standards and Technology, Gaithersburg, MD, USA. values is easy to obtain because the total number of quarks is so large. Many similar models with slightly different numbers of electrochemical quarks also fit within the limits of observation.

These X-ray models could be improved by making some additional requirement for equilibrium. Then electrochemical quark distributions could be further constrained and explained. For more detail about these calculations, please see the X-Rays spreadsheet.

Under Construction

Gamma Rays

Mass

Mass and lifetime are defined. Quark models for nuclear particles like protons and electrons are specified. Quantized gravitons are discussed.

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References
1	Other electrochemical quark distributions are used to model other atoms and molecular bonds.
2	Baryonic and leptonic quarks are not used because we are not considering gamma-rays and nuclear processes.
3	Kramida, A., Ralchenko, Yu., Reader, J., and the NIST ASD Team (2018). NIST Atomic Spectra Database (version. 5.6.1). National Institute of Standards and Technology, Gaithersburg, MD, USA. For this reference, the lines from transitions between levels designated only by principal quantum numbers correspond to observations in which any finer structure is completely unresolved. Reported energy levels are the centre of groups of all fine-structure levels having the same $\rm{n}$ . Their values are based on observations of the Sun.
4	R.D. Deslattes, E.G. Kessler Jr., P. Indelicato, L. de Billy, E. Lindroth, J. Anton, J.S. Coursey, D.J. Schwab, C. Chang, R. Sukumar, K. Olsen, and R.A. Dragoset (2005), X-ray Transition Energies Database (version 1.2). National Institute of Standards and Technology, Gaithersburg, MD, USA.