Mass

Outline

Mass is the most important characteristic for understanding particle mechanics. Many billions of dollars and thousands of physicist-lives have been spent measuring mass because it is such an essential concept in Newtonian mechanics. This applies to particles large and small, from planets to apples. But the richest trove of precise, reproducible data comes from nuclear particles. Mass is their signature quality. Literally hundreds of nuclear particles have been observed and carefully measured at enormous expense. So we start the analysis of mass by considering nuclear particles.

Nuclear Particles

Nuclear events are highly but not perfectly symmetric, like the mollusk shells in this French engraving. — Jean-Baptiste Lamarck (1744-1829), Terebra 402. Tableau Encyclopédique et Méthodique des Trois Règnes de la Nature, Paris 1791-1798. Photograph by D Dunlop.

Events that are highly symmetric combinations of at least sixteen Anaxagorean sensations are called nuclear events. Nuclear events are objectified as nuclear particles. They are modeled by bundles of quarks. Pions need just eight quarks, whereas the Higgs boson requires hundreds.

Nuclear particles must be highly symmetric so that they are stable enough to be measured, but not exactly symmetric. A few small asymmetries are needed or else an event would be indistinct and nondescript. Most nuclear particles are objectified from slightly different numbers of opposing binary sensations so that they have an angular momentum quantum number, baryon number, lepton number, charge and strangeness that are within two or three units of zero.

The way that these small asymmetries are balanced against each other determines a particle’s stability or lifetime. The most stable nuclear particles are the proton and the electron, so they get the most attention and are discussed below. But later, we also show a way of classifying dozens of other more ephemeral particles into 24 families based on their quark content.

Let particle P be characterized by its enthalpy $H$ and the work $W$ required to bring together its component quarks. Then the rest mass of P is defined by

$m \equiv \dfrac{ \sqrt{H^{2}-W^{2}} }{ c^{2} }$

Mass	Definition
heavy	$W^{2} \, \ll \, H^{2}$
material	$W^{2} \, < \, H^{2}$
ethereal	$W^{2} \, = \, H^{2}$
imaginary	$W^{2} \, > \, H^{2}$

where $c$ is a constant. This definition distinguishes several types of particles by their mass. If $m$ is positive then P is a material particle; an ordinary particle of matter, like a coin or a bullet. There is an important special case for material particles when the work required to make them is negligible compared to their enthalpy, then we say they are heavy particles. If the mass is zero then P is ethereal. And if $m^{2} < 0$ then P has an imaginary mass.¹The term imaginary is used here with its mathematical meaning. Particles with an imaginary mass are no more fictitious than any other sort of nuclear particle. They carry momentum and transmit forces like other particles. The main thing about having an imaginary mass is that it puts a particle in a logical category that is different from Newtonian particles. So they are not necessarily required to follow Newtonian laws of motion.

Roughly speaking, the rest mass describes how much energy is leftover after the work of assembling a particle has been completed.

We may use the mass to describe the hardness or density of a particle. Recall that $\left\| \, \overline{\rho} \, \right\|$ is the norm of the radius vector of P. Then the density of P is defined as

$\varrho \equiv \dfrac{ \, m c^{2} }{ \left\| \, \overline{\rho} \, \right\| }$

Particles and anti-particles have the same mass as each other. We have already seen how $H \! ( \mathsf{P} ) = - H \! ( \overline{\mathsf{P}} )$ and $W \! ( \mathsf{P} ) = W \! ( \mathsf{\overline{P}} )$ when conjugate symmetry is assumed. But the mass depends on these quantities squared. So

$m ( \mathsf{P} ) = m ( \mathsf{\overline{P}} )$

Sensory interpretation: Enthalpy characterizes the magnitude of all classes of sensation, whereas the work represents just somatic and visual sensations. The mass is established by their difference, which is mostly due to thermal sensation. So for heavy particles, thermal perceptions are more important than visual sensations. And for particles with an imaginary mass, visual sensations dominate awareness.

Photons are Ethereal

Photons are ethereal because by definition most of the quarks they contain appear in phase anti-symmetric pairs. This is true for all quark types except down-quarks which have $\left| {\Delta}n^{\mathsf{D}} \right| \ge 8 .$ So the electric and magnetic radii of any photon are nil. But the polar radius of a photon is

$\rho_{z} ( \boldsymbol{\gamma} ) = \dfrac{ - \Delta n^{ \mathsf{D}} U^{ \mathsf{D}} }{ k_{\mathsf{F}} }$

This quantity is very small because $U^{ \mathsf{D}} \simeq 0$ . But it is not exactly zero. We use it to assess the work required to assemble the quarks in a photon. In general, this is given by

$\begin{equation*} W = k_{\mathsf{F}} \left( \begin{split} & \; k_{mm} \rho_{m}^{2} + k_{ee} \rho_{e}^{2} + k_{zz} \rho_{z}^{2} \\ & + 2 k_{em} \rho_{e} \rho_{m} + 2k_{mz}\rho_{m} \rho_{z} \\ & \hspace{30px} + 2 k_{ez}\rho_{e} \rho_{z} \; \end{split} \right)^{\frac{1}{2}} \end{equation*}$

But for photons $\rho_{e} = \rho_{m} = 0 .$ And recall that $k_{zz} =1 .$ So the work required to produce a photon is simply

$W \! ( \boldsymbol{\gamma} ) = k_{\mathsf{F}} \left| \, \rho_{z} \, \rule[-3px]{0px}{14px} \right| = \left| \, \Delta n^{ \mathsf{D}} U^{ \mathsf{D}} \, \rule[-3px]{0px}{14px} \right|$

This quantity is small, but not zero. Compare it to the enthalpy which is defined for any particle from the sum

$\displaystyle H \equiv \sum_{\zeta=1}^{16} \Delta n^{\zeta} U^{\zeta}$

For photons, all $\Delta n$ terms are null except for the down quarks, so

$\displaystyle H\! ( \boldsymbol{\gamma} ) = \Delta n^{\mathsf{D}} U^{\mathsf{D}}$

Again, this quantity is small but not zero. However, notice that its modulus has precisely the same value as the work, so $H\! ( \boldsymbol{\gamma} )^{2} = W\! ( \boldsymbol{\gamma} )^{2} .$ Then by definition all photons must have no mass. Not just close to zero, but perfectly zero. Photons are ethereal

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

Lifetime

Recall that the reference sensation of touching ice is used to calibrate the measurement of temperature. So to make good measurements we are getting more precise about what we mean by ice. Accordingly, here is a definition of temperature that is tied to the triple point of water. The thermodynamic temperature is defined by

$T \mathsf{(K)} \; \equiv \; T ( ^{\circ} \mathsf{C} ) + \mathrm{273.15}$

where $T ( ^{\circ} \mathsf{C} )$ is the Celsius temperature. In the following discussion, the symbol $T$ refers to the thermodynamic temperature, in units called kelvins, noted by (K).

Quarks are conserved but compound quarks may decay. Their permanence is characterized by a number called the mean-life. Let particle P be described by its thermodynamic temperature $T$ . The mean life of P is defined as

$\tau \equiv k_{\tau} e^{-T}$

The lifetime of a particle is suggested by this ancient network of threads and spectral figures. — Textile fragment, Chancay people. Pre-Columbian Peru, 50 x 30 cm. Photograph by D Dunlop.

where $e$ notes the exponential function and the constant $k_{\tau} = \mathsf{ 2.6 x 10 }^{\mathsf{56}}$ seconds. Customarily, if P is an atom of hydrogen in its ground-state, then $T=0 \; \mathsf{ (K) }$ and $e^{ 0} = 1$ , so this constant $k_{\tau}$ is called the mean-life of hydrogen²For more detail, see the discussion of stability in the article about atomic hydrogen..

A particle with a negative temperature supposedly has a longer mean-life than hydrogen. But for EthnoPhysics, the only particles like this are some quarks and field quanta which are not given space-time descriptions. All models of observed nuclear particles have a positive thermodynamic temperature.

Particle stability is also characterized by a number called the full width which is noted by $\varGamma$ and defined as

$\varGamma \equiv \dfrac{h}{2 \pi \tau}$

The total number of any specific type of thermodynamic quark does not vary if ordinary-quarks are swapped with anti-quarks of the same type. And with the assumption of conjugate symmetry both kinds of quarks have the same temperature. So

$\tau ( \mathsf{P} ) = \tau ( \overline{\mathsf{P}} )$

and

$\varGamma ( \mathsf{P} ) = \varGamma ( \overline{\mathsf{P}} )$

Particles and their associated anti-particles have the same mean-life and full-width.

The Proton

EthnoPhysics describes the proton by starting with an archetypal chain of events written as

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

The proton is represented by this collage of twentyfour seed icons.

where each repeated cycle $\mathsf{\Omega}$ is a bundle of 24 Anaxagorean sensations. The chain of events $\Psi$ is generically called a history of the proton. To be exact the prototypical sensations are; eight somatic perceptions on the right-side, four on the left, four burning thermal feelings, four freezing, and finally four black visual sensations.

Each Anaxagorean sensation may be objectified to define a seed. And Aristotle refers to the primordial atomic mass³Aristotle, Physics 203 a 21, De Caelo 303 a 16, and De Anima 404 a 4. as πανσπερμία or a seed-aggregate. So to make a seed-aggregate model for the proton we express $\mathsf{\Omega}$ as a bundle of seeds

$\mathsf{\Omega} \! \left( \mathsf{p^{+}} \right) \leftrightarrow \mathrm{4}\mathsf{D} + \mathrm{4}\mathsf{B} + \mathrm{4}\mathsf{T} + \mathrm{8}\mathsf{O} + \mathrm{4}\overline{\mathsf{O}}$

A Quark Model of the Proton

Quarks are defined by pairs of seeds. So the seed-aggregate model of the proton is further developed by associating seeds in pairs to form the following quarks

→

Thus a proton is represented by a bundle of twelve quarks. Here is a mathematical way of expressing the arrangement, along with an iconic image for the model.

$\mathsf{\Omega} \! \left( \mathsf{p^{+}} \right) \leftrightarrow \mathrm{4}\mathsf{d} + \mathrm{4}\mathsf{b} + \mathrm{4}\overline{ \mathsf{t} }$

The proton is represented by this image of twelve quark icons stacked into a parallelepiped.

Using these quarks, the mass of the proton is calculated to be $938.2720460$ (MeV/c²). This is exactly the same as the experimentally observed value because adjustable parameters like quark energies have been carefully chosen⁴The mass of the proton $m$ can be written in terms of the work, enthalpy and quark coefficients. The resulting equation can be solved to find $U^{\mathsf{T}}$ the internal-energy of top-quarks as $U^{\mathsf{T}} = U^{\mathsf{B}} + m c^{2}/4$ . This relationship between top and bottom quarks is then used to constrain the selection of other adjustable parameters. to get this result. For more detail, please see the Nuclear Particles spreadsheet.

A Ground-State Proton Model

In the foregoing model, quark coefficients are all integer multiples of two, and so the image is drawn with the back row of quarks the same as the front row. But we cannot have two identical quarks in the same bundle and still satisfy Pauli’s exclusion principle.

So the quark model is developed further with an additional requirement that the quarks in the front and back rows are out of phase with each other. That is, they are distinguished by the helicity of the reference frame as noted by $\mathsf{F}_{\mdsmwhtcircle}$ or $\mathsf{F}_{\mdsmblkcircle}$ . This satisfies the definition for being in a ground-state and so the new arrangement is called a ground-state model of the proton. It is expressed mathematically as

$\mathsf{\Omega} \! \left( \mathsf{p^{+}} \right) = \left\{ \rule{0px}{14px} \left\{ \rule{0px}{12px} \left\{ \mathsf{d}, \mathsf{b}, \overline{\mathsf{t}} \right\}, \, \mathsf{d}, \, \mathsf{b}, \, \overline{\mathsf{t}}, \, \mathsf{F}_{\mdsmwhtcircle} \right\}, \left\{ \rule{0px}{12px} \left\{ \mathsf{d}, \mathsf{b}, \overline{\mathsf{t}} \right\}, \, \mathsf{d}, \, \mathsf{b}, \, \overline{\mathsf{t}}, \, \mathsf{F}_{\mdsmblkcircle} \right\} \right\}$

To illustrate this model, we show quarks with a background that is dark or bright depending on their phase. Then the image above can be made into a movie that uses shadows, horizons and background brightness to suggest a quark’s relationship with the frame-of-reference.

Proton Stability

The temperature of a proton in its ground-state is easily calculated from the quark average

$\begin{align*} T \! \left( \mathsf{p^{+}} \right) &= \frac{1}{N_{\mathsf{q}} } \sum T_{\mathsf{q}} \\ &= \frac{1}{12} \left( 4T_{\mathsf{d}} + 4T_{\mathsf{b}} + 4T_{\mathsf{t}} \right) \\ &= \mathrm{2.7254885} \, \mathsf{(K)} \rule{0px}{13px} \end{align*}$

This is within experimental uncertainty of the observed⁵Fixsen, D. J., Temperature of the Cosmic Microwave Background, The Astrophysical Journal 707 (2): 916–920 (2009). value of $2.72548 \pm 0.00057$ (K) for the thermal black body spectrum of the microwave background radiation. It is common to describe this background radiation as ‘cosmic’ and to assert that it comes from a ‘big-bang’. But seeing protons, bare naked in their ground-states, might offer another sort of explanation.

The proton temperature corresponds to a calculated mean life of $1.71 \times 10^{55}$ seconds, which is consistent with the observed⁶K.A. Olive et al. Particle Data Group Review of Particle Physics, Chin. Phys. C, 38, 090001 (2014). lower bound of $6.6 \times 10^{36}$ seconds. So the proton is an extremely stable particle. This gives it a starring role in narratives connecting cause and effect.

The Electron

EthnoPhysics describes the electron by starting with an a chain of events written as

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

Here is a collage of quark icons that model an electron.

where each repeated cycle $\mathsf{\Omega}$ is a bundle of 40 Anaxagorean sensations. The chain of events $\Psi$ is generically called a history of the electron. To be exact, the sensations are eight right-side and twelve left-side somatic feelings; two burning, two freezing, two warm and two cool thermal perceptions; four yellow, four blue and four white visual sensations.

Each of these Anaxagorean sensations may be objectified to define a seed. And so to make a seed-aggregate model of the electron we express $\mathsf{\Omega}$ as a bundle of seeds

$\mathsf{\Omega} \! \left( \mathsf{e^{-}} \right) \leftrightarrow \mathrm{4} \mathsf{U} + \mathrm{2} \mathsf{B} + \mathrm{2} \mathsf{T} + \mathrm{2} \mathsf{S} + \mathrm{2} \mathsf{C} + \mathrm{4} \mathsf{G} + \mathrm{4} \mathsf{E} + \mathrm{8} \mathsf{O} + \mathrm{12} \overline{\mathsf{O}}$

A Quark Model of the Electron

Quarks are defined by pairs of seeds. So the seed-aggregate model of the electron is further developed by associating seeds in pairs to form the following quarks

→

Thus an electron is represented by a bundle of twenty quarks. Here is a mathematical way of expressing the arrangement, along with an iconic image for the model.

$\mathsf{\Omega} \! \left( \mathsf{e^{-}} \right) \leftrightarrow \mathrm{4}\overline{\mathsf{u}} + \mathrm{2}\overline{\mathsf{b}} + \mathrm{2}\mathsf{t} + \mathrm{2} \overline{\mathsf{s}} + \mathrm{2}\mathsf{c} + \mathrm{4} \overline{\mathsf{g}} + \mathrm{4}\mathsf{e}$

Using these quarks, the mass of the electron is calculated to be $0.5109989280$ (MeV/c²). This is exactly the same as the experimentally observed value because adjustable parameters like quark energies have been carefully chosen⁷The mass of the electron $m$ can be written in terms of the work, enthalpy and quark coefficients. The resulting quadratic equation can be solved to find $U^{\mathsf{U}}$ the internal-energy of up-quarks as

$U^{\mathsf{U}} = \dfrac{ m^{2} c^{4} - \beta^{2} + 16k_{ee} \left( U^{\mathsf{E}} + U^{\mathsf{G}} \right)^{2} }{ 8\beta -32k_{ez} \left( U^{\mathsf{E}} + U^{\mathsf{G}} \right) }$

where $\beta = -4U^{\mathsf{E}} + 4U^{\mathsf{G}} - 2U^{\mathsf{T}} + 2U^{\mathsf{B}} + 2U^{\mathsf{S}} - 2U^{\mathsf{C}} .$ This relationship is then used to constrain the selection of other adjustable parameters. to get this result. For more detail, please see the Nuclear Particles spreadsheet.

A Ground-State Electron Model

So the quark model is developed further with an additional requirement that the quarks in the front and back rows are out of phase with each other. That is, they are distinguished by the helicity of the reference frame as noted by $\mathsf{F}_{\mdsmwhtcircle}$ or $\mathsf{F}_{\mdsmblkcircle} .$ This satisfies the definition for being in a ground-state and so the new arrangement is called a ground-state model of the electron. It is expressed mathematically as

$\mathsf{\Omega} \! \left( \mathsf{e^{-}} \right) = \left\{ \rule{0px}{14px} \left\{ \left\{ \mathsf{e}, \mathsf{t}, \overline{\mathsf{s}}, \overline{\mathsf{g}}, \overline{\mathsf{u}} \right\}, \, \mathsf{e}, \, \mathsf{c}, \, \overline{\mathsf{b}}, \, \overline{\mathsf{g}}, \, \overline{\mathsf{u}}, \, \mathsf{F}_{\mdsmwhtcircle} \right\}, \left\{ \left\{ \mathsf{e}, \mathsf{t}, \overline{\mathsf{s}}, \overline{\mathsf{g}}, \overline{\mathsf{u}} \right\}, \, \mathsf{e}, \, \mathsf{c}, \, \overline{\mathsf{b}}, \, \overline{\mathsf{g}}, \, \overline{\mathsf{u}}, \, \mathsf{F}_{\mdsmblkcircle} \right\} \right\}$

To illustrate this model, we show quarks with a background that is dark or bright depending on their phase. Then the image above can be made into a short movie that uses shadows, horizons and background brightness to suggest a quark’s relationship with the frame-of-reference.

Gravitons

The union of a photon $\boldsymbol{\gamma}$ and its corresponding anti-photon $\overline{\boldsymbol{\gamma}}$ is noted by the symbol $\mathsf{\Gamma}$ and called a graviton

$\mathsf{\Gamma} \equiv \left\{ \, \boldsymbol{\gamma}, \, \overline{\boldsymbol{\gamma}} \, \rule{0px}{11px} \right\}$

Let $\mathsf{U}$ and $\mathsf{D}$ note the number of up seeds and down seeds in a particle. And recall that seeds are conserved. So the foregoing definition of a graviton implies that

$N^{\mathsf{U}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) + N^{\mathsf{U}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)$

and

$N^{\mathsf{D}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) + N^{\mathsf{D}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)$

Swapping quarks with anti-quarks does not change thermodynamic seed counts, so $N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) = N^{\mathsf{U}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px}\right)$ and $N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) = N^{\mathsf{D}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px}\right)$ . Then the seed coefficients of the graviton are given by

$N^{\mathsf{U}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 2N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right)$

and

$N^{\mathsf{D}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 2N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right)$

Using these values, the angular momentum quantum number of a graviton is found as

$\begin{align*} \textsl{\textsf{J}} \left( \mathsf{\Gamma} \rule{0px}{10px}\right) &\equiv \dfrac{ \, \left| \, N^{\mathsf{U}} \! \left( \mathsf{\Gamma} \rule{0px}{10px}\right) - N^{\mathsf{D}} \! \left(\mathsf{\Gamma} \rule{0px}{10px}\right) \, \right| \, }{8} \\ &= \dfrac{ \, \left| \, 2N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) - 2N^{\mathsf{D}} \! \left(\boldsymbol{\gamma} \rule{0px}{10px}\right) \, \right| \, }{8} \\ &= 2 \, \dfrac{ \, \left| \, N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) - N^{\mathsf{D}} \! \left(\boldsymbol{\gamma} \rule{0px}{10px}\right) \, \right| \, }{8} \\ &= 2 \, \textsl{\textsf{J}} \left( \boldsymbol{\gamma} \right) \rule{0px}{15px} \end{align*}$

But by definition the angular momentum quantum number of any photon is one. So for all gravitons $\textsl{\textsf{J}} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 2.$

Gravitons are defined by the union of a photon with its matching anti-photon. So the net number of any sort of quark in a graviton is zero. This specifically includes down quarks. Photons have mismatched quantities of down-quarks and down anti-quarks. But for gravitons, these imbalances cancel each other perfectly so that $\Delta n = 0$ for all types of quarks. Substituting these conditions into the definitions for charge, strangeness, lepton number, baryon number and enthalpy gives

$q \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0$

$S \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0$

$L \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0$

$B \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0$

and

$H \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0$

Then recall that the lepton-number, baryon-number and charge are conserved, so any particle may freely absorb or emit countless gravitons without altering its own values for these quantum numbers.

Since $\overline{\rho} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = (0, 0, 0)$ no work $W$ is required to assemble the quarks in a graviton

$W \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv k_{\mathsf{F}} \left\| \, \overline{\rho} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \right\| = 0$

This result is combined with the null value for the enthalpy to find the mass of a graviton as

$m \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv \dfrac{1}{c^{2}} \sqrt{ H^{2}-W^{2} \; } = 0$

So gravitons have no mass, they are ethereal.

Examples of Gravitons

Center of Mass

When some particle P absorbs a graviton, we say the graviton is situated at P’s center of mass. To be more exact about this location, let us examine some of the radii we use to make quark models. By definition all gravitons must have $\Delta n = 0$ for all types of quarks, including down-quarks. Therefore, the inner radius of any graviton is nil

$\rho_{in} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv \dfrac{ \, \left| \Delta n^{\mathsf{D}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \right| \, }{8} \sqrt{ \dfrac{hc}{2\pi k_{\mathsf{F}}} \rule{0px}{16px} } = 0$

We can also assess the radius vector of a graviton because $\overline{\rho}$ is defined from sums of quark coefficients, and quarks are conserved. So

$\overline{\rho} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = \overline{\rho} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) + \overline{\rho} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)$

But as discussed earlier, any particle and its anti-particle have symmetrically opposed radius vectors. So for photons

$\overline{\rho} \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) = - \overline{\rho} \left( \overline{\boldsymbol{\gamma}} \right)$

and then

$\overline{\rho} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = (0, 0, 0)$

Thus the graviton is always positioned at the origin of the quark space that frames P. This is consistent with having an inner radius of zero. So the center-of-mass and the origin of quark-space are understood to be the same location.

If P is not electrically or magnetically polarized then we say that it is centered on the polar axis. So for unpolarized particles, the spatial-origin, the center-of-mass, and something like the center-of-P are all in the same place. Furthermore, this same location is later specified as the origin for a Cartesian frame-of-reference centered on P.

So we round-up all these related notions to mathematically designate a center or core of P as follows. Consider some particle Q composed from a subset of the quarks in P. When the norm of Q’s radius vector is smaller than the core radius of P we write

$\left\| \, \overline{\rho}^{\, \sf{Q}} \, \right\| \le \rho_{core} ^{\, \mathsf{P}}$

If this condition is satisfied then we say that Q is in the core of P. Then Q is close to the center of P’s mass. And Q is not free.

Gravitons are objectified from highly symmetric combinations of sensations, as suggested by this bead panel from Borneo. — Baby Carrier panel, Ngaju people. Borneo 20th century, 41 x 26 cm. Photograph by D Dunlop.

Nuclear Particle Classification

Nuclear particles are sorted into 25 different family groups, from baryons to leptons, based on the minimum number of down quarks in their core.

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References
1	The term imaginary is used here with its mathematical meaning. Particles with an imaginary mass are no more fictitious than any other sort of nuclear particle. They carry momentum and transmit forces like other particles. The main thing about having an imaginary mass is that it puts a particle in a logical category that is different from Newtonian particles. So they are not necessarily required to follow Newtonian laws of motion.
2	For more detail, see the discussion of stability in the article about atomic hydrogen.
3	Aristotle, Physics 203 a 21, De Caelo 303 a 16, and De Anima 404 a 4.
4	The mass of the proton $m$ can be written in terms of the work, enthalpy and quark coefficients. The resulting equation can be solved to find $U^{\mathsf{T}}$ the internal-energy of top-quarks as $U^{\mathsf{T}} = U^{\mathsf{B}} + m c^{2}/4$ . This relationship between top and bottom quarks is then used to constrain the selection of other adjustable parameters.
5	Fixsen, D. J., Temperature of the Cosmic Microwave Background, The Astrophysical Journal 707 (2): 916–920 (2009).
6	K.A. Olive et al. Particle Data Group Review of Particle Physics, Chin. Phys. C, 38, 090001 (2014).
7	The mass of the electron $m$ can be written in terms of the work, enthalpy and quark coefficients. The resulting quadratic equation can be solved to find $U^{\mathsf{U}}$ the internal-energy of up-quarks as $U^{\mathsf{U}} = \dfrac{ m^{2} c^{4} - \beta^{2} + 16k_{ee} \left( U^{\mathsf{E}} + U^{\mathsf{G}} \right)^{2} }{ 8\beta -32k_{ez} \left( U^{\mathsf{E}} + U^{\mathsf{G}} \right) }$ where $\beta = -4U^{\mathsf{E}} + 4U^{\mathsf{G}} - 2U^{\mathsf{T}} + 2U^{\mathsf{B}} + 2U^{\mathsf{S}} - 2U^{\mathsf{C}} .$ This relationship is then used to constrain the selection of other adjustable parameters.