Fields
//Bead Panel from a Baby Carrier// (detail). Bahau people. Borneo circa 1900, 36 x 12 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop.
Bead Panel from a Baby Carrier (detail). Bahau people. Borneo circa 1900, 36 x 12 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop.

A particle that is formed entirely from ethereal, imaginary or neutral components can be difficult to characterize and distinguish from other 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 generically noted by the symbol $\mathscr{F}$. For example we may vaguely refer to a set of leptonic seeds as an electromagnetic field, or a collection of gravitons as a gravitational field. But more exactly, a field can be specified by its quark coefficients $\ n$ just like any other compound quark

$\mathscr{F} \leftrightarrow \it{n}^{\sf{e}} \sf{e} + \it{n}^{\bar{\sf{e}}} \bar{\sf{e}} + \it{n}^{\sf{m}} \sf{m} + \it{n}^{\bar{\sf{m}}} \bar{\sf{m}} + \ldots$

If $\Delta n = 0$ for all types of quarks in this field, then by definition its enthalpy $\ H \left( \mathscr{F}\right)$ will be zero. Similarily the definitions for charge, strangeness, lepton number and baryon number imply that

$q \left( \mathscr{F}\right) =0$ $S \left( \mathscr{F}\right) =0$ $L \left( \mathscr{F}\right) =0$ $B \left( \mathscr{F}\right) =0$

The field's radius vector will also be null $\ \overline{\rho} \left( \mathscr{F}\right) = \left( 0, 0, 0 \right)$ and no work is required to bring together a collection of quarks like this. So such a field will also have no mass $\ m \left( \mathscr{F}\right)=0$. But despite all these zero values, $\mathscr{F}$ may still carry a significant amount of energy and momentum. So if part of $\mathscr{F}$ is absorbed by another particle P, the field may change P's temperature and motion. Then we say that P has been excited by an interaction with $\mathscr{F}$.

Fields and Equilibrium

Notice: this page is actively under construction
Notice: this page is actively under construction

We can make a description of how particle P attains dynamic equilibrium by considering that P is surrounded by an enormous field of quark-pairs that are out-of-phase anti-particles to each other. These titbits are like fragments of photons. They are generically written as $\begin{align} \sf { q \overline{q} } \end{align}$ and so $\Delta n = 0$ for all types of quarks in such a field. Equilibrium is achieved through interactions between P and photons made from these $\begin{align} \sf { q \overline{q} } \end{align}$ fragments. The field concept is vague enough so that we can assume there is some chance that any kind of photon will be absorbed or emitted. Then statistical constraints on these chances specify interactions between P and $\mathscr{F}$. These interactions moderate extreme motions, scatter energy values and cause regression toward mean temperatures. Energy and momentum can also be shuffled around between P and other large particles if they are all bathed in a field of photons carrying $\begin{align} \sf { q \overline{q} } \end{align}$ pairs.

More specifically let particle $\sf{P}$ be described by a historically ordered chain of events $\Psi$ and consider some pair of events $\sf{P}_{ \it{i}}$ and $\sf{P}_{ \it{f}}$ from the sequence

$\Psi ^{\sf{P}} = \left( \, \sf{P}_{1}, \sf{P}_{2}, \sf{P}_{3} \ \ldots \ \sf{P}_{ \it{i}} \ \ldots \ \sf{P}_{ \it{f}} \ \ldots \, \right)$

Let $\sf{P}_{ \it{i}}$ mark the absorption of $\sf{\gamma}$ a photon such that $\sf{P}_{ \it{i} \sf{\, -1}} +\sf{\gamma} \to \sf{P}_{ \it{i}}$. And let $\sf{P}_{ \it{f}}$ note the emission of $\sf{\gamma^{\prime}}$ a different photon $\sf{P}_{ \it{f}} \to \sf{P}_{ \it{f} \sf{\, +1}} + \sf{\gamma^{\prime}}$. The differing quark content of the two photons is drawn from the surrounding $\begin{align} \sf { q \overline{q} } \end{align}$ field such that

$\begin{align} \sf{P}_{ \it{f}} - \sf{P}_{ \it{i} } \ \ \leftrightarrow \ \ \frac{ \it{N}^{\, \sf{E}} }{2} \sf{e} \overline{\sf{e}} + \frac{ \it{N}^{\, \sf{M}} }{2} \sf{m} \overline{\sf{m}} + \ldots \end{align}$

Then the interaction, and the approach to equilibrium, is precisely specified by the seed coefficients $\ \it{N}^{\, \sf{E}}$, $\it{N}^{\, \sf{M}}$ etc. For a specific case of using $\begin{align} \sf { q \overline{q} } \end{align}$ pairs to obtain thermodynamic viability, take a look at the difference between the core quarks and the coefficients of all quarks in the models of, for example, pions .

Right.png Next step: particles that are excited by fields.



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