Field Test – EthnoPhysics

Outline

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Fields around Hydrogen

The field of quarks around hydrogen in its ground state is noted by $\mathscr{F} \! \! \left( 1\mathbf{S} \right) .$ It is composed from this force-carrying collection of simple field quanta

$\mathscr{F} \! \! \left( 1\mathbf{S} \right) \leftrightarrow 2\mathsf{m \overline{m}} + 2\mathsf{a \overline{a}} + 2\mathrm{ l \overline{l}} + 2\mathsf{d}\overline{\mathsf{d}}$

An atom of hydrogen is noted by $\mathbf{H} .$ So we write

$\mathbf{H} \! \left( 1\mathbf{S} \right) = \left\{ \, \mathsf{p^{+}} \! , \; \mathsf{e^{-}} \! , \; \mathscr{F} \! \! \left( 1\mathbf{S} \right) \rule{0px}{11px} \right\}$

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Coulomb’s Photon

The remaining simple field quanta around hydrogen are combined to define Coulomb’s photon $\widehat{ \boldsymbol{\gamma} }$ from the collection

$\widehat{\boldsymbol{\gamma}} = 2\mathsf{m \overline{m}} + 2\mathsf{a \overline{a}} + 2 \mathsf{d} \overline{\mathsf{d}} + 2\mathsf{dd}$

Each quark in $\widehat{ \boldsymbol{\gamma} }$ is described by its phase $\delta_{\theta} \hspace{0.5pt} .$ And these phases are used to sort quarks into a pair of sets, noted by $\mathcal{A}_{\mdsmwhtcircle}$ and $\mathcal{A}_{\mdsmblkcircle} \hspace{0.5pt} ,$ such 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 $\widehat{\boldsymbol{\gamma}}$ . We specify Coulomb’s photon in terms of these phase-components as

$\boldsymbol{\widehat{\gamma}} = \left\{ \mathcal{A}_{\mdsmwhtcircle} \hspace{0.5pt} , \, \mathcal{A}_{\mdsmblkcircle} \rule{0px}{12px} \right\}$

$\mathcal{A}_{\mdsmwhtcircle} \! \left( \boldsymbol{\widehat{\gamma}} \right) \equiv \left\{ \rule{0px}{14px} \mathsf{m}_{\mdsmwhtcircle}, \mathsf{d}_{\mdsmwhtcircle} , \left\{ \mathsf{a}_{\mdsmwhtcircle}, \mathsf{d}_{\mdsmwhtcircle} \right\}, \left\{ \rule{0px}{12px} \mathsf{m}_{\mdsmwhtcircle}, \mathsf{d}_{\mdsmwhtcircle}, \left\{ \mathsf{a}_{\mdsmwhtcircle}, \mathsf{d}_{\mdsmwhtcircle} \right\} \right\} \right\}$

$\mathcal{A}_{\mdsmblkcircle} \! \left( \boldsymbol{\widehat{\gamma}} \right) \equiv \left\{ \rule{0px}{14px} \overline{\mathsf{m}}_{\mdsmblkcircle}, \mathsf{d}_{\mdsmblkcircle}, \left\{ \overline{\mathsf{a}}_{\mdsmblkcircle}, \overline{\mathsf{d}}_{\mdsmblkcircle} \right\}, \left\{ \rule{0px}{12px} \overline{\mathsf{m}}_{\mdsmblkcircle}, \mathsf{d}_{\mdsmblkcircle}, \left\{ \overline{\mathsf{a}}_{\mdsmblkcircle}, \overline{\mathsf{d}}_{\mdsmblkcircle} \right\} \right\} \right\}$

Notice that these sets are phase anti-symmetric for all kinds of quarks except down-quarks, so we mark them using the letter $\mathcal{A} \hspace{0.5pt} .$

The total angular momentum quantum number of the photon is $\textsl{\textsf{J}} \, ( \hspace{0.8pt} \boldsymbol{\widehat{\gamma}} \hspace{0.8pt} ) \! = \! 1$ because $N^{\mathsf{D}} ( \hspace{0.8pt} \boldsymbol{\widehat{\gamma}} \hspace{0.8pt} ) \! = \! 8$ and $N^{\mathsf{U}} ( \hspace{0.8pt} \boldsymbol{\widehat{\gamma}} \hspace{0.8pt} ) \! = \! 0 .$

$\left| \Delta n^{\mathsf{D}} \right| = 4.$ So Coulomb’s photon is not free.

For ${\mathcal{A}}_{\mdsmwhtcircle}$ quark coefficients are $\Delta n^{\mathsf{D}} \! = \! -4$ and $\Delta n^{\mathsf{A}} \! = \! \Delta n^{\mathsf{M}} \! = \! -2 .$

For ${\mathcal{A}}_{\mdsmblkcircle}$ quark coefficients are $\Delta n^{\mathsf{D}} \! = \! 0$ and $\Delta n^{\mathsf{A}} \! = \! \Delta n^{\mathsf{M}} \! = \! +2 .$

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Cartesian Hydrogen Model

Then an atom of left-handed, ground-state hydrogen can be expressed as

$\mathbf{H} \! \left( 1\mathbf{S} \right) = \left\{ \rule{0px}{11px} \right.$ $, \, \mathsf{e^{-}} , \, \widehat{\boldsymbol{\gamma}} \hspace{-2px} \left. \rule{0px}{11px} \right\}$

This representation suggests a proton located at the spatial-origin of a Cartesian coordinate-system. The electron is located at some other spatial position in the hydrogen-centered system. And these two particles are presumably bound together by the force of exchanging $\widehat{ \boldsymbol{\gamma} }$ .

𝛄 Core Radius

Recall that the chemical enthalpy $H_{\! chem}$ is defined by a partial sum over just the chemical quarks in any particle. And there are no chemical quarks in Coulomb’s photon. So $\displaystyle H_{\! chem} \! \left( \boldsymbol{\widehat{\gamma}} \right) = 0 .$

Recall that the core radius of any particle is defined by ${\mathcal{R}}_{core} \equiv \dfrac{ H_{\! chem} }{ k_{\mathsf{F}} }$

Then core-radii of the phase-components of $\widehat{\boldsymbol{\gamma}}$ is zero

𝛄 Magnetic Radius

The magnetic radius of any particle is defined by

$\rho_{m} \equiv \dfrac{ \Delta n^{\mathsf{A}} U^{\mathsf{A}} - \Delta n^{\mathsf{M}} U^{\mathsf{M}} }{ k_{\mathsf{F}} }$

where the internal energy of a magnetic quark is $U^{ \mathsf{M}} \! = 1,186$ (MeV). And the internal energy of a amperic quark is hundreds of times smaller at $U^{ \mathsf{A}} \! = 3.122$ (MeV). So the magnetic radii of the Coulomb photon’s phase-components are

$\rho_{m}^{\, \mathcal{A}_{\mdsmblkcircle}} = \dfrac{2}{k_{\mathsf{F}}} \left( U^{\mathsf{A}} - U^{\mathsf{M}} \rule{0px}{11px} \right) = -2,366$ $\left( \mathsf{MeV} \! / k_{\mathsf{F}} \right)$

$\rho_{m}^{\, \mathcal{A}_{\mdsmwhtcircle}} = \dfrac{2}{k_{\mathsf{F}}} \left( U^{\mathsf{A}} - U^{\mathsf{M}} \rule{0px}{11px} \right) = 2,366$ $\left( \mathsf{MeV} \! / k_{\mathsf{F}} \right)$

𝛄 Electric Radius

Coulomb’s photon does not contain any electronic quarks. So $\rho_{e} \, ,$ the electric radius, of its phase-components are all zero.

𝛄 Central Radius

Also the central radius of any particle is defined by

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

and Coulomb’s photon does not contain any up quarks. So $\Delta n^{\mathsf{U}} = 0$ for both phase-components, and the central radius is

$\rho_{z} = \, \mathcal{R}_{core} - \dfrac{ \ U^{\mathsf{D}}}{k_{\mathsf{F}}} \Delta n^{\mathsf{D}}$

Then substituting-in values for the core-radius and quark-coefficients gives

$\rho_{z}^{\, \mathcal{A}_{\mdsmblkcircle}} = \dfrac{2}{k_{\mathsf{F}}} \left( U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} \! - 2U^{\mathsf{D}} \rule{0px}{11px} \right)$

$\rho_{z}^{\, \mathcal{A}_{\mdsmwhtcircle}} = -\dfrac{2}{k_{\mathsf{F}}} \left( U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} \! - 2U^{\mathsf{D}} \rule{0px}{11px} \right)$

The internal energy of a levo quark is $U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} \! = -0.049$ (eV). And the internal energy of a down quark is about a thousand times smaller at $U^{ \mathsf{D}} \! = -27$ (µeV). So the central-radius is approximately the same as the core-radius

$\rho_{z}^{\, \mathcal{A}_{\mdsmblkcircle}} \simeq \mathcal{R}_{core}^{\, \mathcal{A}_{\mdsmblkcircle}}$

$\rho_{z}^{\, \mathcal{A}_{\mdsmwhtcircle}} \simeq \mathcal{R}_{core}^{\, \mathcal{A}_{\mdsmwhtcircle}}$

𝛄 Radius Vector

The radius vector of any particle is defined by

$\overline{\rho} \equiv \left( \rho_{m}, \, \rho_{e}, \, \rho_{z} \right)$

So the radius-vectors of the Coulomb photon’s phase-components are

$\overline{\rho}^{\mathcal{A}_{\mdsmblkcircle}} = \dfrac{2}{k_{\mathsf{F}}} \! \left( \rule{0px}{12px} U^{\mathsf{A}} \! - U^{\mathsf{M}}, \, 0, \, U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} \! - 2U^{\mathsf{D}} \right)$

$\overline{\rho}^{\mathcal{A}_{\mdsmwhtcircle}} = -\dfrac{2}{k_{\mathsf{F}}} \! \left( \rule{0px}{12px} U^{\mathsf{A}} \! - U^{\mathsf{M}}, \, 0, \, U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} \! - 2U^{\mathsf{D}} \right)$

𝛄 Wavelength

Use $W = k_{\mathsf{F}} \left\| \, \overline{\rho} \, \right\|$ to eliminate the norm, and also thereby the dependence on $k_{\mathsf{F}}$

$W^{\! \mathcal{A}} = 2\sqrt{k_{mm}} \, \left| \, \rule{0px}{15px} U^{\mathsf{M}} \! - U^{\mathsf{A}} \right|$

$\lambda_{\mathsf{o}} \! \left( \boldsymbol{\widehat{\gamma}} \right) = \dfrac{hc}{\rule{0px}{11px} \,2 W^{ \! \mathcal{A}} } = \dfrac{hc}{ 4 \sqrt{k_{mm}} \, \left| \, U^{\mathsf{M}} \! - U^{\mathsf{A}} \rule{0px}{12px} \right| \, } \; = \; 0.3 \text{\sf{(fm)}}$

The index of refraction $\eta$ for a photon $\boldsymbol{\gamma}$ in a local environment $\mathtt{f}$ is noted as $\eta \! \left( \boldsymbol{\gamma}, \mathtt{f} \right).$ The refractive index for Coulomb’s photon $\widehat{\boldsymbol{\gamma}}$ when located inside the atomic core of hydrogen $\mathbf{H}$ is written as

$\widehat{\eta} \equiv \eta \! \left( \widehat{\boldsymbol{\gamma}}, \mathbf{H} \right)$

Then

$\lambda \! \left( \boldsymbol{\widehat{\gamma}} \right) = \dfrac{hc}{\rule{0px}{11px} \,2 \widehat{\eta} \hspace{0.5pt} W^{\! \mathcal{A}} } = \dfrac{hc}{ 4 \widehat{\eta} \sqrt{k_{mm}} \, \left| \, U^{\mathsf{M}} \! - U^{\mathsf{A}} \rule{0px}{12px} \right| \, }$

Compare to; radius of atomic hydrogen ~40,000 (fm), radius of proton ~0.8 (fm)

𝛄 Frequency

$\nu \! \left( \boldsymbol{\widehat{\gamma}} \right) \simeq \dfrac{c}{ \, \lambda_{\mathsf{o}} \: } = \dfrac{4}{h} \sqrt{k_{mm}} \, \left| \, U^{\mathsf{M}} \! - U^{\mathsf{A}} \rule{0px}{12px} \right| \; \; \sim 10^{24} \text{\sf{ (Hz) }}$

𝛄 Energy

$E \! \left( \boldsymbol{\widehat{\gamma}} \right) \simeq h \nu = 4 \sqrt{k_{mm}} \, \left| \, U^{\mathsf{M}} \! - U^{\mathsf{A}} \rule{0px}{12px} \right| \; \; \sim 4 \text{\sf{ (GeV) }}$

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Weak Quanta

big electromagnetic bosons?

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.

Strong Quanta

up quarks

Atomic and nuclear interactions are mediated by fields and forces too. Long-distance forces are carried by photons. But short-range forces from collisions, explosions and nuclear reactions can also be accurately modeled using simple $\mathsf{q \overline{q}}$ pairs. We especially consider pairs of baryonic, leptonic, and up-quarks.

These sorts of quark-pairs are an assortment of kicks and torques used to get hydrogen to jump from state to state

For examples, 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 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.

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