Protons in Spacetime – EthnoPhysics

A Spatially Extended Proton

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The foregoing proton models do not contain any leptonic-quarks, so all their electric and magnetic radii are zero. Their central radii are very small, but still finite. So we say these protons are one-dimensional particles.

We can also combine a pointy proton-core with the levo-quarks from $\mathscr{F} \! \! \left( 1\mathbf{S} \right)$ to designate as a extended proton-core

$\leftrightarrow 4\mathsf{b} + 4\overline{\mathsf{t}} + 2 \mathrm{l} \mathrm{\bar{l}}$

$\leftrightarrow$ $+ \; 2 \mathrm{l} \mathrm{\bar{l}}$

$= \left\{ \mathsf{P}_{\mdsmwhtcircle} \hspace{0.8pt} , \, \mathsf{P}_{\mdsmblkcircle} \rule{0px}{12px} \right\}$

$\mathsf{P}_{\! \mdsmwhtcircle} \! \left( \rule{0px}{10px} \right.$ $\left. \rule{0px}{10px} \right) \equiv \left\{ \mathsf{b}_{\mdsmwhtcircle}, \, \overline{\mathsf{t}}_{\hspace{0.8pt} \mdsmwhtcircle}, \, \mathrm{l}_{\hspace{0.8pt} \mdsmwhtcircle}, \left\{ \mathsf{b}_{\mdsmwhtcircle}, \, \overline{\mathsf{t}}_{\hspace{0.8pt}\mdsmwhtcircle}, \, \mathrm{l}_{\hspace{0.8pt}\mdsmwhtcircle} \right\} \rule{0px}{14px} \! \! \right\}$

$\mathsf{P}_{\! \mdsmblkcircle} \! \left( \rule{0px}{10px} \right.$ $\left. \rule{0px}{10px} \right) \equiv \left\{ \mathsf{b}_{\mdsmblkcircle}, \, \overline{\mathsf{t}}_{\hspace{0.8pt} \mdsmblkcircle}, \, \overline{\mathrm{l}}_{\hspace{0.8pt} \mdsmblkcircle}, \left\{ \mathsf{b}_{\mdsmblkcircle}, \, \overline{\mathsf{t}}_{\hspace{0.8pt}\mdsmblkcircle}, \, \overline{\mathrm{l}}_{\hspace{0.8pt}\mdsmblkcircle} \right\} \rule{0px}{14px} \! \! \right\}$

This positively-charged levorotatory kernel can be used to specify the spatial origin for a left-handed Cartesian coordinate system centered on the proton in $\mathbf{H} .$ For a right-handed coordinate system, substitute dextro-quarks for levo-quarks.

Levo quarks are the only sort of chemical quark anywhere in a proton. So the chemical enthalpy of a proton, or of any of its components, is given by

$\displaystyle H_{\! chem} = \Delta n {^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}} \, U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}$

where $U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}$ is the internal energy of a levo-quark. And note that $\Delta n {^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}} ( \mathsf{P}_{\! \mdsmwhtcircle} ) = -2$ and $\Delta n {^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}} ( \mathsf{P}_{\! \mdsmblkcircle} ) = +2 .$ So the chemical-enthalpy of a phase-component is

$H_{chem} \! \left( \mathsf{P}_{\! \mdsmblkcircle} \right) \, = \, - H_{chem} \! \left( \mathsf{P}_{\! \mdsmwhtcircle} \right) \, = \; \pm 2 U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}$

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

$\mathcal{R}_{core} \! \left( \mathsf{P}_{\! \mdsmblkcircle} \right) \, = \, - \mathcal{R}_{core} \! \left( \mathsf{P}_{\! \mdsmwhtcircle} \right) \, = \; \pm \dfrac{\, 2U^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} }{k_{\mathsf{F}}}$

Now let $r_{ \! p}$ note the charge radius [] of a proton. This is experimentally observed [] to be

$r_{ \! p} = 8.4075 \pm 0.0064 \times 10^{-16}$ (m)

$= .84075 \pm 0.00064$ (fm)

Experimentally, $U( \text{lefty stereoisomer} ) \! = -0.1152$ (eV). So

$k_{\mathsf{F}} = 0.13708$ (eV/fm)

$= 1.3708 \times 10^{16}$ (eV/m)

$= 1.3708 \times 10^{10}$ (MeV/m)

$= 13.708$ (MeV/nm)

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Outline

A Ground-State Proton

This article shows a few different ways of making spacetime descriptions of protons. It extends an earlier discussion that presents the proton as a bundle of quarks like this

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

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

Spacetime descriptions require a frame of reference which is noted by $\mathsf{F} .$ The frame is characterized by its chirality which is marked using a subscript like $\mathsf{F}_{\mdsmwhtcircle}$ or $\mathsf{F}_{\mdsmblkcircle} \, .$ The frame’s chirality is used to determine a phase $\delta_{\theta}$ for each quark. And for a ground state model of the proton, quarks in the front and back rows are required to be out of phase with each other. Then the proton has perfect phase symmetry. There are no anti-symmetric quark-pairs, so the proton can be represented by a pair of phase components written as

This diagram shows 2 clumps of quarks as phase-components in different positions that define a polar axis. — A proton in a one-dimensional quark space.

$\mathsf{\Omega} \! \left( \mathsf{p^{+}} \right) = \left\{ \mathcal{S}_{\mdsmwhtcircle} \, , \ \mathcal{S}_{\mdsmblkcircle} \right\}$

where

$\mathcal{S}_{\mdsmwhtcircle} = \left\{ \rule{0px}{11px} \left\{ \mathsf{d}, \mathsf{b}, \overline{\mathsf{t}} \right\}, \, \mathsf{d}, \, \mathsf{b}, \, \overline{\mathsf{t}}, \, \mathsf{F}_{\mdsmwhtcircle} \, \right\}$

$\mathcal{S}_{\mdsmblkcircle} = \left\{ \rule{0px}{11px} \left\{ \mathsf{d}, \mathsf{b}, \overline{\mathsf{t}} \right\}, \, \mathsf{d}, \, \mathsf{b}, \, \overline{\mathsf{t}}, \, \mathsf{F}_{\mdsmblkcircle} \, \right\}$

and

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

The two phase-components are shown separated from each other, and a rod connecting them represents the central axis. Thus a ground-state proton is illustrated in a one-dimensional quark space.

For an isotropic Cartesian spacetime description of the proton, this clump of quarks is presumably whirling around the central axis such that colors and shapes become indistinct. So we make a heuristic picture of a proton in spacetime as a greyish spheroid with a protruding arrow to represent the axis of rotation.

A schematic visual representation of a proton in spacetime. — A spacetime picture of a proton with its angular momentum vector.

There are four down-quarks, and no up-quarks in the ground-state model. So in this configuration, the proton’s chirality is $\delta_{\mathsf{o}} = -1 .$ The total angular momentum quantum number of the proton is

$\textsl{ \textsf{J} } \! \! \left( \mathsf{p^{+}} \right) \equiv \dfrac{ \, \left| \, N^{\mathsf{U}} - N^{\mathsf{D}} \, \right| \, }{8} = \dfrac{1}{2}$

And its principal quantum number is

$\mathrm{n} \! \left( \mathsf{p^{+}} \right) \equiv \dfrac{ \; n^{\mathsf{d}} \! \left( \mathsf{p^{+}} \right) \; }{4} = 1$

Recall that the spatial extent of any particle is characterized by its orbital radius $R ,$ which in-turn depends on the mechanical energy. This energy is observed to be $E = 938.272$ (MeV). So the orbital radius of the proton is

$R \! \left( \mathsf{p^{+}} \right) \equiv \dfrac{hc}{2\pi} \dfrac{ \sqrt{\textsl{\textsf{J}} \; }}{E} = 1.487 \times 10^{-16}$ (m).

This dimension is comparable to observations of the proton’s charge radius at $r_{\mathsf{p}} = 8.414 \times 10^{-16}$ (m). And both values are about a million times smaller than the Bohr radius of a hydrogen atom.

A Proton in 1-Dimensional Space

We mark the central-axis of a proton using the algebraic notation $\hat{z} \equiv (0, 0, 1) .$ This axis is used to define the direction of the angular momentum vector. So in a particle-centered Cartesian frame, ${\mathrm{\overline{J}}} = \left( 0, \ 0, \ {\mathrm{J}}_{z} \right) .$ Following the usual graphic conventions, we draw this axis vertically, with the positive direction pointed towards the top of a page. By definition, the $z$ -component of the angular momentum is

$\mathrm{J}_{z} \equiv \delta_{z} \dfrac{h}{16\pi} \sqrt{ \left( N^{\mathsf{U}}-N^{\mathsf{D}} \right)^{2} + 8 \left| N^{\mathsf{U}}-N^{\mathsf{D}} \right| \; \rule{0px}{12px} }$

This quantity is negative because the chirality of a ground-state proton is $\delta_{\mathsf{o}} = -1 .$ And so

This schematic diagram shows the relationship between the polar axis of a proton and its angular momentum vector. — A proton in a
one-dimensional space.

$\mathrm{\overline{J}} \! \left( \mathsf{p^{+}} \right) = \dfrac{ -\sqrt{3} h }{4\pi} \, \hat{z}$

The spatial character of the proton is described by its inner radius which is defined by

$\rho_{in} \equiv \dfrac{ \, \left| \Delta n^{\mathsf{D}} \rule{0px}{9px} \right| \, }{8} \sqrt{ \dfrac{hc}{2\pi k_{\mathsf{F}}} \rule{0px}{14px} }$

And its outer radius which is

$\rho_{out} \equiv \dfrac{ \, N^{\mathsf{D}} \, }{8} \sqrt{ \dfrac{hc}{2\pi k_{\mathsf{F}}} \rule{0px}{14px} }$

For the unadorned proton $N^{\mathsf{D}} = \left| \Delta n^{\mathsf{D}} \rule{0px}{9px} \right| = 4 ,$ and so $\rho_{in} = \rho_{out} \; .$ But the wavevector of any particle is defined by

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

So for the proton, $\displaystyle \overline{\kappa} ( \mathsf{p^{+}} ) = \left( 0, \, 0, \, 0 \right) .$ The wavenumber is defined by the norm of a wavevector. It is written without an overline as $\kappa \equiv \left\| \, \overline{\kappa} \, \right\|$ . Thus $\kappa ( \mathsf{p^{+}} ) = 0 \, .$ Moreover, the wavelength of any particle is given by

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

So for the proton $\lambda ( \mathsf{p^{+}} ) = 0$ too. We use this wavelength to describe the proton’s motion. If the frame F is inertial, then $\widetilde{\kappa}^{\mathsf{F}} = \left( 0, 0, 0 \right) .$ And for any particle P, the momentum is defined by

$\overline{p} \equiv \dfrac{h}{2\pi} \left( \overline{\kappa}^{\mathsf{P}} \! - N^{\mathsf{P}} \, \widetilde{\kappa}^{\mathsf{F}} \right)$

So the ground-state proton has no momentum. It is stationary, and $p \, ( \mathsf{p^{+}} ) = 0$ in any inertial frame of reference.

History of a 1D Proton

EthnoPhysics describes the history of a proton using an ordered chain of events noted by $\Psi ( \mathsf{p^{+}} ) \, .$ Events are repetitive so that the chain may be written as

$\Psi = \left( \mathsf{\Omega}_{0}, \, \mathsf{\Omega}_{1}, \, \mathsf{\Omega}_{2} \, \ldots \, \mathsf{\Omega}_{k} \, \ldots \, \right)$

where each repeated cycle $\mathsf{\Omega} ( \mathsf{p^{+}} )$ is a bundle of $N$ quarks generically noted by

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

Any displacement of the proton during this history is defined by

$d \! z \equiv \delta_{z} \, \dfrac{\lambda}{\, N \,}$

But as shown above, the wavelength of a ground-state proton is zero. So $d \! z ( \mathsf{p^{+}} ) = 0$ too. The Cartesian coordinate of the $k^{\mathsf{th}}$ event is given by

$\displaystyle z_{k} \equiv z_{\mathsf{o}} + \sum_{i=1}^{k} d \! z_{i}$

So $z_{k} ( \mathsf{p^{+}} ) = z_{\mathsf{o}}$ for any $k.$ And the position of the proton on the central-axis does not change.

The history of a proton may be represented using a Cartesian plane where different bundles are illustrated in different locations along the temporal axis. Here is a diagram showing the $k^{\mathsf{th}}$ event.

A proton in a one-dimensional space with an additional time-axis to show its history.

The period $\widehat{\tau}$ of any particle is defined by

$\widehat{\tau} \equiv \dfrac{h}{\sqrt{ c^{2}p^{2} + m^{2}c^{4} \, \rule{0px}{9px} }}$

A ground-state proton is stationary in any inertial frame, $p = 0 .$ Dp

$\widehat{\tau} \! \left( \mathsf{p^{+}} \right) = \dfrac{h}{mc^{2}} = 4.4 \times 10^{-24}$ (s)

The time of occurrence for any specific event $\mathsf{\Omega}_{k}$ is given by a sum of periods.

$\displaystyle t_{k} \equiv t_{0} + \epsilon_{t} \! \sum_{i=1}^{k} \widehat{\tau}_{i}$

So $\widehat{\tau}$ gives the elapsed time between consecutive events.

This is the finest possible resolution along the temporal axis because spacetime is quantized. Any measurements of $t$ must be at least this big.

$\Psi \! \left( \mathsf{p^{+}} \right) = \left( \mathsf{\Omega}_{0}, \, \mathsf{\Omega}_{1}, \, \mathsf{\Omega}_{2} \; \ldots \; \mathsf{\Omega}_{k} \; \ldots \; \right)$

An Excited Proton

Next we wrap a naked proton with a field additional quarks. The extra stereochemical quarks give the proton a left-handed twist. Some down anti-quarks are included to flatten $\rho_{z} .$ And the muonic quarks give the proton some spatial presence that is transverse to the central-axis. These extra quarks are all paired with each other in simple field quanta. Taken together, they comprise an electromagnetic field that is written as

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

A quark model of a proton with a magnetic field. The dark rod indicates the central-axis. And the light grey rod represents the magnetic axis.

This specific field is also used to model a hydrogen atom in its ground-state which is conventionally written as $\mathbf{H} \! \left( 1\mathbf{S} \right) .$ So the field is noted by $\mathscr{F} \! \! \left( 1\mathbf{S} \right).$

The additional muonic quarks bring red and green chromatic sensations into the description. So in quark-space, the model becomes two-dimensional. A grey rod is used to illustrate this new magnetic axis. Here is a short movie (jump) that gives a quick look around the model.

The quark coefficients of the excited proton give $\Delta n^{\mathsf{D}} ,$ $\Delta n^{\mathsf{M}} ,$ $\Delta n^{\mathsf{A}}$ and $\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}$ all equal to zero. So the enthalpy of the chemical quarks in P is nil, and $\overline{\rho} = (0, 0, 0)$ .

The inner radius of the excited proton is also zero because $\Delta n^{\mathsf{D}} = 0 .$ But $N^{\mathsf{D}} = 8 ,$ so the excited proton’s outer radius is given by

$\rho_{out} = \sqrt{ \dfrac{hc}{2\pi k_{\mathsf{F}}} \rule{0px}{14px} }$

For the excited proton discussed above, $\textsl{ \textsf{J} } = 1 .$ So we say that the proton is rotating. And $N^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} = 4$ while $N^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} = 0$ so the rotation has a left-handed character.

A Proton in 2-Dimensional Space

This schematic diagram shows a proton, in a two-dimensional Cartesian famework, with rotation indicated by stripes on the angular momentum vector. — A proton in a simplified two-dimensional Cartesian framework.

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A Proton in 2-Dimensional SpaceTime

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