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Dense Newtonian Particles

Outline

So far, EthnoPhysics has given extensive deliberation to photons and nuclear particles. But next we consider heftier particles on the way to discussing Newtonian mechanics.

Newtonian Particles are Dense

Let P be a material particle in a steady balance with its environment. It might be emitting and absorbing lots of photons, but not melting or exploding. Then P is presumably steady enough so that we can model it as a sequence of excited states. Let these states be described by  \varrho, their density. And recall that the constant number k_{\mathsf{F}} was introduced earlier to grasp the shape of a particle. Then we say that P is a Newtonian particle if it is so dense that

k_{\mathsf{F}} \ll \varrho

This condition implies that Newtonian particles are heavy, as shown below. And later we also show how interactions between Newtonian particles obey conservation laws for mass and energy.

To be more exact, let P be in an excited state characterized by  \left\| \, \overline{\rho} \, \right\| , the norm of its radius vector, and  m, its rest mass. For Newtonian particles, the energy of the rest-mass is almost the same as the absolute-value of the enthalpy  H. This is because the definition of mass can be rearranged to give

m^{2} c^{4} + W^{2} = H^{2}

Then recall that  W \equiv k_{\mathsf{F}} \left\| \, \overline{\rho} \, \right\| is the work required to assemble the quarks in P, so

m^{2} c^{4} + k_{\mathsf{F}}^{2} \left\| \, \overline{\rho} \, \right\|^{2} = H^{2}

Also, the density is defined such that  \left\|  \, \overline{\rho} \, \right\| = m c^{2} / \varrho. So substituting  \varrho for  \overline{\rho} gives

m^{2} c^{4} \left( 1 + \dfrac{k_{\mathsf{F}}^{2}}{\varrho^{2}} \right) = H^{2}

But the Newtonian condition requires that  k_{\mathsf{F}} \ll \varrho. So the negligible term can be dropped to obtain the approximation m^{2} c^{4} \simeq H^{2} . Then, taking a square root gives

\left| H \right| \simeq m c^{2}

The Newtonian condition also implies that

k_{\mathsf{F}} \left\| \, \overline{\rho} \, \right\| \ll m c^{2}

Eliminating the mass from the last two expressions yields

k_{\mathsf{F}} \left\| \, \overline{\rho} \, \right\| \ll \left| H \right|

But the work required to assemble the quarks in P is W \equiv k_{\mathsf{F}} \left\| \, \overline{\rho} \, \right\| . So for Newtonian particles W \ll \left| H \right| . Then squaring both sides shows that P satisfies the definition for being a heavy particle. That is, W^{2}  \ll  H^{2} \, .

Momentum

Momentum is the modern English word used for translating the phrase “quantity of motion” that Sir Isaac Newton uses on the very first page of his great book, the Principia1Isaac Newton, Mathematical Principles of Natural Philosophy, page 639. Translated by A Motte and F Cajori. University of California Press, 1934.. So to understand motion EthnoPhysics starts by using sensation to define the momentum as follows.

A scalable vector posterized image of Sir Isaac Newton by Mei Zendra.
Sir Isaac Newton. Drawn by Mei Zendra, Sumenep Madura, Indonesia 2022.

Consider some particle P characterized by its wavevector  \overline{ \kappa } and the total number of quarks it contains  N_{\mathsf{q}} \, . Report on any changes relative to a frame of reference F which is characterized using  \widetilde{ \kappa } the average wavevector of the quarks in F. We define the momentum of particle P, in reference frame F, as the ordered set of three numbers

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

where  h and  \pi are constants. The norm of the momentum is marked without an overline as p \equiv \left\| \, \overline{p} \, \right\| . If p=0 we say that P is stationary or at rest in the F-frame. Alternatively, if p \ne 0 then we say that P is moving or in motion.

Momentum is traditionally understood as a product of the mass with a velocity. But the premise of EthnoPhysics questions how we accept spatial ideas like velocity. So instead we start with this sensation-based definition of momentum. Then later, after untangling some entwined concepts, we show how   p \! = \! m \mathsf{v} for many particles and conditions.

XXX

Sensory Interpretation: The momentum is defined by a difference between the wavevector of P and a scaled-down version of the frame’s wavevector. Recall that the wavevector has previously been interpreted as a mathematical representation of somatic and visual sensation. So momentum is like the audio-visual contrast between a particle and its reference frame. This juxtaposition attracts and holds our attention because it is necessary for situational awareness and survival. The EthnoPhysics definition of momentum works by expressing the relevance of reference sensations. Seeing the Sun, seeing blood and seeing gold are vividly pertinent for understanding motivation, movement and motion.

Momentum is Conserved

Recall that quarks are conserved. So if some particles \mathbb{X}, \mathbb{Y} and \mathbb{Z} interact like \mathbb{X} + \mathbb{Y} \leftrightarrow \mathbb{Z} but are otherwise isolated, then  N_{\mathsf{q}} the total number of quarks, is constrained as N_{\mathsf{q}}^{\mathbb{X}} + N_{\mathsf{q}}^{\mathbb{Y}} = N_{\mathsf{q}}^{\mathbb{Z}} . Also, as shown earlier, wavevectors are combined as \overline{\kappa}^{\mathbb{X}} + \overline{\kappa}^{\mathbb{Y}} = \overline{\kappa}^{\mathbb{Z}} if these particles are free. Then by substitution into the definition of momentum

    \begin{align*}  \overline{p}^{\,\mathbb{Z}} &\equiv \dfrac{h}{2\pi} \left[ \overline{\kappa}^{\,\mathbb{Z}} \! - N^{\mathbb{Z}}_{\mathsf{q}} \, \widetilde{\kappa}^{\,\mathsf{F}} \right]  \\   &= \rule{0px}{15px}\dfrac{h}{2\pi} \left[ \left( \overline{\kappa}^{\mathbb{X}} + \overline{\kappa}^{\mathbb{Y}} \right) - \left( N_{\mathsf{q}}^{\mathbb{X}} + N_{\mathsf{q}}^{\mathbb{Y}} \right) \widetilde{\kappa}^{\,\mathsf{F}} \rule{0px}{15px} \right]   \\  &= \dfrac{h}{2\pi} \left[ \left( \overline{\kappa}^{\,\mathbb{X}} \! - N^{\mathbb{X}}_{\mathsf{q}} \, \widetilde{\kappa}^{\,\mathsf{F}}  \right) + \left( \overline{\kappa}^{\,\mathbb{Y}} \! - N^{\mathbb{Y}}_{\mathsf{q}} \, \widetilde{\kappa}^{\,\mathsf{F}} \right)  \rule{0px}{15px} \right]     \\  &= \rule{0px}{15px} \dfrac{h}{2\pi}  \left( \overline{\kappa}^{\,\mathbb{X}} \! - N^{\mathbb{X}}_{\mathsf{q}} \, \widetilde{\kappa}^{\,\mathsf{F}}  \right) + \dfrac{h}{2\pi} \left( \overline{\kappa}^{\,\mathbb{Y}} \! - N^{\mathbb{Y}}_{\mathsf{q}} \, \widetilde{\kappa}^{\,\mathsf{F}} \right)   \\  &= \rule{0px}{15px} \; \overline{p}^{\mathbb{X}} + \overline{p}^{\mathbb{Y}}   \end{align*}

Thus we say that momentum is conserved when compound quarks are formed or decomposed. Newtonian mechanics is built on this relationship. It is important but not unique. Recall that we also have conservation laws for seeds, quarks, charge, lepton number, baryon number and enthalpy. All of these conservation rules follow from the logical requirements of our descriptive method. EthnoPhysics depends on mathematics. Therefore we are constrained by the law of noncontradiction and the associative properties of addition. So any characteristic defined by simple sums of quark coefficients will always be conserved.

De Broglie’s Postulate

In a perfectly inertial frame of reference  \widetilde{ \kappa } ^{ \mathsf{F}}= \left( 0, 0, 0 \right). Then the momentum of P is given by

\overline{p} = \dfrac{h \overline{\kappa}}{2\pi}

And recall that for particles in motion, the wavelength is  \lambda = 2 \pi / \kappa. So taking the norm of the momentum and eliminating the wavenumber obtains Louis de Broglie’s statement about the inverse relationship between momentum and wavelength

p = h / \lambda

Thus de Broglie’s postulate notes a conditional proportionality between  \overline{p} and  \overline{ \kappa } that is just built-in to the EthnoPhysics definitions of these characteristics. Moreover these definitions apply to all sorts of particles, photons as well as Newtonian particles. So De Broglie’s postulate is often used to determine the momentum of a photon since photons are usually described by their wavelength.

Momentum of a Graviton

Gravitons have been defined by the union of a photon and its associated anti-photon. This is written as \mathsf{\Gamma} \equiv \left\{ \, \boldsymbol{\gamma}, \, \overline{\boldsymbol{\gamma}} \, \rule{0px}{9px} \right\}. Also, recall that the wavevector  \overline{\kappa} is defined from sums of quark coefficients, and that quarks are conserved. So the wavevector of a graviton is the sum of the wavevectors of its component photons. But the wavevector of any particle is symmetrically opposed to the wavevector of its matching anti-particle. Thus

    \begin{align*}  \overline{\kappa}  \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)  &=  \overline{\kappa}  \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) +   \overline{\kappa}  \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)   \\   &= \overline{\kappa} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) - \overline{\kappa} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) \rule{0px}{15px} \\  &=  (0, 0, 0)\rule{0px}{14px} \end{align*}

Then, in a frame of reference noted by F, the momentum of a graviton is given by

\overline{p}^{\, \mathsf{\Gamma}}  \! =  - \dfrac{h \widetilde{\kappa}^{\,\mathsf{F}} }{2\pi} \, N_{\mathsf{q}}^{\mathsf{\Gamma}}

where  N_{\mathsf{q}}^{\mathsf{\Gamma}} is the total number of quarks in  \mathsf{\Gamma}, including all types. This expression shows that all of the gravitons in a description have their momenta pointed in the same direction, and this direction is determined by attributes of the reference frame.

Mechanical Energy

Consider a particle P that is described by its mass  m and momentum  p. And please notice that these numbers have been defined by a methodical description of sensation. The mechanical energy of P is defined by

E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \rule{0px}{11px} \; }

where  c is a constant. This statement comes from Max Planck and Paul Dirac . As a special case for material particles, we can divide by  mc^{2} to get

E = mc^{2} \sqrt{ \; 1 + \left( p/mc \right)^{2} \; }

The square root may be expanded in a binomial series as

1 + \dfrac{p^{2}}{2m^{2}c^{2}} - \dfrac{p^{4}}{8m^{4}c^{4}} + \dfrac{3p^{6}}{48m^{6}c^{6}} + \ldots

And if p \ll mc we can ignore the smaller terms to approximate the mechanical energy with the expression

E \simeq mc^{2} \left( 1 + \dfrac{p^{2}}{2m^{2}c^{2}} \right)

The requirement that p \ll mc is called a slow motion condition. An ethereal particle like a photon cannot move slowly because m \! = \! 0 so the condition cannot be satisfied by any value of the momentum.

Here is an example of calculating the mechanical energy for a graviton  \mathsf{\Gamma} . Note that the momentum of a graviton was given above. And the mass of a graviton is zero, so

E^{\mathsf{\Gamma}} = cp^{\mathsf{\Gamma}} = \dfrac{ch}{2\pi} \left\| \,   \widetilde{\kappa}^{\,\mathsf{F}} \right\| N_{\mathsf{q}}^{\mathsf{\Gamma}}

Thus in a perfectly inertial reference frame where \widetilde{\kappa}^{ \mathsf{F}} = (0, 0, 0) gravitons carry no energy or momentum. If we assume that a frame is perfectly inertial, then we are also presuming that gravity can be ignored. For non-inertial frames, both the momentum and energy of a graviton are directly proportional to N_{\mathsf{q}}^{\mathsf{\Gamma}} the total number of quarks it contains.

The Lorentz Factor

The mass and momentum may also be be combined to specify yet another quantity

\gamma \equiv \dfrac{1}{\sqrt{ \; 1 - \left( p/mc \right)^{2} \; \rule{0px}{10px} }}

Hendrik Antoon Lorentz, 1853—1928.

This number  \gamma is called the Lorentz factor after the Dutch physicist Hendrik Lorentz . His original work2H. A. Lorentz, The Theory of Electrons and its Applications to the Phenomena of Light and Radiant Heat, page 225. Published by B. G. Teubner at Leipzig, 1909. expressed  \gamma differently. But later, after discussing the speed of a particle, we will see that the forgoing definition is equivalent. In either case, the Lorentz factor of a moving particle is always greater than one.

We may use the Lorentz factor to classify particles. For example, when the slow motion condition applies, then  \gamma \simeq 1 . But if  \gamma \gg 1 , then we say that a particle is relativistic. To make a useful approximation for the Lorentz factor we expand the square root into a binomial series as

\gamma = 1 + \dfrac{p^{2}}{2m^{2}c^{2}} + \dfrac{3p^{4}}{8m^{4}c^{4}} + \dfrac{15p^{6}}{48m^{6}c^{6}} + \ldots

Note that this is a little different from the previous series used for E, but terms still become progressively smaller. So if motion is not extremely relativistic, the Lorentz factor is taken as just the first two summands. Then we substitute these terms back into the foregoing expression for mechanical energy to obtain

E = \gamma mc^{2}

Measuring Energy

Consider doing a few laboratory experiments to measure  E , the mechanical energy. Here is a review of some terms used to compare theory with observation. Let the measurements be accomplished by any combination of observation and inference whatsoever provided only that they satisfy the professional standards of experimental physicists. For example this means that instruments are painstakingly calibrated. And any new measurement techniques are carefully compared with previous methods so that systematic variations can be evaluated. Ideally experiments are repeated and confirmed by different scientists, working in separate laboratories, located in distant countries.

So overall, measurement is a communal activity that links specific laboratory technique to the globally reproducible report of some number.

In practice, any measurement of a particle requires some sort of interaction that changes the particle’s quark content. The change may be small, ideally even negligible. But nonetheless, there is always a logical distinction between an observed value, versus theories about isolated particles. And despite careful laboratory work, the measurement process itself can introduce new errors and uncertainty.

A customary way of dealing with this issue is to make many observations. So consider a series of  N measurements with results noted by  E^{1}, \, E^{2}, \, E^{3} \ \ldots \ E^{k} \ \ldots \ E^{N} . These observed values are related to  E, the theoretical concept of mechanical energy, by the assertion that

E = \widetilde{E} \pm \sigma_{\! E}

Here  \widetilde{E} is a typical or representative value called the experimental average. The other number  \sigma_{\! E} describes the variation in observed values, it is called the experimental uncertainty. For good measurements  \sigma_{\! E} is small enough so that  E and  \widetilde{E} are interchangeable thus reconciling theory and observation. Usually the experimental average is determined from the arithmetic mean of the set of observations

\displaystyle \widetilde{E} = \dfrac{1}{N} \sum_{k=1}^{N} E^{k}

and the experimental uncertainty is represented by their standard deviation as

\sigma_{\! E} = \sqrt{ \frac{1}{N} \sum_{k=1}^{N} \left( E^{k} - \widetilde{E} \right)^{2} \; }

Thus descriptive statistics can cope with random noise. But there is also a systematic difficulty with energy measurement: Any laboratory experiment that reports  E \! \simeq \! 0 requires perfect isolation to be properly calibrated. This is because the reference sensation of not seeing the Sun was used to grasp the notion of having no energy. So even in principle, we do not have a tangible reference standard for absolute-zero on the energy scale. Furthermore, there are conflicts with theories of dispersion and gravitation which may deny even the possibility of perfect isolation. So this issue is considered in more detail below.

Newtonian particles are dense, somewhat like the imagery in this Indonesian textile with extensive supplementary weft details.
Tampan, Paminggir people. Lampung region of Sumatra, near Semangka Bay, 19th century, 64 x 64 cm. Photograph by D Dunlop.

Binding Energy

An energy associated with the forces that hold a particle together is called its binding energy and written as  \mathcal{U}_{binding} \hspace{1px} . Recall that the positive constant k_{\mathsf{F}} was introduced earlier. Then for a particle described by its polar radius \rho_{z} \hspace{1px} , the binding-energy is defined by

\mathcal{U}_{binding} \equiv -k_{\mathsf{F}} \rho_{z}

Particles and their anti-particles have binding-energies with opposing signs because \rho_{z} (\mathsf{P}) = -\rho_{z} (\overline{\mathsf{P}}) \, . We conventionally assign a positive radius and a negative binding-energy to common particles so that their total-energy is lower when bound together, than if separated. Lower energies are associated with lower temperatures and greater stability. So this convention just confirms that common, everyday particles are composed of ordinary matter. Whereas aggregates of anti-matter are unstable and uncommon.

Anyway, for either sign convention, it is true that \mathcal{U}_{binding}^{\, 2} \! \! \! = k_{\mathsf{F}}^{2} \rho_{z}^{2} \, . Then since the centripetal component of the quark metric k_{zz} is always one, we can write \mathcal{U}_{binding}^{\, 2} \, / k_{\mathsf{F}}^{2} = k_{zz} \, \rho_{z}^{\, 2} and this form is useful for determining the work required to assemble P. We can also express the polar radius \rho_{z} \hspace{1px} in terms of up-quark coefficients  \Delta n ^{\mathsf{U}} , and down-quark coefficients  \Delta n ^{\mathsf{D}} , along with their internal energies  U^{\mathsf{U}} and  U^{\mathsf{D}} . The polar radius also depends on the enthalpy of any chemical quarks in P. In terms quark coefficients, the binding energy can be written as

\displaystyle \mathcal{U}_{binding} \! = \Delta n^{\mathsf{D}} U^{\mathsf{D}} - \Delta n^{\mathsf{U}} U^{\mathsf{U}} - \sum_{\zeta=11}^{16} \Delta n^{\zeta} U^{\zeta}

Thus the binding energy just depends on up-quarks, down-quarks and chemical-quarks. But the internal-energy of an up-quark is billions of times larger than any of these other quarks. So if there are any net up-quarks in P, then they dominate the sum and

\mathcal{U}_{binding} \simeq -\Delta n^{\mathsf{U}} U^{\mathsf{U}}

Kinetic Energy

Consider some Newtonian particle P, described by its mass  m and momentum  p. The kinetic energy of P is defined as

K \equiv \dfrac{\, p^{ 2}}{2m}

Since  m > 0 for material particles,  K is never negative. And recall that for an inertial frame of reference, De Broglie’s postulate states that the momentum is proportional to the wavenumber  \kappa . So  K is proportional to  \kappa ^{2} . But  \kappa is defined by dynamic quarks, not baryonic quarks. So the kinetic energy depends strongly on audio-visual sensations, not thermal sensations.

Potential Energy

Let us also include the mechanical energy  E in the description of P. The difference between  E and the kinetic energy defines another number  \mathcal{U} called the total potential energy

\mathcal{U} \equiv E - K

To evaluate  \mathcal{U} , recall that if P is in slow motion, then the mechanical energy can be approximated as

    \begin{equation*} \begin{split}  E &\simeq mc^{2} \left(1 + \dfrac{p^{2}}{2m^{2}c^{2}} \right) \\ &= \rule{0px}{11px} mc^{2} \left(1 + \dfrac{K}{mc^{2}} \right) \\ &= \rule{0px}{11px} mc^{2} + K \end{split} \end{equation*}

So the total potential energy is approximated as \mathcal{U} \! \simeq \! mc^{2}. And for slowly moving Newtonian particles, the total potential energy depends mostly on the mass. For heavy particles, a sensory interpretation of the mass relates mainly to thermal sensations and baryonic quarks. Thus for Newtonian particles, the total potential energy is also strongly dependent on thermal sensation. However, we are often more concerned with the changes in  \mathcal{U} that arise from interactions with dynamic quarks. We can frequently assume that the rest mass is a constant. And then the following energies are relevant.

Magnetic Potential Energy

An energy associated with the magnetic forces that hold a particle together is called its magnetic potential energy and noted by  \mathcal{U}_{m} \hspace{1px} . It depends on the magnetic polarity which is written as \delta_{\widehat{m}} \! = \! \pm 1 . Recall that the positive constant k_{\mathsf{F}} was introduced earlier. And that k_{mm} marks the magnetic-component of the terrestrial metric. Then for a particle described by its magnetic radius \rho_{m} \hspace{1px} , we define the magnetic potential energy as

\mathcal{U}_{m} \equiv \delta_{\widehat{m}} \, k_{\mathsf{F}}  \sqrt{k_{mm}} \;  \rho_{m}

We can also express the magnetic-radius \rho_{m} \hspace{1px} in terms of the coefficients for southern-quarks  \Delta n ^{\mathsf{A}} , and northern-quarks  \Delta n ^{\mathsf{M}} . Recall that the internal energies for these quarks are  U^{\mathsf{A}} and  U^{\mathsf{M}} . Then

\mathcal{U}_{m} = \delta_{\widehat{m}}  \sqrt{k_{mm}} \left( \Delta n^{\mathsf{A}} U^{\mathsf{A}} - \Delta n^{\mathsf{M}} U^{\mathsf{M}} \right)

Thus the magnetic potential energy of a particle depends only on its muonic quark content. This is mostly due to northern quarks because their lifetime is so much longer, and their internal-energy is so much larger, than southern quarks.

Electric Potential Energy

An energy associated with the electric forces that hold a particle together is called its electric potential energy and noted by  \mathcal{U}_{e} \hspace{1px} . It depends on the electric polarity which is written as \delta_{\widehat{e}} \! = \! \pm 1 . Recall that the positive constant k_{\mathsf{F}} was introduced earlier. And that k_{ee} marks the electric-component of the terrestrial metric. Then for a particle described by its electric radius \rho_{e} \hspace{1px} , we define the electric potential energy as

\mathcal{U}_{e} \equiv \delta_{\widehat{e}} \, k_{\mathsf{F}}  \sqrt{k_{ee}} \;  \rho_{e}

We can also express the electric-radius \rho_{e} \hspace{1px} in terms of the coefficients for good-quarks  \Delta n ^{\mathsf{G}} , and bad-quarks  \Delta n ^{\mathsf{E}} . Recall that the internal energies for these quarks are  U^{\mathsf{G}} and  U^{\mathsf{E}} . Then

\mathcal{U}_{e} = \delta_{\widehat{e}} \, \sqrt{k_{ee}} \left( \Delta n^{\mathsf{G}} U^{\mathsf{G}} - \Delta n^{\mathsf{E}} U^{\mathsf{E}} \right)

Thus the electric potential energy of a particle depends only on its electronic quark content. This is mostly due to good quarks because their lifetime is so much longer, and their internal-energy is so much larger, than bad quarks.

Coulomb Energy

Another sort of energy that accounts for the combined presence of both electronic and muonic quarks is called the Coulomb potential energy in honour of Charles-Augustin de Coulomb . It is defined by

\mathcal{U}_{Coulomb} \equiv  \delta_{\widehat{e}} \, \delta_{\widehat{m}} \,  \sqrt{ \, \mathcal{U}_{e}^{2} + \mathcal{U}_{m}^{2} + \mathcal{U}_{em}^{2} \; }

where the extra electromagnetic contribution is given by

\mathcal{U}_{em} \equiv \delta_{\widehat{e}} \, \delta_{\widehat{m}} \, k_{\mathsf{F}}  \sqrt{\, 2k_{em} \, \rho_{e} \rho_{m} }

Recall that k_{em} notes the electromagnetic component of the terrestrial metric. Particle radii can be eliminated from this expression to put \mathcal{U}_{em} in terms of quark coefficients as

\mathcal{U}_{em} = \delta_{\widehat{e}} \, \delta_{\widehat{m}} \, \sqrt{\, 2k_{em} \left( \Delta n^{\mathsf{G}} U^{\mathsf{G}} - \Delta n^{\mathsf{E}} U^{\mathsf{E}} \right) \left( \Delta n^{\mathsf{A}} U^{\mathsf{A}} - \Delta n^{\mathsf{M}} U^{\mathsf{M}} \right) }

Thus the Coulomb energy of a particle depends on just the leptonic quarks that it contains.

Weak Energies

The energy stored by any weak forces that hold a particle together is called the weak potential energy and noted by  \mathcal{U}_{weak} \hspace{1px} . This term arises from the joint presence of both rotating and leptonic quarks. It is defined as

\mathcal{U}_{weak} \equiv \delta_{\widehat{e}} \, \delta_{\widehat{m}} \, \sqrt{ \, \mathcal{U}_{mz}^{2} + \mathcal{U}_{ez}^{2} \; }

where the electroweak contribution is

\mathcal{U}_{ez} \equiv \delta_{\widehat{e}} \, k_{\mathsf{F}}  \sqrt{2k_{ez} \, \rho_{e} \rho_{z} }

and the magnetoweak term is given by

\mathcal{U}_{mz} \equiv \delta_{\widehat{m}} \, k_{\mathsf{F}}  \sqrt{2k_{mz} \, \rho_{m} \rho_{z} }

The constants k_{ez} and k_{mz} are components of the terrestrial metric. Also remember that particle radii can be stated in terms of quark-coefficients and internal-energies as shown above. Then  \mathcal{U}_{ez} and  \mathcal{U}_{mz} can be expressed as

\mathcal{U}_{ez} = \delta_{\widehat{e}} \sqrt{2k_{ez} \left( \Delta n^{\mathsf{G}} U^{\mathsf{G}} - \Delta n^{\mathsf{E}} U^{\mathsf{E}} \right) \mathcal{U}_{binding} }

and

\mathcal{U}_{mz} = \delta_{\widehat{m}} \sqrt{2k_{mz} \left( \Delta n^{\mathsf{A}} U^{\mathsf{A}} - \Delta n^{\mathsf{M}} U^{\mathsf{M}} \right) \mathcal{U}_{binding} }

Work and Potential Energy

The foregoing potential energies all depend on the electric-polarity, the magnetic-polarity or some other sign convention, that places a factor of   \pm 1 into their definitions. So we can simplify analysis by working with squared terms. For example, the Coulomb and weak energies can be plainly expressed as

\mathcal{U}_{Coulomb}^{2} = \mathcal{U}_{e}^{2} + \mathcal{U}_{em}^{2} + \mathcal{U}_{m}^{2}

and

\mathcal{U}_{weak}^{2} = \mathcal{U}_{mz}^{2} + \mathcal{U}_{ez}^{2}

Also, recall that the binding energy squared can be written as

\mathcal{U}_{binding}^{\, 2} \, / k_{\mathsf{F}}^{2} = k_{zz} \, \rho_{z}^{\, 2}

Similar relationships hold for the other potential energies

\mathcal{U}_{e}^{2} / k_{\mathsf{F}}^{2} = k_{ee} \, \rho_{e}^{\, 2}

\mathcal{U}_{em}^{\, 2} \, / k_{\mathsf{F}}^{2} = 2k_{em} \, \rho_{e}\rho_{m}

\mathcal{U}_{m}^{2} / k_{\mathsf{F}}^{2} = k_{mm} \, \rho_{m}^{\, 2}

\mathcal{U}_{ez}^{\, 2} \, / k_{\mathsf{F}}^{2} = k_{ez} \, \rho_{e} \rho_{z}

and

\mathcal{U}_{mz}^{\, 2} \, / k_{\mathsf{F}}^{2} = k_{mz} \, \rho_{m} \rho_{z}

Now recall that the surface area of P is defined by \widehat{A} \equiv \overline{\rho}^{2} . And this can be written out as

    \begin{equation*} \begin{split}  \widehat{A} & =  k_{mm} \rho_{m}^{2} +  k_{ee} \rho_{e}^{2} +  k_{zz} \rho_{z}^{2}  \\  & \hspace{20px} + 2 k_{em}  \rho_{e}  \rho_{m} + 2k_{mz}\rho_{m}  \rho_{z} \\ & \hspace{40px} + 2 k_{ez}\rho_{e}  \rho_{z} \end{split} \end{equation*}

Then by substitution, the surface area of P can be stated in terms of potential energies as

\widehat{A} = \dfrac{ \mathcal{U}_{binding}^{2} }{k_{\mathsf{F}}^{2}} + \dfrac{ \mathcal{U}_{e}^{2} + \mathcal{U}_{em}^{2} + \mathcal{U}_{m}^{2} }{k_{\mathsf{F}}^{2}} + \dfrac{ \mathcal{U}_{mz}^{2} + \mathcal{U}_{ez}^{2} }{k_{\mathsf{F}}^{2}}

so that

\widehat{A} = \, \mathcal{U}_{binding}^{\, 2} \! + \! \mathcal{U}_{Coulomb}^{\, 2} \! + \! \mathcal{U}_{weak}^{\, 2} \left. \middle/ \rule{0px}{12px} \right. \!  k_{\mathsf{F}}^{2}

Finally recall that  W \! , the work required to assemble the quarks in P, is defined by W \equiv k_{\mathsf{F}} \sqrt{ \widehat{A} \; } so eliminating  \widehat{A} yields

W^{2} = \mathcal{U}_{binding}^{2} + \mathcal{U}_{Coulomb}^{2} + \mathcal{U}_{weak}^{2}

Thus the work required to assemble the quarks in a particle has three components. And later, we discuss a type of force associated with each component. Roughly speaking, we say that to build a particle, work must be done to offset binding, Coulomb and weak forces.

Dynamic Equilibrium

We characterize Newtonian particles as being in some kind of steady balance with their environment. They are presumably interacting with countless photons, bouncing around a lot, and colliding with other particles. But despite much agitation, there is still a central tendency that might loosely be called realistic motion, or perhaps naturalistic movement. Particles that depart too far from this range may be called non-Newtonian, or even unphysical. To be more precise about this fluctuating balance, consider the kinetic energy and the potential energy. These quantities have been defined as

K \equiv \dfrac{\, p^{ 2}}{2m}

and

\mathcal{U} \equiv E - K

We say that a particle is in dynamic equilibrium when its kinetic and potential energies are equal to each other. At equilibrium \mathcal{U} \! = \! K and there is an equal sharing, or equipartition, of energy between kinetic and potential types. So for a particle in dynamic equilibrium, the mechanical-energy is twice the kinetic-energy

E = 2K

This statement is succinct. And equipartition provides an important theoretical connection to traditional ideas about momentum.

But for making measurements, the E = 2K relationship is not much use because it refers to a hypothetical condition of perfect isolation to calibrate the zero-value for  E . Recall that the initial discussion about energy adopted the reference sensation of not seeing the Sun to grasp the notion of having no energy. So we do not have a tangible reference for absolute-zero on the energy scale. Nonetheless, this issue is manageable because physical experience often occurs within distinct regimes that can refer to different ‘zeros’. To be more exact, let us specify \mathcal{U}_{extra} to account for any additional sorts of potential-energy

\mathcal{U}_{extra} \, \equiv \, \mathcal{U} - \mathcal{U}_{binding} \! - \mathcal{U}_{Coulomb} \! - \mathcal{U}_{weak}

This extra energy may be huge. But it can often be ignored if we focus attention on changes that are due to interactions with other particles. We note these energy-differences using the Greek letter  \Delta to write

\Delta \mathcal{U}_{extra} = \Delta \mathcal{U} - \Delta \mathcal{U}_{binding} - \Delta \mathcal{U}_{Coulomb} - \Delta \mathcal{U}_{weak}

Now let us limit consideration to a physical regime where interactions with dynamic quarks are exhaustively described using just the binding energy, the Coulomb energy, and any weak energies. This limitation excludes phenomena like gravity and elasticity. But within this restriction, we can assume that \Delta \! \mathcal{U}_{extra} \! = 0 whenever dynamic quarks are absorbed or emitted. Then for the total potential-energy

\Delta \mathcal{U} = \Delta \mathcal{U}_{binding} + \Delta \mathcal{U}_{Coulomb} + \Delta \mathcal{U}_{weak}

Furthermore, since \mathcal{U} \! \equiv\!  E\!  -\!  K , any changes in the mechanical energy or the kinetic energy are related as \Delta E\!  =\!  \Delta \mathcal{U}\! + \!  \Delta K . So for interactions with dynamic quarks, the variation in mechanical energy is given by

\Delta E = \Delta \mathcal{U}_{binding} + \Delta \mathcal{U}_{Coulomb} + \Delta \mathcal{U}_{weak} + \Delta K

In the laboratory we prefer to measure these energy-differences because a null-value standard for calibration can be selected for experimental convenience. Perfect isolation is not required. And therefore energy-differences are more susceptible of precise observation than absolute values. Results are reported using a slightly different version of the energy which has a shifted origin

E^{\prime} = E + \, \mathsf{an} \, \mathsf{arbitrary} \, \mathsf{constant}

Then \Delta \! {E}^{\prime} \! =  \Delta \!  {E}. But {E}^{\prime} \! \ne 2K and equipartition is inapt for shifted energies.

XXX

Sensory interpretation: As noted above, the kinetic energy characterizes visual stimuli, whereas the potential energy depends more on thermal perception. So there must be a balanced experience of both thermal and visual sensation for events to be objectified as particles in dynamic equilibrium. This requirement for eyes-open visual sensation means that, for example, a dream about flying while asleep cannot meet equilibrium conditions. And neither can watching cartoons on TV, because television only transmits audio-visual sensations, not thermal sensations. So dynamic equilibrium is more like experiencing ordinary daily circumstances in our classrooms and laboratories on Earth. Unlike many movies, dreams and hallucinations.

Dynamic equilibrium results from balanced energies, as suggested by the writhing dragons in this Indonesian weaving.
Tampan, Paminggir people. Lampung region of Sumatra, 19th century, 77 x 70 cm. Photograph by D Dunlop.

Conservation of Energy

Newtonian particles are dense and heavy. And we regularly assume that they are in dynamic equilibrium with their surroundings. Then as discussed earlier, their enthalpy  H , is related to their mass  m , by the approximation

\left| H \right| \simeq mc^{2}

But the mechanical energy  E of any material particle is approximately E \simeq \gamma m c^{2} where  \gamma is the Lorentz factor. Then

E \simeq \gamma \left| H \right|

For particles in slow motion  \gamma \simeq 1 so that E \simeq \left| H \right|. But the absolute-value signs can usually be ignored because ordinary particles are composed from electrons, neutrons and protons which all have positive enthalpy. So if we exclude anti-particles and processes like annihilation, then we usually have

H \simeq E \simeq mc^{2}

Thus the mechanical energy and the enthalpy are almost interchangeable for slow Newtonian particles made of ordinary matter. But enthalpy is conserved for all particles and conditions. So the energy and mass must also be approximately conserved for slow Newtonian particles too. This idea is honoured as an energy conservation law because it is so important for classical mechanics. Moreover, a conservation law for mass is a basic principle in benchtop chemistry. These excellent approximations are used everyday. Together with the routine assumption of dynamic equilibrium they typify Newtonian particles.

Refraction

Consider some particle P characterized by its wavevector  \overline{\kappa} and the total number of quarks it contains  N_{\mathsf{q}} \, . Report on any changes relative to a frame of reference F which is characterized using  \widetilde{ \kappa } the average wavevector of the quarks in F. The momentum of P in the F-frame is defined as

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

Let the frame of reference be formed from a large component \mathbb{G} and a smaller part \mathbb{S} that surrounds P so that

\mathsf{F} = \left\{ \mathbb{G}, \, \mathbb{S} \rule{0px}{10px} \right\}

and

\overline{\kappa}^{\mathsf{F}} = \overline{\kappa}^{\mathbb{G}} + \overline{\kappa}^{\mathbb{S}}

Let the large part of F be responsible for any gravitons that interact with P so that when gravitational effects are completely negligible \overline{\kappa}^{\mathbb{G}} = (0, \, 0, \, 0) \, . Then P’s momentum is given by

\overline{p} = \dfrac{h}{2\pi} \left( \overline{\kappa}^{\mathsf{P}} - \dfrac{N^{\mathsf{P}}}{N^{\mathsf{F}}} \, \overline{\kappa}^{\mathbb{S}} \right)

From de Broglie’s postulate we have \lambda = h / p so

\lambda = \dfrac{2\pi}{ \left\| \, \overline{\kappa}^{\mathsf{P}} - \dfrac{N^{\mathsf{P}}}{N^{\mathsf{F}}} \overline{\kappa}^{\mathbb{S}} \right\| }

To simplify, set \overline{\kappa}^{\mathbb{S}} = (0, 0, 0) in the expression above to define  \lambda_{\mathsf{o}} as the wavelength of P in a non-dispersive surrounding medium

\lambda_{\mathsf{o}} \equiv \dfrac{2\pi}{ \left\| \, \overline{\kappa}^{\mathsf{P}} \right\| }

Many different combinations of photons and media are usefully characterized by defining the ratio

\eta \equiv  \lambda_{\mathsf{o}} / \lambda

This number  \eta is called the index of refraction. Then in environments where there is no dispersion, \lambda \!  = \!  \lambda_{\mathsf{o}} and the motion of a photon may be described with a statement that  \eta \! = \! 1 .

Refraction describes the shimmering iridescence evoked by this lustrous silken textile from Indonesia.
Tampan, Paminggir people. Sumatra 19th century, 58 x 61 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop.
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EthnoPhysics faviconTime

Time is grasped by counting days and heartbeats. Planck's postulate is discussed. Cause, effect and stability are analyzed. Phase angle is defined.
References
1Isaac Newton, Mathematical Principles of Natural Philosophy, page 639. Translated by A Motte and F Cajori. University of California Press, 1934.
2H. A. Lorentz, The Theory of Electrons and its Applications to the Phenomena of Light and Radiant Heat, page 225. Published by B. G. Teubner at Leipzig, 1909.