Solid Newtonian Particles • EthnoPhysics

Outline

Newtonian Particles are Dense

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

Let P be a material particle in a stable balance with its environment. It might be emitting and absorbing lots of photons, but not melting or breaking into pieces. Then P is presumably steady enough so that we can model it using 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 mechanical 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 $m^{2} c^{4} \simeq H^{2} .$ Finally, taking a square root gives

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

Newtonian Particles are Heavy

Let particle P, with a rest mass $m$ , be in an excited state characterized by the norm of its radius vector $\left\| \, \overline{\rho} \, \right\|$ . The density $\varrho$ is defined by $\varrho \equiv m c^{2} / \left\| \, \overline{\rho} \, \right\|$ . So the Newtonian condition that $k_{\mathsf{F}} \ll \varrho$ implies that

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

From the discussion above about enthalpy, we have $mc^{2} \simeq \left| H \right| .$ Then, eliminating the mass from the last two equations 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, $W^{2} \ll H^{2} \, .$

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.

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 $N$ . 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

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

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 in motion.

Momentum is traditionally understood as a product of the mass with a velocity. But the premise of EthnoPhysics frowns on the naive acceptance of spatial concepts. And we require some discussion of ideas like dynamic equilibrium before getting to velocity. But later, after untangling some entwined concepts, we ultimately show how the foregoing sensation-based definition of momentum is equivalent to the traditional understanding.

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. The EthnoPhysics definition of momentum works by expressing the relevance of reference sensations for understanding human experience. Seeing the Sun, seeing blood and seeing gold are vividly pertinent for almost everyone.

Here is an example of determining the momentum for a graviton $\mathsf{\Gamma} .$ Recall that the wavevector $\overline{\kappa}$ is defined from sums of quark coefficients, and 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

$\overline{p} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv \dfrac{h}{2\pi} \! \left[ \, \overline{\kappa} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) - N \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \widetilde{\kappa}^{ \mathsf{F}} \rule{0px}{14px} \right] = - \dfrac{h \widetilde{\kappa}^{\mathsf{F}} }{2\pi} N \hspace{-4px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right)$

where $N ( \mathsf{\Gamma} )$ 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 the character of the reference frame.

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

$\lambda = \dfrac{h}{p}$

Thus de Broglie’s postulate notes a conditional proportionality between $\overline{p}$ and $\overline{ \kappa }$ that is just built-in to the definitions for these characteristics.

Momentum is Conserved

Recall that quarks are conserved. So if some free particals $\mathbb{X}$ , $\mathbb{Y}$ and $\mathbb{Z}$ interact like $\mathbb{X} + \mathbb{Y} \leftrightarrow \mathbb{Z}$ and are otherwise isolated, then

$N^{\mathbb{X}} + N^{\mathbb{Y}} = N^{\mathbb{Z}}$

Also, as shown earlier, wavevectors are combined as $\overline{\kappa}^{\mathbb{X}} + \overline{\kappa}^{\mathbb{Y}} = \overline{\kappa}^{\mathbb{Z}}$ . Then by the foregoing definition, momenta are related as

$\overline{p}^{\mathbb{X}} + \overline{p}^{\mathbb{Y}} = \overline{p}^{\mathbb{Z}}$

Thus we say that momentum is conserved when compound quarks are formed or decomposed. Newtonian mechanics is based 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 necessarily be conserved.

Momentum conservation is like counting the beads in this textile from Kalimantan. — Baby Carrier panel, Basap people. Borneo 19th century, 39 x 20 cm. Photograph by D Dunlop.

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 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 determining the 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

$\begin{align*} E \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) &\equiv \sqrt{c^{2}p^{2} + m^{2}c^{4} \, } \\ &= cp \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \rule{0px}{14px} \\ &= \frac{ch}{2\pi} \left\| \, \widetilde{\kappa}^{\mathsf{F}} \right\| \, N \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \rule{0px}{16px} \end{align*}$

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 the total number of quarks it contains.

The Lorentz Factor

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

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

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 velocity, we will see that the forgoing definition is equivalent.

The Lorentz factor is used 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

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

Note that this is a little different from the previous series. But for non-relativistic motion, terms become progressively smaller, and the Lorentz factor is roughly given by just the first two summands. Then we can substitute these terms back into the mechanical energy approximation to obtain

$E \simeq \gamma mc^{2}$

Energy Measurements

Consider some laboratory experiments to measure $E$ and let us review some terms often used to compare these observations with theory. Let the experiment be accomplished by any combination of observation and inference whatsoever provided only that it satisfies 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 any systematic variations can be evaluated. Ideally experiments are repeated and confirmed by different scientists in other laboratories. So overall; measurement is a communal activity, with ancient roots, that links specific laboratory practice to the reproducible report of some number. The twentieth century has left us with an outstanding legacy of data about nuclear particles that come from measurements like this.

Any measurement of a particle presumably involves some sort of interaction that changes its quark content. The change may be small, maybe even negligible, but nonetheless there is always a logical distinction between an observed value and the theoretical concept of the energy of an isolated particle.

The customary way of assessing this 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

$E = \widetilde{E} \pm \delta \! E$

where $\widetilde{E}$ is a typical or representative value called the experimental average. The other number $\delta \! E$ describes the variation in observed values, it is called the experimental uncertainty. For good measurements $\delta \! E$ 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

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

Another important number is the coefficient of variation in the data which is defined by the ratio $\delta \! E / \widetilde{E} .$ The inverse of this quantity is known as the signal to noise ratio

$\varsigma = 10 \log{ \left(\widetilde{E} / \delta \! E \right) }$ (dB)

expressed on a logarithmic scale in units of decibels .

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.

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 be called realistic motion, or perhaps naturalistic movement. Particles that depart too far from this balance may be called non-Newtonian, or even unphysical. To be more exact about this we define the kinetic and potential energy.

Kinetic Energy

Consider a material 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 in an inertial frame, momentum is proportional to the wavenumber $\kappa$ . So $K$ is proportional to $\kappa ^{2}$ . Then recall that the wavenumber depends only on the coefficients of dynamic quarks. So the kinetic energy depends strongly on P’s dynamic quark content. Dynamic quarks are objectified from somatic and visual sensations. So the kinetic energy depends strongly on these audio-visual sensations too.

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 potential energy

$\mathcal{U} \equiv E - K$

To evaluate $\mathcal{U}$ , recall that if P is in slow motion, then

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

So the potential energy is approximated by $\mathcal{U} \simeq mc^{2}$ . Thus for slowly moving Newtonian particles, the potential energy depends strongly on the mass. Then remember that for heavy particles, a sensory interpretation of the mass relates mainly to baryonic quarks and thermal sensations. And so for Newtonian particles, the potential energy is mostly associated with thermal sensations too.

What is Dynamic Equilibrium?

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 mechanical energy between kinetic and potential types. But the potential energy is defined above by $E$ less $K$ . So for a particle in dynamic equilibrium

$E= \mathcal{U} + K = 2K$

This account of dynamic equilibrium is succinct. And equipartition provides an important theoretical linkage to traditional notions of momentum. But it is not very helpful in the laboratory. Difficulties arise from using a hypothetical condition of perfect isolation to set the zero-value for energy measurements. Recall that our initial discussion of internal energy adopted the reference sensation of not seeing the Sun to grasp the notion of having no energy. So even in principle, there is no tangible reference standard for absolute-zero on the energy scale. And this is noticeable now that measurements can be made to a few parts in a trillion.

Moreover, there are conflicts with theories of dispersion and gravitation which may deny the possibility of perfect isolation. Anyway, these difficulties are manageable because physical phenomena often occur within distinct energy regimes which have different ‘zeros’. In the laboratory we measure energy changes, that are related to each other by $\Delta E = \Delta \mathcal{U} + \Delta K$ . Then a null-value standard for calibrated energy-difference measurements can be selected for experimental convenience. Results are reported using a slightly different version of the energy with a shifted origin

$E = E^{\prime} + \; \text{an arbitrary constant}$

Then $\Delta {E}^{\prime} = \Delta {E}$ . These energy-differences are more susceptible of precise laboratory observation than absolute values. But ${E}^{\prime} \! \ne 2K$ and equipartition is inapt for shifted energies.

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.

Energy Conservation

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 $N$ . 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

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

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

$\eta \equiv \dfrac{ \, \lambda_{\mathsf{o}}}{\lambda}$

This number $\eta$ is called the index of refraction. Then in environments where there is no dispersion, $\lambda = \lambda_{\mathsf{o}}$ and motion is described by $\eta =1 .$

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.

Time

Time is grasped by counting days and heartbeats. Planck's postulate is discussed. Cause, effect and stability are analyzed. Phase angle is defined.

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References
1	Isaac Newton, Mathematical Principles of Natural Philosophy, page 639. Translated by A Motte and F Cajori. University of California Press, 1934.
2	H. 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.