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.

Sir Isaac Newton. Drawn by Mei Zendra, Sumenep Madura, Indonesia 2022.

We can use the mass to describe the hardness or density of P. Recall that $\left\| \, \overline{\rho} \, \right\|$ is the norm of a radius vector. Then the radial energy density of P is defined by

$\varrho \equiv \dfrac{ \, m c^{2} }{ \left\| \, \overline{\rho} \, \right\| }$

We presume that P is steady enough so that we can model it as a sequence of excited states. Let these states be described by $\varrho$ , their radial energy density. And recall that the constant number $k_{\mathsf{F}}$ was introduced earlier.

We say that P is a Newtonian particle if it is so dense that

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

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 $W \ll \left| H \right| .$

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The Momentum

Definition of Momentum

Consider some particle P in a space that has a reference frame noted by F. Let P and F be characterized by their type of matter $\delta_{\mathfrak{m}}$ and their quantity of motion in vacuo $p_{\mathsf{o}} .$ Assume that F is not centered on P. And perhaps not even aligned with P. Then the central-axis unit-vectors of P and F may be different from each other, they are marked as $\widehat{z}^{\, \mathsf{P}}$ and $\widehat{z}^{\, \mathsf{F}} .$ We define the linear momentum of P in the F-frame by

$\overline{p}^{\,\mathsf{P}}_{\,\mathsf{F}} \hspace{2px} \equiv \hspace{2px} \delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} \widehat{z}^{\, \mathsf{P}} - \, \delta_{\mathfrak{m}}^{\, \mathsf{F}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{F}} \hspace{2px} \widehat{z}^{\, \mathsf{F}}$

We mark the norm of the momentum by $p \equiv \left\| \hspace{1px} \overline{p} \hspace{1px} \right\| .$ Then if $p \! = \! 0$ we say that P is stationary or at rest in the F-frame. Otherwise we may say that P is moving or in motion relative to F.

Momentum in a Grounded Frame

Any reference-frame is composite quark like other particles, so consider that F may be in its ground state. Then we say that F is a grounded frame. Recall that the ground-state of any particle is defined by perfect phase symmetry. This causes the quark-flux vector and wavevector in vacuo to both be null. Then ultimately, the quantity of motion for F is also nil. So $p_{\mathsf{o}}^{\hspace{1px} \mathsf{F}} \! = \! 0$ and in a grounded-frame the momentum is simply

$\overline{p}^{\,\mathsf{P}}_{\,\mathsf{F}} \hspace{2px} = \hspace{2px} \delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} \widehat{z}^{\, \mathsf{P}}$

Recall that particles and their conjugate twins are always made of opposite types of matter. So $\delta_{\mathfrak{m}} ( \mathsf{P} ) \! = \! - \delta_{\mathfrak{m}} ( \mathsf{\overline{P}} ) .$ But the quantity-of-motion does not vary between conjugate-twins. And neither do frame characteristics like $\widehat{z} .$ So in a grounded-frame, the momentum of particles and their conjugate-twins are related as

$\overline{p} \hspace{1px} ( \mathsf{P} ) \! = \! - \overline{p} \hspace{1px} ( \mathsf{\overline{P}} )$

If P is a graviton then $\delta_{\mathfrak{m}}^{\,\mathsf{P}} \! = \! 0$ and $\overline{p}^{\,\mathsf{P}}_{\,\mathsf{F}} \! = \! \left( 0, 0, 0 \right).$ We examine this case later. But for now let P be a particle of ordinary-matter or anti-matter. That is, let $\delta_{\mathfrak{m}}^{\,\mathsf{P}} \! = \! \pm 1 .$ Also, remember that the quantity-of-motion is never negative, so $p_{\mathsf{o}} \! \ge \! 0 .$ And recall that $\widehat{z}$ marks a unit-vector, so $\| \hspace{1px} \widehat{z} \hspace{1px} \| \! = \! 1 .$ Then in a grounded-frame, the norm of a momentum-vector $p \hspace{1px} ,$ is given by

$p \equiv \left\| \hspace{1px} \overline{p} \hspace{1px} \right\| = \left\| \hspace{1px} \delta_{\mathfrak{m}} \hspace{1px} p_{\mathsf{o}} \hspace{2px} \widehat{z} \hspace{1px} \right\| = p_{\mathsf{o}}$

Thus a careful choice of symbols makes this statement seem tautological. But sometimes we have to remember that it depends on the presumption of a grounded-frame. More attention may be required when comparing momenta in different frames.

Momentum in Other Frames

Let particle P be described in a reference-frame H, that is different from F. Characterize H by its own matter-type $\delta_{\mathfrak{m}}^{\, \mathsf{H}} \, ,$ its own quantity of motion $p_{\mathsf{o}}^{\, \mathsf{H}}$ and its own central-axis as noted by the unit-vector $\widehat{z}^{\, \mathsf{H}} .$ Then the momentum of P in the H-frame is

$\overline{p}^{\,\mathsf{P}}_{\mathsf{H}} \equiv \hspace{2px} \delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} \widehat{z}^{\, \mathsf{P}} - \, \delta_{\mathfrak{m}}^{\, \mathsf{H}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{H}} \hspace{2px} \widehat{z}^{\, \mathsf{H}}$

This statement always true by definition. But next we consider an example that is useful for understanding the helicity. To compare F and H we make some restrictive conditions: Let all central-axes be aligned with F, and let F be grounded. Then $\widehat{z}^{\, \mathsf{F}} \! = \widehat{z}^{\, \mathsf{H}} \! = \widehat{z}^{\, \mathsf{P}}$ and momenta can be written as

$\overline{p}^{\,\mathsf{P}}_{\mathsf{F}} = \delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} \widehat{z}^{\, \mathsf{F}}$

and

$\overline{p}^{\,\mathsf{P}}_{\mathsf{H}} = \delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} \widehat{z}^{\, \mathsf{F}} \! - \delta_{\mathfrak{m}}^{\, \mathsf{H}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{H}} \hspace{2px} \widehat{z}^{\, \mathsf{F}}$

Moreover, let the frame H be made of ordinary matter so that $\delta_{\mathfrak{m}}^{\, \mathsf{H}} \! = \! 1 .$ And let H have a quantity-of-motion that is twice as much as F so that $p_{\mathsf{o}}^{\mathsf{H}} \! = \! 2\delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} .$ Then the momentum of P in the H-frame is $\overline{p}^{\,\mathsf{P}}_{\mathsf{H}} \! = - \delta_{\mathfrak{m}}^{\, \mathsf{P}} \hspace{1px} p_{\mathsf{o}}^{\mathsf{P}} \hspace{2px} \widehat{z}^{\, \mathsf{F}} .$ So for this example

$\overline{p}^{\,\mathsf{P}}_{\,\mathsf{F}} \hspace{2px} = - \hspace{1px} \overline{p}^{\,\mathsf{P}}_{\,\mathsf{H}}$

This expression shows that the direction of P’s momentum can change sign depending on the frame of reference. The example is narrow, and there are limitations. If P is a photon then its speed in any frame is a constant noted by $c .$ But frames are always material particles. So their speed is limited to being less than $c .$ It is not possible for H to move faster than P. So the momentum of a photon cannot reversed by shifting the description to another frame.

De Broglie’s Postulate

For particles that are moving, the definition of the wavelength in vacuo gives

$\lambda_{\mathsf{o}} = 2\pi / \kappa_{\mathsf{o}}$

Substituting into the definition of quantity-of-motion

$p_{\mathsf{o}} \equiv \dfrac{h}{2 \pi} \kappa_{\mathsf{o}}$

gives

$p_{\mathsf{o}} = h / \lambda_{\mathsf{o}}$

This applies to all sorts of moving particles, both photons as well as material particles

Conservation of Momentum

In an achiral frame, the momentum of any particle P can be written as

$\overline{p} = \delta_{\mathsf{o}} \hspace{0.5px} p \hspace{3px} \widehat{z}$

Quarks are conserved when particles are formed or when they decay. So for any sort of quark noted by $\mathsf{q} \hspace{1px} ,$ and for any generic particles $\mathbb{A} ,$ $\mathbb{B}$ and $\mathbb{C} ,$ if $\left\{ \mathbb{A} , \mathbb{B} \rule{0px}{10px} \right\} \equiv \mathbb{C}$ then

Quarks are conserved so $n_{\mathsf{q}}^{\mathbb{A}} + n_{\mathsf{q}}^{\mathbb{B}} = n_{\mathsf{q}}^{\mathbb{C}}$

Radii are conserved so $\overline{\rho}^{\mathbb{A}} + \overline{\rho}^{\mathbb{B}} = \overline{\rho}^{\mathbb{C}}$

By the law of cosines $\left\| \, \overline{p}^{\mathbb{C}} \right\| = \, \left\| \, \overline{p}^{\mathbb{A}} + \overline{p}^{\mathbb{B}} \right\| = \sqrt{ \; \left( \overline{p}^{\mathbb{A}} \right)^{2} + \left( \overline{p}^{\mathbb{B}}\right)^{2} + 2 \left( \overline{p}^{\mathbb{A}} \cdot \overline{p}^{\mathbb{B}} \right) \; }$

Consider the momentum vector of $\mathbb{A}$ in an achiral frame of reference noted by F.

$\overline{p}^{\, \mathbb{A}} = \delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \hspace{3px} \widehat{z}^{\, \mathbb{A}}$

In the F-frame, the unit-vector $\widehat{z}^{\, \mathbb{A}}$ may point in any direction. So its F-frame components are noted by

$\widehat{z}^{\, \mathbb{A}} = \left( \, \varrho_{x}^{\, \mathbb{A}}, \, \varrho_{y}^{\, \mathbb{A}}, \, \varrho_{z}^{\, \mathbb{A}} \, \rule{0px}{12px} \right)$

The values of $\varrho$ are not fixed. But $\widehat{z}^{\, \mathbb{A}}$ is still a unit-vector so $\varrho_{x}^{2} + \varrho_{y}^{2} + \varrho_{z}^{2} = 1 .$

Then the momentum of $\mathbb{A}$ in F is

$\overline{p}^{\, \mathbb{A}} = \delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \left( \, \varrho_{x}^{\, \mathbb{A}}, \, \varrho_{y}^{\, \mathbb{A}}, \, \varrho_{z}^{\, \mathbb{A}} \, \rule{0px}{12px} \right)$

Similar reasoning for particles $\mathbb{B}$ and $\mathbb{C}$ gives

$\overline{p}^{\, \mathbb{B}} = \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \left( \, \varrho_{x}^{\, \mathbb{B}}, \, \varrho_{y}^{\, \mathbb{B}}, \, \varrho_{z}^{\, \mathbb{B}} \, \rule{0px}{12px} \right)$

$\overline{p}^{\, \mathbb{C}} = \delta_{\mathsf{o}}^{\, \mathbb{C}} p^{\, \mathbb{C}} \left( \, \varrho_{x}^{\, \mathbb{C}}, \, \varrho_{y}^{\, \mathbb{C}}, \, \varrho_{z}^{\, \mathbb{C}} \, \rule{0px}{12px} \right)$

Then, by the usual rules of linear algebra, conservation of momentum requires that

$\delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \varrho_{x}^{\, \mathbb{A}} + \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \varrho_{x}^{\, \mathbb{B}} = \delta_{\mathsf{o}}^{\, \mathbb{C}} p^{\, \mathbb{C}} \varrho_{x}^{\, \mathbb{C}}$

$\delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \varrho_{y}^{\, \mathbb{A}} + \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \varrho_{y}^{\, \mathbb{B}} = \delta_{\mathsf{o}}^{\, \mathbb{C}} p^{\, \mathbb{C}} \varrho_{y}^{\, \mathbb{C}}$

$\delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \varrho_{z}^{\, \mathbb{A}} + \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \varrho_{z}^{\, \mathbb{B}} = \delta_{\mathsf{o}}^{\, \mathbb{C}} p^{\, \mathbb{C}} \varrho_{z}^{\, \mathbb{C}}$

To simplify calculations without loss of generality, choose the frame of reference so that $\widehat{z}^{\, \mathbb{C}}$ is positively aligned with $\widehat{z}^{\, \mathsf{F}} .$ Then

$\delta_{\mathsf{o}}^{\, \mathbb{C}} \! = \! 1$ and $\widehat{z}^{\, \mathbb{C}} \! = \left( 0, 0, 1 \right)$

For this special case, conservation of momentum requires that

$\delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \varrho_{x}^{\, \mathbb{A}} + \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \varrho_{x}^{\, \mathbb{B}} = 0$

$\delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \varrho_{y}^{\, \mathbb{A}} + \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \varrho_{y}^{\, \mathbb{B}} = 0$

$\delta_{\mathsf{o}}^{\, \mathbb{A}} p^{\, \mathbb{A}} \varrho_{z}^{\, \mathbb{A}} + \delta_{\mathsf{o}}^{\, \mathbb{B}} p^{\, \mathbb{B}} \varrho_{z}^{\, \mathbb{B}} = p^{\mathbb{C}}$

Momentum of a Graviton

To do: all graviton momenta are directed like the frame, and proportional to the number of quarks

Mechanical Energy

actually Total mechanical energy, but don’t change links or slugs

Definition of Mechanical Energy

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

where $c$ is a constant.

This statement comes from Max Planck and Paul Dirac .

By this definition, the mechanical energy is never negative, $E \! \ge \! 0 .$

Please notice that these numbers have been defined by a methodical description of sensation.

Mechanical Energy and Gross Energies

Consider a particle P that is described by its gross field energy $\mathtt{E}_{\, \mathsf{field}} \hspace{2px} ,$ and its gross core energy $\mathtt{E}_{\, \mathsf{core}} \hspace{2px} .$

Now assume that $\mathtt{E}_{\, \mathsf{field}} \sim \mathtt{E} \! \left( \boldsymbol{\gamma} \right) .$ That is, let the gross field-energy be represented by the gross photonic energy which is defined by

$\mathtt{E} \! \left( \boldsymbol{\gamma} \right) \equiv \dfrac{hc}{2 \pi} \kappa_{\mathsf{o}}$

where $\kappa_{\mathsf{o}}$ is the wavenumber. Then substituting in the definition of the quantity of motion and eliminating the wavenumber gives

$\mathtt{E} \! \left( \boldsymbol{\gamma} \right) = cp$

Furthermore, assume that $\mathtt{E}_{\, \mathsf{core}} \sim \mathtt{E} \! \left( m \right) .$ That is, let the gross core-energy be manifest as the gross material energy which is defined by

$\mathtt{E} \! \left( m \right) \, \equiv \, mc^{2}$

where $m$ is P’s mass. Then

$E^{2} = \mathtt{E}_{\, \mathsf{core}}^{\, 2} \! + \mathtt{E}_{\, \mathsf{field}}^{\, 2}$

So $E$ is like the gross-energy for the whole of particle P, both core and field together.

Slow Motion

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 quantity-of-motion.

The Lorentz Factor

The mass and quantity-of-motion may also be be combined to specify yet another quantity

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

A photo of H. Lorentz. — Hendrik Antoon Lorentz, 1853—1928.

This number $\gamma$ is called the Lorentz factor after the Dutch physicist Hendrik Lorentz . His original work¹H. 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

$E = \gamma mc^{2}$

Mechanical Energy of a Graviton

Here is an example of calculating the mechanical energy for a graviton $\mathsf{\Gamma} .$

Note that the wavevector of a graviton is … ?

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 quantity-of-motion. 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 quantity-of-motion and energy of a graviton are directly proportional to $N_{\mathsf{q}}^{\mathsf{\Gamma}}$ the total number of quarks it contains.

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.

Kinetic Energy

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

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

Since $m \ge 0$ for material particles, $K$ is never negative. And recall that for an inertial frame of reference, De Broglie’s postulate states that the quantity-of-motion 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.

Total 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 material 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.

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.

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.

Conservation of Energy

Newtonian particles are dense. 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.

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.

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	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.