# Mass, Energy and Momentum

In this article we will show how quantum-geometry dynamics explains mass, energy, momentum.

As discussed in earlier articles, quantum-geometry dynamics proposes that there be only two fundamental particles: The preon(-), which dimensionalizes space, and the preon(+), the fundamental particle of matter. What distinguishes quantum-geometry dynamics from dominant theories of fundamental physics, collectively known as the standard model, is that all properties of elementary particles are intrinsic. Essentially, this means that no additional particles and their properties are necessary to explain interactions not only at the fundamental scale but at all scales.

According to QGD, every particle or material structure is made of preons(+). That includes, without exception, all particles we currently consider to be elementary. Electrons, positrons, neutrinos and even photons are thus composed of preons(+).

Preons(+) possess two intrinsic properties. The first is that they are strictly kinetic. Preons(+) are always in motion. Their speed is constant, hence their kinetic energy is constant and equal to $c$. The speed of preons(+) is constant whether they are free or bound within a composite particle or structure. The constancy of the speed of preons(+) is a direct consequence of the quantum-geometrical structure of space. Preons(+) move by leaping from preons(-) to preons(-). The preonic leap is fundamental unit of distance. Since there is no smaller distance than the preonic leap, there is no faster motion than that of a preon(+).

The second intrinsic property of preons(+) is that they interact with each other through p-gravity, which is an attractive force acting between them. The p-gravity interaction between two preons(+) is the fundamental unit of p-gravity or $\displaystyle {{g}^{+}}$ .

### Mass

According the QGD model, mass is an intrinsic property of matter. Since preons(+) are the fundamental particle of matter, their mass must be the fundamental unit of mass. It follows that the mass of an object should be understood as being equal to the number of preons(+) it contains. So when we write that ${{m}_{a}}$is the mass of a particle $a$ we mean that it contains $m$ number of preons(+).

### Energy

Like mass, energy is an intrinsic property of preon(+). The fundamental unit of energy corresponds to the kinetic energy of the preon(+) and is equal to its mass times its speed, which is equal to $c$ , that is ${{E}_{{{p}^{\left\langle + \right\rangle }}}}={{m}_{{{p}^{\left\langle + \right\rangle }}}}c$ or, since ${{m}_{{{p}^{\left\langle + \right\rangle }}}}=1$ , we have ${{E}_{{{p}^{\left\langle + \right\rangle }}}}=c$. So the kinetic energy of a preon(+) is mathematically equivalent to its speed.

The same formula applies to composite particles or structures. So if $a$ is a particle or structure with mass ${{m}_{a}}$ , then its energy must correspond to the added energies of the preons(+) that compose it. In mathematical terms the relationship between energy and mass is expressed by the formula $\displaystyle {{E}_{a}}={{m}_{a}}c$; the number of preons(+) multiplied by the energy of each preon(+).

Thus quantum-geometry dynamics provides a simple and elegant explanation of the relationship between the mass of a body and its energy. It is important to note though QGD’s equation appears to be similar to Einstein’s mass-energy equivalence equation, it differs in important ways.

First, ${{E}_{a}}={{m}_{a}}c$ is not an equivalence relation between mass and energy. Mass corresponds to the number of preons(+) of a body and energy corresponds to product of mass by the fundamental energy of the preon(+). So the QGD equation expresses a proportionality relation between mass and energy, not an equivalence relation. So according to the QGD model, mass can never be converted to energy, nor energy be converted to mass. Mass and energy are two intrinsic but distinct properties of preons(+).

At first glance this may appear to contradict observations. For instance, nuclear reactions result in a certain amount of mass being transformed into energy. What the QGD model suggests is that during a nuclear reaction, mass is not transformed into energy, but rather, bound particles become free from the structures they were bound to and carry with them their momentum. There is no conversion of mass into energy, but only the release of particles having momentum. We will see in the next section how kinetic energy is the only kind of energy that exists.

### Momentum

That mass and energy are intrinsic properties of preons(+) implies that unless a composite particle or massive structure loses or acquires preons(+), its mass and energy remains constant regardless of its speed. This is in disagreement with dominant physics theories which define the energy of an object has the sum of its intrinsic energy, also known as its energy at rest, and its kinetic energy or momentum; a relation that is expressed by the equation $E=m{{c}^{2}}+mv$ where $m{{c}^{2}}$ is the energy of the body at rest, $c$ the speed of light, and $mv$,the product of the its mass by its speed $v$,its kinetic energy or momentum.

The accepted definition of the energy of a body is logically correct. The reasoning behind it is simple. In order to accelerate an object, one needs to impart energy to it. Thus it makes perfect sense that the accelerated object carries this energy, which we call kinetic energy, with it. So the assumption that the total energy of body must be equal to the energy it had prior to acceleration plus the kinetic energy that is imparted to it is perfectly logical. However, we will show here that using the axioms of QGD we arrive at different descriptions and explanations which, though in disagreement with dominant physics theories, are consistent with observations.

According to QGD, there is only one kind of energy known as kinetic energy or momentum.

As explained earlier, the speed of a preon(+) is constant, hence its momentum is constant and equal to $c$. It follows that the total energy of a particle or massive structure must be equal to the sum energy of its constituents. This is described by the formula ${{E}_{a}}={{m}_{a}}c$.

From this point we will use vectors to represent momentum and direction. We will, for example, associate the vector $\vec{c}$ to the momentum of a preon(+) so that $c=\left\| {\vec{c}} \right\|$ . The energy of a preon(+) is thus the magnitude of its momentum vector. This distinction allows us to define ${{E}_{a}}$ and ${{P}_{a}}$, respectively the energy and momentum of a particle $a$ in the following way.

$\displaystyle {{E}_{a}}=\sum\limits_{i=1}^{{{m}_{a}}}{\left\| {{{\vec{c}}}_{i}} \right\|}={{m}_{a}}c$

${{P}_{a}}=\left\| \sum\limits_{i=1}^{{{m}_{a}}}{{{{\vec{c}}}_{i}}} \right\|$

To illustrate this, let’s consider the simple particle made of two bound preons(+) as shown in the following figure 1.

Here the particle, which we’ll denote $a$ , is composed of two preons(+), $p_{1}^{\left\langle + \right\rangle }$ and$p_{2}^{\left\langle + \right\rangle }$, bound by gravitational interaction (the resulting effect of p-gravity and n-gravity). The purple arrows represent their trajectories in quantum-geometrical space. The yellow vectors represent their momentum at a certain point before taking into account the bounding force. This magnitude of this vector is equal to $c$.

In this example, the interaction between the preons(+) is strong enough to deviate them from what would be their free trajectories, which would coincide with their momentum vector.

If the composite particle in our example is not subjected to any other force, then ${{E}_{a}}=\sum\limits_{i=1}^{{{m}_{a}}}{\left\| {{{\vec{c}}}_{i}} \right\|}={{m}_{a}}c=2c$, and its momentum is ${{P}_{a}}=\left\| \sum\limits_{i=1}^{{{m}_{a}}}{{{{\vec{c}}}_{i}}} \right\|=\left\| {{{\vec{c}}}_{1}}+{{{\vec{c}}}_{2}} \right\|=0$. The zero value indicates that particle is at rest relative to quantum-geometrical space.

However, if $a$ interacts with a massive structure $b$ (see figure 2), then the trajectories of the preons(+) of $p_{1}^{\left\langle + \right\rangle }$ and $p_{2}^{\left\langle + \right\rangle }$ will change and affect the momentum of $a$. We see that ${{E}_{a}}=2c$ but that ${{P}_{a}}=\left\| {{{\vec{c}}}_{1}}+{{{\vec{c}}}_{2}} \right\|=2v$ where $v$ is the speed of the particle.

This example shows how the energy of a particle does not change as a function of its speed. The momentum of a particle changes but is already taken into account in the formula ${{E}_{a}}={{m}_{a}}c$. The same applies to material structures at all scales.

We should address here the issue of the maximum possible speed of massive structure. From what we have seen so far, the speed of any structure is $\displaystyle \frac{{{P}_{a}}}{{{m}_{a}}}=\left\| \sum\limits_{i=1}^{{{m}_{a}}}{{{{\vec{c}}}_{i}}} \right\|=v$ . Since $\displaystyle \max \left\| \sum\limits_{i=1}^{{{m}_{a}}}{{{{\vec{c}}}_{i}}} \right\|=\sum\limits_{i=1}^{{{m}_{a}}}{_{i}}\left\| {\vec{c}} \right\|$ , the maximum speed of a particle or structure is achieved when the momentum of a particle is equal to its energy or ${{P}_{a}}={{E}_{a}}$ . This is evidently the case for light, but that shouldn’t be confused with current idea that light is pure energy. The energy of a photon is still that of its component preons(+). The same applies to a structure when the trajectories of all its component preons(+)are parallel and oriented in the same direction. In such case we have ${{P}_{a}}=\sum\limits_{i=1}^{{{m}_{a}}}{_{i}}\left\| {\vec{c}} \right\|={{m}_{a}}c$ so that ${{v}_{a}}=\frac{{{P}_{a}}}{{{m}_{a}}}=\frac{{{m}_{a}}c}{{{m}_{a}}}=c$. Particles for which energy and momentum are always equal include photons and neutrinos but theoretically any structure regardless of its mass can achieve $c$ if it is submitted to a strong enough force.

 Slowing Down of Clocks The conservation of the speed of preons(+) implies that any increase in the speed of a structure translates in a reduction of the transversal speeds of the component preons(+), hence a decrease in the speeds of the component substructures. In other words, all internal motion must slow down as v increases. This means that all internal motion of any structure will slow down to a stop when v=c. The slowing down of internal motion explains why clocks slow down without the necessity of introducing the mechanism of time dilation. Thus, QGD provides a non-relativistic explanation of the slowing down of clocks due to an increase in speed or gravitation. Therefore, the slowing down of clocks (or any internal motion of a particle or system) is consistent with QGD’s assumption that time is not physical property of reality, but a purely relational concept allowing us to compare events to cyclic periodic systems; in other words, clocks. It is clocks that slow down, not time. Note that this explains observations showing that fast muons decay at a much slower rate than slow muons. Whatever internal process cause muons to decay will slow as its speed increases. This is explained in detail in the article The Slowing Down of Clocks as Explained by QGD (see link).

## Calculating Direction of a Bound Preon(+) and Structures

When a preon(+), denoted ${{p}^{\left\langle + \right\rangle }}$, is subjected to an interaction with an object $a$ , expressed by force vector $\overset{\leftrightarrow }{\mathop{G}}\,\left( {{p}^{\left\langle + \right\rangle }};a \right)$ , its direction changes while its speed, hence its energy remains constant and equal to $\left\| {\vec{c}} \right\|$. The momentum of a preon(+) is conserved under change in direction.

If $\vec{c}$ is the momentum vector of a preon(+) and $\vec{c}'$ is its momentum vector resulting from a change in direction cause by attractive force, then $\left\| {\vec{c}} \right\|=\left\| \vec{c}' \right\|=c$ . The momentum is conserved in accordance to the formula $\displaystyle \vec{c}'=\frac{c}{\left\| \vec{c}+\vec{G} \right\|}\vec{c}+\vec{G}$ for a single preon(+) influenced by a single force $\vec{G}$ .

When a preon(+) is subjected to $n$ forces we have $\displaystyle \vec{c}'=\frac{c}{\left\| \vec{c}+\sum\limits_{i=1}^{n}{{{{\vec{G}}}_{i}}} \right\|}\vec{c}+\sum\limits_{i=1}^{n}{{{{\vec{G}}}_{i}}}$. Also, because the momentum vector of a particle or massive structure is the resultant of the sum of the momentum vectors of all its components, that is ${{P}_{a}}={{m}_{a}}\left\| \sum\limits_{i=1}^{{{m}_{a}}}{_{i}}\vec{c} \right\|={{m}_{a}}{{v}_{a}}$, and we can threat the particle as a whole when it is subjected to a force. Then using the momentum vector $\displaystyle {{\vec{P}}_{a}}$ for the particle $a$ can have $\displaystyle {{\vec{P}}_{a}}'={{\vec{P}}_{a}}+\vec{G}\left( a;b \right)$ where $\displaystyle \vec{G}\left( a;b \right)$ is the gravitational interaction between $a$ and $b$ and $\displaystyle \vec{G}\left( a;b \right)=\frac{\overset{\leftrightarrow }{\mathop{G}}\,\left( a;b \right)}{{{m}_{a}}}$.

In conclusion, change in speed and the corresponding change in momentum of a composite particle or massive structure is caused by discrete changes in the trajectories of their component preons(+). The kinetic energy of a composite particle or structure is thus a directional component of the total energy along the axis of motion. As such, the kinetic energy of body is already included in the intrinsic energy of an object, itself the sum of the kinetic energy of its preons(+). At the most fundamental scale, the preonic scale, since preons(+) are kinetic, there is no such thing as energy at rest. But there can be, as we saw earlier, non-fundamental structures in which the momentum vectors cancel out resulting in a null net momentum.

As the reader can see, the equation $E=mc$ naturally emerges from the first principles of quantum-geometry dynamics.

Understanding how energy is conserved under acceleration is a simple shift with important impact in physics. In cosmology, for example, a universe which undergoes an accelerated expansion does not violate the law of conversion of energy. The energy of the galaxies undergoing acceleration does not change (except from the variations of their masses). This also implies that the mass and energy of the Universe are conserved.

Note: We will see in the coming article about gravity how the effect attributed to dark energy can be attributed to negative gravitational force (where n-gravity interactions exceed p-gravity interactions). We will also show why dark energy being gravitational, it cannot be detected by instruments.

## Heat, Temperature and Entropy

Using the concepts we have introduced so far, we will now derive quantum-geometrical explanation of the properties of heat, temperature and entropy.

Given a system $S$ having $n$ unbound particles, the heat of the system is equal to $\displaystyle \sum\limits_{i=1}^{n}{{{P}_{i}}}$ , where ${{P}_{i}}$ is the momentum of the ${{i}^{nt}}$ particle and its temperature is $\frac{\sum\limits_{i=1}^{n}{{{P}_{i}}}}{Vo{{l}_{S}}}$.where $Vo{{l}_{S}}$ is the volume of the system measured in preon(-), the fundamental and discrete particle which forms and dimensionalizes quantum-geometrical space. The total energy of the system being equal to $\sum\limits_{i=1}^{n}{{{m}_{i}}c}$ , it follows that if we define entropy in the classical sense, then the entropy of$S$ is $\sum\limits_{i=1}^{n}{{{m}_{i}}c}-\sum\limits_{i=1}^{n}{{{P}_{i}}}$.

##### Application to Exothermic Reactions within a System

The QGD definitions can be used to describe the changes in heat and temperature resulting from chemical or nuclear reactions. The particles involved are different, as are the reaction mechanisms, and the reactions occur at different scales, but both result in changes in the structure and number of bound particles.

Consider $\displaystyle {{S}_{1}}\to {{S}_{2}}$ where ${{S}_{1}}$ is a dynamic system containing ${{n}_{1}}$ unbound particles (or structures) some of which reacting with each other, and ${{S}_{2}}$ the resulting system containing $\displaystyle {{n}_{2}}$ unbound particles, if ${{n}_{2}}>{{n}_{1}}$ then $\displaystyle \sum\limits_{i=1}^{n}{{{P}_{i}}}<\sum\limits_{i=1}^{n}{{{P}_{i}}}'$ and change in heat of the system is$\Delta H=\sum\limits_{i=1}^{{{n}_{2}}}{{{P}_{i}}}'-\sum\limits_{i=1}^{{{n}_{1}}}{{{P}_{i}}}$ is positive.

For example, let say the system contains only a particle ${{e}^{-}}$ and a particle ${{e}^{+}}$ which annihilate to give $n$ photons ($\gamma$ ), then $\displaystyle \Delta H=\sum\limits_{i=1}^{n}{{{m}_{{{\gamma }_{i}}}}c}-\left( {{v}_{{{e}^{-}}}}{{m}_{{{e}^{-}}}}+{{v}_{{{e}^{+}}}}{{m}_{e+}} \right)$. Here, the difference in heat depends on the speed of interacting electrons and is at the lowest when electrons achieve the speed of light; in which case $\Delta H=0$.Note that from the QGD model, when electrons achieve$c$, internal motion stops, so that component preons(+) move in parallel trajectories. Note that QGD predicts that electrons accelerated to $c$become indistinguishable from neutrinos and become electrically neutral. The electrical charge of a particle is caused by internal motion of its component preon(+) which interact with the preonic field (the free preons(+) populating quantum-geometrical space). Since all internal motion stop at speed$c$, the electron moving at that speed must lose their electric charge.

Also worth nothing is that $\displaystyle \sum\limits_{i=1}^{n}{{{m}_{{{\gamma }_{i}}}}}={{m}_{{{e}^{-}}}}+{{m}_{e+}}$ which implies that ${{E}_{{{S}_{1}}}}={{E}_{{{S}_{2}}}}$,that is; mass and energy are conserved. This holds for any closed system. So though it is believe that a nuclear reaction results in the conversion of mass into energy, it is, according to QGD, it results in the freeing of bound particles which carry with them momentum, hence increase the heat of the system. Aside from the reaction mechanism, the only difference between exothermic chemical and nuclear reactions is in the type of particles that become free. For chemical reactions these particles are molecules, atoms and photons, and for nuclear reactions they nuclei and other subatomic particles.

##### Application to Cosmology

In the initial state of the Universe, QGD theorizes that all preons(+) were free. That means that the energy of the Universe was equal to its heat. So if that its entropy was equal to zero. That is:$\displaystyle {{m}_{U}}c-\sum\limits_{i=1}^{n}{{{P}_{i}}}=0$, where ${{m}_{U}}$ is the masse of the Universe in preons(+) and since all preons(+) are free ${{m}_{U}}=n$. It follows that the temperature of Universe in its initial state was $T_{0}^{U}=\frac{{{m}_{U}}c}{Vo{{l}_{U}}}$.

Though the Universe as evolved, its total energy remains ${{m}_{U}}c$, but as particles and structures are formed its heat decreases resulting in an increase in entropy according. In formal terms we have $\displaystyle {{m}_{U}}c-\sum\limits_{i=1}^{n}{\left\| {{P}_{i}} \right\|}>0$).

For those interested in reading further, the chapter on cosmology in Introduction to Quantum-Geometry Dynamics explains that the temperature of the Universe in its initial state and to the only two constants of QGD; $c$ the kinetic energy of the preon(+) and $k$ the proportionality constant between the n-gravity and p-gravity, are related by the equation ${{T}_{0}}=\frac{c}{\sqrt{k}}$.

The next article discusses how gravity emerges from the interactions between preons.