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3.5.7 Motion within the Energy Continuum

In the geometry of the energy continuum, its hyper-spherical shape ensures that each point within its curved volume is directly adjacent to the void into which the continuum is expanding. This is analogous to the way in which the curved surface area of an expanding sphere is adjacent to the volume inside and outside the sphere. It is our inability to visualize this curved volume, that presents the major problems in analyzing motion within the continuum.

For a small volume of the continuum, its inherent curvature may be assumed to be zero, so that an analysis can be simplified to the situation of a volume of flat continuum moving through the void at expansion velocity, \(v_x\), which is also the speed of light, \(v_x=c_0\). In classical physics this situation is the subject of Newton's laws of motion. The energy continuum is the Ether which, in classical physics, permeates the whole of space, and expansion velocity is ignored because it cannot be detected. The question to be asked and answered is this; does expansion velocity invalidate Newton's laws.

To include expansion velocity in a classical analysis, it is necessary to take into account that this velocity is perpendicular to each of the three coordinate axes of the flat continuum. This requires that the three orthogonal axes of Newtonian, Euclidean space be augmented by a fourth axis in the direction of expansion and orthogonal to Newtonian space. Hence, each point in this coordinate system is specified by four coordinates, \(R, J(ix, jy, kz)\), where the symbol \(J=\sqrt{-1}\) denotes perpendicularity to the three unit vectors \((i, j, k)\) of Euclidean space.

In the analysis below, we consider movement of energy within the continuum itself, on scales of time and space compatible with human experience. We show that, when velocity and momentum are acknowledged to extend in the direction of expansion, Newton's laws of motion, which are valid in three dimensions, still hold true in this four dimensional frame work. However, to ensure this we need to establish definitions of both mass and momentum appropriate to four dimensions.

We consider the movement of energy within a gravity free volume of the continuum. We assume that this volume is small enough so that the inherent curvature of the universe within it is so small as to be negligible. Expansion of this volume, too, can be safely ignored since the exponent of exponential volume expansion, \(3Ht\), is extremely small. However, the expansion velocity, \(c_o\), - the speed of light - is very large and must be taken into account since it endows each point within the continuum with momentum. This situation is a simplification of particle motion, and its results will be used in the following chapter on particle structure.

As before, we imagine a small element of constant volume so that its angular momentum content, \(A\), is constant. In this case, the energy content of this volume element will not remain constant since it is moving within the continuum and will acquire kinetic energy due to its velocity, \(v\). We assume that kinetic energy may be attributed to an increase in energy density, \(\rho\), which in turn is enabled by increases in expansion velocity, \(c\), and curvature \(\kappa\), as per the energy constraint equation,
\begin{align}
Energy\ content\ = Volume\ \rho = A\ c\ \kappa \nonumber
\end{align}

Since we regard the local continuum as flat and not expanding, we can represent points within it using three dimensional Cartesian coordinates. However, the movement of this whole Cartesian frame of reference into the forth dimension with expansion velocity \(c_o\) must be accounted for by adding \(c_o\) to velocities within the continuum as an orthogonal vector component.

This constant volume element is moving with velocity \(v\) within the local continuum, and also in the direction of expansion with velocity, \(c_o\). The total velocity, \(c\), therefore, is the vector sum of \(v\) and \(c_o\). We can express the orthogonality of \(c_o\) by use of the imaginary operator \(J=\sqrt{-1}\). Even though the velocity \(v\) is itself a three dimensional vector within the Cartesian frame of reference, so long as \(v\) remains in a constant direction it can be treated as a scalar quantity. Hence the total velocity of the energy element is now the vector, \(c= v+Jc_o\), which has a modulus given by \(|c|^2 = c_o^2+v^2\).

The momentum of the volume element within the continuum, \(P_v\), is the conventional, Newtonian momentum, \(p=mv\), where \(m\) is the mass equivalent of the energy content of the constant volume element. We may then write,

\begin{align}
mv &= E \frac{v}{c^2} \nonumber \\
mc &=\frac{E}{c} \nonumber
\end{align}
where \(c = \sqrt{v^2+c_o^2}\).

This leads to the result \(E=mc^2\), which has no definite meaning since both \(m\) and \(E\) are not defined in the general case where \(v\ne0\). However, setting \(v=0\), we can establish a relationship between rest mass, \(m_o\) and rest energy, \(E_o\), as \(E_o=m_oc_o^2\), which is the Relativistic relationship.

To determine how mass and energy vary when \(v\ne0\), it is necessary to impose an additional constraint. This can be done by insisting that the modulus of total momentum is a constant. That is,
\begin{align}
|P| = E (v^2+c_o^2)^{-\frac{1}{2}} =\ constant\ \nonumber
\end{align}
Setting \(v=0\), the constant, \(|P|\), may be evaluated as \(m_o c_o =E_o/c_o\).

This means that the energy content of the volume element, \(E\), is given by
\begin{align}
E= \frac{E_o}{c_o} (v^2+c_o^2)^{\frac{1}{2}} = E_o\sqrt{1+v^2/c_o^2} \nonumber
\end{align}

Alternatively, we can write
\begin{align}
E= \frac{E_o}{c_o} (v^2+c_o^2)^{\frac{1}{2}} &= m_oc_o^2\sqrt{1+v^2/c_o^2} \nonumber \\
&=m_o c_o c \nonumber \\
&= (m_o \frac{c}{c_o}) c_o^2 \nonumber
\end{align}
If a definition of moving mass is required we can write, \(m=m_o c/c_o\), which will give the expression for total energy of the moving element as \(E= mc_o^2\). But this is only complicates the definition of mass, and since mass is a convenient parameter and not a substance in its own right, it makes sense to stick with rest mass \(m_o\).

However, in Special Relativity, mass is said to increase with velocity. Taking the above tentative expression for moving mass, \(m=m_oc/c_o\), we can derive the following.
\begin{align}
m &= m_o \frac{c}{c_o} \nonumber \\
&=m_o \frac{c}{\sqrt{c^2 -v^2}}\nonumber \\
&= \frac{m_o}{\sqrt{1 -v^2/c^2}}\nonumber
\end{align}
Since Special Relativity does not distinguish between \(c\) and \(c_o\), the above is the same as the Relativistic expression for moving mass.

With this definition of energy in terms of velocity, we may now find the relationship between kinetic energy \(E_k\), and velocity, \(v\). We have that,
\begin{align}
E_k &= E - E_o \nonumber \\
&= E_o\sqrt{1+v^2/c_o^2} -E_o \nonumber
\end{align}

The square root term may be represented by its binomial expansion, that is,

\(\sqrt{1+v^2/c_o^2} = 1 + v^2/2c_o^2 - v^4/3c_o^4+ higher\ order\ terms\)

In the real world, the ratio \(v^2/c_o^2\) is very small. For example, at the very high speed of 30,000 meters per second, it is about \(10^{-8}\), and the third term is of the order \(10^{-16}\).

Hence, for use not involving high speed particles, we can truncate the binomial expansion to its first two terms, \(1+v^2/2c_o^2\). This reduces the expression for kinetic energy to
\begin{align} E_k = E_o(1+\frac{1}{2}\frac{v^2}{c_o^2} - 1) &= m_oc_o^2\frac{1}{2}\frac{v^2}{c_o^2} \nonumber \\
&= m_o\frac{1}{2}v^2 \nonumber
\end{align}
This is the expression for kinetic energy in both classical and Relativistic physics.

Finally, we note that the expression for total energy of the moving volume element may be rearranged and expanded as follows.
\begin{align}
E &= |P| (v^2+c_o^2)^{\frac{1}{2}} \nonumber \\
squaring\ gives\ E^2 &= |P|^2 (v^2+c_o^2) \nonumber \\
and\ E^2 &= |P|^2 v^2+|P|^2c_o^2 \nonumber
\end{align}

Writing \(|P|=m_o c_o\), we have
\begin{align}
E^2 &= (m_o c_o)^2 v^2+(m_o c_o)^2c_o^2 \nonumber \\
E^2 &= p^2 c_o^2 + m_o^2 c_o^4 \nonumber
\end{align}
where \(p=m_ov\), the classical expression for the momentum of a mass within three dimensions.

This is the same expression for total energy derived via Special Relativity.

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