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3.3.4 Momentum

In his equations of motion, Newton gives priority to the concept of momentum over that of mass. He defines momentum as, \(p = mv\), where \(m\) is mass and \(v\) is its velocity within three dimensional, Euclidean space. In extending this concept to a particle moving in four dimensions, an appropriate definition for total momentum is \(P=E/c\), where \(E\) is the energy content of the particle and \(c\) is the total velocity \(c=(c_o+Jv)\). The symbol \(J=\sqrt{-1}\) represents the 90 degree rotation in four dimensions between the energy continuum and the direction of expansion. The four dimensional coordinate system becomes, in effect, a 'complex volume', and momentum becomes \(P=P_0+JP_v\).

We note that since \(c\) is the vector \(c=(c_o+Jv)\), total momentum, \(P\), becomes a vector in the four dimensional coordinate system, even though, \(v\) is a vector within the continuum.
\begin{align}
P= \frac{E}{c_o+Jv} &= \frac{E(c_o-Jv)}{(c_o^2+v^2)} \nonumber \\
&= E(\frac{c_0}{c_v^2}) - JE(\frac{v}{c_v^2}) \nonumber \\
&= P_0 + J P_v \nonumber
\end{align}
where, \(c_v\), given by \(c_v^2=(c_0^2+v^2)\), is the total velocity of the particle in four dimensions.

If the part of total momentum within the energy continuum is \(P_v\) and the momentum into the fourth dimension is \(P_0\), then the modulus of total momentum is given by, \(|P|^2 =P_v^2 + P_0^2\). We have then
\begin{align}
P_v= E(\frac{v}{c_v^2})\ \ and\ P_0 = E(\frac{c_o}{c_v^2})\ hence\ |P|^2 =&\ E^2 \ \frac{v^2+c_0^2}{c_v^4} \nonumber \\
and\ modulus\ |P|= E (\frac{\sqrt{v^2+c_o^2}}{c_v^2}) \ =&\ \frac{E}{c_v} \nonumber
\end{align}

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