Consider conservation of energy and momentum
\(mc^2=\cfrac{1}{2}mc^2+\cfrac{1}{2}mc^2\)
\(mc^2=\cfrac{1}{2}(\cfrac{m}{2})(\sqrt{2}c)^2+\cfrac{1}{2}(\cfrac{m}{2})(\sqrt{2}c)^2\)
after \(n\) split, each particle will have energy,
\(E_p=\left\{\cfrac{1}{2}\right\}^nmc^2\) ---(1)
and its mass is,
\(m_p=\cfrac{m}{2^n}\)
And the total number of particles after \(n\) split is,
\(2^n\)
\(0=\cfrac{m}{2}\sqrt{2}c+\cfrac{m}{2}\sqrt{2}c\) --- (1)
Subsequently, each particle starts with a momentum of
\(P_p=\cfrac{m}{2}\sqrt{2}c \)
the second split results in
\(\cfrac{m}{2}\sqrt{2}c=\cfrac{m}{4}\sqrt{2}c+\cfrac{m}{4}\sqrt{2}c=\cfrac{m}{2}\cfrac{1}{2}\sqrt{2}c+\cfrac{m}{2}\cfrac{1}{2}\sqrt{2}c\)
after \(n\) split, each particle will have momentum,
\(P_p=\cfrac{1}{2}\left\{\cfrac{1}{2}\right\}^{n-1}m\sqrt{2}c=\left\{\cfrac{1}{2}\right\}^{n}m\sqrt{2}c\)
Consider the total energy of the particle,
\(E_p=PE_p+KE_p\)
But since we started with \(PE_p=0\),
\(E_p=KE_p=\cfrac{1}{2}mv^2=\cfrac{1}{2}mv.v=\cfrac{1}{2}P_p.v\)
\(\cfrac{1}{2}P_p.v=E_p=\left\{\cfrac{1}{2}\right\}^nmc^2=\cfrac{1}{2}.\cfrac{1}{2}\left\{\cfrac{1}{2}\right\}^{n-1}m\sqrt{2}c.v\)
\(v=\sqrt{2}c\)
Up to here, if \(E=mc^2\) is our kinetic energy in time that denotes our existence, where \(c\) is our time speed along the time axis, then it is possible to reach speed in space of up to \(\sqrt{2}c\). And since the derivation up to this point, has not included any losses (both conservation laws apply), this is the max speed by which we can attained in space (a higher value of \(v\) implies \(E>mc^2\), a lower value would imply loss, for which there is none). So, the maximum speed in free space is \(\sqrt{2}c\) and not \(c\).
