One more time to get it right:
The Minkowski equation of motion is
$$ K = m A $$
with K the Minkowski four-force and A the four-acceleration. The Minkowski three-force is then related to the non-relativistic force as
$$ \vec{K} = \gamma \vec{F} $$
In traditional relativity, and with primes indicating traditional-relativity-unique energy quantities here and following, the kinetic energy is obtained as
$$
\begin{eqnarray}
E'_k \equiv \int_0^u \vec{F} \cdot d \vec{r} = \int_0^u \frac{1}{\gamma} \vec{K} \cdot d \vec{r}
\nonumber \\ = \int_0^u d \vec{u} \left[ m \gamma^3\right] \cdot \vec{u} \nonumber \\ = \int_0^u m \gamma^3 u du \nonumber \\ = \int_0^u m c^2 d \gamma \nonumber \\ = m c^2 \gamma(u) - m c^2 \gamma(0) \nonumber \\
= m c^2 \gamma - m c^2 \nonumber \\
= m c^2 \left[\gamma - 1\right].
\end{eqnarray}
$$
Thus, in traditional relativity the classical kinetic energy relation $$E_k = p^2/2m$$ is an approximation valid only at low velocity. However, the Einstein relativistic energy E' can be related to the time component of the four-momentum as $$p_0 \equiv m\gamma c = E'/c$$. This gives rise to the ``energy-momentum'' four-vector of traditional relativistic mechanics. In Einstein relativity, the time component of four-momentum is
$$
\begin{eqnarray}
p_0 \equiv \gamma m c = \frac{\gamma m c^2}{c} = \frac{E'}{c},
\end{eqnarray}
$$
where E' is the Einstein relativistic energy. The temporal momentum thus cannot be negative unless the particle energy is negative.
Noting that
$$d\gamma/d\beta = \beta \gamma^3$$ so $$ d \gamma = u \gamma^3 d\beta / c = u \gamma^3 du / c^2 $$, and with $$ \gamma^2 -1 = (1 - \beta^2)^{-1} - (1 - \beta^2)/(1 - \beta^2) = \beta^2/(1 - \beta^2) = \beta^2 \gamma^2$$, the kinetic energy may be evaluated from Eq. (\ref{EkInt1}) as
$$
\begin{eqnarray}
E_k = \int_0^u d\vec{u} \left[ m \gamma^4\right] \cdot \vec{u} \nonumber \\ = \int_0^u m \gamma^4 u du \nonumber \\ = \int_0^u m c^2 \gamma d \gamma \nonumber \\ = \frac{1}{2} m c^2 \gamma^2(u) - \frac{1}{2} m c^2 \gamma^2(0) \nonumber \\
= E(u) - E(0) \nonumber \\
= \frac{1}{2} m c^2 \gamma^2 - \frac{1}{2} m c^2 \nonumber \\
= \frac{1}{2} m c^2 \left[\gamma^2 - 1\right] \nonumber \\
= \frac{1}{2} m \gamma^2 u^2 \nonumber \\
= \frac{p^2}{2m},
\end{eqnarray}
$$
where$$p = m \gamma u$$ is the relativistic momentum.
In Osiak relativity, as in Einstein relativity, the kinetic energy is the total energy minus the rest energy, \emph{ i.e.}, $$E_k = E(u) - E(0)$$. Using the Osiak total and rest energy formulas as obtained, $$E = \gamma^2 m c^2/2$$ and $$E_0 = m c^2/2$$ obtains the familiar relation from non-relativistic classical physics, $$E_k = p^2/2m$$, but p here is the relativistic momentum.[/tex]
