Newton's momentum and power from the 4-force or Minkowski force How can I get Newton's momentum and power from the time and space component of the 4-force at low speeds?
 A: Consider $\,\mathbf f\boldsymbol{=}\left(f_1,f_2,f_3\right)\,$ to be a 3-vector representing a force applied on a particle of rest mass $\,m_{0}\,$ moving with velocity 3-vector $\,\mathbf u\boldsymbol{=}\left(u_1,u_2,u_3\right)$. If this force is preserving the rest mass $\,m_{0}\,$ of the particle then we can $''$build$''$ a Lorentz 4-vector $\,\mathbf F\,$ as follows
\begin{equation}
\mathbf F \boldsymbol{=}\left(\gamma_{\mathrm u}\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c},\gamma_{\mathrm u}\mathbf f \right)\boldsymbol{=}\left(F_0,F_1,F_2,F_3\right)
\tag{01}\label{01}
\end{equation}
where $\,\gamma_{\mathrm u} \boldsymbol{=}1\bigg{/}\sqrt{1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}}\,$.
For the power we have
\begin{equation}
\mathbf f\boldsymbol{\cdot}\mathbf u\boldsymbol{=}\gamma^{\boldsymbol{-}1}_{\mathrm u}\,c\,F_0\boldsymbol{=}\sqrt{1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}} c\,F_0
\tag{02}\label{02}
\end{equation}
For low speeds $\,\gamma^{\boldsymbol{-}1}_{\mathrm u}\boldsymbol{=}\sqrt{1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}}\boldsymbol{\approx}1 $
so
\begin{equation}
\boxed{\:\:\texttt{power}\boldsymbol{=}\mathbf f\boldsymbol{\cdot}\mathbf u\boldsymbol{\approx}c\,F_0\vphantom{\dfrac{a}{b}}\:\:}
\tag{03}\label{03}
\end{equation}
Note that for the low speed approximations we must take the limit $\:\mathrm u/c \longrightarrow 0\:$ and not $\:\mathrm c \longrightarrow\infty$. The latter corresponds to the non-relativistic approximations.
The 4-vector $\,\mathbf F\,$ of equation \eqref{01} is built to be a Lorentz 4-vector from its very definition as the differential of the 4-momentum Lorentz 4-vector $\,\mathbf P\,$ with respect to the proper time $\,\tau$, a Lorentz invariant scalar
\begin{equation}
\mathbf F \stackrel{\rm def}{\boldsymbol{\equiv\!\!\!\equiv}}\dfrac{\mathrm d\mathbf P}{\mathrm d\tau} \boldsymbol{=}\dfrac{\mathrm d\left(m_{0} \mathbf U\right)}{\mathrm d\tau}\boldsymbol{=}m_{0} \dfrac{\mathrm d\mathbf U}{\mathrm d\tau}\boldsymbol{=}m_{0} \dfrac{\mathrm d\left(\gamma_{\mathrm u}\,c,\gamma_{\mathrm u}\mathbf u\right)}{\mathrm d\tau}\boldsymbol{=}m_{0}\gamma_{\mathrm u} \dfrac{\mathrm d\left(\gamma_{\mathrm u}\,c,\gamma_{\mathrm u}\mathbf u\right)}{\mathrm d t}
\tag{04}\label{04}
\end{equation}
We see that for the  4-momentum vector $\,\mathbf P\boldsymbol{=}m_{0} \mathbf U\boldsymbol{=}m_{0}\left(\gamma_{\mathrm u}\,c,\gamma_{\mathrm u}\mathbf u\right)\,$ to be a Lorentz 4-vector we need the rest mass $\,m_{0}\,$ to be invariant.
The space component of equation \eqref{04} yields
\begin{equation}
\left(F_1,F_2,F_3\right) \boldsymbol{=}m_{0}\gamma_{\mathrm u} \dfrac{\mathrm d\left(\gamma_{\mathrm u}\mathbf u\right)}{\mathrm d t}
\tag{05}\label{05}
\end{equation}
But
\begin{equation}
\dfrac{\mathrm d\left(\gamma_{\mathrm u}\mathbf u\right)}{\mathrm d t} \boldsymbol{=}\left(\dfrac{\mathrm d\gamma_{\mathrm u}}{\mathrm d t}\right)\mathbf u\boldsymbol{+}\gamma_{\mathrm u}\dfrac{\mathrm d\mathbf u}{\mathrm d t}\boldsymbol{=}\left(\gamma^3_{\mathrm u}\dfrac{\mathrm u}{c^2}\dfrac{\mathrm d\mathrm u}{\mathrm d t}\right)\dfrac{\mathbf p}{m_{0}}\boldsymbol{+}\dfrac{\gamma_{\mathrm u}}{m_{0}}\dfrac{\mathrm d\mathbf p}{\mathrm d t}
\tag{06}\label{06}
\end{equation}
So
\begin{equation}
\left(F_1,F_2,F_3\right) \boldsymbol{=}\left(\gamma^4_{\mathrm u}\dfrac{\mathrm u}{c^2}\dfrac{\mathrm d\mathrm u}{\mathrm d t}\right)\mathbf p\boldsymbol{+}\gamma^2_{\mathrm u}\dfrac{\mathrm d\mathbf p}{\mathrm d t}
\tag{07}\label{07}
\end{equation}
For the low speed approximation $\:\mathrm u/c \boldsymbol{\approx} 0\:$, $\:\gamma_{\mathrm u}\boldsymbol{\approx}1\:$ and \eqref{07} gives
\begin{equation}
\boxed{\:\:\mathbf f\boldsymbol{=}\dfrac{\mathrm d\mathbf p}{\mathrm d t}\boldsymbol{=}\left(\dfrac{\mathrm d p_1}{\mathrm d t},\dfrac{\mathrm d p_2}{\mathrm d t},\dfrac{\mathrm d p_3}{\mathrm d t}\right)\boldsymbol{\approx}\left(F_1,F_2,F_3\right)\vphantom{\dfrac{a}{b}}\:\:}
\tag{08}\label{08}
\end{equation}
Of course equation \eqref{08} could be extracted directly from \eqref{01} and the low speed approximation since
\begin{equation}
\left(F_1,F_2,F_3\right)\boldsymbol{=}\gamma_{\mathrm u}\mathbf f\boldsymbol{=}\gamma_{\mathrm u}\left(f_1,f_2,f_3\right)\boldsymbol{=}\gamma_{\mathrm u}\left(\dfrac{\mathrm d p_1}{\mathrm d t},\dfrac{\mathrm d p_2}{\mathrm d t},\dfrac{\mathrm d p_3}{\mathrm d t}\right)\boldsymbol{=}\gamma_{\mathrm u}\dfrac{\mathrm d\mathbf p}{\mathrm d t}
\tag{09}\label{09}
\end{equation}
A: At low speeds the spacial components of Minkowski's four-vectors become practically equal to the quantities Newton cared about. In particular, if $(F_0,F_1,F_2,F_3)$ is four-force then Newton's force is $(F_1,F_2,F_3)\,.$ In short: you can ignore the time component. Same for momentum. Power is neither a four-vector nor a three-vector.
