Both classically and relativistically, the force on an object is what causes a time rate of change of its momentum: $\mathit{F}\mathbf{=}\mathit{d}\mathit{p}\mathbf{/}\mathit{d}\mathit{t}\mathbf{.}$

(a) using the relativistically correct expression for momentum, show that

$\mathit{F}\mathbf{=}{{\mathit{\gamma }}_{\mathbf{u}}}^{\mathbf{3}}\mathit{m}\frac{du}{dt}$

(b) Under what conditions does the classical equation $\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$hold?

(c) Assuming a constant force and that the speed is zero at $\mathit{t}\mathbf{=}\mathbf{0}$, separate t and u, then integrate to show that

$\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{+}{\left(Ft/mc\right)}^{\mathbf{2}}}}\frac{\mathbf{F}}{\mathbf{m}}\mathit{t}$

(d) Plot $\mathit{u}$verses$\mathit{t}$. What happens to the velocity of an object when a constant force is applied for an indefinite length of time?

Force on an object is what causes a time rate of change in its momentum

$F=\frac{dp}{dt}$

Using relativistic momentum expression in the above equation,

$$\begin{gathered} F=\frac{d}{dt}\left({\gamma }_{u}mu\right) \\ =m\left[{\gamma }_u\frac{du}{dt}+u\frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)\right] \\ =m\left[{\gamma }_u\frac{du}{dt}+u\left(\frac{-1}{2}{\left(1-\frac{u^2}{c^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(-\frac{2u}{c^2}\frac{du}{dt}\right)\right)\right] \\ =m\left[{\gamma }_u\frac{du}{dt}+{\gamma }_u\frac{du}{dt}\frac{c^2}{u^2}{\left(1-\frac{u^2}{c^2}\right)}^{-1}\right] \end{gathered}$$
Taking ${\gamma }_u\frac{du}{dt}$out of the bracket and solving will give the final expression as follows.

$F={\gamma }_u^{3}m\frac{du}{dt}$

From part (a), we know the relativistic relation for force.

$$\begin{gathered} F={\gamma }_u^{3}m\frac{du}{dt} \\ ={\gamma }_u^{3}ma \\ =m{\left(\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\right)}^3 \\ =ma{\left[1-\frac{u^2}{c^2}\right]}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Using the binomial expression, $(1-x{)}^n=1+nx+\frac{n(n+1)}{|2}{x}^2+...$

$\begin{array}{l}F=ma{\left[1-\frac{u^2}{c^2}\right]}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =ma\left[1+\frac{3}{2}\frac{u^2}{c^2}+\frac{15}{8}\frac{u^4}{c^4}+.....\right]\end{array}$

For velocities close to zero, the higher-order terms vanish.Therefore,

$F=ma$

Therefore, this equation holds for objects moving at very low speed.

Now, if we consider the relativistic equation of force from part (a), that is,

$F={\gamma }_u^{3}m\frac{du}{dt}$

And separating t and u and integrating,

$$\begin{gathered} F=\frac{m}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{du}{dt} \\ \frac{d}{{\left(1{\displaystyle \raisebox{1ex}{$-u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{F}{m}d{t}^{} \\ \int \frac{du}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\int \frac{F}{m}dt \\ \frac{u}{\left(\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}\right)}=\frac{Ft}{m} \end{gathered}$$

Solving further,

$\begin{array}{l}u=\left(\frac{Ft}{m}\right)(1-\raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.{)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ u^2={\left(\frac{Ft}{m}\right)}^2\left(1-\raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.\right)\end{array}$

Rearranging the above expression,

$u=\frac{1}{{\sqrt{1+\left({\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$mc$}\right.}\right)}}^2}\frac{Ft}{m}$

Here,$\raisebox{1ex}{$Ft$}\!\left/ \!\raisebox{-1ex}{$mc$}\right.$is taken on the x-axis, $\raisebox{1ex}{$u$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$
and is taken on the y-axis.

As the time increases, the velocity of the object reaches the speed of light