Question

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?

Short answer

(a)The final expression is $F={\gamma }_u^{3}m\frac{du}{dt}$,

(b) The equation holds for objects moving at very low speeds .

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

(d) The graph is shown below:

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

Step-by-step solution

Step 1:Determine a relativistic expression for Newton’s second law

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.

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

Deduce the above equation in the classical form

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.

Determine the expression for the relativistic velocity of an accelerating object

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$}\right.}\right)}}^2}\frac{Ft}{m}$

Plot uvs t

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