Question

Classically, the net work done on an initially stationary object equals the final kinetic energy of the object. Verify that this also holds relativistically. Consider only one-dimension motion. It will be helpful to use the expression for p as a function of u in the following:

$W=\int Fdx=\int \frac{dp}{dt}dx=\int \frac{dx}{dt}dp=\int udp$

Short answer

Determining the differential change in momentum and integrating the above expression with respect to momentum will prove that relativistically,the net work done on an initially stationary object is equal to the final kinetic energy of the object.

Step-by-step solution

Determine the differential change in momentum

The one-dimensional motion net work is given by the following.

$$\begin{gathered} W={\int }_0^{u_f}udp \\ ={\int }_0^{u_f}ud\left(\frac{mu}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right) \end{gathered}$$

Let’s get the expression for $dp$first.

$$\begin{gathered} dp=d\left(\frac{mu}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right) \\ =m\left[\frac{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}du-u\left({\displaystyle \frac{1}{2}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(-{\displaystyle \frac{2udu}{c^2}}\right)\right)}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \\ =m\left[\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}+\frac{u^{2}du}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ =m\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\left[1+\frac{{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \end{gathered}$$$$\begin{gathered} dp=d\left(\frac{mu}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right) \\ =m\left[\frac{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}du-u\left({\displaystyle \frac{1}{2}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(-{\displaystyle \frac{2udu}{c^2}}\right)\right)}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \\ =m\left[\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}+\frac{u^{2}du}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ =m\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\left[1+\frac{{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \end{gathered}$$

Hence, the final expression for the differential change in momentum is

$dp=\frac{mdu}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Determine net work done on a stationary object.

Putting the expression for $dp$in expression of net work done.

$$\begin{gathered} W={\int }_0^{U_f}u\left(\frac{mdu}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right) \\ =m{\int }_0^{u_f}\frac{udu}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ ={\left[\frac{m{c}^2}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right]}_0^{u_f} \\ ={\left[{\gamma }_{u}m{c}^2\right]}_0^{u_f} \end{gathered}$$

Hence, the final expression for net work done on the stationary object is

$W=\left({\gamma }_{u_f}-1\right)m{c}^2$

$W=\left({\gamma }_{u_f-1}\right)m{c}^2$

Thus, it is proved that relativistically, the net work done on an initially stationary object is equal to the final kinetic energy of the object.