A New Look at Energy — Question 2

Question 2

Work over a known distance determines speed without solving the trajectory

A constant laboratory force F accelerates a particle from rest through distance L along the force, with $FL=mc^2$ and no losses. Use $E^2=p^2c^2+m^2c^4$, $E=\gamma mc^2$, $K=E-mc^2$, and $\mathbf v=c^2\mathbf p/E$.

1. Use work-energy balance to find the final speed and momentum.

2. Find the elapsed laboratory time using impulse and compare with the classical final-speed estimate.

Question 2 - Solution

Work over a known distance determines speed without solving the trajectory

(a) Method and calculation

$$K=FL=mc^2,\quad E=2mc^2\quad\Rightarrow\quad\boxed{\gamma=2}.$$$$\boxed{v=(\sqrt3/2)c},\qquad\boxed{p=\sqrt3\,mc}.$$

(b) Interpretation and further calculation

$$Ft=p\quad\Rightarrow\quad\boxed{t=\sqrt3\,mc/F}.$$$$\tfrac12mv_{\mathrm{cl}}^2=mc^2\quad\Rightarrow\quad v_{\mathrm{cl}}=\sqrt2c\text{ is invalid}.$$

Physical check

The classical kinetic-energy formula has been used outside its low-speed domain; relativistic work-energy balance keeps the speed below c.