Answer to Question #98258 in Mechanics | Relativity for AbdulRehman

Question

. What happens in the relativistic addition of velocities formula (derived
in problem (10) when the two velocities are both very small? Try it for
vBA = .01c and vA = .01c. How much does the relativistic result differ
from the Newtonian result (1.2) in this case?

Answer 1

When the two velocities are both much smaller than the velocity of light, then $v_A v_{BA} / c^2 \ll 1$, and this quantity can be neglected compared to unity in the denominator of equation (3.4) for relativistic addition of velocities. In this case, we obtain the Newtonian formula of addition $v_B \approx v_A + v_{BA}$. For $v_A = v_{BA} = 0.01 c$, the Newtonian formula gives $v_B \approx 0.02 c$, while the relativistic formula gives $v_B \approx 0.019998 c$, with the difference $\Delta v_B \approx 2 \times 10^{-6} c$, or relative difference $\Delta v_B / v_B \approx 2 \times 10^{-6} / 0.02 = 10^{-4}$, which is rather small.