A particle of rest mass m0 travels at a speed v = 0.18c. At what speed will its momentum be quadrupled?

Answer 1

Let p = mv; where p is the momentum for the effective mass m when v = .18 c

Then P = 4p = MV; where P is quadruple the momentum for effective mass M when V is some velocity > v.

Take the ratio P/p = 4 = MV/mv; so that V = 4mv/M and V^2 = 16 m^2 v^2//M^2 = 16 m0^2/l^2 v^2//m0^2/L^2 = 16 (1/l^2) v^2//(1/L^2); where l^2 = (1 - (v/c)^2) and L^2 = (1 - (V/c)^2) which are the Lorentz transforms at v and V for the rest mass m0 (which cancels out in the ratio). Combining all the factors, we have V^2 = 16 (L^2/l^2) v^2 where l^2 = .97 when v = .18c.

As l^2 = .97 ~ 1.00, we simplify V^2 = 16 L^2 v^2 = 16 (1 - (V/c)^2) v^2 = 16 (1 - V^2/c^2) v^2 = (16 - 16V^2/c^2) (.18c)^2 = 16(.18c)^2 - 16V^2(.18)^2 c^2/c^2 = 16(.18)^2 (c^2) - 16(.18)^2 V^2

Bring all the V^2 factors to the LHS we have V^2 + .51 V^2 = .51 c^2; so that 1.5 V^2 = .5 c^2 and V = c sqrt(1/3) ~ .57 c.

If I've done the arithmetic right, the momentum will quadruple at about .57 light speed.

The important physics lesson, even if my arithmetic is wrong, is that momentum = p = mv; where m is not the rest mass and it increases as v increases. So we have both m and v increasing which makes the momentum increase non linear with increasing velocity.

Answer 2

use p = mc*(gamma^2 - 1)^(0.5), solve for gamma, solve for v