- 天同Lv 76 years agoFavorite Answer
It is just a matter of differentiation.
From momentum definition: p = ymv
where m is the rest mass, v is the velocity, y = (1-v^2/c^2)^(-1/2)
c is the speed of light.
Force F = dp/dt = m[d(yv)/dt] = my(dv/dt) + mv(dy/dt)
But my(dv/dt) = mya where a is the acceleration, which equals to dv/dt
and dy/dt = (-1/2).y^3.(-2v/c^2).(dv/dt) = (y^3).(v/c^2).a
i.e. mv(dy/dt) = m.y^3(v^2/c^2)a
Therefore, F = mya + m.y^3.(v^2/c^2).a
If v << c, then v^2/c^2 << 1, and y is more or less equal to 1. The equation reduces to F = ma, the same form as in Newtonian mechanics.