Relativistic Equation of Motion

Suppose that the particle discussed in the previous section has a mass $m_0$ in its instantaneous rest frame. Given that the particle's acceleration in its instantaneous rest frame is $a_0\,{\bf e}_x$, the particle is clearly subject to a force ${\bf f} = f\,{\bf e}_x$, where

$$f = m_0\,a_0.$$ (3.158)

Thus, according to Equation (3.153), the particle's equation of motion in an inertial reference frame in which its instantaneous velocity is ${\bf v} = v\,{\bf e}_x$ is

$$\frac{dv}{dt} = \frac{f}{m_0}\left(1-\frac{v^2}{c^2}\right)^{3/2}$$ (3.159)

However, the previous equation can be rearranged to give

$$f = \frac{d}{dt}\!\left[\frac{m_0\,v}{(1-v^2/c^2)^{1/2}}\right] = \frac{d(\gamma\,m_0\,v)}{dt}.$$ (3.160)

Let us define the relativistic mass of the particle as

$$m= \gamma\,m_0,$$ (3.161)

and its relativistic momentum as

$${\bf p} = m\,{\bf v}.$$ (3.162)

Thus, Equation (3.160) implies that the relativistic equation of motion of the particle is

$${\bf f} = \frac{d{\bf p}}{dt},$$ (3.163)

which is analogous in form to Newton's second law of motion, (1.17). Thus, we conclude that the reason that a particle of rest mass (i.e., mass in its instantaneous rest frame) $m_0$, subject to a constant force ${\bf f}$, never achieves a speed greater than the speed of light is that the particle's relativistic mass, $\gamma\,m_0$, increases as it moves faster, and tends to infinity as its speed approaches the speed of light.