Relativistic Particle Dynamics

Consider a particle that, in its instantaneous rest frame $S_0$ , has mass $m_0$ and constant acceleration in the $x$ -direction $a_0$ . Let us transform to a frame $S$ , in the standard configuration with respect to $S_0$ , in which the particle's instantaneous velocity is $u$ . What is the value of $a$ , the particle's instantaneous $x$ -acceleration, in S?

The easiest way in which to answer this question is to consider the acceleration 4-vector [see Equation (1726)]

$$A^{\,\mu} = \gamma\left(\frac{d\gamma}{dt}\,{\bf u} + \gamma \,{\bf a},\, c\,\frac{d\gamma}{dt}\right).$$ (1836)

Using the standard transformation, (1694)-(1697), for 4-vectors, we obtain

$$a_0 = \gamma^{\,3}\, a,$$ (1837)

$$\frac{d\gamma}{dt} = \frac{u\,a_0}{c^{\,2}}.$$ (1838)

Equation (1839) can be written

$$f = m_0\,\gamma^{\,3}\,\frac{du}{dt},$$ (1839)

where $f=m_0 \,a_0$ is the constant force (in the $x$ -direction) acting on the particle in $S_0$ .

Equation (1841) is equivalent to

$$f = \frac{d(m\,u)}{dt},$$ (1840)

where

$$m = \gamma \,m_0.$$ (1841)

Thus, we can account for the ever decreasing acceleration of a particle subject to a constant force [see Equation (1839)] by supposing that the inertial mass of the particle increases with its velocity according to the rule (1843). Henceforth, $m_0$ is termed the rest mass, and $m$ the inertial mass.

The rate of increase of the particle's energy $E$ satisfies

$$\frac{dE}{dt} = f\,u = m_0\, \gamma^{\,3}\, u\, \frac{du}{dt}.$$ (1842)

This equation can be written

$$\frac{dE}{dt} = \frac{d (m\,c^{\,2})}{dt},$$ (1843)

which can be integrated to yield Einstein's famous formula

$$E = m \,c^{\,2}.$$ (1844)

The 3-momentum of a particle is defined

$${\bf p} = m \,{\bf u},$$ (1845)

where ${\bf u}$ is its 3-velocity. Thus, by analogy with Equation (1842), Newton's law of motion can be written

$${\bf f} = \frac{d{\bf p}}{dt},$$ (1846)

where ${\bf f}$ is the 3-force acting on the particle.

The 4-momentum of a particle is defined

$$P^{\,\mu} = m_0\, U^{\,\mu} = \gamma\, m_0\, ({\bf u},\, c) = ({\bf p},\, E/c),$$ (1847)

where $U^{\,\mu}$ is its 4-velocity. The 4-force acting on the particle obeys

$${\cal F}^{\,\mu} = \frac{dP^{\,\mu}}{d\tau} = m_0\, A^{\,\mu},$$ (1848)

where $A^{\,\mu}$ is its 4-acceleration. It is easily demonstrated that

$${\cal F}^{\,\mu} = \gamma\left({\bf f}, \,c \,\frac{dm}{dt}\right) = \gamma \left( {\bf f}, \,\frac{{\bf f}\cdot{\bf u}}{c}\right),$$ (1849)

because

$$\frac{d E}{dt} = {\bf f}\cdot{\bf u}.$$ (1850)