Relativistic particle dynamics

Consider a particle which, 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 Eq. (1429)]

$$A^\mu = \gamma\left(\frac{d\gamma}{dt} {\bf u} + \gamma {\bf a}, c \frac{d\gamma}{dt}\right).$$ (1541)

Using the standard transformation, (1397)-(1400), for 4-vectors, we obtain

$$a_0 = \gamma^3 a,$$ (1542)

$$\frac{d\gamma}{dt} = \frac{u a_0}{c^2}.$$ (1543)

Equation (1542) can be written

$$f = m_0 \gamma^3 \frac{du}{dt},$$ (1544)

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

Equation (1544) is equivalent to

$$f = \frac{d(m u)}{dt},$$ (1545)

where

$$m = \gamma m_0.$$ (1546)

Thus, we can account for the ever decreasing acceleration of a particle subject to a constant force [see Eq. (1542)] by supposing that the inertial mass of the particle increases with its velocity according to the rule (1546). Henceforth, $m_0$ is termed the rest mass, and $m$ the inertial mass.

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

$$\frac{dE}{dt} = f u = m_0 \gamma^3 u \frac{du}{dt}.$$ (1547)

This equation can be written

$$\frac{dE}{dt} = \frac{d (m c^2)}{dt},$$ (1548)

which can be integrated to yield Einstein's famous formula

$$E = m c^2.$$ (1549)

The 3-momentum of a particle is defined

$${\bf p} = m {\bf u},$$ (1550)

where ${\bf u}$ is its 3-velocity. Thus, by analogy with Eq. (1545), Newton's law of motion can be written

$${\bf f} = \frac{d{\bf p}}{dt},$$ (1551)

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

The 4-momentum of a particle is defined

$$P^\mu = m_0 U^\mu = \gamma m_0 ({\bf u}, c) = ({\bf p}, E/c),$$ (1552)

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

$${\cal F}^\mu = \frac{dP^\mu}{d\tau} = m_0 A^\mu,$$ (1553)

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

$${\cal F}^\mu = \gamma\left({\bf f}, c \frac{dm}{dt}\right)... ...amma \left( {\bf f}, \frac{{\bf f}\!\cdot\!{\bf u}}{c}\right),$$ (1554)

since

$$\frac{d E}{dt} = {\bf f}\!\cdot\!{\bf u}.$$ (1555)