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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)]

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

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

$\displaystyle a_0$ $\displaystyle = \gamma^{\,3}\, a,$ (1837)
$\displaystyle \frac{d\gamma}{dt}$ $\displaystyle = \frac{u\,a_0}{c^{\,2}}.$ (1838)

Equation (1839) can be written

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

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

where

$\displaystyle m = \gamma \,m_0.$ (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

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

This equation can be written

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

which can be integrated to yield Einstein's famous formula

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

The 3-momentum of a particle is defined

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

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

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

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

The 4-momentum of a particle is defined

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

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

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

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

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

because

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


next up previous
Next: Force on a Moving Up: Relativity and Electromagnetism Previous: Field Due to a
Richard Fitzpatrick 2014-06-27