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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. (2.85))

\begin{displaymath}
A^\mu = \gamma\left(\frac{d\gamma}{dt}\,{\bfm u} + \gamma {\bfm a},
c\,\frac{d\gamma}{dt}\right).
\end{displaymath} (268)

Using the standard transformation (2.57) for 4-vectors, we obtain
$\displaystyle \frac{d\gamma}{dt}\,u + \gamma a$ $\textstyle =$ $\displaystyle a_0,$ (269)
$\displaystyle \frac{d\gamma}{dt}$ $\textstyle =$ $\displaystyle \frac{u\,a_0}{c^2}.$ (270)

It follows that
\begin{displaymath}
a = \frac{a_0}{\gamma^3}.
\end{displaymath} (271)

The above equation can be written
\begin{displaymath}
f = m_0\gamma^3\,\frac{du}{dt},
\end{displaymath} (272)

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

Equation (2.214) is equivalent to

\begin{displaymath}
f = \frac{d(mu)}{dt},
\end{displaymath} (273)

where
\begin{displaymath}
m = \gamma m_0.
\end{displaymath} (274)

Thus, we can account for the ever decreasing acceleration of a particle subject to a constant force (see Eq. (2.213)) by supposing that the inertial mass of the particle increases with its velocity according to the rule (2.216). Henceforth, $m_0$ is termed the rest mass, and $m$ the inertial mass.

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

\begin{displaymath}
\frac{dE}{dt} = fu = m_0 \gamma^3\, u\, \frac{du}{dt}.
\end{displaymath} (275)

This equation can be written
\begin{displaymath}
\frac{dE}{dt} = \frac{d (mc^2)}{dt},
\end{displaymath} (276)

which can be integrated to yield Einstein's famous formula
\begin{displaymath}
E = m c^2.
\end{displaymath} (277)

The 3-momentum of a particle is defined

\begin{displaymath}
{\bfm p} = m {\bfm u},
\end{displaymath} (278)

where ${\bfm u}$ is its 3-velocity. Thus, by analogy with Eq. (2.215), Newton's law of motion can be written
\begin{displaymath}
{\bfm f} = \frac{d{\bfm p}}{dt},
\end{displaymath} (279)

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

The 4-momentum of a particle is defined

\begin{displaymath}
P^\mu = m_0 U^\mu = \gamma m_0 ({\bfm u}, c) = ({\bfm p}, E/c),
\end{displaymath} (280)

where $U^\mu$ is its 4-velocity. The 4-force acting on the particle obeys
\begin{displaymath}
{\cal F}^\mu = \frac{dP^\mu}{d\tau} = m_0 A^\mu,
\end{displaymath} (281)

where $A^\mu$ is its 4-acceleration. It is easily demonstrated that
\begin{displaymath}
{\cal F}^\mu = \gamma\left({\bfm f}, c \,\frac{dm}{dt}\right...
...a \left( {\bfm f}, \frac{{\bfm f}\!\cdot\!{\bfm u}}{c}\right),
\end{displaymath} (282)

since
\begin{displaymath}
\frac{d E}{dt} = {\bfm f}\!\cdot\!{\bfm u}.
\end{displaymath} (283)


next up previous
Next: The force on a Up: Relativity and electromagnetism Previous: The electromagnetic field due
Richard Fitzpatrick 2002-05-18