## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 6.8

A relativistic particle P, initially at rest at the origin in frame  $\psi$,  is moving along a straight line under a constant force  $\mathbf{F}_0$.  Determine the relativistic speed and the distance traveled by P as functions of time.

## Verified Solution

The equation of motion for the relativistic particle is given by (5.34) in which  $\mathbf{F}(P, t)=\mathbf{F}_0$  is a constant force and (6.9) is to be used. Hence, separation of the variables and integration of  $\mathbf{F}_0 d t=d(m \mathbf{v})=d\left(\gamma m_0 \mathbf{v}\right)$,  with the initial values  $\mathbf{v}(P, 0)=\mathbf{0}$  and  $\gamma=1$,  yields  $m \mathbf{v}=\mathbf{F}_0 t$.  Thus, recalling (6.9) and noting that  $\mathbf{v}=v \mathbf{t} \text { and } \mathbf{F}_0=F_0 \mathbf{t}$  are parallel vectors, we have only one nontrivial component equation:  $m_0 v /\left(1 – v^2 / c^2\right)^{1 / 2}=F_0 t$.  This scalar equation yields the rectilinear, relativistic speed

$\mathbf{F}(P, t)=\frac{d \mathbf{p}(P, t)}{d t}=\frac{d}{d t}[m(P) \mathbf{v}(P, t)]$                    (5.34)

$m=\gamma m_0=\frac{m_0}{\sqrt{1 – \beta^2}} \quad \text { with } \quad \beta \equiv \frac{\dot{s}}{c} \text {. }$             (6.9)

$v(P, t)=\frac{c k t}{\sqrt{1 + (k t)^2}} \quad \text { with } \quad k \equiv \frac{F_0}{m_0 c} \text {. }$                 (6.25a)

Introducing  $v=\dot{s}$  into (6.25a) , separating the variables, and integrating  $d s=v d t$  with the initial value  $s(0)=0$,  we obtain the rectilinear distance traveled by P:

$s(P, t)=\frac{c}{k}\left(\sqrt{1 + (k t)^2} – 1\right) .$             (6.25b)

Notice in (6.25a) that  $v / c<1$  for all  t, and  $v / c \rightarrow 1 \text { as } t \rightarrow \infty$;  that is, under a constant force, the relativistic particle speed cannot exceed the speed of light c. This  result is quite different from the corresponding speed  $v=F_0 t / m_0$  described by (6.22) for a Newtonian particle of mass  $m=m_0$  initially at rest and subject to a constant force  $F_0$;  in this case  $v \rightarrow \infty$  with t. If $m_0 c$  is large compared with  $F_0 t$  so that  $k t \ll 1$,  then (6.25a) and (6.25b) reduce approximately to

$\mathbf{v}(P, t)=\frac{\mathbf{F}_0}{m} t + \mathbf{v}_0$           (6.22)

$v(P, t)=c k t=\frac{F_0}{m_0} t, \quad s(P, t)=\frac{1}{2} c k t^2=\frac{F_0}{2 m_0} t^2$            (6.25c)

The se are the Newtonian formulas described by (6.22) and (6.23) for the  corresponding rectilinear motion of a particle of mass $m_0$  initially at rest at the origin and acted upon by a constant force  $F_0$.  In the present relativistic approximation , however, these results are valid for only a sufficiently small  time for which  $v / c=k t \ll 1$.

$\mathbf{x}(P, t)=\frac{\mathbf{F}_0}{2 m} t^2 + \mathbf{v}_0 t + \mathbf{x}_0$                    (6.23)