I'm doing some special relativity exercises. I have to find $x(t)$ and $v(t)$ of a charged particle left at rest in $t=0$ in an external constant uniform electric field $\vec{E}=E_{0} \hat{i}$, then with that velocity I should find the Liénard–Wiechert radiated power.

I will show you what I did but I feel that it is wrong.

We should solve the equation of motion given by

$$ \tag{1}\frac{dp^{\mu}}{d\tau} = \frac{q}{c} F^{\mu \nu}u_{\nu} $$

The four-velocity is given by

$$ u^{\mu} = (u^{0},u^{1},u^{2},u^{3}) = \gamma (c,v^{1},v^{2},v^{3}) $$

where $v^{\alpha}$ are the components of the three-velocity. The four-momentum is

$$ p^{\mu} = mu^{\mu} $$

This will give us four equtions where two of them will give a constant velocities and the other two are

$$ \tag{2}\frac{d\gamma}{d\tau} = -\frac{qE_{0}}{mc^{2}}\gamma v_{1} $$

$$ \tag{3}\frac{d\gamma}{d\tau} v_{1} + \gamma \frac{dv_{1}}{d\tau} = \frac{qE_{0}}{m} \gamma $$

Replacing $(2)$ in $(3)$ gives

$$ \tag{4}\frac{dv_{1}}{d\tau} = -\frac{qE_{0}}{mc^{2}} (v_{1})^{2} + \frac{qE_{0}}{m} $$

The solution of the ODE $(4)$ gives something like

$$ \tag{5}v_{1}(\tau) = A\tanh{(B\tau)} $$

This component of the three-velocity is in terms of the proper time $\tau$ and the problem ask me to find the velocity in terms of the time $t$. So my attempt was to solve

$$ \tag{6}\frac{dt}{d\tau} = \gamma (\tau) = \frac{1}{\sqrt{1 - \frac{(v_{1}(\tau))^{2}}{c^{2}}}} $$

and then replacing this solution for $\tau$ in $(5)$. But the solution of $(6)$ is this. Which doesn't make any sense to me.

I think that I'm misunderstanding something or missing something that will give me a easier solution to this problem. I thought it because in the Liénard–Wiechert radiated power I sould do $dv_{1}/dt$ which is almost impossible to do it without WolframAlpha.

  • $\begingroup$ I've added the homework-and-exercises tag. Please use that tag on homework problems. $\endgroup$
    – user4552
    Commented Sep 19, 2017 at 15:40

1 Answer 1


You need to match the initial conditions,

\begin{align*} a_{0} &= \frac{qE_{0}}{m} \\ v &= c\tanh \frac{a_{0} \tau}{c} \\ \gamma &= \cosh \frac{a_{0} \tau}{c} \\ &= \frac{dt}{d\tau} \\ t &= \int_{0}^{\tau} \cosh \frac{a_{0} \tau}{c} \, d\tau \\ &= \frac{c}{a_{0}} \sinh \frac{a_{0} \tau}{c} \\ \frac{a_{0} t}{c} &= \sinh \frac{a_{0} \tau}{c} \\ v &= \frac{a_{0} t}{\sqrt{1+\left( \dfrac{a_{0}t}{c} \right)^{2}}} \\ \end{align*}

  • $\begingroup$ Oh, what are you saying is that I forgotten to use equation $(2)$ to recover what $\gamma$ looks like right? $\endgroup$ Commented Mar 4, 2017 at 19:57
  • $\begingroup$ Closely, sometimes it's useful to check your results with the classical limit and relativistic limit. For $t\approx 0$, $v\approx a_{0} t$ whereas $t\to \infty$, $v\to c$. $\endgroup$ Commented Mar 4, 2017 at 20:02
  • $\begingroup$ So the Lorentz factor $\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ is only true when the velocity is a constant? $\endgroup$ Commented Mar 4, 2017 at 20:25
  • $\begingroup$ What I mean is try to fit the integration constants $A$ and $B$ by looking at $\tau \to 0$, $v\to AB\tau$ and $\tau \to \infty$, $v\to A$ you immediately get the result. Actually, I'd rather formulate directly as $qE_{0}=\gamma^3 ma \,$ so that $$\frac{qE_{0}t}{m}=\int_{0}^{v} \frac{dv}{\left( 1-\frac{v^2}{c^2}\right)^{3/2}}=\frac{v}{\sqrt{1-\dfrac{v^2}{c^2}}}$$ $\endgroup$ Commented Mar 4, 2017 at 20:45

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