I'm doing some special relativity exercises. I have to find $$x(t), v(t)$$ 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.(adsbygoogle = window.adsbygoogle || []).push({});

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 http://www.wolframalpha.com/input/?i=integrate 1/sqrt(1 - a^2*tanh(bx)^2). 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.

Thanks for the read.

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# I Relativistic charged particle in a constant uniform electric field

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