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I Relativistic charged particle in a constant uniform electric field

  1. Mar 4, 2017 #1
    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.

    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.
  2. jcsd
  3. Mar 4, 2017 #2
    Your equation (5) makes look like a rapidity; could you use that fact to transform to the frame you want?
  4. Mar 4, 2017 #3


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    Well, it seems to me that a charged particle in a uniform electric field should be equivalent to the well-known "relativistic rocket". See for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html.

    That article (the old sci.physics.FAQ article) gives $$v = c \, \tanh (a \tau/c)$$, which seems to match your answer - and my recollections.

    If you want the solution for velocity in terms of time t and not ##\tau##, the article gives the less-well-known answer:

    $$v = \frac{at}{\sqrt{1 + \left( \frac{at}{c} \right) ^2}}$$

    which might be helpful. It remians to calculate the Lamour radiation, something I haven't done.

    Also the FAQ article doesn't give the derivation of the relativistic rocket equation - you can find that in for instance MTW's "Gravitation", I think wiki has some, but since you seem to be getting the right answer, I'm not sure a reference is really needed.
  5. Mar 4, 2017 #4
    Hello, thanks for that article. Yes, this is correct and the way to recover that velocity in terms of t is to use the equation (2) in the OP which gives you the functional form of $$\gamma (\tau)$$ and then solve $$dt/d\tau = \gamma (\tau)$$.

    Thanks, now my problem is complete.
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