I Relativistic charged particle in a constant uniform electric field

[notdroid]

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.

 

[John Park]

Your equation (5) makes Bτ look like a rapidity; could you use that fact to transform to the frame you want?

 

[pervect]

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.

 

[notdroid]

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.

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.