## Lienard-Wiechert potentials. To solve an equation.

Hi, I have a doubt about a problem of classical electrodynamics (specifically for calculating the Lienard-Wiechert potentials).

(t_r is the retarded time, and t the time).

The position that has a particle is given by: x (t_r) = e cos (w t_r).

The squared modulus of the relative position vector is: R ^ 2 = r ^ 2 + x ^ 2 - 2rx

On the other hand, we know that: R ^ 2 = c ^ 2 (t-t_r) ^ 2

Equating the two expressions:

r ^ 2 + x ^ 2 - 2rx = c ^ 2 (t-t_r) ^ 2

Then, I have to replace x (t_r) and solve the equation for t_r, but I don't know how to solve it...maybe there's any trick for doing it.

Any idea?

Thank you.
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 Hello lailola, such equation is difficult to solve exactly. Often it is sufficient to find approximate solution. One way is as follows. - reexpress the equation as a formula for t_r: t_r = f(t_r) - guess first approximation t_r(1) = r/c and insert to the right-hand side - evaluate the right-hand side. This is the second approximation; should be better than the first one. - repeat how many times is necessary Often the first approximation t_r = r/c is sufficient; but if r ~ x, it will not be. Then you have to either iterate the above procedure many times or find another approach.

 Quote by Jano L. Hello lailola, such equation is difficult to solve exactly. Often it is sufficient to find approximate solution. One way is as follows. - reexpress the equation as a formula for t_r: t_r = f(t_r) - guess first approximation t_r(1) = r/c and insert to the right-hand side - evaluate the right-hand side. This is the second approximation; should be better than the first one. - repeat how many times is necessary Often the first approximation t_r = r/c is sufficient; but if r ~ x, it will not be. Then you have to either iterate the above procedure many times or find another approach.
Hi Jano, thanks for your help. But why can I set t_r(1)=r/c?

## Lienard-Wiechert potentials. To solve an equation.

The retarded time is exactly t_r = t - R/c.

R can be (if r>>x) approximated by r, so

t_r ~ t - r/c.

 Quote by Jano L. I made a mistake. The retarded time is exactly t_r = t - R/c. R can be (if r>>x) approximated by r, so t_r ~ t - r/c.
Ok! But, in that case, I don't use x for anything. That's strange, isn't it?
 Well, it is an approximation, good as long as x is much smaller than r. The second approximation will contain x.
 How would it be the second aproximation? I was thinking about Taylor series but I don't know how to apply that here. And, another question, would it be reasonable (supossing w is small) to aproximate cos(wt) by 1-(wt)^2/2? Thank you!
 The Taylor series will be accurate only for times much less than period. But in such problems we are usually interested in long periodic motion, so Taylor probably won't be a good idea. It would be much better if you can post the full assignment.
 I have to find the Lienard-Wiechert potentials $\vec{A}=\frac{q\vec{v}}{cR-\vec{R}\vec{v}}$ $\phi=\frac{qc}{cR-\vec{R}\vec{v}}$ (both evaluated in t_r) with $\vec{R}=\vec{r}-\vec{x}(t_r)$. $\vec{x}(t)=Acos(wt) \hat{z}$ is the trajectory of the particle. Then I have to find E, B and S.
 OK, what we wrote applies, you have equations $$t_r = t - R/c$$ $$R^2(t_r) = c^2(t-t_r)^2$$ The last equation can be written as $$t_r = t - R(t_r)/c~~(*)$$ The zeroth approximation (in x) is as if the particle was at x = 0: $$t_{r0} \approx t - r/c ~(0th~ approx.)$$ The first approximation in x is obtained from * by substituting the first approximation to the right-hand side: $$t_{r1} \approx t - R(t_{r0})/c~(1st ~approx.)$$ just plug-in the expression for R(t_{r0}). Higher approximations are derived in the same spirit.

 Quote by Jano L. OK, what we wrote applies, you have equations $$t_r = t - R/c$$ $$R^2(t_r) = c^2(t-t_r)^2$$ The last equation can be written as $$t_r = t - R(t_r)/c~~(*)$$ The zeroth approximation (in x) is as if the particle was at x = 0: $$t_{r0} \approx t - r/c ~(0th~ approx.)$$ The first approximation in x is obtained from * by substituting the first approximation to the right-hand side: $$t_{r1} \approx t - R(t_{r0})/c~(1st ~approx.)$$ just plug-in the expression for R(t_{r0}). Higher approximations are derived in the same spirit.
Ok, I'm going to work the problem that way. If I have any problem, I'll come back!

Thank you a lot, Jano.