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Lienard-Wiechert potentials. To solve an equation.

  1. Aug 19, 2012 #1
    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.
  2. jcsd
  3. Aug 19, 2012 #2

    Jano L.

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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.
  4. Aug 19, 2012 #3
    Hi Jano, thanks for your help. But why can I set t_r(1)=r/c?
  5. Aug 19, 2012 #4

    Jano L.

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    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.
  6. Aug 19, 2012 #5
    Ok! But, in that case, I don't use x for anything. That's strange, isn't it?
  7. Aug 20, 2012 #6

    Jano L.

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    Well, it is an approximation, good as long as x is much smaller than r. The second approximation will contain x.
  8. Aug 20, 2012 #7
    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!
  9. Aug 20, 2012 #8

    Jano L.

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    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.
  10. Aug 20, 2012 #9
    I have to find the Lienard-Wiechert potentials



    (both evaluated in t_r)

    with [itex]\vec{R}=\vec{r}-\vec{x}(t_r)[/itex]. [itex]\vec{x}(t)=Acos(wt) \hat{z}[/itex] is the trajectory of the particle.

    Then I have to find E, B and S.
    Last edited: Aug 20, 2012
  11. Aug 20, 2012 #10

    Jano L.

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    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.
  12. Aug 20, 2012 #11
    Ok, I'm going to work the problem that way. If I have any problem, I'll come back!

    Thank you a lot, Jano.
  13. Aug 21, 2012 #12

    Jano L.

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    Glad to help.
  14. Aug 21, 2012 #13
    Another thing! If I want to know the results from an inertial frame which is moving with velocity v on the z axis, how can I do it? Just with Lorentz transformation? Or do I have to solve the problem from the start?
  15. Aug 21, 2012 #14

    Jano L.

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    You can find the fields and motion of the particles in any inertial frame, the results are connected via Lorentz transformation. In your assignment you have prescribed motion of the particle x =... in lab frame, so it makes sense to solve everything in this frame and then transform into another if you need to.
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