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Potentials and Poynting Conservation Theorem

  1. Aug 20, 2012 #1
    1. The problem statement, all variables and given/known data

    I have to find the Lienard-Wiechert potentials

    [itex]\vec{A}=\frac{\vec{qv}}{R-\vec{R}\vec{v}/c}[/itex]

    [itex]\phi=\frac{q}{R-\vec{R}\vec{v}/c}[/itex]

    (both evaluated in t_r)

    with [itex]\vec{R}=\vec{r}-\vec{x}(t_r)[/itex]. [itex]\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}[/itex] is the vector asociated to the observation point, [itex]\vec{x}(t)=\vec{a}+ct \hat{z}[/itex] is the trajectory of the particle.

    Then I have to find E, B and verify the conservation theorem for the energy density.

    Finally, I have to compare my results with the results obtained in a frame that is moving with velocity [itex]\vec{v}=v\hat{z}[/itex]

    2. Relevant equations

    [itex]R=c(t-t_r)[/itex]
    [itex]\vec{v}=\frac{dx(t_r)}{dt_r}=c \hat{z}[/itex]


    3. The attempt at a solution

    If I calculate the denominator:

    [itex]R-\vec{R}\vec{v}/c = c(t-t_r)-(zc-ac-c^2t_r)/c= ct-z-a[/itex]

    Substituting in the expressions I get A and [itex]\phi[/itex].

    For calculating E and B I use:

    [itex]\vec{E}=-grad(\phi)-\frac{1}{c}\frac{\partial\vec{A}}{\partial t}[/itex]

    [iex]\vec{B}=rot(\vec{A})[/itex]

    I find B=0, so [itex]\vec{S}=0[/itex]

    The energy density is [itex]\varepsilon=\frac{c}{4\pi}(E^2+B^2)=\frac{c}{4\pi}(E^2)[/itex]

    Finally, the conservation theorem says:

    [itex]grad(\vec{S})+\frac{\partial \varepsilon}{\partial t}= 0[/itex]

    One of my problems is that this last part is not satisfied, and I can't find the error.

    The other problem is to solve the problem in the new inertial frame. I don't know how to do it.


    Tthanks for any help.
     
    Last edited: Aug 20, 2012
  2. jcsd
  3. Aug 24, 2012 #2

    gabbagabbahey

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    Are you told that [itex]\vec{a}[/itex] points in the [itex]z-[/itex]direction?

    This may be related to the well-known radiation problem of a charge in hyperbolic motion. See J. Cohn, Am. J. Phys. 46, 225 (1976) and references therein.

    I assume you are supposed to apply the Lorentz transform to [itex]\vec{x}(t)[/itex] and then recalculate using the same method (You could also use the transformation equations for the E- & B-fields if you know them)
     
  4. Aug 24, 2012 #3

    TSny

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    Do you really want the particle to be traveling at the speed of light? I suspect that you should replace c by v here.
     
  5. Aug 24, 2012 #4
    [itex]\vec{a}[/itex] is just an arbitrary constant vector with three components. I don't know anything about it.

    Si, if I apply Lorentz transform to [itex]\vec{x}(t)[/itex] and I solve the problem again, I'll obtain the same results as if I apply Lorentz transf. to E and B. Is that right?

    Thanks!
     
  6. Aug 24, 2012 #5
    I know it may be strange but it's c, not v.
     
  7. Aug 24, 2012 #6
    Another question. In this same problem, what would be the radiated power through the plane z=b?

    I know that [itex]P=∫\frac{dP}{d\Omega} d\Omega[/itex] but I don't know how to calculate the power through that plane.
     
  8. Aug 24, 2012 #7

    TSny

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    For a charged particle moving at the speed of light we can expect some singular behavior of the potentials and fields. I believe the following is true, but I welcome corrections.

    In the case of a particle moving at light speed, the denominators of the Lienard-Wiechert potentials are just equal to the difference, Δz, in the z coordinate of the present location of the particle and the z coordinate of the observation point . (This is what you already essentially stated.) So, we have a singularity in the potentials when Δz = 0. The point on the trajectory corresponding to Δz = 0 is the point of closest approach of the particle to the observation point. Call this point of the trajectory O.

    It is important to note that if the particle at the present time t has not yet made it to O, then there does not exist a retarded time tr associated with this present time t. Thus, for a particle moving at the speed of light, you can only find retarded times for those present times for which the particle has already passed O!

    So, the retarded Lienard-Wiechert potentials for a particle moving at light speed are only defined at the observation point for present times t for which the particle has already passed O.

    If you use the potentials to find E and B for the times for which the potentials are defined, you will find E = 0 and B = 0 at those times. But, the retarded potentials cannot tell you what the fields are at the observation point at times when the particle has not yet made it to O or when the particle is at O.

    You could use the so-called “advanced” Lienard-Wiechert potentials to find E and B for times when the particle has not yet made it to O. Again, I believe you would find E = B = 0. But these advanced potentials will be undefined at the instant the particle passes O and for all times after which the particle passes O. So even if we introduce both advanced and retarded potentials, we cannot get the fields at the observation point at the instant the particle passes O.

    I think it would be better to begin with the speed v of the particle less than c. Then the retarded potentials are defined for all times and you can work out the E and B fields at the observation point for all times (as done in many textbooks). You can then take the limit of these fields as v approaches c. You will find in this limit that E = B = 0 at the observation point for all times except the instant when the particle passes O. At the instant the particle passes O, E and B at the observing point diverge to infinity as v approaches c.

    All of this corresponds to the fact that the faster a charged particle moves at constant velocity, the stronger the fields become in a plane containing the particle and oriented perpendicular to the trajectory. For a particle moving at near light speed, the fields are essentially confined to this plane and have a very large magnitude.
     
    Last edited: Aug 24, 2012
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