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Retarded Potential and Energy

  1. Sep 18, 2013 #1
    Hi All,

    I've been reading Griffiths E&M and Feynman Lectures (vol 2, E&M), and it made me think about a gedanken that I'm trying to resolve. I'm pretty sure once I really peruse the chapters on the relativistic formulation it'll make sense, but I'm impatient :biggrin:

    I have two identical positive charges, [itex]q_0[/itex] and [itex]q_1[/itex], on the x-axis.

    [itex]q_a[/itex] is at [itex]-ax[/itex], travelling toward the origin at [itex]v_0[/itex].

    [itex]q_b[/itex] is at [itex]+bx[/itex], travelling toward the origin at [itex]-v_0[/itex].

    Since they are both positive charges, they will repulse each other due to the Coulomb field. But, from what I understand, the repulsive force is less than what they would experience in a static case, since at each moment in time it is the retarded (further away) position of the particles that influence each other. (I think the field in the x-axis is reduced by a factor of [itex]\gamma = \frac{1}{\sqrt{1-v^{2}_{0}/c^2}}[/itex])

    This is what I don't understand: The faster the charges zoom toward each other, the less work they impart into the field potential energy (since it is reduced in the x-direction). If they return to their previous positions at a slower velocity, the return trip will result in them having more kinetic energy than they started with?
  2. jcsd
  3. Sep 18, 2013 #2
    That's not how it works. Look up Liénard–Wiechert potential which gives the relativistically correct electric potential produced by a moving charge
  4. Sep 19, 2013 #3
    Griffiths derives the [itex]\mathbf{E}[/itex] field of a point charge moving with constant velocity from the Liénard–Wiechert potentials in chapter 10.3:

    \mathbf{E}(\mathbf{r},t) = \frac{q}{4\pi\epsilon_{0}} \frac{1-v^{2}/c^{2}}{\left(1-v^{2}\sin^{2}{\theta/c^{2}}\right)}\frac{\hat{\mathbf{R}}}{R^{2}}

    [itex]\mathbf{R} \equiv \mathbf{r} - \mathbf{v}t[/itex]

    [itex]\theta[/itex] is the angle between [itex]\mathbf{R} [/itex] and [itex]\mathbf{v}[/itex]

    He then says:

    I was wrong about the reduction factor, but it is still present. So i'm still curious: if the repulsive electric fields of two positive charges zooming toward each other is mutually reduced, where does the energy go/come from?
    Last edited: Sep 19, 2013
  5. Sep 21, 2013 #4
    Ok, I think I have a piece of the puzzle. I was not considering the Lorentz force on each particle relativistically. The relativistic Lorentz force is actually:

    \mathbf{F} = \gamma q(\mathbf{E} \mathbf{v} \times \mathbf{B})

    Where [itex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/itex]

    But, since [itex]\mathbf{E}[/itex] and [itex]\mathbf{F}[/itex] are directly proportional, it still means that the repulsive force on each particle is still reduced by a factor of[itex]\sqrt{1 - \frac{v^2}{c^2}}[/itex]...
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