How Does Retarded Potential Affect Energy in Moving Charges?

[Smacal1072]

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

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], traveling toward the origin at [itex]v_0[/itex].

[itex]q_b[/itex] is at [itex]+bx[/itex], traveling 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?

 

[dauto]

That's not how it works. Look up Liénard–Wiechert potential which gives the relativistically correct electric potential produced by a moving charge

 

[Smacal1072]

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:

[tex]
\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}}
[/tex]

[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:

Because of the [itex]\sin^{2}{\theta}[/itex] in the denominator, the field of a fast-moving charge is flattened out like a pancake in the direction perpendicular to the motion. In the forward and backward directions [itex]\mathbf{E}[/itex] is reduced by a factor [itex](1-v^{2}/c^{2})[/itex] relative to the field of a charge at rest...

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?

 

[Smacal1072]

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:

[tex]
\mathbf{F} = \gamma q(\mathbf{E} \mathbf{v} \times \mathbf{B})
[/tex]

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]...

 

[PJC01]

Hi there,

It sounds like you are on the right track with your understanding of retarded potential and energy. In this scenario, the charges are indeed experiencing less repulsive force due to the Coulomb field being reduced in the x-direction. This is because the Coulomb field takes time to propagate, and by the time it reaches the other charge, it is further away from its original position. This results in a reduced force between the two charges.

As for the potential energy and kinetic energy, it is important to remember that energy is always conserved. In this scenario, the charges may have less potential energy due to the reduced repulsive force, but they will also have less kinetic energy since they are traveling at a slower velocity. As they return to their original positions, the potential energy will increase, but the kinetic energy will decrease, resulting in the same total energy as before.

I hope this helps clarify your understanding of retarded potential and energy. Keep exploring and asking questions, and you will continue to deepen your understanding of these concepts.