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, q_0 and q_1, on the x-axis.

q_a is at -ax, traveling toward the origin at v_0.

q_b is at +bx, traveling toward the origin at -v_0.

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 \gamma = \frac{1}{\sqrt{1-v^{2}_{0}/c^2}})

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 \mathbf{E} field of a point charge moving with constant velocity from the Liénard–Wiechert potentials in chapter 10.3:

<br /> \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}}<br />

\mathbf{R} \equiv \mathbf{r} - \mathbf{v}t

\theta is the angle between \mathbf{R} and \mathbf{v}

He then says:

Because of the \sin^{2}{\theta} 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 \mathbf{E} is reduced by a factor (1-v^{2}/c^{2}) 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:

<br /> \mathbf{F} = \gamma q(\mathbf{E} \mathbf{v} \times \mathbf{B})<br />

Where \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

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