# Retarded Potential and Energy

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 $-ax$, travelling toward the origin at $v_0$.

$q_b$ is at $+bx$, travelling 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

1 person
Smacal1072
Griffiths derives the $\mathbf{E}$ field of a point charge moving with constant velocity from the Liénard–Wiechert potentials in chapter 10.3:

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

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

$\theta$ 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?

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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:

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

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$\sqrt{1 - \frac{v^2}{c^2}}$...