Electrodynamics/Lienard-Wiechert Potentials

Suppose there is a point charge q traveling with location at all times given by $$\mathbf{\gamma}(t)$$. We will use the formulas developed in the previous section to find the potentials and the fields.

The retarded time is given by $$t_{ret}=t-\frac{||\mathbf{r}-\mathbf{\gamma}(t_{ret})||}{c}$$ (since, when we measure go back, the particle wasn't where it is now, it is at $$ \gamma(t_{ret})$$) Note that at most one previous instance of the particle is generating the field at the point, since the particle is moving at subluminal speeds. Thus, $$\mathbf{\eta}=\mathbf{r}-\mathbf{\gamma}(t_{ret})$$ is a unique vector.

Thus, the scalar potential is


 * $$\phi(\mathbf{r},t)=\frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\mathbf{r}-\mathbf{r'},t_{ret})}{||\mathbf{\eta}||} dV = \frac{1}{4 \pi \epsilon_0 ||\mathbf{\eta}||} \int \rho(\mathbf{r}-\mathbf{r'},t_{ret})dV$$

You might think that equals
 * $$\frac{q}{4 \pi \epsilon_0 ||\mathbf{\eta}||}$$

but that is wrong! The reason is very subtle: for the integral to be equal to the total charge, the charge distribution has to be taken at a specific time. However, we are obliged to evaluate the distribution at different times for each point! Thus, the charged particle is "smeared" out! Even though we are considering a point particle, the formula is still wrong, since the correction factor doesn't depend on geometric size!

Suppose the particle is a box of length a and is moving towards us. However, we will observe the particle to have length b, because the light that is simultaneously reaching our eyes from the front and back of the box originated from different times. In the time that light from the back of the box travels the extra distance b, the car traveled distance $$b-a$$, so
 * $$\frac{b}{c}=\frac{b-a}{v}$$, and $$b=\frac{a}{1-v/c}$$; the box is stretched by factor $$\frac{1}{1-v/c}$$ that has nothing to do with the size of the object. This is not an effect of length contraction; this is rather more similar to the Doppler shift.

The correct formula for the scalar potential is
 * $$\phi=\left(\frac{1}{1-\mathbf{\hat{\eta}} \cdot \mathbf{w}'(t_{ret})/c} \right)\frac{q}{4 \pi \epsilon_0 ||\mathbf{\eta}||}$$

where $$\mathbf{w}'(t_{ret})$$ is the particle velocity at time $$t_{ret}$$

The current density is $$\rho \mathbf{v}$$, and by similar arguments,
 * $$\mathbf{A}=\left(\frac{1}{1-\mathbf{\hat{\eta}} \cdot \mathbf{w}'(t_{ret})/c} \right)\frac{\mu_0 q \mathbf{v}}{4 \pi ||\mathbf{\eta}||}=\frac{\mathbf{v}}{c}\phi

$$

These are the Lienard-Wiechert Potentials for a moving charge. The correction factors are for the components of velocity pointing to the point we are measuring the potentials at.