# How to calculate the retarded time

1. Aug 29, 2012

### lailola

1. The problem statement, all variables and given/known data

A charged particle is moving along the x-axis and its position is given by: $\vec{r}'(t)=\sqrt{a^2+c^2t^2}\vec{e_x}$

I have to calculate the Lienard-Wiechert potentials, the electric and magnetic fields and the Poynting vector.

2. Relevant equations

$\vec{A}=\frac{q\vec{v}}{cR-\vec{R}\vec{v}}$

$\phi=\frac{qc}{cR-\vec{R}\vec{v}}$

(both evaluated in t_r)

with $\vec{R}=\vec{r}-\vec{r}'(t_r)$.

R=c(t-tr)

3. The attempt at a solution

I have to find the retarded time tr to calculate the denominator of the potentials, and that is my doubt. I do:

$R^2=(x-\sqrt{a^2+c^2t^2})^2+y^2+z^2$
$R^2=c^2t_r^2+c^2t^2-2c^2tt_r$

Equating these two expressions I get tr but when I do it I get a horrible thing. It's an exam question so I think there will be another way to do this. Any idea?

Thank you

2. Aug 29, 2012

### gabbagabbahey

Shouldn't you have $t_r$ in there instead of $t$? Other than that, it looks fine.

3. Aug 29, 2012

### lailola

Yes, it's tr. But solving for tr is still horrible.

4. Aug 29, 2012

### gabbagabbahey

Horrible is a relative concept (relative to one's own perspective). You end up with a quadratic equation for $t_r$, which I'm sure you know how to solve (despite the fact that some of the coefficients are rather unpleasant), and you can select the correct root by looking at the case where $t=0$.

There may be a better way, but I can't think of it off hand.

5. Aug 30, 2012

### lailola

Ok. When i solve the equation it appears an 'x^2' in the denominator. Should I consider separately the two cases (x=0,x≠0 )?

And, when I set t=0, does tr have to be negative?

Thanks

6. Aug 30, 2012

### gabbagabbahey

Depends on how thorough you want to be. I doubt you instructor will care too much about x=0.