# How to calculate the retarded time

lailola

## Homework Statement

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.

## Homework 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)

## 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

Homework Helper
Gold Member
$R^2=(x-\sqrt{a^2+c^2t^2})^2+y^2+z^2$

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

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

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

Homework Helper
Gold Member
Yes, it's tr. But solving for tr is still horrible.

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.

lailola
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.

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

Homework Helper
Gold Member
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 )?

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

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