How to Find the Retarded Time for a Moving Charged Particle?

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Homework Help Overview

The discussion revolves around calculating the retarded time for a moving charged particle, specifically in the context of Lienard-Wiechert potentials, electric and magnetic fields, and the Poynting vector. The original poster presents a position function for the particle and expresses uncertainty regarding the calculation of the retarded time.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the equation for retarded time, questioning whether to include the retarded time in certain expressions. There is mention of deriving a quadratic equation for retarded time, with concerns about the complexity of the resulting expressions.

Discussion Status

Participants are exploring different interpretations of the problem and discussing the implications of their findings. Some guidance is offered regarding the nature of the quadratic equation and considerations for specific cases, but there is no explicit consensus on a single approach.

Contextual Notes

There is a mention of the potential complexity of the solution and the possibility of needing to consider different cases for the variable x. Additionally, the discussion touches on the implications of setting time to zero in relation to the retarded time.

lailola
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Homework Statement



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

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

Homework Equations



[itex]\vec{A}=\frac{q\vec{v}}{cR-\vec{R}\vec{v}}[/itex]

[itex]\phi=\frac{qc}{cR-\vec{R}\vec{v}}[/itex]

(both evaluated in t_r)

with [itex]\vec{R}=\vec{r}-\vec{r}'(t_r)[/itex].

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:

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

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
 
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lailola said:
[itex]R^2=(x-\sqrt{a^2+c^2t^2})^2+y^2+z^2[/itex]

Shouldn't you have [itex]t_r[/itex] in there instead of [itex]t[/itex]? Other than that, it looks fine.
 
gabbagabbahey said:
Shouldn't you have [itex]t_r[/itex] in there instead of [itex]t[/itex]? Other than that, it looks fine.

Yes, it's tr. But solving for tr is still horrible.
 
lailola said:
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 [itex]t_r[/itex], 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 [itex]t=0[/itex].

There may be a better way, but I can't think of it off hand.
 
gabbagabbahey said:
Horrible is a relative concept (relative to one's own perspective). You end up with a quadratic equation for [itex]t_r[/itex], 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 [itex]t=0[/itex].

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
 
lailola said:
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?

It had better be, otherwise you are using the advanced time instead of the retarded time.
 
Ok, thank you!
 

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