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How to calculate the retarded time

  • Thread starter lailola
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  • #1
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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
 

Answers and Replies

  • #2
gabbagabbahey
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[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.
 
  • #3
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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.
 
  • #4
gabbagabbahey
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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.
 
  • #5
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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
 
  • #6
gabbagabbahey
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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.
 
  • #7
46
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Ok, thank you!
 

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