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Retarded time

by dipole
Tags: retarded, time
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dipole
#1
Apr25-12, 05:09 PM
P: 440
I have a quick question about the retarded time when dealing with moving charges.

The retarded time is:

[itex] t' = t - \frac{r}{c}[/itex]

where [itex] r [/itex] is the distance between the point of observation and the position of the charge.

My question is very simple, is [itex] r [/itex] a function of the normal time [itex] t [/itex], or the retarded time [itex] t' [/itex]?

That is, which equation is correct?

1. [itex] t' = t - \frac{r(t)}{c}[/itex]

2. [itex] t' = t - \frac{r(t')}{c}[/itex]

Thanks.
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K^2
#2
Apr26-12, 01:05 AM
Sci Advisor
P: 2,470
Depends on which one is moving. If the observation point stays still, and the charge is moving, then you are interested in r(t'), the distance at time the signal was emitted.

More generally, suppose both the charge and observation point are moving, with rc and ro being positions of charge and observation point respectively relative to some fixed origin. In that case, the distance traveled by the wave will be function of both.

[tex]t' = t - \frac{||\vec{r}_c(t')-\vec{r}_o(t)||}{c}[/tex]


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