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I am reading Laudau-Lifshitz The Classical Theory of Field (4th edition). In (63.2), it says

In the system of reference in which the particle is at rest at time $t^\prime,$ the potential at the point of observation at time t is just the Coulomb potential:

$$\varphi = \frac e{R(t^\prime)} \tag{63.2}$$

This $R(t^\prime)$ is the distance between the moving point charge and the point of observation in the observation frame. I thought if the frame is changed to that in which the particle is at rest, there should be Lorentz Transformation (contraction) of $R(t^\prime).$ Why does it use the same $R(t^\prime)$ in that equation?

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  • $\begingroup$ Hi. I do not understand where the change of reference comes from. If I get correctly the part from the book, you have set up a certain reference frame where the particle is at rest for a moment t' so that, then, the potential at a point for this certain reference frame is φ. $\endgroup$ – Constantine Black Sep 19 '16 at 17:29
  • $\begingroup$ Thank you! It tries to find the field at the point of observation P(r) at time t, which is "determined by the state of motion of the charge at the earlier time t'. So in the frame of observation, the distance between the charge and observation at t is R(t) = r - r0(t'), where r is the location of the observation and ro(t') is the location of the charge at t'. So R(t') = c(t-t'). R(t') is the distance in the observation frame (?). Then it considers the rest frame when the charge particle is at rest (the situation in my question), why it doesn't have to do Lorentz transformation for R? $\endgroup$ – HYW Sep 20 '16 at 13:35
  • $\begingroup$ Hi, I think I understand what's going on now. I misunderstood the sentence (maybe this is what you meant already, Constantine?). It only says at t', it is at rest in the system of reference (of observation). It didn't say there is change of frame. And since it is at rest, of course, there is no length contraction. After that, the author tried to come up a general equation so that it coincides with the case in which it is at rest at t'. Sorry for the confusing questions. But I only understand after going through this "posting exercise". Thanks a lot! $\endgroup$ – HYW Sep 20 '16 at 14:33
  • $\begingroup$ Hi again. No problem. Glad I helped. $\endgroup$ – Constantine Black Sep 21 '16 at 8:10

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