# Retarded field interaction

I was doing the following thought experiment: imagine a particle that moves at a certain velocity. Imagine also that the particle generates a field that propagates at velocity $v_f$. Well, if the particle's velocity is higher than $v_f$ then it will interact with its own field if for some reason the particle decelerates (imagine a plane that goes Mach 2 and then "brakes" to zero, it will hear its own sound after a while). But what if the particles's velocity is exactly $v_f$?

My guess is that it will interact with its own field... is this correct?

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The interaction of a particle with its own field depends of the position and retardation depends of the distance between the source particle and the test particle. In principle the field is present at the particle position and the retardation is exactly zero in this one-particle case, therefore the particle always interact with its own field with independence of its velocity

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Thank you very much for nice answer! –  PML Oct 17 '12 at 21:09
You are welcome –  juanrga Oct 18 '12 at 15:27