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## Potential due to a moving charge

Suppose that a particle carrying a charge moves with uniform velocity through a frame . Let us evaluate the vector potential, , and the scalar potential, , due to this charge at a given event in .

Let us choose coordinates in so that and . Let be that frame in the standard configuration with respect to in which the charge is (permanently) at rest at (say) the point . In , the potential at is the usual potential due to a stationary charge,

 (1518) (1519)

where . Let us now transform these equations directly into the frame . Since is a contravariant 4-vector, its components transform according to the standard rules (1397)-(1400). Thus,
 (1520) (1521) (1522) (1523)

since in this case. It remains to express the quantity in terms of quantities measured in . The most physically meaningful way of doing this is to express in terms of retarded values in . Consider the retarded event at the charge for which, by definition, and . Using the standard Lorentz transformation, (1346)-(1349), we find that
 (1524)

where denotes the radial velocity of the change in . We can now rewrite Eqs. (1520)-(1523) in the form
 (1525) (1526)

where the square brackets, as usual, indicate that the enclosed quantities must be retarded. For a uniformly moving charge, the retardation of is, of course, superfluous. However, since
 (1527)

it is clear that the potentials depend only on the (retarded) velocity of the charge, and not on its acceleration. Consequently, the expressions (1525) and (1526) give the correct potentials for an arbitrarily moving charge. They are known as the Liénard-Wiechert potentials.

Next: Fields due to a Up: Relativity and electromagnetism Previous: Transformation of fields
Richard Fitzpatrick 2006-02-02