Re: LienardWiechert potential
 From: D.McAnally@i'm_a_gnu.uq.net.au (David McAnally)
 Date: Sun, 1 Jan 2006 02:00:29 +0000 (UTC)
D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:
<snip>
>Panofsky and Phillips include a derivation of the EM field corresponding
>to the LienardWiechert potential, which is therefore the EM field
>generated by the charge. The EM field is given by
>E = q/(4 \pi e_0) {(1/s^3) S (1u^2/c^2) + (1/(c^2 s^3)) [R,[S,du/dt]]},
>B = [R,E]/(R c),
>where R and u are as before,
i.e. where R is such that the charge is a position rR at time tR/c,
and u is the velocity of the charge at that event.
>S = R  R u/c, s = R  u.R/c, du/dt is
>the acceleration of the charge when it is at rR at time tR/c, and for
>vectors v and w, [v,w] denotes the cross product of v and w.
This can be written in covariant form. The conventions will be adopted
that the metric in Minkowski space satisfies g_00 = 1, g_11 = g_22 = g_33
=  1/c^2, and the antisymmetric tensor F_{uv} representing the
electromagnetic field is normalized so that its components are:
F_{01} =  F_{10} =  E_x,
F_{02} =  F_{20} =  E_y,
F_{03} =  F_{30} =  E_z,
F_{23} =  F_{32} = B_x,
F_{31} =  F_{13} = B_y,
F_{12} =  F_{21} = B_z.
It follows that
F^{01} =  F^{10} = c^2 E_x,
F^{02} =  F^{20} = c^2 E_y,
F^{03} =  F^{30} = c^2 E_z,
F^{23} =  F^{32} = c^4 B_x,
F^{31} =  F^{13} = c^4 B_y,
F^{12} =  F^{21} = c^4 B_z.
If, at a given event P, the past light cone of P does not intersect the
worldline of the charge, then the contribution to the EM field at P due
to the charge is
F_{uv} = 0.
For an event P for which the past light cone of P intersects the worldline
of the charge, let T be the displacement 4vector of P relative to the
intersection of the past light cone and the worldline (so T.T = 0 and
T^0 > 0  since the charge is travelling at less than the speed of light,
then T is unique). Let U be the 4velocity of the charge at the
intersection between the past light cone and the worldline, and let W be
the 4acceleration of the charge at the same event (so U.U = 1 and
U.W = 0). The contribution to the EM field at P due to the charge is
F^{uv} = q (U^u T^v  U^v T^u)/(4 \pi e_0 c (T.U)^3)
+ q (T.W) (T^u U^v  T^v U^u)/(4 \pi e_0 c (T.U)^3)
+ q (W^u T^v  W^v T^u)/(4 \pi e_0 c (T.U)^2).

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