# Re: Lienard-Wiechert 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 Lienard-Wiechert 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 (1-u^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 r-R at time t-|R|/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 r-R at time t-|R|/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 4-vector 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 4-velocity of the charge at the
intersection between the past light cone and the worldline, and let W be
the 4-acceleration 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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