Single-Particle Electrodynamics - Assassination Science

6(c 1·c 3 ) + 4c 2 2 = 0,
8(c 1·c 4 ) + 12(c 2·c 3 ) = 0,
10(c 1·c 5 ) + 16(c 2·c 4 ) + 9c 2 3 = 0,
12(c 1·c 6 ) + 20(c 2·c 5 ) + 24(c 3·c 4 ) = 0. (2.89)
We now divide the temporal component d τ z 0 of (2.86) into the spatial components
d τ z, to obtain
v(τ) ≡ d t z(τ)
≡ d τz(τ)
d τ t(τ)
≡ c 1 + 2c 2 τ + 3c 3 τ 2 + 4c 4 τ 3 + 5c 5 τ 4 + 6c 6 τ 5 + O(τ 6 )
c 0 1 + 2c 0 2τ + 3c 0 3τ 2 + 4c 0 4τ 3 + 5c 0 5τ 4 + 6c 0 6τ 5 + O(τ 6 ) . (2.90)
To proceed from here, one needs to perform the division (2.90) term-by-term;
at each step, one needs to make use of the identities (2.89), and then replace τ
wherever it appears in favour of t (by reverting the expression already found,
to the preceding order, for t as a function of τ). The result at each step is
compared to the parametrisation (2.87), to replace the c α i by the quantities
˙v, ¨v, v ... and .... v . The procedure is straightforward, but tedious; the author
will spare the reader the gory details.
parametrisation is
One finally finds that the correct
t(τ) = τ + 1 6 ˙v2 τ 3 + 1 8 ( ˙v·¨v)τ 4 + 1 {
13 ˙v 4 + 3¨v 2 + 4( ˙v·... v) } τ 5
120
+ 1 {
( ˙v·....
v) + 2(¨v·... v) + 27 ˙v 2 ( ˙v·¨v) } τ 6 + O(τ 7 ), (2.91)
144
z(τ) = 1 2 ˙vτ 2 + 1 6 ¨vτ 3 + 1 { ...
v + 4 ˙v 2 ˙v } τ 4
24
+ 1 { ....
v + 10 ˙v 2¨v + 15( ˙v·¨v) ˙v } τ 5 + O(τ 6 ). (2.92)
120
(It will be noted that (2.91) contains terms up to sixth order in τ, whereas
(2.92) only contains terms up to fifth order. This order of expansion has been
96