Single-Particle Electrodynamics - Assassination Science

Our second piece of verification evidence lies with the observation that γ(τ)
may alternatively be computed via γ(τ) ≡ d τ t(τ); this gives
γ(τ) = 1 + 1 2 τ 2 ˙v 2 + 1 2 τ 3 ( ˙v·¨v) + 13
24 τ 4 ˙v 4 + 1 8 τ 4¨v 2 + 1 6 τ 4 ( ˙v·... v)
+ 1
24 τ 5 ( ˙v·.... v) + 1
12 τ 5 (¨v·... v) + 9 8 τ 5 ˙v 2 ( ˙v·¨v) + O(τ 6 ), (G.17)
which is identical to (G.16). Now, upon reversion of t(τ), one finds
τ(t) = t − 1 6 t3 ˙v 2 − 1 8 t4 ( ˙v·¨v) − 1 40 t5 ˙v 4 − 1
40 t5¨v 2 − 1
30 t5 ( ˙v·... v)
− 1
144 t6 ( ˙v·.... v) − 1 t6 (¨v·...
v) − 1 72 24 t6 ˙v 2 ( ˙v·¨v) + O(t 7 ). (G.18)
Using (G.18) in (G.15), we thus find
v(t) = t ˙v + 1 2 t2¨v + 1 6 t3 ...
v + 1 24 t4 ....
v + O(t 5 ),
which of course defines ˙v, ¨v, ... v and .... v; this completes the verification of the
input expressions.
G.6.3
Trajectories of rigid body constituents
We now compute the trajectory of the constituent r. This is given by
z α r (τ) = z α (τ) + ∆z α r (τ),
(G.19)
where
where u(0) = r, and
∆t r (τ) = γ(τ) ( u(τ)·v(τ) ) ,
∆z r (τ) = u(τ) +
γ2 (τ) ( )
u(τ)·v(τ) v(τ), (G.20)
γ(τ) + 1
d τ u(τ) =
γ3 (τ)
γ(τ) + 1 u(τ)×( v(τ)× ˙v(τ) ) .
(G.21)
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