A particle moves with uniform acceleration if its acceleration measured in the instantaneous rest-frame of the rocket is constant. We parameterise the path by proper time τ . Then we can write the speed u in frame S as dx ± dt u= dτ dτ
Differentiating x0 = γ(x − vt), gives
dx0 dτ
and
dt0 dτ
µ =γ µ
=γ
t0 = γ(t − vx/c2 ).
dx dt −v dτ dτ
dt v dx − 2 dτ c dτ
¶
¶
(∗)
dt = γ(u − v) dτ
³ uv ´ dt =γ 1− 2 . c dτ
Then we divide the results to give the usual velocity addition law u−v 0 u = 1 − uv/c2
Now we differentiate u0 =
u−v : 1−uv/c2
du/dτ du0 (u − v)(v/c2 )du/dτ 1 − v 2 /c2 du = . + = 2 2 2 2 2 dτ 1 − uv/c (1 − uv/c ) (1 − uv/c ) dτ The acceleration, a, in S is by definition du/dt and similarly for S 0 so Á 0 Á 0 0 2 2 du du dt 1 − v /c du 0 2 dt a = 0 = = γ(1 − uv/c ) 2 2 dt dτ dτ (1 − uv/c ) dτ dτ 3 2 2 v /c ) 2
(1 − = a. 2 3 (1 − uv/c )
If S 0 is the instantanous rest frame of the accelerating observer, so that u0 = 0 and u = v, and the acceleration a0 in this frame is constant, then 3 2 2 v /c ) 2
(1 − a = a. 2 3 (1 − uv/c ) 0
becomes 2
2
3 2
a = (1 − u /c ) a0
Now
du a= dτ
Á
dt dt 2 2 − 12 and = (1 − u /c ) dτ dτ
so we can find the parameterised equation of the world line by integrating du = (1 − u2 /c2 )a0 . dτ This gives u = c tanh(a0 τ /c),
γ = cosh(a0 τ /c).
(choosing τ so that u = 0 when τ = 0)
Then from dt/dτ = γ, we find that t = c/a0 sinh(a0 τ /c), where we have assumed (or chosen the origin of t) such that t = 0 when τ = 0. Finally, dx dt dx = = uγ = c sinh(a0 τ /c) =⇒ x = c2 /a0 cosh(a0 τ /c) dτ dt dτ where we have assumed that x = c2 /a0 when t = 0. Uniformly accelerated particles therefore move on rectangular hyperbolas of the form x2 − (ct)2 = (c2 /a0 )2 .
The space-time diagram for an accelerated observer. The thick hyperbola is the observer’s world line. An observer ‘below’ the dashed lines could in principle send a message to the observer marked as a heavy dot; other observers could not.
