This is a short explanation of how one finds the distance and time travelled in both the reference frame of the rocket and that of the "fixed stars", in terms of how much of the initial mass is fuel, what the properties of the fuel are, and what the acceleration is.
The usual way you transform things like momentum and energy between reference frames is through the Lorentz transformation, in which the velocity, v, is a parameter. But here you do a "continuous boost": at each instant in time the rocket is moving at velocity v relative to the "fixed stars". It burns an infinitesimal quantity of fuel, changing its momentum and therefore its velocity by an infinitesimal amount, dv, in the process. You then "boost" onto the new rest frame at velocity v+dv by Lorentz transformation from the old one at velocity v, and so on...
Consider a fuel which self-reacts in some way, converting a fraction f of its rest energy, m_f c^2 ("mass"), into kinetic energy and the rest into exhaust mass. Of the kinetic energy, another fraction g is lost and the rest converted into directed exhaust momentum. Denote the fuel rest energy by E_f = m_f c^2 and the exhaust total energy by E_x; then the difference is just the energy lost, g times f times E_f. The energy of the exhaust, E_x, is given by its rest energy, m_x c^2, and its directed momentum, p, through the relation
(E_x)^2 = (m_x c^2)^2 + (p c)^2,
the exhaust mass is given in terms of the energy liberated,
m_x c^2 = (1 - f) m_f c^2,
which now gives the total energy of the exhaust:
E_x = E_f - g f E_f = (1 - g f) E_f
(note that if g is 1 then E_x is all rest energy). The first line then reads
(1 - g f)^2 E_f^2 = (1 - f)^2 E_f^2 + (p c)^2,
and then one obtains p in dimensionless form:
p
beta_f = ----- = sqrt [ 2 (1 - g) f - (1 - g^2) f^2 ],
m_f c
for which the best one could do would be beta_f = 1 for complete annihilation and perfect direction of the exhaust energy. By comparison, for hydrogen fusion and 50 per cent losses, beta_f is about 0.06 (this excludes proton-proton fusion, which is a weak interaction).
The acceleration proceeds through successive infinitesimal additions of momentum, dp, to the rocket in its rest frame, causing infinitesimal changes in the velocity, dv, measured relative to the "fixed stars". The change to the rocket's velocity in its rest frame, Dv, due to the burning of an infinitesimal amount of fuel, dm, is given by
dp = M Dv = beta_f dm c,
where M is the total mass, payload plus fuel at the give instant. To get the sign right, denote M by M_0 - m, where M_0 is the total initial mass, and m is the mass of the fuel on board.
The boost to the new rocket frame involves adding the velocity Dv to the current velocity v. Conversely, the change in v is given by the difference between this boosted velocity and v:
v + Dv
dv = ------------ - v.
1 + v Dv/c^2
This is reduced to first order in the changes by multiplying through by the denominator and dropping the dv Dv term on the left side. Expressing v in dimensionless form, beta = v/c, find
d beta dv Dv beta_f ------ = -- = -- (1 - beta^2) = ------- (1 - beta^2), dm dm dm M_0 - m
and then combining terms,
d beta dm ---------- = beta_f -------. 1 - beta^2 M_0 - m
This is integrated from 0 to beta on the left side and 0 to f_m = m/M_0 on the right side to obtain
1 1 - beta
log (1 - f_m) = -------- log --------,
2 beta_f 1 + beta
or the fuel mass as a fraction of total mass (f_m) needed to propel the rocket to the given velocity, v = beta c. Note that for a one-way trip both an acceleration and a deceleration leg are needed, so the payload mass is
M_P = (1 - f_m)^2 M_0.
There are two formulas that relate the small changes in velocity and time between the two reference frames of the rocket and the fixed stars. Changes in time are given by the Lorentz transformation:
dT -- = sqrt(1 - beta^2), dt
where T is the proper time of the rocket, t is the proper time of the fixed stars, and beta = v/c, with v the velocity of the rocket relative to the fixed stars. For each small acceleration of the rocket in its own reference frame, one must take into account this change, and "boost" the rocket frame such that the velocity of the rocket in its own frame stays zero. This is called the "continuous boost" and is found by application of the velocity addition rule:
v + Dv
dv = ------------ - v,
1 + v Dv/c^2
where Dv is the change of the rocket's velocity due to the acceleration within dT (that is, Dv = a dT), and dv is the change in the rocket's velocity relative to the fixed stars. For simplicity, let "a" be kept constant, and define the dimensionless variable
alpha = a t/c.
The above formulas can be combined into the differential equation,
d beta ------- = (1 - beta^2)^(3/2) d alpha
(that is, one minus beta squared, all to the 3/2 power), which has the solution
beta a t a T ------------ = alpha, hence --- = sinh ---. (1 - beta^2) c c
from these and the velocity definition v = dx/dt, one can find
(x + c^2/a)^2 - c^2 t^2 = c^4/a^2,
the well-known hyperbola law for constant acceleration in space-time.
One can then turn this around to give the proper time, T, in terms of the distance, x, and acceleration, a:
T = (c/a) inv cosh [(a x/c^2) + 1]
and from this the necessary fuel.
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