Sebastiaan said:
I though the vessel would accerate less the closer it neared the speed of light
"Accelerate" is ambiguous. The
proper acceleration--the acceleration actually felt by the spacecraft and its crew--is constant; it is 600 kN divided by the rest mass of the spacecraft . The
coordinate acceleration in the outside observer's frame--the rate at which the spacecraft 's speed in that frame changes with time in that frame--does decrease as the vessel speeds up. See below.
Sebastiaan said:
How can the Force be actualy be higher the faster you travel?
"Force" is also ambiguous. The force felt by the spacecraft is constant--600 kN, the thrust exerted by the engines. But you were asking about the force as measured by the outside observer. It turns out that I was mistaken in saying that increases; it is also constant! I had misinterpreted the formula in the Usenet Physics FAQ article you linked to; it was using ##a## to mean ##dv/dt##, the rate of change of the vessel's speed with respect to time in the outside observer's frame. So let's re-calculate.
We have the speed ##v## of the vessel in the outside observer's frame given by ##v = \tanh w##. We also have, from the formula I gave earlier for ##t##, ##w = \sinh^{-1} a t##. (These formulas aren't affected by my misinterpretation of the article that I just referred to.) Putting these together gives (note that I am here using units where ##c = 1##, as I did in previous posts; at the end I'll show what happens when we put the factors of ##c## back in):
$$
v = \frac{at}{\sqrt{1 + a^2 t^2}}
$$
We then simply take the derivative of this with respect to ##t## to get, after some algebra:
$$
\frac{dv}{dt} = \frac{a}{\left( 1 + a^2 t^2 \right)^{3/2}}
$$
where ##dv / dt## is the vessel's coordinate acceleration in the outside observer's frame. Notice that this decreases as ##t## increases.
We can go further and rewrite the above in terms of ##\gamma##, by observing that ##\gamma = \cosh w = \sqrt{1 + \sinh^2 w} = \sqrt{1 + a^2 t^2}##. This means that
$$
v = \frac{at}{\gamma}
$$
and
$$
\frac{dv}{dt} = \frac{a}{\gamma^3}
$$
If we then substitute this into the (properly interpreted) formula from the article, which is ##F = \gamma^3 m dv/dt##, we get ##F = m a##! That is, the force as measured by the outside observer is constant, and
the same as the force felt by the spacecraft crew. (Here I am using ##a## to mean the proper acceleration, and ##dv/dt## to mean the coordinate acceleration, to avoid confusion.) What is going on here is that the decrease in ##dv/dt## as the ship speeds up is exactly compensated by the increase in mass of the ship as seen by the outside observer, plus the effects of time dilation. (See the discussion of Isp below.) Sorry for the mixup on my part in misstating this before; it's been a while since I worked through this from the outside observer's viewpoint.
We can also see that
$$
\frac{d\gamma}{dt} = \frac{a^2 t}{\sqrt{1 + a^2 t^2}} = a v
$$
Putting this together with the equation for ##dv/dt## tells us that
$$
\frac{d \left( \gamma v \right)}{dt} = v \frac{d\gamma}{dt} + \gamma \frac{dv}{dt} = a v^2 + \frac{a}{\gamma^2} = a
$$
So we can rewrite our equation for ##F## as ##F = m d\left( \gamma v \right) / dt = m dp/dt##, where ##p## is the momentum of the spacecraft in the outside observer's frame. This formula is useful because it generalizes, without any modification, to the case where ##F## and ##p## are vectors, i.e., to the case where the force is not parallel to the velocity.
Finally, what about Isp? Unlike the force above, that
is different for the outside observer vs. the spacecraft and its crew. The formula I gave for ##t## above applies: for ##\tau = 1,000,000##, and assuming a spacecraft rest mass of 60 tons, so ##a = 10##, we have (adding factors of ##c## where needed to make the units come out right--I was using units where ##c = 1## in the previous formulas):
$$
t = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right) = \frac{1}{10} 299,792,458 \sinh \left( \frac{10,000,000}{299,792,458} \right) = 1,000,185
$$
So the ship's fuel burn will last an extra 185 seconds according to the outside observer. Also, the ship's speed when the burn finishes, according to the outside observer, will be (again adding back the factors of ##c##):
$$
\frac{v}{c} = \frac{at}{\sqrt{c^2 + a^2 t^2}} = 0.033344
$$
So the ship, starting from rest, would indeed reach (and exceed) 1% of the speed of light during the burn.
