Ender wrote:Which point was that at? I know at one point I dropped a "*" between kg and m/s simply to make the end numbers line up right.
The seconds should be squared.
Ender wrote:If I go to 50% the speed of light and back down ten times, I will never exceed the speed of light, but my total delta V will still be 3*10^9 m/s.
This makes sense in Newtonian mechanics, if a bit unusual, but in relativity, you need to be looking at change in momentum instead. If you say Δv = 1.90c, for example, it makes quite a lot of difference whether it was an accelerated to 0.950c and back to zero or two such operations to 0.475c. Burning fuel changes the rocket's momentum in more or less orthodox manner, with only a slight difference if the exhaust is relativistic, but the way this change of momentum corresponds to a change of velocity is highly nonlinear if the current velocity is a sizable fraction of c.
Ender wrote:Honestly, I'm not following what you are saying here. However, I do remember the thread in question, and I later found out that the answer given there (that I would need a seperate equation to compensate for relativity) was incorrect.
But you do. With apologies to ClaysGhost, I did make a mistake in having one too many factors of γ at that time, but that only changes the interpretation of u. Otherwise, the calculation works out to the same result. If your exhaust velocity is u, then for a mass change Δm, the corresponding momentum change is Δp = uΔm. The rest of the derivation is the same. The local acceleration α = [dp/dτ]/m = u dm/m transforms with a factor of Γ³ = [1-v²]^{-3/2} as before; the dτ/dt = Γ takes away one factor of Γ, giving dv/(1-v²) = u[dm/m], or atanh(v) = -u log R. This is equivalent to v = [1-R^{2u}]/[1+R^{2u}].
Ender wrote:Again, not quite following what you are saying here, but if I understand you right I'm really confused - the ejecta is totally seperate from the reactants, so why would I be trying to convert the ejecta into energy?
Because that's how rockets work. Let's say you observe some amount exhaust of mass M and velocity u. In the process of expelling it, the rocket mass is decreased by both the rest mass M
and the mass-equivalent of the kinetic energy of the exhaust. Normally, this energy is taken from chemical potential energy, which is so small as to allow one to ignore its mass-equivalent with impunity. It is no longer ignorable if the exhaust is relativistic. Try looking at it backwards. In burning some amount of fuel Δm and using it for exhaust with some fraction ε going into kinetic energy (K = εΔm), the rest mass of the exhaust will be ΔM = (1-ε)Δm. The kinetic energy is Κ = (γ-1)[(1-ε)Δm] = εΔm, solving which gives γ = 1/[1-ε] and u = [γ²-1]^{1/2}/γ = [2ε-ε²]^{1/2}. Note that then Δp = γuΔM = uΔm, as above.
Ender wrote:If the engine provided both thrust and power like some real life engine designs do that would make sense (and it would make a lot of sense for starfighters, but alas that idea is contradicted by canon), but the reactor and engines are totally seperate in SW designs.
It doesn't matter. The reactors take energy from somewhere, and thus the mass of some material is decreased. If the reactor works by converting mass to energy, then there you go. If it works by fusion, then the mass of fusion products is less than the reactants, in according to E = mc². If it works by combustion, then this mass decrease corresponds to the change in chemical potential energy. The exact mechanics of the reactor are irrelevant. No matter where the energy comes from, it is taken out of the mass of the ship in some fashion.