Yay! Everything you said makes sense to my enthusiast knowledge. Except that bit. Rather than sailing right through it (and by it, we're talking it's relative speed from the original reference frame), wouldn't it just edge closer and closer to C, become increasingly more massive, and there's some length contraction which I'm aware of, but never really understood.
In most instances, raising momentum means raising counter momentum - a la Newton's 3rd law (N3) - and this is the fundamental reason why kinetic energy scales as the half square of velocity.
For example, following the standard mechanical energy term KE = 1/2 mV^2, it costs 1/2 a Joule to accelerate a 1 kg mass, from stationary, up to an initial velocity of 1 meter / sec.
But to raise its velocity by a further meter / sec is more expensive. The 1/2 J per kg-m^2 special introductory offer doesn't roll over to the next kg-m^2 - rather, its cost rises fourfold, to 2 J.
And each successive kg-m/s of momentum we buy thereafter costs ever more energy to purchase. A third unit costs 4.5 J. A fourth, 8 J, and a fifth, 12.5 J and so on.
Each time we're buying the same amount of momentum - 1 kg-meter / second (or equivalently, 1 kg-meter squared), but the more we buy, the more expensive it gets.
This is because to accelerate a mass, we have to apply a a force, over a distance - following the work / energy equivalence principle, work = force times displacement.
And since force = mass times acceleration, and acceleration is rate of change of rate of change of location with respect to time, the faster a mass is moving, the more force is required, or else the more displacement over which a constant force must be applied.
So the ultimate cost of momentum is determined by the practicalities of being dependent upon reaction mass.
If however we had a notional violation of Newton's 3rd law (an "N3 symmetry break" or just N3 break for short), then we could arrange a perfectly elastic collision (ie. having no dissipative losses), with an asymmetric resulting distribution of momentum.
When such an interaction is cycled successively, the net system momentum rises.
This is obviously not what happens in normal elastic mechanical interactions - the division of momentum is usually equal and opposite in sign, and so cycling the interaction does not change the net system momentum.
Internally however, the masses in an asymmetric interaction are only paying for their own, relative displacements, not that of the net system.
Inertial motors are generally considered impossible - until Shawyer's EM drive came along, few scientists would even consider such a possibility. No mechanical example has ever been found, as far as i know.
But
if an N3 break were possible, then the device would not be paying for its rise in net system momentum resulting from the accumulating asymmetric distribution of momentum.
Internally, if it moved say 1 kg back and forth by 1 meter each cycle, that cost is 1 J per cycle, however assuming it was fully elastic it's only 1 J to set it in motion, which then continues indefinitely until friction or some other force is applied. Externally however, from a stationary reference frame, the system would be accumulating more momentum than could be explained by the 1 J / cycle on-board expenditure.
Since energy is conserved we could not assume this was free energy, but rather supplied by whatever elementary field is ultimately responsible for the N3 break, or simply just inertia itself (ie. the Higgs field).
I raised this same objection on the forum of the NASA team replicating these EM drives - reactionless thrust is only half the controversey; the other half is the implied free energy. Any effective example of an N3 break is also over-unity. A working EM drive accelerating through space would pass some threshold velocity at which it had more kinetic energy than could have been supplied to it or carried by it in terms of potential energy stores. It would only be paying for its internal workload in manifesting the N3 break. The rise in KE of the net system, with its accumulating momentum and diverging reference frame, would be anomolous and so not amenable to calorimetry for example.
In principle however, with a mechanical N3 break, whatever its unit cost of anomolous momentum per cycle, it would remain constant as that net momentum built up - and not rising as a function of velocity.
So if our first kg-m/s cost 1/2 a Joule, so would the second, third and fourth and so on, forever.
If we could accelerate 1 kg up to lightspeed using a perfectly-efficient external energy source, the final kg-m/s of momentum that tipped us from 299792457 m/s up to 299792458 m/s (C) would cost a whopping 299792500 Joules.
If however it was purchased via an N3 break, its minimum cost could remain as low as 1/2 J - the same cost as the first kg-m/s off the launchpad, and a total expenditure of 149896229 Joules - a nice saving of 44937758487152171 Joules over the standard ticket price.
But whatever the unit energy cost of momentum sourced via an N3 break - even if it was a megajoule per kg-m/s - it would in principle remain constant per cycle, regardless of rising net system momentum, and so provided we could buy enough of it, we would eventually pass some threshold velocity beyond which we were over-unity, as measured from a static external frame.
If you think about the practical limitations on accelerating to lightspeed, it's reaction mass, and on-board potential energy.
But an N3 break is velocity-agnostic. It has no external reference for its own velocity, and is effectively its own independent, unique reference frame.
Provided you had the on-board energy to keep cycling it, it would have no speed limit.
We can only speculate whether matter as we know it can even exist above C. But assuming the craft didn't somehow evaporate, there'd be no practical constraints upon exceeding C.
Obviously, we can already see objects in the sky exceeding C as a function of relative velocity (to us observers) - and all of the universe beyond the red-shift horizon is effectively already at superluminal speed relative to us, due to Lambda (the constant of accelerating expansion).
Similarly, a craft above C on a parallel vector would be invisible, but on an orthogonal vector would remain visible, although we'd be seeing where it was, not where it is.
On-board, you'd have no sensation of increased mass - 1 kg would still register as such on a scales under 9.81 m/s of acceleration. Likewise, there'd be no change in your rate of time, or metric of distance.
Again, one can only speculate what superluminal travel might be like, but with an effective N3 violation, C would no longer present any barrier in purely mechanical terms.
In reality (by that i mean here, not ED), at any point when thrusting is stopped, the ship is relatively stationary. So at any point thrusting again provides the same artificial sense of gravity. To an outside observer that remains stationary from the "starting point" of the test, the ship will never go faster than the speed of light. But relativity will have time slowing down for the ship; From the ships reference frame wouldn't the acceleration be constant and thus the artificial sense of gravity constant as well, no matter how close to the speed of light it got in the starting reference frame?
