But as already noted, gyroscopes (rotating bodies generally) and the magnetic field (as evinced by Faraday's spinning inductor paradox) also have their own unique frames (classically, the "outermost sphere of fixed stars") - and they're not warp bubbles!?
A warp bubble is specifically a contortion of spacetime. Granted such a craft would need no reaction mass, but then it has no need to accelerate, either, since it is space itself that is translating. A reactionless drive would probably take a good while to reach anything approaching "warp" speeds though. If it even had the energy reserves. Although there is one potential reactionless system that could perhaps give a warp drive a run for its money.. Suppose we had a Bessler wheel! The classic "gravity mill" - say, applying an effective N3 break to gain energy from gravity. Gravity is equivalent to an acceleration, so if we attach such a system to an accelerating spacecraft, it'll be powered by its own inertia to that acceleration. Use the energy to power an EM drive and you have a positive closed feedback loop - the harder it accelerates, the more energy the Bessler mechanism is able to supply to the motor, and so on... to infinity and beyond, presumably.. It'd be the last word on thrill rides (infinite exponential acceleration!), albeit lethal (the craft will quickly exceed its structural limits)..and hence in practice, still slower than warp.. On Mon, Mar 14, 2016 at 3:59 PM, Russ George <russ.geo...@gmail.com> wrote: > Great vibrations but where you say, “Without physical reaction mass, such > a system has its own unique reference frame - from within which, energy may > be conserved, but which from without, cannot be.” Might we start using > common terminology…. Your description is of course more popularly known as > the ‘warp bubble’;) > > > > > > *From:* Vibrator ! [mailto:mrvibrat...@gmail.com] > *Sent:* Monday, March 14, 2016 4:04 AM > *To:* vortex-l@eskimo.com > *Subject:* Re: [Vo]:Re: EM Drive(s) > > > > Yes, and this is why KE = 1/2 MV^2 - ie., why the acceleration unit cost > escalates; a given force has to be applied over an ever-greater distance as > velocity (time rate of change of position) increases. Alternatively, we > could hold displacement constant and progressively raise the force > magnitude. > > Yet Craig still seems to have a point - without some kind of corporeal > reaction mass, what is an EM drive's velocity actually relative to? What's > its reference frame, if not the thing it's pushing against? > > To illustrate the conundrum, suppose i have an EM drive aboard a train, > and you the observer are standing on the platform as the train passes > through the station: I fire the engine, and it accelerates by 1 meter / > sec. > > Suppose the engine weighs 10 kg. From my perspective, its KE has > increased by 5 Joules - ie. it's perrformed 5 J of mechanical work, > regardless of how much more energy may have been wasted to heat. > > But if the train was already travelling at 10 m/s, and the drive > accelerated in the same direction, then from your stationary perspective > the drive has accelerated up from 10 to 11 m/s - and for a 10 kg mass > that's a workload of 105 J - bringing its KE up from 500 J to 605 J. > > So, has the drive burned 5 J or 105 J? > > If i cheated - the drive doesn't really work, and i just gave it a > surreptitious shove - this same paradox is resolved by a corresponding > deceleration of the train - ie. if i accelerate a small mass against the > inertia of a larger mass, the latter is decelerated and net momentum is > conserved. > > Except here, the drive ISN'T pushing against the train. Yet it still > benefits from its ambient velocity. Net momentum is NOT conserved, and > neither is energy. > > And so the question arises, how does the EM drive "know" what its > reference frame is? Shawyer claims (or seems to imply) that the unit cost > of acceleration increases as we would normally expect (distance over which > a given force is applied keeps rising) - but how does it measure > "distance"? Relative to what, exactly? Without physical reaction mass, > such a system has its own unique reference frame - from within which, > energy may be conserved, but which from without, cannot be. > > I mean this not as a crtitique against the plausibility of such systems, > and share the prevailing cautious optimism. But if they do work, then we > also have an energy anomaly. > > In the many years i've been researching classical symmetry breaks, one > thing has become clear - the only way to explain away a real symmetry break > is to invoke another somewhere else up or downstream (it's a standard > recourse for pseudoskeptics). As much as i'd welcome free energy, momentum > and FTL travel, and despite Shawyer's assurances everything's classically > consistent, these enigmatic implications remain.. for me, at least. > > > > On Mon, Mar 14, 2016 at 4:17 AM, <mix...@bigpond.com> wrote: > > In reply to Craig Haynie's message of Sun, 13 Mar 2016 21:08:43 -0400: > Hi, > [snip] > > Note the use of the word "acceleration". > > Acceleration produces a force. Force times distance = energy. > > >This doesn't make any sense: > > > >"For a given acceleration period, the higher the mean velocity, the > >longer the distance travelled, hence the higher the energy lost by the > >engine." > > > >Since we're not talking about relativistic speeds, then the idea that a > >device will consume more energy, over a given period of time, simply > >because it's moving, would violate Einstein's Special Relativity which > >says there's no preferred frame of reference. The moving object cannot > >be said to be moving at all. > > > >Craig > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > > >
