Offering the implied presence of classical symmetry breaks as evidence of their impossibility - ie. "it can't be right because it'd break the laws of physics" - is surely redundant; the claim is explicitly a classical symmetry break, that's its whole prospective value, and reason for our interest.
It is of course trivial that linear momentum can be converted to angular momentum, so the distinction in this sense is also redundant - this is one classical CoM violation, not two discrete ones. Plus the classical CoE violation. It's important to keep that "classical" bit in the description of the phenomenon in question - evidence of the non-conservation of momentum or energy is intrinsically impossible - akin to positing evidence of a causality violation; an inherently oxymoronic position.. We can only rationally intepret an apparent example of non-conservation as evidence that a system usually thermodynamically closed is somehow open. So we don't simply halt at the prospect, presume we've run out of road and pack up to go home.. we park up our expectations, grab a few essentials and continue following the trail that's lead us here.. Both the CoM and CoE anomalies are eminently resolvable from first principles. To simplify the system, let's presume we have a passively superconducting frustum. Because I^2 * R * T is the only disssipative loss mechanism involved, eliminating it provides a perfectly-efficient means of applying a momentum asymmetry... but NOT thrust! The thrust is, in turn, produced by the momentum asymmetry. This is the key distinction. Our input energy is only performing a constant internal workload - specifically, generating an asymmetry in the ratios of positive to negative ambient momenta exchanged between the virtual photon field and opposite ends of the frustum. In principle, merely rendering a field costs nothing, aside from a small but notionally conserved input energy. Only if that field then performs work, is that workload applied to the field-source energy supply, via Newton's 3rd law or its EM counterpart, Lenz's law. Furthermore, if that workload is merely mass displacement at constant velocity then all energy invested in the acceleration remains conserved in that momentum, and can likewise be recouped to the field source. However the acceleration corresponding to a cyclical (ie. closed-loop) momentum asymmetry represents work performed by the field itself - or, more precisely, the quantum exchanges that constitute the substance of the force in question. The input energy only needs to generate and sustain the imbalance of opposing momenta. To clarify this, let's eliminate the EM component entirely, and examine a hypothetical but purely mechanical CoM violation: - suppose we can 'turn off' mass at will, accelerating a body with zero inertia, before switching it back on again, and so generating free momentum and energy ..so in this scenario, the Higgs field has output mechanical work for us, we've netted ambient momentum and energy directly from the vacuum potential, and the only cost to us was that required to momentarilly disable the mass / inertia. But crucially, we input no work against the Higgs.. However even if that input energy was far greater than the output KE netted, provided we can keep repeating the interaction, building up successively more net system momentum, we eventually pass a threshold wherein the net system KE as measured from the non-inertial frame exceeds the net input energy as measured from within the accelerating system. If the EM-drive principle is viable, then its subsequent gain in energy and momentum is a direct linear function of its internal momentum asymmetry, not its input energy, and its internal efficiency or CoP never exceeds unity... The confusion arises when we assume that a passively-superconducting EM drive would only be spending energy on its own mechanical acceleration - after all if we've eliminated heat loss then acceleration seems the only remaining workload... ..however here, the net acceleration is the sum of not just the applied positive momenta, but also the circumvented or negated negative momenta! IE. it's a function of the momentum disunity, which can be anything from 100% (ie. full cancellation of counter-force) down to any non-zero value. So the net system energy is augmented by the KE value of its diverging reference frame. The clearest handle on this is given by consideration of just why KE=1/2mV^2 in the first place - why does KE have different dimensions to momentum (P=mV)..? The reasons are obvious when you realise them, but subtle enough that resolution of this "vis viva" debate was surely THE most fundamental breakthrough of the enlightenment in classical mechanics... Energy evolves as the half square of inertia times velocity as a direct consequence of the constraint of Newton's 3rd law - we must inevitably accelerate reaction mass, and the faster our reaction mass is moving away from us, the more force * displacement = work we need to apply in order to keep applying the same amount of momentum. If we plug some simple numbers into P=mV and KE=1/2mV^2, we see that an initial 1 kg-m/s unit of momentum has an energy cost of 1/2 a Joule. However, this interaction has now raised the relative velocity between the subject mass and its reaction mass. Because F=mA, if we want to generate another identical acceleration, we need to account for this velocity delta, raising the applied force or displacement, or both, to compensate the speed difference.. and because of this, the cost of a second identical 1 kg-m/s unit of momentum, on top of the first, rises four-fold to 2 Joules. Likewise, a third identical acceleration costs 4.5 J (1/2 * 1 kg * 3 meters/sec sqaured), then rising to 8 J for a fourth kg-m/s, 12.5 J for a fifth and so on.. So, while the first 1 kg-m/s off the launchpad only costs 1/2 a Joule, the final kg-m/s up to lightspeed costs 299,792,500 Joules. An effective N3 violation reduces or eliminates this practical accumulator on the rising unit energy cost of momentum. So, whether a given amount of momentum costs 1/2 J or 300 MJ, it will remain constant over successive input cycles, not rising with climbing velocity. But the system remains fully internally conservative, since the rising net system momentum presents no load upon the on-board energy source, and inertia alone is not velocity dependent. The energy asymmetry only becomes apparent in relation to a stationary external frame, and is due to the acceleration of the inertial frame itself, and our adding of its velocity onto that of the internally-performed work. For example, if Alice is on a train passing Bob on the platform, and she acclelerates a 1 kg mass forwards in the direction of the train's travel by 1 meter/sec, then she has only spent 1/2 a Joule... yet Bob sees the mass's KE rise by 1 meter/sec PLUS that of the train's own velocity... so if the train's moving at say 10 meter/sec through the station, Bob sees an acceleration from 10 m/s up to 11 m/s, and applying KE=1/2mV^2, accelerating 1 kg from 10 up to 11 m/s costs 10.5 J - 20 times more energy than Alice has spent... Of course the train also incurs a 1 kg-m/s counter-acceleration, imperceptible divided by its greater mass, but conserving the net system momentum. This is where the analogy with an EM drive breaks down.. there's no apparent counterforce applied back to the net system, and so that 20-fold gain becomes folding wedge.. the cost of generating the discount remains constant, but the discount actually returned increases by the half-square of rising momentum. There can be no mystery as to the source and sink when it is right there, spelled out in the terms of the interaction - we're applying an effective classical N3 exception to buy mechanical momentum on the cheap. It really is that simple. We 'gain' energy that hasn't needed to be input in the first place... and need never be output again either, provided we decelerate the same way we accelerated - sinking the momentum and KE back where it came from. If however we keep it here in the mechancial / thermodynamic domain, then the vacuum potential is left out of pocket. Notwithstanding that it is the same reserve likely underwriting all classical energy anyway.. Asymmetric classical interactions are open thermodynamic systems, and their existence is implicit within the terms of Noether's theorem, whereby equality of input vs output force * displacement integrals depends upon temporal invariance of the respective fields - in other words an 'over unity' (or under unity) interaction from a closed-loop displacement requires and implies passively time-varying forces. Thus a working EM drive is, by definition, exploiting a passive force/time delta between opposing ends of the photoelectric interaction with the frustum. Energy and momentum are conserved, and cannot be created or destroyed ex nihilo. However, the source of the gain (or loss) is defined by the form of its terms - it is raw mechanical momentum, which is a mass's resistance to velocity changes multiplied by its velocity, and caused by massive particles exchanging quantum momentum in signed units of h-bar with the vacuum potential via the Higgs interaction. That is, were we to harvest that gain, for a "perpetual motion machine" or whatever usage, the form of the energy we're harnessing is simply mechanical KE, manifested by a mass's resistance to deceleration - normal, symmetrical every-day interactions are all vacuum interaction, since all fundamental forces are nothing more or less. We input some work against the vacuum via a field's force, and it duly outputs some back at us. When we get more out than we've put in, it hasn't suddenly 'come from nowhere' - it's still the same force, a manifestation of the same gauge bosons mediating the same positive and negative packets of momentum between charges or masses. Ditto for non-dissipative losses. Inequality of input to output integrals for a closed-loop cycle means a time-variant interaction has occurred. CoE does not and cannot be applied to a system that is by definition not thermodynamically closed (ie. time-variant systems). It's become an open thermodynamic system. These same points apply to the relativistic mass increase - which again only arises from the non-inertial frame - inertia remains velocity-invariant within the accelerating frame, so relativistic mass increase presents no more challenge to mass constancy than it normally does. So, there's an intriguing thought to end on - if an EM-driven spacecraft subsequently decelerates again by simply performing a 180° rotation and continuing to apply constant thrust, all of the 'anomolous' momentum and energy is neatly returned to source. If however we were to aerobrake around a gas giant or something, we'd have dissipated the gained momentum and energy mechanically, causing a net change in the balance of 'classical thermodynamic energy' (that finite amount the universe was born with and which is ever-diminishing to 2LoT) relative to the vacuum potential. Ultimately, the thermodynamic outcome depends upon whether the pilot decides to employ non-dissipative deceleration (ie. reactionless braking) or not.. On Wed, Dec 28, 2016 at 6:43 PM, Stephen A. Lawrence <sa...@pobox.com> wrote: > Just to point something out -- the EM drive *obviously* doesn't need to > be outside the craft to work, since it doesn't eject mass. > > Furthermore (and consequently), it violates conservation of momentum, > conservation of angular momentum, conservation of energy, and conservation > of mass. While data trumps theory, this doesn't seem like the most likely > explanation of the effect to me. > > Gedanken: Put an EM drive in a box. Attach it to a wire. Attach the > other end of the wire to a pivot (like one of those old gas powered toy > planes people used to have before the days of radio control). Let the box > with the EM drive go. It will accelerate in a circle, around the pivot > point. > > Power consumption inside the box is presumably constant. Power generated > varies in proportion to the speed of the box (power = force * velocity). > So, at some point it'll be generating more power than it's consuming. And > there's the violation of CoE. (With a bit of cleverness you can turn it > into a Type I perpetual motion machine.) > > Meanwhile it's going lickety split around the pivot, with increasing > angular momentum; with no mass ejection there's no compensating decrease > anywhere else. There's the violation of conservation of angular momentum. > > And as its velocity increases, its mass increases as gamma*m. There's the > violation of conservation of mass. > > And violation of linear momentum is obvious. > > On the other hand if it doesn't work, then all that's being violated is > the assumption that the handful of extremely delicate high precision > experiments that have been done to show the effect were not somehow botched. > > I'm not holding my breath on this one. > > > On 12/28/2016 02:02 AM, David Roberson wrote: > > Russ, > > Can you verify that the Chinese actually have a functioning EM drive on > their space station. Also, how much thrust are they claiming? Finally, is > that device or group of devices capable of maintaining all of the > orientation required for the station? > > Dave > > > > -----Original Message----- > From: Russ George <russ.geo...@gmail.com> <russ.geo...@gmail.com> > To: vortex-l <vortex-l@eskimo.com> <vortex-l@eskimo.com> > Sent: Tue, Dec 27, 2016 3:45 pm > Subject: [Vo]:EM Drive need not be outside the spacecraft > > A curious facet of the EM drive, such as the one now operating on the > Chinese space station is that it need not be on the outside of the > spacecraft, it’s thrust is independent of the position and surrounding > matter. This enables all manner of interesting spacecraft geometries. > > >
