But Robin, there is no kinetic energy associated with a moving object in its own reference frame. It will know that it accelerated because it can directly sense that operation, but once the drive cuts out it is at rest in its latest frame.
I suppose a spaceman on the EM Drive powered ship could calculate the force causing his acceleration and thus know how much internal mass he has converted into energy to make it happen. And, someone residing within the space ships original frame could likewise make the same calculation. The problem would show up if the space ship then reversed his maneuver and returned back to that original frame. Both people would again calculate approximately the same mass conversion that occurs due to the force generation. So at the end of the trip, the ship would be depleted of some of its mass for no net apparent change in velocity or position. Where did the energy go? Dave -----Original Message----- From: mixent <mix...@bigpond.com> To: vortex-l <vortex-l@eskimo.com> Sent: Mon, Mar 14, 2016 5:06 pm Subject: Re: [Vo]:Re: EM Drive(s) In reply to Vibrator !'s message of Mon, 14 Mar 2016 11:03:43 +0000: Hi, [snip] >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. [snip] Is the energy anomaly resolved if it pushes against the mass of the universe (i.e. against space-time itself)? In which case it would indeed be just like a train on rails. In short, momentum is conserved, and all the energy ends up with the moving object. I suspect that this is the basis of Shawyers argument. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
