Quote from: deltaMass on 07/10/2015 12:19 AMQuote from: TheTraveller on 07/09/2015 10:38 PMPhase velocity + group velocity = 2c. WrongGroup velocity X phase velocity = 2c.

Quote from: TheTraveller on 07/09/2015 10:38 PMPhase velocity + group velocity = 2c. Wrong

Phase velocity + group velocity = 2c.

Quote from: TheTraveller on 07/10/2015 12:20 AMQuote from: deltaMass on 07/10/2015 12:19 AMQuote from: TheTraveller on 07/09/2015 10:38 PMPhase velocity + group velocity = 2c. WrongGroup velocity X phase velocity = 2c.Also wrong

Quote from: TheTraveller on 07/09/2015 10:38 PMQuote from: WarpTech on 07/09/2015 09:12 PMQuote from: deltaMass on 07/09/2015 07:55 PMQuote from: WarpTech on 07/09/2015 07:31 PM1. For a phase velocity >> c, the group velocity is << c. So F * v_{g} will have a lot more Force with a lot less Power than a photon rocket. So yes, u >> c is very important.The important thing is, Pin = Pout. The vehicle will accelerate dependent on the available input power until relativistic effects start to change the parameters of the problem. It never goes over-unity. That is simply a mis-inerpretation of what "break-even" means.ToddI believe, based on what I wrote above, that you have not made the case for "no power breakeven until v = c". Indeed, we agree that this breakeven speed is substantially less than c if we take your formula for k to be correct and we take the phase velocity to be superluminal. The question which intrigues me is what the value of 'b' might be (u = b*c). Do you have a handle on quantifying this phase velocity value?What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. The phase velocity depends on the dimensions of the frustum and the frequency driving it. There will be more details on that in my forthcoming paper.ToddPhase velocity + group velocity = 2c. In a waveguide, group velocity is below 1c and phase velocity is therefore above 1c. As such nothing in a waveguide moves at phase velocity or greater than 1c. Everything in a waveguide moves at group velocity or below 1c.Outside a waveguide group velocity = phase velocity = 1c.That's incorrect. v_{p} * v_{g} = c^{2}Inside a waveguide, the phase of the wave moves with the phase velocity, the "energy" moves with the group velocity. IMO, the phase is not "nothing", it has a physical meaning even if it's not a physically measurable quantity, the effects of phase-interference are.Todd

Quote from: WarpTech on 07/09/2015 09:12 PMQuote from: deltaMass on 07/09/2015 07:55 PMQuote from: WarpTech on 07/09/2015 07:31 PM1. For a phase velocity >> c, the group velocity is << c. So F * v_{g} will have a lot more Force with a lot less Power than a photon rocket. So yes, u >> c is very important.The important thing is, Pin = Pout. The vehicle will accelerate dependent on the available input power until relativistic effects start to change the parameters of the problem. It never goes over-unity. That is simply a mis-inerpretation of what "break-even" means.ToddI believe, based on what I wrote above, that you have not made the case for "no power breakeven until v = c". Indeed, we agree that this breakeven speed is substantially less than c if we take your formula for k to be correct and we take the phase velocity to be superluminal. The question which intrigues me is what the value of 'b' might be (u = b*c). Do you have a handle on quantifying this phase velocity value?What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. The phase velocity depends on the dimensions of the frustum and the frequency driving it. There will be more details on that in my forthcoming paper.ToddPhase velocity + group velocity = 2c. In a waveguide, group velocity is below 1c and phase velocity is therefore above 1c. As such nothing in a waveguide moves at phase velocity or greater than 1c. Everything in a waveguide moves at group velocity or below 1c.Outside a waveguide group velocity = phase velocity = 1c.

Quote from: deltaMass on 07/09/2015 07:55 PMQuote from: WarpTech on 07/09/2015 07:31 PM1. For a phase velocity >> c, the group velocity is << c. So F * v_{g} will have a lot more Force with a lot less Power than a photon rocket. So yes, u >> c is very important.The important thing is, Pin = Pout. The vehicle will accelerate dependent on the available input power until relativistic effects start to change the parameters of the problem. It never goes over-unity. That is simply a mis-inerpretation of what "break-even" means.ToddI believe, based on what I wrote above, that you have not made the case for "no power breakeven until v = c". Indeed, we agree that this breakeven speed is substantially less than c if we take your formula for k to be correct and we take the phase velocity to be superluminal. The question which intrigues me is what the value of 'b' might be (u = b*c). Do you have a handle on quantifying this phase velocity value?What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. The phase velocity depends on the dimensions of the frustum and the frequency driving it. There will be more details on that in my forthcoming paper.Todd

Quote from: WarpTech on 07/09/2015 07:31 PM1. For a phase velocity >> c, the group velocity is << c. So F * v_{g} will have a lot more Force with a lot less Power than a photon rocket. So yes, u >> c is very important.The important thing is, Pin = Pout. The vehicle will accelerate dependent on the available input power until relativistic effects start to change the parameters of the problem. It never goes over-unity. That is simply a mis-inerpretation of what "break-even" means.ToddI believe, based on what I wrote above, that you have not made the case for "no power breakeven until v = c". Indeed, we agree that this breakeven speed is substantially less than c if we take your formula for k to be correct and we take the phase velocity to be superluminal. The question which intrigues me is what the value of 'b' might be (u = b*c). Do you have a handle on quantifying this phase velocity value?

1. For a phase velocity >> c, the group velocity is << c. So F * v_{g} will have a lot more Force with a lot less Power than a photon rocket. So yes, u >> c is very important.The important thing is, Pin = Pout. The vehicle will accelerate dependent on the available input power until relativistic effects start to change the parameters of the problem. It never goes over-unity. That is simply a mis-inerpretation of what "break-even" means.Todd

Quote from: Rodal on 07/09/2015 07:26 PMQuote from: WarpTech on 07/09/2015 07:22 PMQuote from: aero on 07/09/2015 06:59 PMThere was some question about growth in field strength in the resonant cavity. This shouldn't be a surprise. This is a resonant cavity, an energy storage device, with some value of Q which may be modest or very high. It doesn't matter in this instance because we are working with only 32 cycles from the start at zero energy. 32 cycles is not enough input energy to build up the stored energy in the cavity to its potential. After 32 cycles the stored energy cannot be be close to saturation. We are expecting the stored energy to reach 1000 times that value. The energy (amplitude) within the cavity is increasing due to the constant energy input. I guess this does underline the need to look at some much longer meep runs.We may be overlooking the significance that I've mentioned several times already. It should only thrust when the thing is charging or discharging. Not when it is in steady state. In this charging phase, we are seeing a Poynting vector doing exactly what it is expected to do. Have a strong DC offset and increasing in amplitude, both of which imply there is thrust. Once it reaches steady state, please don't make the mistake of seeing an oscillating Poynting vector and no DC offset and saying "See, there should not be any thrust!". That would be a huge mistake, IMO.ToddExcellent thinking!Yes, although the analysis may not be exactly what is going on (since we have not yet arrived at a formal proof that is universally accepted) the transient response is more important to understand than the standing wave response which is easier to understand and can be obtained from exact solutions.Just another point of attention.The resulting electromagnetic force, with the fields confined inside cavity, calculated from the poyting vector, is F=-(1/c^2).d(integral_vol(S))/dt, where S=ExHSo a constant, at principle, make no difference to resulting force on whole cavity.Of course, the graph is about one spacial point, and is a extrapolation of few temporal samples, but the resulting force depends of all volume and the distribuition of the fields.If one argument about a net force resulting of vibrations, changing the total volume/format of the cavity, will have to demonstrate it, and it will be very hard because the deformations are very tiny, cyclical and field dependent, resulting a zero net force at principle.

Quote from: WarpTech on 07/09/2015 07:22 PMQuote from: aero on 07/09/2015 06:59 PMThere was some question about growth in field strength in the resonant cavity. This shouldn't be a surprise. This is a resonant cavity, an energy storage device, with some value of Q which may be modest or very high. It doesn't matter in this instance because we are working with only 32 cycles from the start at zero energy. 32 cycles is not enough input energy to build up the stored energy in the cavity to its potential. After 32 cycles the stored energy cannot be be close to saturation. We are expecting the stored energy to reach 1000 times that value. The energy (amplitude) within the cavity is increasing due to the constant energy input. I guess this does underline the need to look at some much longer meep runs.We may be overlooking the significance that I've mentioned several times already. It should only thrust when the thing is charging or discharging. Not when it is in steady state. In this charging phase, we are seeing a Poynting vector doing exactly what it is expected to do. Have a strong DC offset and increasing in amplitude, both of which imply there is thrust. Once it reaches steady state, please don't make the mistake of seeing an oscillating Poynting vector and no DC offset and saying "See, there should not be any thrust!". That would be a huge mistake, IMO.ToddExcellent thinking!Yes, although the analysis may not be exactly what is going on (since we have not yet arrived at a formal proof that is universally accepted) the transient response is more important to understand than the standing wave response which is easier to understand and can be obtained from exact solutions.

Quote from: aero on 07/09/2015 06:59 PMThere was some question about growth in field strength in the resonant cavity. This shouldn't be a surprise. This is a resonant cavity, an energy storage device, with some value of Q which may be modest or very high. It doesn't matter in this instance because we are working with only 32 cycles from the start at zero energy. 32 cycles is not enough input energy to build up the stored energy in the cavity to its potential. After 32 cycles the stored energy cannot be be close to saturation. We are expecting the stored energy to reach 1000 times that value. The energy (amplitude) within the cavity is increasing due to the constant energy input. I guess this does underline the need to look at some much longer meep runs.We may be overlooking the significance that I've mentioned several times already. It should only thrust when the thing is charging or discharging. Not when it is in steady state. In this charging phase, we are seeing a Poynting vector doing exactly what it is expected to do. Have a strong DC offset and increasing in amplitude, both of which imply there is thrust. Once it reaches steady state, please don't make the mistake of seeing an oscillating Poynting vector and no DC offset and saying "See, there should not be any thrust!". That would be a huge mistake, IMO.Todd

There was some question about growth in field strength in the resonant cavity. This shouldn't be a surprise. This is a resonant cavity, an energy storage device, with some value of Q which may be modest or very high. It doesn't matter in this instance because we are working with only 32 cycles from the start at zero energy. 32 cycles is not enough input energy to build up the stored energy in the cavity to its potential. After 32 cycles the stored energy cannot be be close to saturation. We are expecting the stored energy to reach 1000 times that value. The energy (amplitude) within the cavity is increasing due to the constant energy input. I guess this does underline the need to look at some much longer meep runs.

A proofhttp://www.dpedtech.com/VelocityEquation.pdfpage 3

...We are not going to see an EMdrive effect unless some other "teeming streaming background field" is put in that picture of energy_repartition_end minus energy_repartition_start in a closed system.

...but switching off the process we will see the energy tail leave the antenna a little bit before it reaches the walls : during this transient the kick is opposite to the initial one, and we are back at the same inertial rest frame as initially (albeit a little bit translated). Overall, on the on/off cycle we have gained a delta X but not a delta V.

I went outside for a walk and I found out that actually:(Group velocity*Electric Permitivity of Free space)*(Phase velocity*Magnetic Permeability of Free space) = 1(all these multiplied together give exactly one)How about that?

Quote from: Ricvil on 07/09/2015 08:34 PMQuote from: Rodal on 07/09/2015 07:26 PMQuote from: WarpTech on 07/09/2015 07:22 PMQuote from: aero on 07/09/2015 06:59 PMThere was some question about growth in field strength in the resonant cavity. This shouldn't be a surprise. This is a resonant cavity, an energy storage device, with some value of Q which may be modest or very high. It doesn't matter in this instance because we are working with only 32 cycles from the start at zero energy. 32 cycles is not enough input energy to build up the stored energy in the cavity to its potential. After 32 cycles the stored energy cannot be be close to saturation. We are expecting the stored energy to reach 1000 times that value. The energy (amplitude) within the cavity is increasing due to the constant energy input. I guess this does underline the need to look at some much longer meep runs.We may be overlooking the significance that I've mentioned several times already. It should only thrust when the thing is charging or discharging. Not when it is in steady state. In this charging phase, we are seeing a Poynting vector doing exactly what it is expected to do. Have a strong DC offset and increasing in amplitude, both of which imply there is thrust. Once it reaches steady state, please don't make the mistake of seeing an oscillating Poynting vector and no DC offset and saying "See, there should not be any thrust!". That would be a huge mistake, IMO.ToddExcellent thinking!Yes, although the analysis may not be exactly what is going on (since we have not yet arrived at a formal proof that is universally accepted) the transient response is more important to understand than the standing wave response which is easier to understand and can be obtained from exact solutions.Just another point of attention.The resulting electromagnetic force, with the fields confined inside cavity, calculated from the poyting vector, is F=-(1/c^2).d(integral_vol(S))/dt, where S=ExHSo a constant, at principle, make no difference to resulting force on whole cavity.Of course, the graph is about one spacial point, and is a extrapolation of few temporal samples, but the resulting force depends of all volume and the distribuition of the fields.If one argument about a net force resulting of vibrations, changing the total volume/format of the cavity, will have to demonstrate it, and it will be very hard because the deformations are very tiny, cyclical and field dependent, resulting a zero net force at principle.Since a consensus seems to appear that a steady state (spatially integrated) Poynting vector imply no thrust, that asymmetric (integrated) radiation pressure (=apparent net force=thrust) will only occur when such Poynting vector magnitude is increasing or decreasing, and since such a transient (by definition) can't last forever, the question that naturally arise is then : isn't the apparent thrust occurring on the rising transient (between switch on and live steady state) of same magnitude and opposite to the decay transient (between switch off and dead steady state), with a total gained momentum of 0 after end of such on/off cycle (and only a very modest gained delta X) ? I'm not talking here of the long term destiny of the energy that was transferred from on board battery to some wall of cavity, as heat will radiate away sooner or later, and can impart some real net momentum when leaving, as IR waste, but obviously at photon rocket efficiency at most. Argument goes as follow : Poynting vector is flow of energy (power) (per surface). From dead steady state to live steady flow, we are starting to move some mass_energy from one place (on board battery) to another (absorbing cavity walls) through RF injection port. Initially, when power on energy front just left the antenna and is coasting to destination, we see thrust. The energy front doesn't even know yet that the other end of the cavity is closed, if it were open this initial thrust would be gained "for good". But since it is closed, it is absorbed and opposite thrust compensate the initial kick. We have reached live steady state, a constant flow of energy from places A (battery) to places B (cavity walls, some patches more, some less). During this phase, the process is not thrusting, but it is coasting, at (very small) constant velocity that was given by the (very short lived) initial transient kick. We would like to continue on this acquired velocity at no ongoing energetic flow (constant power) cost, but switching off the process we will see the energy tail leave the antenna a little bit before it reaches the walls : during this transient the kick is opposite to the initial one, and we are back at the same inertial rest frame as initially (albeit a little bit translated). Overall, on the on/off cycle we have gained a delta X but not a delta V.I'm assuming here absorbing walls, I know the whole point of the EM drive cavity is to resonate and the walls have low absorbivity (?), that complicates the argument a little bit but don't change the following conclusion (I think). Take whatever resonance, whispering or evanescent modes or real or virtual gradients of refraction, linear or nonlinear effects, point is at the end of an on/off cycle what has changed (from start to end) is the repartition of density of energy within the volume of the hull of a closed spacecraft system (battery is lighter, hot walls are heavier). Not only this will give no long lasting delta V (as would a normal thruster, for which acquired velocity is not lost after on/off cycle) but the delta X gained is limited to the size of the hull : a 100m long star cruiser of 100 tons will move 90m at most when it has blasted its 90 tons of matter+antimatter stored in its nose and have sent the massive photon flux to crash on the inside of its (indestructible and perfectly thermally isolated) rear plate, making it 90 tons heavier by heat alone. For a final acquired deltaV of ... 0 (until the heat is released and thrusting for good, at photon rocket efficiency)For a 2m long spacecraft of 2kg, a lithium battery of 1kg storing E=1MJ in front, discharging in a cavity (whatever shape, or a resistor, would be as good) at rear, is a move of E/c²=10^-11 kg by 2m backward, allowing a move of 2kg by 10^-11 m. The spacecraft can go forward by a tenth of an angstrom, and stop. Just shooting 1MJ as collimated photons (photon rocket) would yield an acquired momentum 3.3*10-3 kg*m/s => 1.6 mm/s for said spacecraft. It takes 6ns to travel 10^-11m at 1.6mm/s : in the closed case, this is the time of free coasting of energy during the transit from initial position front battery to final destination rear plate.We are not going to see an EMdrive effect unless some other "teeming streaming background field" is put in that picture of energy_repartition_end minus energy_repartition_start in a closed system.

Quote from: WarpTech on 07/09/2015 09:12 PM[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important

[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]

Quote from: deltaMass on 07/10/2015 12:04 AMQuote from: WarpTech on 07/09/2015 09:12 PM[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important It's a good thing I'm going to bring this debate to a close. The answer is embarrassingly simple. It embarrasses me, you and everyone else who's been gawking at it and writing about this paradox. Not to forget to mention everyone of us who "thinks" we know rocket science! In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.So then I decided to look at it relativistically rather than solve for an incorrect speed... That's when I said "OUCH!" This is going to hurt.... but you said yourself, you're power source is a battery on board the moving vehicle. Convinced now?Todd

Quote from: Rodal on 07/09/2015 07:26 PMQuote from: WarpTech on 07/09/2015 07:22 PMQuote from: aero on 07/09/2015 06:59 PM.........Just another point of attention.The resulting electromagnetic force, with the fields confined inside cavity, calculated from the poyting vector, is F=-(1/c^2).d(integral_vol(S))/dt, where S=ExHSo a constant, at principle, make no difference to resulting force on whole cavity....

Quote from: WarpTech on 07/09/2015 07:22 PMQuote from: aero on 07/09/2015 06:59 PM.........

Quote from: aero on 07/09/2015 06:59 PM......

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