Mach Effect progress

[hanelyp]

The claim that energy conservation is the result of using propellant is hand-waving.

On the contrary, accounting for reaction mass (in his example propellant) is just doing it right.

There is nothing in his analysis to show a propellantless system will violate conservation.

While his example didn't directly address a reactionless thruster (as opposed to a thruster with distant reaction mass), I am throughly convinced that a reactionless thruster is unable to conserve energy-momentum. A thruster that pushes on external mass has no such disability. But note that the relative velocity of the external mass impacts the power required for a given thrust.

Similarly, I am satisfied that a device that modifies inertial mass without direct side effect is non-conservative in a relativistic universe. On the other hand, a device might be conservative that allows mass to be borrowed between a local matter and distant matter, momentum being carried along with the mass being loaned or repaid. Such transfer of momentum with borrowed mass has strong implications on the performance of a device based on the effect.

Also, if the lack of propellant would yield a conservation violation, we would see this in the case of electric motors and we do not.

Lack of reaction mass would violate conservation. Where the reaction mass is located matters only to the device carrying the force.

[GIThruster]

Well you can believe what you want. Just noting, Jim says that use to Noether's Theorem guarantees conservation, and there are no physicists in almost 2 decades who have made this conservation objection. Given no one here has ever shown a salient distinction between a propellant based and propellantless thruster, I can't agree.

I am throughly convinced that a reactionless thruster is unable to conserve energy-momentum.

M-E thrusters are not reactionless. Have you read the book?

[williatw]

I wish I could say more on this forum, but as GI-Thruster (Ron) points out, I'm now covered by a number of Non-Disclosure Agreements (NDA) that make divulging technical details on what the Eagleworks Lab is currently working on problematic at best. I can say though that we've not had any earth shattering experimental results to date, but have pursued several interesting test campaigns over the last year that have produced data that could support either Dr. Woodward's Mach Effect (M-E) or Dr. White's Quantum Vacuum Fluctuation (QVF) thruster conjectures. I think though that our upcoming home-grown 2.45 GHz resonant cavity test series we are just now starting has the potential of showing which one of these thruster conjectures is the more inclusive one. However as in all things experimental, once you have your data, someone always comes along and points out what you didn't do to clinch the deal either way.

Thank you for the update, sure most of us here are awaiting you and Dr. White's results with bated breath.


http://www.huffingtonpost.com/2013/09/2 ... 12175.html


Number Of Alien Planets Confirmed Beyond Our Solar System Nears 1,000, Data Shows

...And it's becoming more and more apparent that this diversity is stunning. Scientists have found exoplanets as light and airy as Styrofoam, for example, and others as dense as iron. They've also discovered a number of worlds that appear to orbit in their stars' habitable zone — that just-right range of distances that could support the existence of liquid water and thus, perhaps, life as we know it.
But the search continues for possibly the biggest exoplanet prize: the first true alien Earth. Kepler was designed to determine how frequently Earth-like exoplanets occur throughout the Milky Way, and mission scientists have expressed confidence that they can still achieve that primary goal. So some Earth analogs likely lurk in Kepler's data, just waiting to be pulled out...

[93143]

Remember, this is a maximum. In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.

If that were true, then this case would not apply in the case of a thruster on a swing arm, and it does; so obviously the propellant is not the salient explanation.

If you put the whole propellant tank on the end of the arm with the thruster, what I said above would apply. In the case where the propellant is continuously fed in from somewhere else, the mechanism is a bit different, but the principle is the same if you think about it, and it gives you essentially the same answer.

Okay, time for Part 2:

Take a pair of thrusters (to avoid undesirable bending moments on the shaft) mounted tangentially on a flywheel which rotates at a rate such that the speed of the outer edge is V. Propellant is fed to the thrusters continuously from an external source. They have an exhaust velocity v_exh and a combined mass flow rate mdot, resulting in a jet power (the effective input to this mechanical system) of

  P_jet = 0.5*mdot*v_exh^2

(Note that this is after inefficiencies in the thrusters; ie: P_jet = η*P_in, where P_in could be electrical, chemical, nuclear thermal...)

The outputs are the exhaust energy rate and shaft power:

  P_exh = 0.5*mdot*(V-v_exh)^2 = 0.5*mdot*(V^2 - 2*V*v_exh + v_exh^2)
  P_sh = F*V = mdot*V*v_exh

This results in a total output power of

  P_exh + P_sh = 0.5*mdot*(V^2 - 2*V*v_exh + v_exh^2) + mdot*V*v_exh = 0.5*mdot*(V^2 + v_exh^2)

which looks like a violation of conservation. In fact, for V > 0.5*v_exh, P_sh > P_jet and the device looks like a free energy machine.

But wait - where's mdot coming from? Of course it has to be pumped up to the kinetic energy it has when fed into the thrusters. So the pumping power is

  P_pump = 0.5*mdot*V^2

This is the theoretical minimum (no losses) and comes directly out of the useful work produced by the device, in this case the shaft power:

  P_sh - P_pump = mdot*V*v_exh - 0.5*mdot*V^2

which is quadratic in V, exhibiting a maximum at

  mdot*v_exh - mdot*V = 0

or V = v_exh. Sound familiar?

The maximum in question is of course

  P_sh[@V=v_exh] = mdot*v_exh^2 - 0.5*mdot*v_exh^2 = 0.5*mdot*v_exh^2 = P_jet

So you cannot actually get more power out of this system than you put in. It is at most (for non-electric thrusters) a silly-looking heat engine.

The total power output, including the energy of the exhaust and with pumping power accounted for, is

  P_out = P_exh + P_sh - P_pump = 0.5*mdot*(V^2 - 2*V*v_exh + v_exh^2) + mdot*V*v_exh - 0.5*mdot*V^2
  P_out = 0.5*mdot*v_exh^2

or exactly the same as the input power.

So what we have here is that the kinetic energy lost by expending propellant in the spacecraft case is replaced by the pumping power necessary to supply propellant in the flywheel case. It's the same thing in two different forms.

If you had used the thrust/power FOM for a stationary ion thruster

This sounds distressingly like something you said earlier, that was wrong:

If you take any thrust efficiency figure of merit in force/input power (N/W) and allow the thruster to accelerate at any rate over an arbitrarily long period of time, you can show that all thrusters will appear to violate conservation at some time in the future. This is because thrust efficiencies are stationary. Energy and thus power are not invariant so GR requires anyone doing this calculation will seem to uncover a conservation violation. I think it was chris who stated the issue quite well, as well as one of the folks over at NBF (the physicist there, not GoatGuy), that power in is linear, but power out would be quadratic, so this MUST yield a conservation violation. The remedy is to note that the dynamical relationship for stationary thrust can only be used while the thruster is stationary.

This is not true, as I demonstrated in the previous example. I used "stationary" values for input power and thrust: η*P = 0.5*mdot*v_exh^2 and F = mdot*v_exh. Those are invariant. And I got conservation.

Okay, only the four-force is Lorentz invariant. But the three-force is Galilean invariant, and for a model problem in Newtonian mechanics that's all you need.

Now, it is possible to use an accelerating reference frame and get the right answer (you have to assume a fictitious force, which naturally enough does fictitious work), but the fact that the thruster itself is accelerating has nothing to do with this... as long as the acceleration is insignificant compared with that of the reaction mass being accelerated in the thruster, which is generally a pretty good assumption. Relax that assumption and you just get a slight acceleration dependence of the thruster performance parameters, which complicates the math greatly but gives you basically the same answer - energy conservation.

And you don't need any form of relativity to do the math in the non-inertial reference frame. It's still Newtonian. You never need GR to get energy conservation in a model problem. It's built into Newtonian mechanics and will always be observed if you use the equations properly.

BTW, if you really feel up to it, you can do a very similar calculation using the active mass of an M-E thruster, and calculating the action of the mass using it as propellant. After all, M-E thrusters are not technically "propellantless" but rather they use a "recycled propellant" -- the active mass of the thruster.

No, that's exactly the mistake GoatGuy made. His control volume was too small.

Do you understand the concept of a "black box"? It doesn't matter what's going on inside; it's what it does that matters. In the case of an M-E thruster, you can turn it on, then off again, and you've got a spacecraft of exactly the same mass as before moving at a different velocity, meaning its kinetic energy is (in general) different. No part of the spacecraft can serve as a 'sink' for that energy difference because it's all exactly the same as it was before the thruster was used, with the exception of whatever energy transfer happened to operate the thruster. That energy transfer is invariant (e.g. chemical energy from a fuel cell) and thus cannot be matched with the spacecraft's kinetic energy change except by arbitrarily selecting a reference frame.

To get conservation with an M-E thruster, you need to consider the distant mass that it is ultimately reacting against. Without that you really do get an apparent conservation violation no matter what you do.

Also note, that if the exclusion of propellant were an explanation for a real conservation violation, then both linear and circular electric motors would go overunity and they do not.

You can't just consider the motor and its payload; you have to account for everything that might experience a force as a result of its operation.

In the case of an electric car, the reaction mass is the Earth. And it's true - without accounting for the energy gained/lost by the Earth, you will get an apparent conservation violation in the general case. The fact that you can usually analyze an electric car without reference to the Earth's kinetic energy is due to the Earth being much more massive than the car, so its change in velocity (and thus kinetic energy) is negligible during the operation of the car and its local surface velocity can be used as a reference frame for the calculation. This is only almost right, but it's close enough. Use a different reference frame and you do need to account for Earth's motion - for instance, if you're driving alongside a car and it begins to brake, it appears to be accelerating by doing negative work. What's actually happening is that the Earth is doing work on it - from your perspective.

The claim that energy conservation is the result of using propellant is hand-waving.

No, it's called Newton's Third Law. And it's a beautiful thing, when you finally get it...

More to come, when time allows.

[hanelyp]

To get conservation with an M-E thruster, you need to consider the distant mass that it is ultimately reacting against. Without that you really do get an apparent conservation violation no matter what you do.

Yup.

Also note, that if the exclusion of propellant were an explanation for a real conservation violation, then both linear and circular electric motors would go overunity and they do not.

You can't just consider the motor and its payload; you have to account for everything that might experience a force as a result of its operation.

In the case of an electric car, the reaction mass is the Earth. And it's true - without accounting for the energy gained/lost by the Earth, you will get an apparent conservation violation in the general case. The fact that you can usually analyze an electric car without reference to the Earth's kinetic energy is due to the Earth being much more massive than the car, so its change in velocity (and thus kinetic energy) is negligible during the operation of the car and its local surface velocity can be used as a reference frame for the calculation.

A good model for distant reaction mass. It's worth noting that the power needed for a given thrust/acceleration is proportionate to the velocity relative to the reaction mass, which is not constant in this model.

On a deeper level, mass fluctuations are non-conservative unless accompanied by something like mass/momentum coupling to distant mass. One consequence I see for such a coupling is that an element driven with oscillating mass, as in an ME thruster, would experience (to borrow a startrek term) "continuum drag" as momentum is exchanged between the active mass element and the distant mass. Taking the total mass of the system as defining a reference frame, when the active mass increased it would slow, when the active mass decreased the remaining mass would speed up.

[93143]

The M-E equation appears to obey Lorentz covariance. It does not contain any direct reference to any velocity state other than that of the local active mass. This is to be expected from an equation derived from a model of inertia, since inertia does not reference any absolute velocity (only its time rate of change, if Woodward and Sciama are correct about Mach's principle).

Thus I would expect that an M-E device would be observed to operate in the same way regardless of its velocity with respect to the bulk of the distant universe. Experiments seem to bear this out - AFAIK the lab results don't seem to depend on the device's orientation, or the time of day, or the season. There would be better evidence if the thrust efficiency were higher...

But the distant universe is supposed to be what the M-E thruster ultimately pushes on. So how does the above square with using the thruster's Hubble sphere (cf. Paul March) as reaction mass?

Part 3:

Consider a Mach-effect device attached to a spacecraft, the combined vehicle having a mass m and traveling at velocity v_v. At time t = 0 and vehicle velocity v_v = v0_v, the device is switched on and begins to produce a constant force F. For clarity, and without loss of generality, we assume that F and v_v are in the same axis. The vehicle's rate of increase in kinetic energy is

  P_v = F*v_v = F*(v0_v + F*t/m)

and the vehicle's kinetic energy is

  E_v = 0.5*m*v0_v^2 + F*v0_v*t + 0.5*F^2*t^2/m = E0 + F*(v0_v*t + 0.5*F*t^2/m)

This is the infamous quadratic kinetic energy curve, to be contrasted with the linear increase in energy consumed resulting from a presumed constant power input to the device (indeed, by Galilean invariance, the power input cannot be dependent on an observer's velocity, as it would have to be to match this output). Certain critics have made a big deal of this, as if this were where the story ended.

But per Newton's Third Law, the thruster also exerts a force -F on the far-off active mass (FOAM). If the FOAM has a mean velocity v_u, its total rate of change of kinetic energy is

  P_u = -F·v_u

The total kinetic power resulting from the operation of the drive at a given moment is thus

  P = F·v_v - F·v_u

As with the previous calculations, for clarity and without loss of generality we simplify this so that F, v_v and v_u are all on the same axis:

  P = F*v_v - F*v_u = F*(v_v - v_u)

This is the minimum power requirement for an ideal M-E thruster, assuming energy is conserved. Note that while it is linear with the difference between the velocity of the vehicle and the velocity of the FOAM in the axis of thrust at a given moment, it is independent of any absolute value of velocity, so it respects Galilean invariance and can in principle result in energy conservation. Since v_v is changing, we get

  P = F*(v0_v + F*t/m - v_u)

which is dependent on both time and initial vehicle velocity.

But we established above that the Mach effect equation appears to obey Lorentz invariance. So P should be totally independent of both v_v and v_u. There is only one solution:

  v_u = v_v - B in all cases, where B is a constant.

If the behaviour of the thruster is independent of velocity or orientation, that implies that the effective mean relative velocity of the far-off active mass must be directly tied to that of the thruster no matter what. How exactly this is compatible with the idea that the rest of the Hubble volume is the reaction mass, I'm not totally sure - I have a half-formed idea or two, but I'd have to study the matter more closely before I'd be confident in articulating anything specific.

Note that this means

  P = F*(v_v - v_v + B) = F*B

resulting in a system kinetic energy of

  E = E0 + F*B*t

Which is linear, with a time-dependent term that corresponds exactly with the cumulative energy productively expended in the operation of the thruster. In other words, energy is conserved. Please note that this is not the only case where energy conservation is respected by a Mach-effect thruster; it is merely the only case where conservation is respected by a Mach-effect thruster with invariant input power and thrust.

(Notice that if we set B = 0 as seems reasonable, there is no net increase in kinetic energy, and thus (barring non-obvious GR or QM weirdness or something like that) no upper limit on the thrust efficiency of an ideal Mach-effect device...)

[hanelyp]

If the behaviour of the thruster is independent of velocity or orientation, that implies that the effective mean relative velocity of the far-off active mass must be directly tied to that of the thruster no matter what. How exactly this is compatible with the idea that the rest of the Hubble volume is the reaction mass, I'm not totally sure ...

I've come to the same conclusion. I see some possible takeoffs from this:

- the behavior of the thruster is not invariant with velocity relative to mass of the hubble volume. This would seem to have problems with experimental results. If it operated as postulated in my above post, there would be an effect on capacitors carrying AC that I'd expect would be noticed.

Or

- The effect preferentially couples with mass by relative velocity. This is at odds with how ME is commonly described. But it would allow for the described performance projections without great difficulty in momentum-energy.

This all presumes ME works.

[AcesHigh]

93143, some good very "complete" posts from you lately, including lots of equations. You should post those on Nasa SpaceFlight forums, as well as NextBigFuture when they post something about Mach Effect again and GoatGuy starts dissing ME as pseudoscience.

[TallDave]

To get conservation with an M-E thruster, you need to consider the distant mass that it is ultimately reacting against. Without that you really do get an apparent conservation violation no matter what you do.

Apparently we're never going to stop explaining that.

[Spudd86]

I have a question. If we assume that the Mach Effect thusters work as Prof Woodward claims could we actually build one that gives significantly higher thrust than his test thrusters with current materials, if he is in fact right about them would we likely end up like the space elevator, we know roughly how to build it and what is needed to do so but we don't have the right stuff to actually make it out of?

[GIThruster]

There are several advanced materials that are on the table and as Woodward just recently posted, one of the other physicists involved at Fullerton is looking at what would be a good new next step past pure PMN.

In addition to these materials is the complication that each of them have complex trades based upon whether they're sintered crystal or nano-crystal, whether they're thin or thick film from one of half a dozen deposition methods, or if they're single crystal, which is the most expensive form but has important benefits. It also matters what frequency each of these would be run at. For large arrays, the dissipation factor matters when designing active thermal management like heatpipes, etc. Probably we don't know what the first commercial iterations will look like but there are plenty of options.

If we can see much greater thrust than mass of the thruster and power system, I doubt there will ever be an interest in designing any sort of transport system that requires expensive infrastructure. Individual ships are a better solution in many, many ways. For example, since they're not throw-away, they're relatively cheap compared to single-use spacecraft. Ships that don't follow a ballistic trajectory like rockets don't require the same sort of mission control and are more like commercial aircraft. MET's have no exhaust so they can fly from any parking lot large enough to hold them. And since there are no dangerous fuels to handle, the infrastructure for a MET spacecraft should actually be less than for commercial aircraft. High thrust/mass ratio METs make things like the Millenium Falcon and other flying bricks extremely practical.

However, given much more modest early commercial prototype success than a high thrust/mass ratio, meaning a thruster that with its power system weighs more than the thrust it generates; a first commercial application is likely to be something like satellite station keeping. Satellites that require thrusters to keep them on orbit like ISS, would be an excellent first application. NASA currently spends 500 million dollars each year to keep ISS in place, and if it were equipped with modest thrust MET's, that cost could be completely eliminated. Likewise, all commercial sats would so benefit, especially those on lower orbits where they strike more atmosphere. New kinds of sats like sat tenders, or tugs could be designed and launched with chemical rockets into low orbit and tow older sats to higher orbits and so extend their lifetimes by decades. Tugs could be used to clean the orbital pathways of junk. And of course deep space probes that would otherwise use Ion propulsion would greatly benefit from low thrust MET's. In fact, with a tiny space reactor and some low thrust MET's, one could replace the original multi-billion dollar JIMO design with a cheap replacement, since even now METs require so much less power than Ion propulsion. We could see a single Falcon 9 launch half a dozen deep space probes sent to do a detailed survey of the asteroids for example. And done so cheaply, we could see dozens of such launches to get lots of detailed info. Look for early commercial ventures to take such an interest:

http://www.planetaryresources.com/

[vnbt4]

I was looking for some warp drive news and I found this talk by Harold White. It is from November.
https://www.youtube.com/watch?v=9M8yht_ofHc

[AcesHigh]

thanks for the video. Very interesting. Specially the part of Q-Thrusters, where although I am somewhat skeptical, Dr White claims they got some pretty good results in N/W.

(the warp drive part itself is just too early yet so it´s pure theory that won´t see application in the next 100 years)

[djolds1]

thanks for the video. Very interesting. Specially the part of Q-Thrusters, where although I am somewhat skeptical, Dr White claims they got some pretty good results in N/W.

(the warp drive part itself is just too early yet so it´s pure theory that won´t see application in the next 100 years)

If the theory works we should be able to see concrete results in 40-50 years, maximum (special relativity/ e=mc2 of 1905 to Trinity of 1945, 40 years).

If its dragging beyond that we're dealing with metaphysics as we have been in cosmology for the last 40 years, not theory showing actual and validated results.

[AcesHigh]

If the Q-Thrusters work, I really don´t care that much about warp drive in the next 50-100 years. The solar system is pretty huge. There is a lot of space to colonize, mine, build bases, etc, before we run out of resources, etc.