A FINAL WORD on the thermodynamics of 'Mach Thrusters'.

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chrismb
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A FINAL WORD on the thermodynamics of 'Mach Thrusters'.

Post by chrismb »

CaptainBeowulf wrote:Is there any way that as the M-E ship accelerates, Q would grow for it, and Q on the distant matter would grow too, resulting in the two summing to zero?
Is there an alternative mathematical description that makes sense for such a hypothetical M-E thruster? Or do you get to the same over-unity situation that GoatGuy did for years?
CaptainBeowulf wrote:Note: Of course if you were just observing a M-E thruster accelerate without ejecting any propellant it would appear to reach the point where it would go over-unity, because you wouldn't know what it was pushing against. There would be no conservation of momentum.

I ask about an alternative mathematical model because there is, for example, no "m kg/s of material" being ejected - but there should (according to the theory) be force exerted on the distant matter.
This has arisen in another thread analysing the general case of a thruster based on propellant:

viewtopic.php?t=4190

In that thread, a full set of equations were prepared from first principles which, unsurprisingly, came down to showing that the energy expended was the kinetic energy of the propellant, frame-independent.

Capt B asks the next obvious question to ask - how does this shape up for hypothesised propellantless systems.

It seems to look like this;

Take a test mass, M kg, travelling at velocity Q. By means unknown, somehow it reacts with the rest of the universe, ROTU, so as to accelerate the test mass by an accelerating force, F, whilst the ROTU of mass U at velocity V is caused to accelerate in the opposite direction by the 'remote' reaction force, -F. Now consider the energy required for this in a time sample of Δt at t=0.

Impulse Equations (Force = rate of change of momentum):
Force on rest-of-the-universe = -F = -d(U.V)/dt
Force on test mass = F = d(MQ)/dt. For a small Δt; ΔQ=F/M=-U.V.Δt/M

Conservation of momentum:
-U.ΔV = M.ΔQ


Do the maths!...

A: Power, by rate of change of kinetic energy, at t=0;
Test Mass---
ΔKE = ½ M.(Q+ΔQ)² - ½ M.Q = +M.Q.ΔQ + ½ M ΔQ²
lim ΔQ->0; ΔQ²=0
So ΔKE = M.Q.ΔQ = M.Q.U.V.Δt/M = Q.U.V.Δt
Rest Of The Universe---
ΔKE = ½ {(V+ΔV)² – V²}.U = V.ΔV.U + ½ .ΔV².U
lim ΔV->0; ΔV²=0
ΔKE = V.ΔV.U = V.U.(ΔV/ΔQ).ΔQ = -V.U.(M/U).U.V.Δt/M = -V².U.Δt

TOTAL ΔKE = U.V.Δt.(Q-V)



B: Power, by rate of expended work;
Test Mass---
Rate of Work = U.V.Δt.(Q+ΔQ) = Q.U.V.Δt + Q.U.V.Δt.ΔQ
lim Δt->0; ΔQ.Δt=0
So Work = Q.U.V.Δt (as per consideration by KE)
ROTU
Rate of Work = -U.V.Δt.(V+ΔV) = -V².U.Δt (as per consideration by KE)

TOTAL Rate of Work = U.V.Δt.(Q-V)



The physical meaning of these equations are that the rate of change of KE of 'the rest of the universe' is its rate of change of KE (!) [average a change from a datum zero to -V².U over a segment of time and it is -½.V².U]. Also that the rate of change of KE of the propelled vehicle (from its initial frame) is proportional to its velocity in that frame.

These conclusions are trivial. However, they do allude to a simple analysis that is more 'familiar'.

In the proposed 'propellantless' scenario, some force is generated between the universe (largest mass) and the thrusted vehicle (relatively ~zero mass).

It is a common and well-understood outcome in physics that when two bodies of very dissimilar mass interact, that one does all the 'changes' and the other does little. Only when the masses are similar do more complex interactions occur.

e.g. a marble hitting a bowling ball, the bowling ball experiences almost no KE change whilst the marble reverses its direction.

The fraction of the 'output' KE going to one or other is defined in m/M+m and M/M+m. So the light one comes away with almost all the change in KE, and the heavy one undergoes an insignificant change.

This is easy to consider when thinking of the momentum versus the KE - as the mass increases so the velocity of the heavier one decreases (for a given force - force is the change of momentum and momentum is the product of mass and velocity). But as the velocity comes down, so the square of the velocity comes down more quickly. So as the mass tends to 'very large', the fraction of KE changes going on in the system becomes 'very small' for the massive object.

A virtually insignificant mass will therefore receive essentially all of the energy expended in the system as an increase of its KE. The massive object will not 'benefit' from an increase of KE from the energy expended.

OK, so the obvious analogy to a hypothesised propellantless craft is a motor car driving along a road. As it accelerates from its originating inertial frame, almost all the mechanical traction applied to the road is converted to the car's KE (some of that is then lost in other processes - discount the losses in this analysis, for comparison with hypothesised thrusters). It would be untrue to say that the KE of the Earth does not change, but it is truly insignificant. It is so close to unity that one would actually calculate the mechanical and thermal efficiency of the car by how much of the chemical energy of the fuel goes into increasing the car's KE. No-one would dream of writing a conservation-of-momentum equation for the Earth against which the car pushed, because the fraction of the energy expended that went to the Earth's KE would be <<ppb of that expended by the car's engine.

So here one can see exactly the same scenario with speculated propellantless thrusters, just as a motor car is thrust along without propellant but instead uses a force to thrust against a much larger mass.

Where does this leave an analysis of speculated propellantless thrusters? Simply, that because all the energy expended by the thruster would go into its own KE, so it can be seen that the work done on the thruster with a constant thrust would be, simply, proportional to its distance travelled (work=force x distance).

So this is real-simple! The consequent behaviour of the Universe can be ignored to all practical purpose in this theorised scenario when considering energy expended, because any change of momentum of it will come along with a diminishingly small velocity (thus a KE change that can be disregarded).

Example; a hypothesised propellantless thruster, that starts at rest wrt ROTU, generates 1N of thrust will generate 1J of ΔKE for each meter it travels.

A hypothesised propellantless thruster that generates 100N and travels 10 billion kilometres from ROTU-stationary will have generated 1 PJ of ΔKE. It would be trivial to then work out its velocity at the end of that 10 billion km trip to the edge of the solar system;

Say it has a 100kg mass; 1^15J = ½.100.V^2
So, V= ~4,472 km/s.

(No need to accommodate relativity at these speeds.)

So the actual mechanics of hypothesised propellantless thrusters turns out to be really trivial and there are no immediately apparent violations!

HOWEVER!!!

The above acceleration, to V= ~4,472 km/s, takes place from a 'standing start' relative to the ROTU.

The question is; what is the actual relative velocity difference between a thruster and ROTU, and does it make a difference?

Here's the problem; what cannot be done is to attempt to treat the thrust of hypothesised propellantless thrusters as being like a conventional rocket, because it would actually be more like the grip of a car tyre. The propellant in a conventional rocket has a much higher capacity to do work once it is at speed (this is that extra term that cancels out, in the other post), this is the key difference.

Now, because the hypothesised propellantless thruster is reacting with the ROTU the physics relate to the inertial frame of the ROTU. Imagine a thruster moving at 1,000m/s wrt ROTU which can somehow generate a 1N propellantless thrust reacted against the ROTU. This is made bold to highlight that in a normal rocket system the reaction is against gases that are already moving in the same frame as the rocket. THIS IS A HUGE DIFFERENCE!!! What it means is that per second that 1N thrust will generate 1kJ of energy in the ROTU frame.

This shows up in the equations above – see how the power expended is equal to the PRODUCT of the magnitude of the rate of change of momentum (of either ROTU or the thrusted mass) AND the difference in velocities of the two parts.

Don't believe it? OK, take a look at the motor-car-on-a-road scenario, which matches the hypothesised propellantless thruster scenario (big mass pushes small mass, thrust without propellant, intermediate force between tyres and road).

Now imagine that this wheel-driven car (in a vacuum and with no rolling losses) is rolling along at 1,000m/s. No power is required and it will carry on, just like a space vehicle. Now it goes to accelerate from 1,000m/s. It accelerates from 1,000m/s to 1,010m/s in 1 second. The question is – does this take MORE power from the engines to accelerate in this interval than it does from zero to 10m/s?

Answer – NO: It takes A HECK of a lot more! Let's say that the torque required on a driven wheel to accomplish this acceleration is 100Nm. When stationary up to 10m/s let's say the tyre goes from zero rad/s to 20 rads/s. Power on a rotating axle is its rotational velocity times torque, so as it accelerates to 10m/s it hits a power input to the axle of 2,000W. Averaged from zero power at zero speed, the average power from 0 to 10m/s is 1kW. 1kJ total.

Now, at 1,000 m/s the tyre is now rotating at 2,000rad/s. It still needs 100Nm to achieve the same acceleration rate, because this is F=ma at the tyre/road interface. But at 1,000m/s the engine has to develop 200kW and at 1,010m/s it is generating 202kW, average 200.5kW. 2005kJ total.

The tyre grip on the road would be analogous to 'inertial engagement with ROTU'.

So that result above, TOTAL Rate of Work = U.V.Δt.(Q-V), is all that is needed to understand that such a hypothesised thruster is not frame-independent, and it would not work at all the same way as a conventional rocket. It should ALSO be noted that Q-V has a sign, and it will reverse (viz. will generate power) if the hypothesised thruster slows down, wrt ROTU.

Revisit the hypothesised propellantless thruster that generates 100N and travels 10 billion kilometres:

Now assume that when the hypothesised thruster started its journey that it was already travelling at 4,472km/s wrt ROTU. How does this affect the power consumption?

If the thruster accelerates so that its relative speed wrt ROTU increases, then it will need to generate an additional x3 power expended (x2^2 – 1) than if it were starting from stationary, viz. it would have to generate 3 PJ during its 52 day trip. It would have to generate progressively more power. At the start of its trip it'd need ~500kW, and at the end of its trip it would have to be generating 100N x 8944m/s = ~1MW.

However, if it were to 'accelerate' to its 4,472km/s by decelerating wrt ROTU, then it would have to dissipate 1 PJ of energy during the trip (~-500kW to start with, dropping to zero once it has reached 4,472km/s relative to its starting frame). The analogy with the car on the road is that instead of applying +100Nm from the engines, that -100Nm of braking is used. The brakes will have to dissipate the car's KE.

So these are the physical ramifications if a hypothesised propellantless thruster mechanism can be created.

To generate 'free energy' here on Earth then all that is needed is to accelerate a mass wrt Earth but that slows down wrt ROTU. Energy will somehow have to be dissipated through the coupling mechanism that connects it to ROTU. Energy could be extracted in this process. This mass now has a positive KE wrt Earth, so it can then be 'slowed down' relative to Earth's velocity by means that, again, extracts energy from that process.

This is not yet a thermodynamic violation because the net of what is happening is that the Earth will have been slowed down towards the ROTU velocity, and the energy (in both of those steps) came from the reduction of the Earth's KE wrt ROTU.

Better still, miss out a moving thruster altogether and bolt it to the Earth itself. The inertial coupling mechanism can then be activated so as to decelerate the Earth wrt ROTU and generate enormous amounts of power (at the risk of knocking the Earth out of orbit!).

So, would such an inertial coupling be a violation of thermodynamics? Despite first appearances, it looks like it would NOT NECESSARILY be a violation of thermodynamics!

...Does it feel 'non-physical' that these thrusters could start dissipating petawatts of power if they were bolted directly to the Earth, so that they simply slow the Earth down directly, wrt ROTU, as a means to generate power?

… hmmmmm …..

GIThruster
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Post by GIThruster »

I have been extremely specific with you chris, that M-E thrusters create constant stationary thrust, just like any thruster, and that therefore they do not cause constant acceleration. They act just like any other thruster.

You proved it yourself with your example of the car. Not much left to say, chris.

I will however reiterate, that in the case of exotic matter with negative inertia, the maths do seem to indicate a true "overunity" condition will arise for reasons completely unrelated to all what you've been considering. If you want to look up the work of Dr. Robert Forward you'll find better info on that issue.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

chrismb
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Post by chrismb »

ME thrusters have not yet been proven to produce any sort of thrust. GIT should resist reifications in this thread. It is an attempt to put at least some degree of formalisation on what an ME thruster would look like, if one were to ever be created.

There is no indication in the above formalisation of any 'over-unity' circumstances wherein the input energy would be less than the KE gain in the ROTU frame.

GIThruster wrote:... M-E thrusters.. just like any thruster
The above proves an ME thruster would be frame-dependent on the inertial frame of the ROTU, and the required power input would be increasingly positive as it accelerates wrt ROTU, and would produce an energy excess (i.e. would need to dissipate power) if it decelerates wrt ROTU.

Maybe this is what GIT is saying, but it is difficult to understand if GIT is hoping that 'stationary thrust' relates to 'stationary wrt ROTU', which is true, or to an arbitrary frame, which is false.

'Any other thruster' is frame-independent. Propellantless thrust would be ROTU dependent.

GIThruster
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Post by GIThruster »

simply hopeless. . .
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

chrismb
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Post by chrismb »

GIThruster wrote:simply hopeless. . .
GIT to explain the logic of this statement?

GIThruster
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Post by GIThruster »

chris, you're the only guy I've ever met who can give fair warning that he's going to post some Earth-shaking, ground-breaking analysis and come up with this sort of wishy-washy nonsense. I'm just skimming it and it's not worth the time for a detailed read, but honestly, if you're AGREEING that M-E theory does not necessarily entail a violation of conservation, then I have to ask what was the point in the post?

If you want to understand M-E theory, you need to get conversant on Mach's Principle and then on all the papers. Elementary analysis by someone wholy unqualifed and deliberatly ignorant is not worth anyone's time. How is it you still don't get it? One more time:

You're not a physicist.

There really isn't any reason I can see from your post that anyone should take the time to read it. No value added, chris. Sheesh. Just saying, all the physicists are satisfied with the back-reaction explanation supplied by Mach's Principle and Wheeler-Feynman Absorber Theory. So why would anyone take an interest in high school algebra poseted up by an habitual complainer and troll? I don't think you understand much the way you come across online, chris.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

kcdodd
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Post by kcdodd »

He's demonstrated more physics here than you have, GIT.
Carter

chrismb
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Post by chrismb »

GIThruster wrote:There really isn't any reason I can see from your post that anyone should take the time to read it. No value added, chris. Sheesh.
So, does GIT agree with all the conclusions, or which conclusions does he not agree with?

MSimon
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Post by MSimon »

kcdodd wrote:He's demonstrated more physics here than you have, GIT.
I'd have to agree with your assessment. So far.

Chris,

Loved your pointing out the relative mass question and the simple way you have of explaining what it means.

Thrust without propellant would be rather nice. But you still need energy.

And in fact we can come rather close to that already. Ion engines.
Engineering is the art of making what you want from what you can get at a profit.

CaptainBeowulf
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Post by CaptainBeowulf »

Thanks for posting a detailed reply Chris. I'm pretty busy right now so I didn't even notice it at first, and don't really have much to add at present.

I will note that Andrew Palfreyman and some of the others who have worked with M-E effects in the past went and set up a Yahoo forum last October, generated 535 posts in about 2.5 months, and then apparently lost interest and went quiet. However, they've posted some mathematical analysis similar to what Chris did (and some similar analogies too - instead of cars, Oxfordian and Cantabrigian carts). Similarly, they seem to have thought that if a M-E device works it would appear locally to violate conservation of energy (and be used as a power source) but would in fact conserve it, and not violate thermodynamics, with regard to the universe as a whole.

There are some differences though, and a great deal of talk about the rocket/M-E device strapped to a flywheel idea. Finally, there are a couple of interesting oddball experimental attempts I hadn't heard about before, with a replication of one apparently planned until the group went quiet. I found them last month and read many of their posts before work got busy again. The site is here:

http://tech.groups.yahoo.com/group/spacedrives/

If you find the subject interesting Chris, you might get something out of reading their posts and math and cross-checking, falsifying or verifying some of it.

CaptainBeowulf
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Post by CaptainBeowulf »

Main difference between Chris' interpretation and that of Andrew et al., so far as I can tell without reviewing again properly:

Andrew et al. still seem to think that the idea of strapping a M-E device to a flywheel could be used to generate energy locally by reacting against the rest of the universe.

Chris' interpretation seems more nuanced in arriving at the position that to generate 'free energy' on Earth what is needed is to accelerate a mass with regard to Earth but that slows down wrt rest of the universe.

Seeing as how I can't do the GR Lorentz transformations for myself in a manner I would feel remotely confident in, I'll just have to accept that on the one hand we have some physicists who believe that any rocket can be made to appear to go over-unity unless you apply GR .

Then, we have Andrew et al. and Chris, both parties which present reasonable mathematical models (though simpler than GR models) which don't appear to result in over-unity with a conventional rocket.

Andrew et al. and Chris then come up with somewhat different models of how you could extract energy reacting against the rest of the universe using a M-E device.

The overall positive I see here is that, no matter how wrong we consider one interpretation or another to be, each group discussing M-E in detail seems to find that there are not necessarily thermodynamics violations.

So... I'll definitely keep watching to see if Woodward can generate a stronger signal with better materials and then build a self-contained device. If a M-E device does turn out to work, practical experimentation could be done to start to clarify some of these other points of contention surrounding the theory...

chrismb
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Post by chrismb »

There would be no way to recover energy from a circular system given the mechanics equations above.

Just think of the whole situation as being in a car. The tyre is this speculated connection with the mass of the universe. One way or another, creating a 'thrust' (torque on the tyre) achieves nothing in a cyclic system, it performs a conversion of some energy to KE by creating linear momentum.

But it'd be simpler than anything complicated - just make the whole of planet Earth a 'Mach Thruster' and decelerate it. Energy should then be recoverable.

Just imagine if a group of intelligent beings lived on a car rolling along an infinitely long road with no rolling resistances, but they had ran out of energy and wanted a means to generate power to carry on their lives in this [bizarre!] car. If they developed a means to contact the road surface with a generator, say with a wheel attached to the spindle, they could hold it out of the car window and hold it in contact with the road. Then they'd get energy out of it (at the expense of slowing the car down).

In fact, this mechanics description analysis might well lead to a very clear and distinct outcome that would give an absolutely clarion-clear signal in an ME experiment that any observed thrust is from such a mechanism, rather than any erroneous signals....

...can anyone spot what it is?.. (it'd be a bigger 'oh yeah!' realisation if one figures it out for oneself...)

kcdodd
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Post by kcdodd »

I was ok with the conclusion until I just realized something. The problem is two parts. First, no thruster can change the total momentum and energy of the universe from an inertial reference frame, right? Hopfully that is agreed. Second, if a thruster acts on the whole universe, then the whole universe is in a non-inertial reference frame, INCLUDING any observers. So, the total energy and momentum of the universe will NOT appear to be constant to physical observers, because you are measuring the total energy in different reference frames as the thruster operates. It would only appear to be constant for some observer "outside the universe" where the thruster does not react against anything.
Carter

GIThruster
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Post by GIThruster »

I think Carter is correct. Too I would note that although chris hasn't said what he has in mind, he seems to be assuming 2 things that are both incorrect. The first is that the M-E thruster is a transducer, when in fact it is a transistor. The second is the transducer can have its action reversed like turning an electric motor into a generator. Neither of these things are true. Of course one can only guess what chris is thinking so we'll wait and see what he proposes next.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

chrismb
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Post by chrismb »

GIThruster wrote:The first is that the M-E thruster is a transducer, when in fact it is a transistor. The second is the transducer can have its action reversed like turning an electric motor into a generator. Neither of these things are true. Of course one can only guess what chris is thinking so we'll wait and see what he proposes next.
This appears to be just an unsupported ejection of 'thought-juice', without any supportive substantiation.

The above is a Newtonian mechanical analysis based simply on the one starting proposition that a thrust can be applied to the rest-of-the-universe. All else, from that proposition, follows logically therefrom. That was the intent of this thread - start with a stated proposition, then look at the logical inferences.

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