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?

This has arisen in another thread analysing the general case of a thruster based on propellant: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.

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

**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.**

*reacted against the ROTU.*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.

**, 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.**

*It still needs 100Nm to achieve the same acceleration rate*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 …..