Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1799073 times)

Online SeeShells

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...Than I'm kind of confused at what I see here Dr.
http://emdrive.com/dynamictests.html
Shell, @frobnicat and I were discussing the only EM Drive tests reported having been performed in a partial vacuum.
Correct me if I am wrong but wasn't the test flawed with a leaky cap?
what test being flawed? are you referring to the NASA Eagleworks test in partial vacuum ?
Somewhere I read or heard someone saying that the capacitors were failing and were even replaced once. No? Did I glean something else. I've read so much lately that it's kind of melding together. ;)

Online SeeShells

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Quote
What can be done (and you have done an excellent job showing this  :) ) is to show what an EM Drive cannot do in Space. (For example: it cannot provide constant acceleration at constant power)
I'd never say cannot as we are not dealing with a black and white situation. It may not provide constant acceleration at constant power, we simply don't know.
If it provides constant steady acceleration at constant power, then it can be used to provide free energy, I think that Todd agrees with that too.
It may provide nothing.

Offline Rodal

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...Somewhere I read or heard someone saying that the capacitors were failing and were even replaced once. No? Did I glean something else. I've read so much lately that it's kind of melding together. ;)
In the initial Brady et.al report they couldn't perform any vacuum tests because of such capacitor problems.

Paul March addressed those issues (replaced capacitors with ones that can operate in the vacuum of Space, etc.) and since December 2014 he reported a number of tests in partial vacuum without those issues.

The issue it has however is that according to the reports the average thrust is significantly lower than what @frobnicat considers because when turned around 180 degrees the thrust was significantly smaller.
« Last Edit: 06/07/2015 06:08 PM by Rodal »

Offline WarpTech

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To revisit the rocket staging thought experiment, consider two scenarios: EMdrive 1 accelerating itself to its "limiting velocity", say 100 km/s, and a chemical rocket carrying EMdrive 2 accelerating itself to 100 km/s by conventional means. The discussion upthread indicates that if COE is conserved, EMdrive 1, having brought itself up to 100 km/s will not be able to accelerate further. But if the rocket propelling EMdrive 2 "stages" away, releasing its EMdrive payload, would EMdrive 2 be able to accelerate to a higher velocity than EMdrive 1 relative to the "real" inertial reference frame EM1 obeys at their mutual starting point? Where you place the "real reference frame" matters.

This strikes me as a qualitative problem with Traveller's theory. How does EMdrive 1 "know" that if it continues to accelerate, it will violate COE? Your description of some sort of subatomic stress left me totally unconvinced, as it has no empirical backing. If you are going to rely on such a thing, please provide a reference to a peer reviewed journal with an impact factor greater than 1.

Quote from: Traveller
This physical change in the properties of matter alters atomic and sub-atomic spacing,
and energies, identical to how gravity contracts matter falling into a gravity well.

Why do I need a peer reviewed paper to prove that the momentum of a sub-atomic particle such as an electron is, p = h/lambda, relative to the rest frame? Where h is Planck's constant. This is basic QM. If a particle is accelerated to a higher momentum and h is a constant, the wavelength must be decreased. It's not rocket science.

The fact that the amount the wavelength contracts is equal to the amount predicted by Lorentz Contraction, is mere coincidence?

Matter stores inertia, simply put, the longitudinal mass will increase by gamma^3 and the transverse mass will increase by gamma. This shortens the wavelengths of matter waves, that is where the inertia is stored.

See: http://www.fourmilab.ch/etexts/einstein/specrel/www/
ON THE ELECTRODYNAMICS OF MOVING BODIES
By A. Einstein
June 30, 1905

To answer your question, if the rocket engine and the EM Drive have the same thrust-to-power ratio, then switching off the rocket and switching on the EM Drive will make no difference. Just because EM Drive 2 was only along for the ride makes no difference. Every particle of matter in both devices still absorbed the increased inertia when it was accelerated by the rocket engine to the new potential, (v - v0)^2. Note, this is the "change" in velocity, relative to where it started from, relative to its rest frame. The rest frame it started in is a preferred frame for that object, but it was not at rest in any "absolute" sense of the word.

Todd


Offline WarpTech

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Todd -

I commend you on your efforts to develop a theory that can help to resolve the CoE issue of the emdrive.  I have a comment and a question.

First, a comment:
Quote
The physics is similar in nature to hovering in a Newtonian gravitational field, where Special Relativity
does not apply. The Newtonian gravitational potential Φ has units of (m / s)2
, such that the gradient derivative yields an acceleration vector. It represents the potential energy per unit mass and may be treated identically to the velocity squared in Newtonian kinetic energy, v2 = 2E / m .
   

It appears to me that you have confused the gravitational potential, Φgravity=GM/r, with an actual evaluation of the gravitational potential energy.  The gravitational potential is work done by gravity accelerating a unit mass from infinity to the point of evaluation (http://en.wikipedia.org/wiki/Gravitational_potential).  If you actually wanted the gravitational potential energy, you are missing some mathematical mechanics.  I don't think this changes too much, but it is necessary.

Now for the actual issue:  This theory is a completely testable hypothesis with a trivial experiment.

Take a ball and hold it  above the ground.  It has gravitational potential energy equal to m*g*h.  There is a force between the ball and the earth equal to m*g.  At t =0 release this ball. 

Then:
Thrust to power ratio = mg/(d/dt(mgh))
                                =mg/(mgv),         dh/dt=v
                                =1/v

So at t=0 when we release the ball, the thrust to power ratio for the ball is given by: 1/v = 1/0 = inf.

Therefore, the limiting velocity of the ball is 0!  This ball should not be able to move (by my understanding of your theory), but clearly balls do move.  How can this be resolved?

You're forgetting the gradient in the thrust-to-power ratio is required to get g. In your example, the power is the energy exchanged between the ball and the gravitational field. If it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass. When you let go of the ball, the gradient is then free to act. Note, this is Newtonian mechanics, not GR. It's explained differently in GR, because the ball is now in geodesic motion, which is in fact described by the derivatives of the metric potentials.
« Last Edit: 06/07/2015 06:29 PM by WarpTech »

Offline WarpTech

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...
I guess I have a question myself.  I was wondering what this velocity is with respect to.  Normally when I see a velocity I think that V=V1-V2 where it is a comparison between two frames with different velocity.  My question is if this velocity is a comparison with us and the accelerated frame of space time?  That is we are stationary but the space flowing into the earth is not so we can assign it a relative velocity.  This relative velocity dilates space and time.  Lower to the surface of the earth clocks run slower and space is more contracted because space is flowing faster.  Higher from the surface space is flowing slower with respect to us so clocks are running faster.  So is this velocity a comparison between our frame and the flow of space time's frame?  In that sense in a gravitational field the ball would move?  Further away from gravitational fields where the relative velocity with respect to space time is zero the ball would not accelerate? 

Hmm, be warned.  I think my hypothesis about free space might be flawed.  The reason being if you are falling faster than free space then its resistance would slow you down.  This might make sense for momentum resisting acceleration and assuming free space catches up to you but in the sense of assuming a velocity for in falling space I am not sure it makes sense.  Unless some ones local space moving towards the earth at near light speed could fall faster than the surrounding space.  Do super fluids behave this way? 

The problem with what you said is that, you can't measure "space moving". You can only measure objects in space moving. To answer your question, the velocity is wrt the rest-frame the object started from, just as in SR. This rest frame defines the initial inertial mass = m0*|g_11|^3/2, where g_11 is the gravitational potential at the location of the object. At this inertial mass, the object's speed is defined as v0=0. The speed defined in (P/F) = v, is the speed it was accelerated to, relative to the rest-frame it started from (v - v0) = v. Sorry if that was not obvious.

Todd

Offline wallofwolfstreet

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Todd -

I commend you on your efforts to develop a theory that can help to resolve the CoE issue of the emdrive.  I have a comment and a question.

First, a comment:
Quote
The physics is similar in nature to hovering in a Newtonian gravitational field, where Special Relativity
does not apply. The Newtonian gravitational potential Φ has units of (m / s)2
, such that the gradient derivative yields an acceleration vector. It represents the potential energy per unit mass and may be treated identically to the velocity squared in Newtonian kinetic energy, v2 = 2E / m .
   

It appears to me that you have confused the gravitational potential, Φgravity=GM/r, with an actual evaluation of the gravitational potential energy.  The gravitational potential is work done by gravity accelerating a unit mass from infinity to the point of evaluation (http://en.wikipedia.org/wiki/Gravitational_potential).  If you actually wanted the gravitational potential energy, you are missing some mathematical mechanics.  I don't think this changes too much, but it is necessary.

Now for the actual issue:  This theory is a completely testable hypothesis with a trivial experiment.

Take a ball and hold it  above the ground.  It has gravitational potential energy equal to m*g*h.  There is a force between the ball and the earth equal to m*g.  At t =0 release this ball. 

Then:
Thrust to power ratio = mg/(d/dt(mgh))
                                =mg/(mgv),         dh/dt=v
                                =1/v

So at t=0 when we release the ball, the thrust to power ratio for the ball is given by: 1/v = 1/0 = inf.

Therefore, the limiting velocity of the ball is 0!  This ball should not be able to move (by my understanding of your theory), but clearly balls do move.  How can this be resolved?

You're forgetting the gradient in the thrust-to-power ratio is required to get g. In your example, the power is the energy exchanged between the ball and the gravitational field. If it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass. When you let go of the ball, the gradient is then free to act. Note, this is Newtonian mechanics, not GR. It's explained differently in GR, because the ball is now in geodesic motion, which is in fact described by the derivatives of the metric potentials.

You know, I understand all those words you used, but in that arrangement they make absolutely no sense to me.

Also, I'm not unfamiliar with SR and GR, or physics and math in general. That is not where the issue in my understanding lies.

Quote
You're forgetting the gradient in the thrust-to-power ratio is required to get g.

What does this mean?  What gradient in the thrust-to-power ratio?  The thrust-to-power ratio is location-independent, it is a constant throughout space (in fact it isn't even uniquely defined through space, because it is velocity dependent.)  It has zero gradient (more accurately, it's gradient is completely undefined.  There is no such thing as the gradient in the thrust-to-power ratio).  And how does g play into it?

Quote
If it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass.

Once again, what does it mean for a force to overcome a gradient in a ratio?  This "gradient" has units of s/m2 because the ratio of thrust-to-power is in s/m.  What equation relates them such that one can "overcome" the other? 

Quote
When you let go of the ball, the gradient is then free to act.
 

The gradient (if we are still talking about the gradient in thrust-to-power) isn't a force.  What do you mean by "the gradient is free to act"?  Act on what?

I know I'm coming off here a little aggressive, but there is nothing for it.  The majority of people reading aren't commenting.  If they are to understand correctly, I suppose it is up to us commenters to ask the tough questions.         
« Last Edit: 06/07/2015 07:20 PM by wallofwolfstreet »

Offline deltaMass

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Consider the domain of the following to be a field-free region of flat spacetime in which all measured velocities in some arbitrary inertial observer frame are severely subrelativistic. Thus Galilean transformations and Newton's laws suffice for an analysis to first order in this domain. We also consider the motion to be confined to one dimension - i.e. there exist no transverse forces.

It is stated that "F = P/v" is an accurate description of the dynamics of a propellantless space drive, such as the EmDrive. What are the consequences of this assertion?

The measured acceleration of the drive will be the same in any inertial frame, since the measured value of deltaV is the same in any inertial frame.
[deltaV = v1 - v0 = (v1 + V) - (v0 + V), where V is the relative velocity of any two inertial frames]

In the limit of small deltas, the measured acceleration of the drive is directly deducible from deltaV
[a = deltaV/deltaT] and as shown is not dependent on which inertial frame is used to measure it.

Since the mass is an invariant of the motion, we can via observation therefore directly deduce the value of the thrust [F = m a, where acceleration is deduced as above].

The power input P to the device is taken to be a constant and is also known accurately. The device carries its own power supply on board, which delivers constant input power by design.

Putting all the above together, it is clear that we are able to directly test the validity of the assertion "F = P/v" from any inertial frame, and that the measured values of F and P will be identical in any and all of these frames. Clearly, however, the measured value of v will be different in different inertial frames.

This leads to a contradiction:
a) The measured value of F/P is the same in all different inertial frames
b) The measured value of v is different in all different inertial frames

The conclusion is that "F = P/v" cannot be true for an EmDrive in space.

Now, for a car accelerating along a flat road, we know that F = P/v is a true description of the situation. Given the above, how can that be?

The answer is that in the case of the car, there is indeed a preferred frame in respect of the car, and that is the road itself. The car's tyres are in physical contact with the road and thus 'v' has an absolute significance as far as the car's engine is concerned. It is doing work by pushing off against the road and the elementary mechanics of the situation shows that, for constant P, the thrust F will decrease with increasing v, and F = P/v will hold. Equal and opposite momentum is transferred to the road, which of course is not something we notice, given the mass of the Earth.

But for an EmDrive in space, there is no road, and there are no tyres. The drive is oblivious to the value 'v' because there is no preferred reference frame, because there is no solid connection to one.

Only if a preferred reference frame exists can "F = P/v" be true for an EmDrive in space.
But the existence of a preferred reference frame is a direct violation of a core principle of physics - that the physics is the same in all different inertial frames in a flat, field-free spacetime. Therefore, for an EmDrive in space, "F = P/v" can only be true if this core tenet is violated.

I rest my case.

« Last Edit: 06/07/2015 07:31 PM by deltaMass »

Offline frobnicat

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...
What can be done (and you have done an excellent job showing this  :) ) is to show what an EM Drive cannot do in Space. (For example: it cannot provide constant acceleration at constant power)

Thank you for the sensible reply.

I must (you know I must) only add, for sake of precision, to the last remark, that what me and a number of sceptics are trying to show is that it cannot provide averaged stationary thrust (be it operated while free to accelerate, as per Shawyer ideas, or operated when not accelerating, as per what results on static balances roughly hint as being possible, should inertial frame invariance hold) at an averaged stationary power such that thrust/power (averaged as a limit of any stationary process, be it constant, periodic, or chaotic) > 1/c, unless in deep space vacuum a frame invariant proper source/reserve of energy exists and can be harnessed.

Trying to be concise but accurate.

If the effect as measured on lab balances is real (not a force exchanged with earthly surroundings), I find constant_thrust/constant_power +  frame invariant proper source/reserve of energy in vacuum actually quite less unbelievable than other proposed schemes positing a span/duration/velocity limit so far. Just an opinion.

Offline birchoff

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Consider the domain of the following to be a field-free region of flat spacetime in which all measured velocities in some arbitrary inertial observer frame are severely subrelativistic. Thus Galilean transformations and Newton's laws suffice for an analysis to first order in this domain. We also consider the motion to be confined to one dimension - i.e. there exist no transverse forces.

...


Is this assumption true for the universe as we currently understand it?

...

But for an EmDrive in space, there is no road, and there are no tyres. The drive is oblivious to the value 'v' because there is no preferred reference frame, because there is no solid connection to one.

Only if a preferred reference frame exists can "F = P/v" be true for an EmDrive in space.
But the existence of a preferred reference frame is a direct violation of a core principle of physics - that the physics is the same in all different inertial frames in a flat, field-free spacetime. Therefore, for an EmDrive in space, "F = P/v" can only be true if this core tenet is violated.

I rest my case.

How is this conclusion valid outside the constraints you initially defined; "Consider the domain of the following to be a field-free region of flat spacetime in which all measured velocities in some arbitrary inertial observer frame are severely subrelativistic."

Offline deltaMass

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If this spherical chicken was good enough for the physics greats to conduct their own thought experiments, then you can rest assured that it's good enough for me.

Try thinking less like an engineer and more like a physicist. There is such a thing as an acceptable approximation, and there is such a thing as "to first order".

Offline WarpTech

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You know, I understand all those words you used, but in that arrangement they make absolutely no sense to me.

Also, I'm not unfamiliar with SR and GR, or physics and math in general. That is not where the issue in my understanding lies.

Quote
You're forgetting the gradient in the thrust-to-power ratio is required to get g.

What does this mean?  What gradient in the thrust-to-power ratio?  The thrust-to-power ratio is location-independent, it is a constant throughout space (in fact it isn't even uniquely defined through space, because it is velocity dependent.)  It has zero gradient (more accurately, it's gradient is completely undefined.  There is no such thing as the gradient in the thrust-to-power ratio).  And how does g play into it?

Quote
If it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass.

Once again, what does it mean for a force to overcome a gradient in a ratio?  This "gradient" has units of s/m2 because the ratio of thrust-to-power is in s/m.  What equation relates them such that one can "overcome" the other? 

Quote
When you let go of the ball, the gradient is then free to act.
 

The gradient (if we are still talking about the gradient in thrust-to-power) isn't a force.  What do you mean by "the gradient is free to act"?  Act on what?

I know I'm coming off here a little aggressive, but there is nothing for it.  The majority of people reading aren't commenting.  If they are to understand correctly, I suppose it is up to us commenters to ask the tough questions.       

If you read the paper, please understand I'm referring to the gradient of equation (5);

(P/F)^2 = (v - v0)^2,

Sorry, I see now I was being sloppy. I hadn't finished my first cup of coffee yet this morning.  :-[ It happens on forums, I'm not as formal with my language as I should be and my thoughts are not always as clearly stated as I think they are at the time. In haste, I may say things incorrectly or abbreviate my thoughts in ways that are not clear to others. Please, on this forum, taken what I say with a grain of salt. If I publish a paper, then that is what I have given thought to and where I attempted to express them clearly.

It is not the gradient of F/P, as I mistakenly said above. Equation 5 is correct in the paper we are discussing, refer to that for the Math please.

The gradient of (P/F)^2 = v^2 is not a force, it is an acceleration vector that opposes the force when you are accelerating an object, and results in the Newtonian force of gravity when it is the gravitational acceleration;

GM/r = (v - v0)^2

d/dr GM/r = -GM/r^2 = g  m/s^2

d/dr (P/F)^2 = -a   m/s^2,

Where the ratio P/F is only constant at a constant r, but is a variable wrt r. This gradient opposes the thrust. When the force of the thruster acting on the mass is offset by the gradient (-a), the object can no l longer increase its inertial mass, it has reached equilibrium with its own gravity, due to its increased inertial mass. Pushing on the ground at 1g will not move the earth.

Todd
« Last Edit: 06/07/2015 08:03 PM by WarpTech »

Offline WarpTech

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Consider the domain of the following to be a field-free region of flat spacetime...

No such domain exists. The EM and Gravitational fields span the entire universe. There is no way to "eliminate" them for convenience and have a meaningful discussion of this topic. Sorry!

Please read my paper and tell me where I made the mistake in the Math, rather than another long analogy in a nonexistent universe with no fields to interact with. The problem cannot be solved without them, not even at rest.

Thank you.

Todd
Quote
...
I rest my case.

Offline vulture4

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If we accept frame invariance (and it is hard not to) then the reactionless thrust cannot vary with velocity.  But that means that the engine power output increases without limit. So it's hard to avoid breaking conservation of energy.

As to the varying energy of the microwave photons as they pass through the waveguide, here is my question.

The energy of a photon is proportional to its frequency. E=hf where h is Planck's constant

So if photons approaching the small end of the waveguide have lower energy, they must have lower frequency. That should be easily measured to a precision of 1 in 10^6 or greater. There is nothing in the literature that I can find about frequency varying with waveguide diameter. In other words, the group velocity of a photon may vary with the medium through which it propogates (e,g, free space, waveguide, dielectric) but photon energy (and hence radiation pressure) does not vary with group velocity in the manner that the recoil force of a billiard ball varies with the velocity with which it propagates across a pool table.
https://en.wikipedia.org/wiki/Radiation_pressure#Radiation_pressure_by_particle_model:_photons
« Last Edit: 06/07/2015 08:34 PM by vulture4 »

Offline not_a_physicist

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Every particle of matter in both devices still absorbed the increased inertia when it was accelerated by the rocket engine to the new potential, (v - v0)^2. Note, this is the "change" in velocity, relative to where it started from, relative to its rest frame. The rest frame it started in is a preferred frame for that object, but it was not at rest in any "absolute" sense of the word.

If you and the EM drive were aboard the spaceship after the conventional rocket finished accelerating it, would you back able to take apart and inspect the EM drive to tell that it was "spent" (as in already at its maximum velocity)? Presumably (if I am understanding your predictions correctly), you could tell it was spent by running it and seeing if it accelerated your ship further, but could you tell by looking at its particles?

Thanks for all your patience with us!

Offline Tetrakis

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To revisit the rocket staging thought experiment, consider two scenarios: EMdrive 1 accelerating itself to its "limiting velocity", say 100 km/s, and a chemical rocket carrying EMdrive 2 accelerating itself to 100 km/s by conventional means. The discussion upthread indicates that if COE is conserved, EMdrive 1, having brought itself up to 100 km/s will not be able to accelerate further. But if the rocket propelling EMdrive 2 "stages" away, releasing its EMdrive payload, would EMdrive 2 be able to accelerate to a higher velocity than EMdrive 1 relative to the "real" inertial reference frame EM1 obeys at their mutual starting point? Where you place the "real reference frame" matters.

This strikes me as a qualitative problem with Traveller's theory. How does EMdrive 1 "know" that if it continues to accelerate, it will violate COE? Your description of some sort of subatomic stress left me totally unconvinced, as it has no empirical backing. If you are going to rely on such a thing, please provide a reference to a peer reviewed journal with an impact factor greater than 1.


To answer your question, if the rocket engine and the EM Drive have the same thrust-to-power ratio, then switching off the rocket and switching on the EM Drive will make no difference. Just because EM Drive 2 was only along for the ride makes no difference. Every particle of matter in both devices still absorbed the increased inertia when it was accelerated by the rocket engine to the new potential, (v - v0)^2. Note, this is the "change" in velocity, relative to where it started from, relative to its rest frame. The rest frame it started in is a preferred frame for that object, but it was not at rest in any "absolute" sense of the word.

So to be absolutely clear, you agree and have yourself now stated that an absolute rest frame is required for your theory to work (when you postulate that COE is forbidden, but that the EMdrive may still create net thrust). Do you understand what this means? If the starting point for my thought experiment was  at the surface of the moon, that point is in motion relative to Texas. That point is in motion relative to the center of the earth. That point is in motion relative to the sun, and so on and so on. Who decides where velocity is zero? You? Who is to say that the velocity of the frame at staging is or is not a valid rest frame? As was explained many pages ago, an EMdrive working according to your theory would also act as a universal spedometer anyone could use to determine their current velocity relative to the "universal preferred frame of reference".

I'm just glad that we agree on something: if the EMdrive works as advertised, there must be either a COE violation or there must be an absolute rest frame and therefore universal isotropy is broken. To me this indicates that the "principles" used to design the experimental EMdrives are baseless, and therefore any observed effect will be the result of luck. Might as well build a thousand EMdrives of different dimensions and specifications and test them all on spacecraft, that would give you more insight than the current regime.
« Last Edit: 06/07/2015 08:41 PM by Tetrakis »

Offline wallofwolfstreet

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If you read the paper, please understand I'm referring to the gradient of equation (5);

(P/F)^2 = (v - v0)^2,

Sorry, I see now I was being sloppy. I hadn't finished my first cup of coffee yet this morning.  :-[ It happens on forums, I'm not as formal with my language as I should be and my thoughts are not always as clearly stated as I think they are at the time. In haste, I may say things incorrectly or abbreviate my thoughts in ways that are not clear to others. Please, on this forum, taken what I say with a grain of salt. If I publish a paper, then that is what I have given thought to and where I attempted to express them clearly.

It is not the gradient of F/P, as I mistakenly said above. Equation 5 is correct in the paper we are discussing, refer to that for the Math please.

The gradient of (P/F)^2 = v^2 is not a force, it is an acceleration vector that opposes the force when you are accelerating an object, and results in the Newtonian force of gravity when it is the gravitational acceleration;

GM/r = (v - v0)^2

d/dr GM/r = -GM/r^2 = g  m/s^2

d/dr (P/F)^2 = -a   m/s^2,

Where the ratio P/F is only constant at a constant r, but is a variable wrt r. This gradient opposes the thrust. When the force of the thruster acting on the mass is offset by the gradient (-a), the object can no l longer increase its inertial mass, it has reached equilibrium with its own gravity, due to its increased inertial mass. Pushing on the ground at 1g will not move the earth.

Todd

Okay.... but how does that solve the problem?  Instead of referring to the gradient of (F/P), you are referring to the gradient of (P/F)2, correct?

So in my example, wtih a ball just sitting in a gravitational field, the F/P ratio is just given by 1/v.  There is no spatial dependence.  It is not uniquely defined.

There is no such thing as the gradient of (P/F)^2 for this case.

So is there something special in my hypothetical case that causes you theory to be inapplicable?

Offline vulture4

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Quote from: vulture4
If we accept frame invariance (and it is hard not to) then the reactionless thrust cannot vary with velocity.
Precisely. That is all I am saying. Todd and Rodal, however, seem to think differently.
So what do you think about the momentum argument, i.e. that radiation pressure exerted by a photon depends only on frequency, and not on group velocity?

Offline Rodal

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...

The power input P to the device is taken to be a constant and is also known accurately. The device carries its own power supply on board, which delivers constant input power by design.

Putting all the above together, it is clear that we are able to directly test the validity of the assertion "F = P/v" from any inertial frame, and that the measured values of F and P will be identical in any and all of these frames. Clearly, however, the measured value of v will be different in different inertial frames.

...
How can it be stated on one hand that

Quote
The power input P to the device is taken to be a constant and is also known accurately.
and that on the other hand
Quote
the measured value of v will be different in different inertial frames

Power is the time rate of energy, power is Force*velocity.  If velocity is frame dependent, so is Power frame dependent, where Power = Force * velocity

...
Now, for a car accelerating along a flat road, we know that F = P/v is a true description of the situation. Given the above, how can that be? The answer is that in the case of the car, there is indeed a preferred frame in respect of the car, and that is the road itself. The car's tyres are in physical contact with the road and thus 'v' has an absolute significance as far as the car's engine is concerned. It is doing work by pushing off against the road and the elementary mechanics of the situation shows that, for constant P, the thrust F will decrease with increasing v, and F = P/v will hold. Equal and opposite momentum is transferred to the road, which of course is not something we notice, given the mass of the Earth....

Since the distance covered while applying a force to an object depends on the inertial frame of reference, so does the work done.  The amount of work is frame-dependent.

In order to resolve this one has to use Newton's third law ( http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion ) , to find the reaction force that does work depending on the inertial frame of reference in an opposite way. Then, (when using Newton's reaction force dependent on the inertial frame) the total work done becomes independent of the inertial frame of reference.

The kinetic energy, and also the change in this energy due to a change in velocity, depends on the inertial frame of reference. The total kinetic energy of an isolated system also depends on the inertial frame of reference: it is the sum of the total kinetic energy in a center of momentum frame and the kinetic energy the total mass would have if it were concentrated in the center of mass. Due to the conservation of momentum the latter does not change with time, so changes with time of the total kinetic energy do not depend on the inertial frame of reference.


As an aside, furthermore, the relativistic energy is the addition of the rest energy plus the kinetic energy, where kinetic energy is dependent on velocity.  A body having an inertial rest mass possesses a definite amount of energy proportional to its mass m, this amount being not small in comparison with its kinetic energy for v << c
« Last Edit: 06/07/2015 09:05 PM by Rodal »

Offline deltaMass

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Quote from: vulture4
If we accept frame invariance (and it is hard not to) then the reactionless thrust cannot vary with velocity.
Precisely. That is all I am saying. Todd and Rodal, however, seem to think differently.
So what do you think about the momentum argument, i.e. that radiation pressure exerted by a photon depends only on frequency, and not on group velocity?
Radiation pressure from a beam of photons depends solely on P. When one is talking about beamed power like that, and when the relative velocity of source and mirror becomes relativistic, it gets complex. I have worked through the relativistic issues on StackExchange, but they are not relevant here, since the relative velocity of source and cavity is zero.

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