Author Topic: Woodward's effect  (Read 802990 times)

Offline aceshigh

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Re: Woodward's effect
« Reply #1380 on: 04/17/2018 05:05 am »
a bit short on time to watch 7 hours of video lol

but at 6:25:00, the man talking tells "it's a bit intimidating having Lawrence Krauss sitting right there, and I felt a little sorry for Dr Sonny White explaining his propulsion concept"

I suppose Lawrence Krauss gave Dr Sonny White a hard time? Has anyone seen Dr Sonny White's talk?

ok, watched it... apparently, Krauss basically grilled Dr Sonny White and said Sonny White should look for other things to do (rather than physics field, based on how grumpy he was, including saying Dr White was fooling himself)


Zubrin was also on the conference and also gave some hard time to Dr Sonny White.


It seems none of them addressed directly Dr Heidi Fern, although Krauss did argue a little at that panel at the end, basically saying people shouldnīt pay attention to anything that disagrees with what physics has already understood.

7:07:00
Krauss: "the difference is that when something disagrees with experiment, it's wrong"

Jeff Greason from Tau Zero: "yes, when something disagrees with experiment, it's wrong. When something disagrees with our interpretation of the experiments, there is room for argument"

But even Marc Millis, from Tau Zero Foundation (partner of Jeff Greason) had some flak to give to Dr Sonny White.

Krauss doesnīt even look at the panelists, when he stops talking angrily, he turns kind of sideways, like a defensive posture "I donīt care to read your replies".

« Last Edit: 04/17/2018 06:29 am by aceshigh »

Offline WarpTech

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Re: Woodward's effect
« Reply #1381 on: 04/17/2018 09:25 pm »
a bit short on time to watch 7 hours of video lol

but at 6:25:00, the man talking tells "it's a bit intimidating having Lawrence Krauss sitting right there, and I felt a little sorry for Dr Sonny White explaining his propulsion concept"

I suppose Lawrence Krauss gave Dr Sonny White a hard time? Has anyone seen Dr Sonny White's talk?

ok, watched it... apparently, Krauss basically grilled Dr Sonny White and said Sonny White should look for other things to do (rather than physics field, based on how grumpy he was, including saying Dr White was fooling himself)


Zubrin was also on the conference and also gave some hard time to Dr Sonny White.


It seems none of them addressed directly Dr Heidi Fern, although Krauss did argue a little at that panel at the end, basically saying people shouldnīt pay attention to anything that disagrees with what physics has already understood.

7:07:00
Krauss: "the difference is that when something disagrees with experiment, it's wrong"

Jeff Greason from Tau Zero: "yes, when something disagrees with experiment, it's wrong. When something disagrees with our interpretation of the experiments, there is room for argument"

But even Marc Millis, from Tau Zero Foundation (partner of Jeff Greason) had some flak to give to Dr Sonny White.

Krauss doesnīt even look at the panelists, when he stops talking angrily, he turns kind of sideways, like a defensive posture "I donīt care to read your replies".

Prof. Krauss says some pretty uninformed things. "One is, that the quantum vacuum has no (rest) frame. There is nothing you can push against."  The latter is NOT necessarily implied by the former statement. Even though the QV has no rest frame, when "pushing" the goal is that it is no longer an "inertial frame" anyway. It is an accelerated frame, in which case there is Unruh radiation and we can push against it by exchanging momentum with this elevated "thermal field". Krauss is ignoring the details in those statements.

In addition, even if Dr. White's conjecture about the density of e-p pairs in the vacuum is wrong. It is not wrong to expect such particle densities "in the vicinity" of charged particles, such as electrons and nuclei. As one approaches the center of charge, there is a limit regarding how close you can get because when the distance is on the order of the Compton wavelength, pair-creation can and does occur. So the field Dr. White is referring to is KNOWN to exist, but only in the vicinity of charge. It is not observed in free space in the way he conjectures.

Krauss: "the difference is that when something disagrees with experiment, it's wrong"

He may eat those words if it turns out the Em Drive actually does produce thrust. It would imply that conventional physics does not agree with experiment! IMO, it does but there is some exchange of momentum going on with the environment that we are not aware of yet.
« Last Edit: 04/17/2018 09:27 pm by WarpTech »

Offline Bob Woods

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Re: Woodward's effect
« Reply #1382 on: 04/18/2018 03:48 pm »
Krauss doesnt even look at the panelists, when he stops talking angrily, he turns kind of sideways, like a defensive posture "I donīt care to read your replies".
I watched the live stream as it happened, and Krauss was cranky and too dismissive. I hope it might be something like low blood sugar, because I get cranky when mine is low, but it seemed more like overarching arrogance. Challenging a position is good, dismissing the person is not.


There was a time where many great minds were absolutely certain that the world was flat. It was flat from the reference frame that they lived their lives in, it just didn't hold when their perspective expanded.

Offline Jim Davis

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Re: Woodward's effect
« Reply #1383 on: 04/18/2018 04:23 pm »
There was a time where many great minds were absolutely certain that the world was flat. It was flat from the reference frame that they lived their lives in, it just didn't hold when their perspective expanded.

This is fascinating. Who were these great minds and when did they live?

Offline ppnl

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Re: Woodward's effect
« Reply #1384 on: 04/18/2018 05:18 pm »
"One is, that the quantum vacuum has no (rest) frame. There is nothing you can push against."  The latter is NOT necessarily implied by the former statement. 





It would imply that conventional physics does not agree with experiment! IMO, it does but there is some exchange of momentum going on with the environment that we are not aware of yet.


Yeah, I don't think you are thinking it through.

Think of a propeller aircraft pushing against the air. How much power does it take to get a given acceleration? Well that depends on how fast the craft is traveling currently. The faster it is going the more power is needed to maintain the current acceleration. This is necessary for conservation of energy since energy goes with the square of velocity. And it is possible because there is a special frame in which the air is motionless.

Now imagine a craft pushing against the quantum vacuum. The problem is that the QV has no frame in which it is motionless. The amount of power needed for a constant acceleration is a constant. This immediately causes a violation of conservation of energy. Exchanging momentum with parts of the environment does not help unless you can give that part a special frame where it is motionless.  The QV does not allow this.

Any attempt to get around this will necessarily involve physics beyond the conventional. And I would add that often the whole point of doing experiments is to find places where experiment disagrees with conventional physics. Finding such a thing is good because it may mean you just won the Nobel.

 

Offline Bob Woods

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Re: Woodward's effect
« Reply #1385 on: 04/18/2018 05:48 pm »
There was a time where many great minds were absolutely certain that the world was flat. It was flat from the reference frame that they lived their lives in, it just didn't hold when their perspective expanded.

This is fascinating. Who were these great minds and when did they live?
https://en.wikipedia.org/wiki/Flat_Earth


If you wish to take issue with "many great minds", fine.


The article cites both sides, and notes that the Chinese did not adopt a spherical view until the 17th century. I think most of us grew up with the belief that certainty did not reign globally until after Columbus and Magellan.

Offline Jim Davis

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Re: Woodward's effect
« Reply #1386 on: 04/18/2018 08:28 pm »
The article cites both sides, and notes that the Chinese did not adopt a spherical view until the 17th century.

That's fine, Bob, your point is made. But I think it is grossly unfair to compare someone in the 21st century who thinks there is something to conservation of momentum and conservation of energy to people in the 17th who thought the earth was flat.

I don't think the evidence in favor of propellant propulsion, such as it is, of whatever flavor, is so overwhelming that doubters can be dismissed as irrational traditionalists.

Offline aceshigh

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Re: Woodward's effect
« Reply #1387 on: 04/18/2018 10:59 pm »
The article cites both sides, and notes that the Chinese did not adopt a spherical view until the 17th century. I think most of us grew up with the belief that certainty did not reign globally until after Columbus and Magellan.

we grew up with that belief due to urban myth. Everyone in Columbus time knew the Earth was round.

Columbus had trouble convincing people to sponsor his trip exactly because he was wrong. He thought the Earth was smaller than it really was, and thus, that Japan was much nearer to western Europe than it really is.

That's why Columbus thought he had reached the Indias when he landed in the Caribbean. No navigator would make such mistake, as they could already measure longitude at that time... unless said navigator thought the distance to east Asia was smaller.

Columbus was incredibly lucky. If the Americas did not exist, everyone would die from thirst and starvation, 1/3 to 1/4 of the way to East Asia.

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Krauss was cranky and too dismissive. I hope it might be something like low blood sugar,

MAYBE his problems with sexual harassment accusations in February? Arizona State University even put him in paid leave. I would also be cranky if those allegations were true. And EVEN MORE CRANKY if they were a lie.
https://en.wikipedia.org/wiki/Lawrence_M._Krauss

Quote
I watched the live stream as it happened

did Krass made any comment or questions directly to Dr Heidi Fern? Or anything at all related to Woodward Effect?

What about Zubrin or others?
« Last Edit: 04/18/2018 11:02 pm by aceshigh »

Offline WarpTech

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Re: Woodward's effect
« Reply #1388 on: 04/19/2018 12:49 am »
"One is, that the quantum vacuum has no (rest) frame. There is nothing you can push against."  The latter is NOT necessarily implied by the former statement. 


It would imply that conventional physics does not agree with experiment! IMO, it does but there is some exchange of momentum going on with the environment that we are not aware of yet.


Yeah, I don't think you are thinking it through.

Think of a propeller aircraft pushing against the air. How much power does it take to get a given acceleration? Well that depends on how fast the craft is traveling currently. The faster it is going the more power is needed to maintain the current acceleration. This is necessary for conservation of energy since energy goes with the square of velocity. And it is possible because there is a special frame in which the air is motionless.

Now imagine a craft pushing against the quantum vacuum. The problem is that the QV has no frame in which it is motionless. The amount of power needed for a constant acceleration is a constant. This immediately causes a violation of conservation of energy. Exchanging momentum with parts of the environment does not help unless you can give that part a special frame where it is motionless.  The QV does not allow this.

Any attempt to get around this will necessarily involve physics beyond the conventional. And I would add that often the whole point of doing experiments is to find places where experiment disagrees with conventional physics. Finding such a thing is good because it may mean you just won the Nobel.

I don't think you realize what you just said. First, the propeller-driven aeroplane in the atmosphere is a poor example. Instead, think of a photon rocket.

1. The amount of power required for constant acceleration is not constant.
2. There is no violation of conservation of energy.
3. It exchanges momentum simply by emitting EM waves that carry momentum, which can be quantized as photons.
4. The QV allows this.
5. There may not be new physics, just new understanding or interpretations of what we already know.

If you think it through, you will find that the QV spectrum is Lorentz invariant, so it is identical in any inertial reference frame. Pushing on it is a non-inertial action, which results in a temperature gradient due to the Unruh effect being asymmetrical. Similar to the Casimir effect, if a flat plate pushes on the QV, the EM modes on the leading side are blue-shifted and on the back side are red-shifted. This results in a pressure imbalance that allows some "push" off the QV. However, the effect is only as good as a photon rocket whose output is the difference between the two sides of the plate. Unfortunately, the highest energy photons of the QV pass right through the plate, so we are limited by the hard X-ray transparency of matter.

In the case of a MEGA drive, I conjecture that the electric charge on the PZT stack has asymmetrical radiation due to its asymmetrical acceleration. Thereby, improving the coupling by increasing the number of photons emitted in one direction over the other. Similarly, gravity is the coupling of matter to the QV. Matter and QV are constantly exchanging photons at resonance. Radiative damping due to other influences, such as a nearby massive planet, causes a gradient in this process which results in matter drifting toward the source of the damping because it results in a lower energy state of matter.
« Last Edit: 04/19/2018 12:55 am by WarpTech »

Offline SeeShells

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Re: Woodward's effect
« Reply #1389 on: 04/19/2018 02:08 am »
In this video from today's meeting at Stanford, Prof. Heidi Fearn has two minutes or so beginning at ~2:43:43 where she talks about the Mach Effect drive and uses a simple analogy to describe it.

https://www.youtube.com/watch?time_continue=3&v=3GiN-tWAV_k

Here's a direct link for people to jump to the spot:

https://www.youtube.com/watch?v=3GiN-tWAV_k#t=2h43m44s


(Btw, does her oil drop analogy sound similar to the one made for Pilot Wave theory? Interesting how that was also used for rationalizing EMdrive)

Her oil drop (silicone) experiment is precisely what Couder et al did in France. it would seem she is laying the groundwork for connecting QM and GR with at least quasiparticles (QFT) and possibly negative vacuum.

Couder Video


https://en.wikipedia.org/wiki/Pilot_wave_theory

See also
Couder
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.97.154101
https://arxiv.org/abs/1405.0466
This video is what got me thinking about the possability of Pilot Wave Theory.

Shell

Offline ppnl

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Re: Woodward's effect
« Reply #1390 on: 04/19/2018 04:23 am »

I don't think you realize what you just said. First, the propeller-driven aeroplane in the atmosphere is a poor example. Instead, think of a photon rocket.

1. The amount of power required for constant acceleration is not constant.

For a rocket in the accelerating frame of the rocket it is. A rocket using fuel at a constant rate thus producing energy at a constant rate will have a constant acceleration neglecting the change of weight due to fuel depletion. Factor in fuel depletion and it gets even better. This seems like a good thing until you factor in the cost of carrying all that fuel. That's what the rocket equation is all about.

An airplane or car does not expel any reaction mass. It instead reacts against an external medium such as the air or ground. That means it needs ever increasing amounts of power to maintain constant acceleration. That seems like a bad thing. But actually it will outperform a rocket because it does not have to carry its reaction mass with it and does not have to deal with the rocket equation.

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2. There is no violation of conservation of energy.

Certainly not in a photon rocket or any other rocket. As long as you are converting something internal to the rocket into photons and expelling them out the back you can have constant acceleration with constant power. Or better. But the rocket equation apples and you will run out of fuel. For a photon rocket this is bad because you need a megawatt of power just to get three newtons of force. Nothing less than the energy density of matter/antimatter will do. And if your target velocity is anything short of a massive fraction of the speed of light you are wasting massive amounts of energy. Almost anything else would work better.
 
Quote
3. It exchanges momentum simply by emitting EM waves that carry momentum, which can be quantized as photons.

But how many photons? A photon rocket of any size would need to produce a laser beam more powerful than any laser ever produced on earth. Remember a megawatt for just three newtons. A photon rocket that could accelerate at one g would burn a hole deep into the Earth.


Quote
4. The QV allows this.

No. It. Does. Not.

You still need to push against a massive amount of photons. Megawatts of them for barely measurable thrust. If the source of the photons is from energy stored on the ship then this is possible if impractical. And you would run out of photon fuel before getting very far.

But you seem to want to push against the QV which is outside the ship. The power you need for a given acceleration then depends on your velocity with respect to the QV the same way the power a car needs for a given acceleration depends on how fast the car is going with respect to the road.

The quantum vacuum has no preferred frame of reference. You cannot calculate your velocity with respect to it. If you could react against it you would either create a preferred frame of reference for the universe or violate conservation of energy. Or both. Depending on just what your engine did.   



Quote
In the case of a MEGA drive, I conjecture that the electric charge on the PZT stack has asymmetrical radiation due to its asymmetrical acceleration.

You cannot have asymmetrical acceleration. Ever. This is the very definition of violating conservation of momentum. For anything accelerating in one direction there must be something with equal acceleration in the other. Because they must push each other equally. And if that thing is photons then for any reasonable thrust you would need enough of them to burn down a city.

Unless you introduce some very new physics. 

Offline WarpTech

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Re: Woodward's effect
« Reply #1391 on: 04/19/2018 05:09 am »

3. It exchanges momentum simply by emitting EM waves that carry momentum, which can be quantized as photons.

But how many photons? A photon rocket of any size would need to produce a laser beam more powerful than any laser ever produced on earth. Remember a megawatt for just three newtons. A photon rocket that could accelerate at one g would burn a hole deep into the Earth.

Not true! If gravity is acting on a mass and it is accelerating relative to the Earth at 1g, there must be an exchange of momentum between the mass and the Earth. That exchange has to be either gravitons or photons, but either way it is the same amount of power we are talking about. I don't see falling objects burning holes through the Earth, so there must be a way to replicate this, and I believe the key is the Mach effect that I'm working on.

4. The QV allows this.
No. It. Does. Not.

You still need to push against a massive amount of photons. Megawatts of them for barely measurable thrust. If the source of the photons is from energy stored on the ship then this is possible if impractical. And you would run out of photon fuel before getting very far.

But you seem to want to push against the QV which is outside the ship. The power you need for a given acceleration then depends on your velocity with respect to the QV the same way the power a car needs for a given acceleration depends on how fast the car is going with respect to the road.

The quantum vacuum has no preferred frame of reference. You cannot calculate your velocity with respect to it. If you could react against it you would either create a preferred frame of reference for the universe or violate conservation of energy. Or both. Depending on just what your engine did.   

You contradicted yourself. The QV is Lorentz invariant, therefore there is no "drag" force dependent on velocity like you have with air, or wrt "the road". You can't measure your velocity wrt the vacuum because there is no drag force relative to it. I'm not the one saying there is.

There IS a drag force when the time derivative of "acceleration" is non-zero. The 3rd derivative of the position will result in a drag force on an atom, but this is not an inertial reference frame. It is "Radiation Reaction", and it's proportional to the Larmour radiation from the accelerating charges.

In the case of a MEGA drive, I conjecture that the electric charge on the PZT stack has asymmetrical radiation due to its asymmetrical acceleration.
You cannot have asymmetrical acceleration. Ever. This is the very definition of violating conservation of momentum. For anything accelerating in one direction there must be something with equal acceleration in the other. Because they must push each other equally. And if that thing is photons then for any reasonable thrust you would need enough of them to burn down a city.

Unless you introduce some very new physics.

Not true, relative to the center of mass. Take 2 unequal masses, M1 and M2 attached to the ends of a coiled Spring of length "L" and constant "k", and calculate the acceleration of each mass, relative to the center of mass. They are not the same at all! The force is the same, but not the acceleration.

M1*a1 = -M2*a2, but a1 =/= -a2 if M1 =/= M2.

Larmour radiation depends on the square of the acceleration and the square of the charge. So if the masses are charged equally, like a capacitor, then one end will radiate more than the other because the accelerations are not equal.
« Last Edit: 04/19/2018 05:15 am by WarpTech »

Offline ppnl

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Re: Woodward's effect
« Reply #1392 on: 04/19/2018 09:46 am »

Not true! If gravity is acting on a mass and it is accelerating relative to the Earth at 1g, there must be an exchange of momentum between the mass and the Earth. That exchange has to be either gravitons or photons, but either way it is the same amount of power we are talking about.


Gravity is a field not a rocket motor. It does not work by emitting an exhaust of gravitons. There are no actual gravitons involved at all. They are only virtual particles introduced to do a calculation. In order to have an actual gravitron you would need to furnish the energy to create it. Nobody has managed to do that yet. Gravity also obeys conservation of momentum and energy. No orbiting configuration of masses will accelerate anywhere without a real exhaust.

There is a similar field called an electric field that is mediated by the photon. There are no actual photons there as again it is only virtual particles. To see a real photon you have to furnish the energy to create it. Fortunately these can be created by smacking an electric field up side the head. The electric field also obeys conservation of momentum and energy. No configuration of interacting electric fields will accelerate anywhere without a real exhaust.

Quote
You contradicted yourself. The QV is Lorentz invariant, therefore there is no "drag" force dependent on velocity like you have with air, or wrt "the road".

Aiiiiiiii....

Drag has nothing to do with it. The fact that you need four times the energy to go twice as fast is not caused by drag and has nothing to do with drag. Ke=1/2mv^2 even in a frictionless system. A car with a lossless transmission and zero rolling friction will still need ever increassing power to maintain constant acceleration. This because energy goes with the square of velocity.

The problem isn't that there is no drag force with the QV. The problem is that there can be no process at all that lets you know how fast you are going with respect to the QV. This means that a drive reacting against the QV must produce constant acceleration with constant power or otherwise you can measure how fast you are going with respect to the QV. But constant acceleration with constant power is a violation of conservation of energy because energy goes with the square of velocity.

Quote
Not true, relative to the center of mass. Take 2 unequal masses, M1 and M2...

Well yes if the masses are different. But so what? The forces are equal and momentum and energy is conserved as it must be.

Quote
Larmour radiation depends on the square of the acceleration and the square of the charge. So if the masses are charged equally, like a capacitor, then one end will radiate more than the other because the accelerations are not equal.

Uh, no.

Say you have unequal masses tied together with a rubber band. They bounce to and fro exchanging energy and momentum. But energy and momentum is conserved so the center of mass never changes. It ain't goin' anywhere.

Now say you give the masses a charge. Now when they bounce to and fro they emit photons. That's what happens when you smack an electric field around. But however many photons you have going in one direction you have an equal number going in the opposite direction and in fact most of the photons go out in a ring perpendicular to the direction of acceleration. This is true of each individual mass so it is true of the combination of masses no matter how they pull on each other. In any case all the momentum cancels out and you still ain't goin' anywhere.

In any case these are real photons that you have to supply the energy to create. To get a large number of photons you have to smack the electric field really hard. That means a gigawatt of power would at most give you three newtons of force even if you could get around the fact that all the momentum cancels out.

Offline bad_astra

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Re: Woodward's effect
« Reply #1393 on: 04/19/2018 03:45 pm »
The article cites both sides, and notes that the Chinese did not adopt a spherical view until the 17th century. I think most of us grew up with the belief that certainty did not reign globally until after Columbus and Magellan.

Columbus had trouble convincing people to sponsor his trip exactly because he was wrong. He thought the Earth was smaller than it really was, and thus, that Japan was much nearer to western Europe than it really is.

That's why Columbus thought he had reached the Indias when he landed in the Caribbean. No navigator would make such mistake, as they could already measure longitude at that time... unless said navigator thought the distance to east Asia was smaller.

Columbus was incredibly lucky. If the Americas did not exist, everyone would die from thirst and starvation, 1/3 to 1/4 of the way to East Asia.
Or Columbus was as good a navigator as his past and later history proved him to be, and he devised his silly voyage proposal precisely because he knew from certain fisherman who were already secretly fishing off the grand banks (as well as possibly Greenlander records that were less than 200 years old at that point) that SOMETHING was out there. I know this has nothing to do with this thread, but over the years I think the oddest idea of Columbus proposal was his misconception of how small the world was, and I don't buy it anymore. He knew. But he needed a reason to get his ships west. It's no mistake that Caboto followed Columbus only a few years later but this time hit the economic bulls eye right in the Grand Banks. By then Columbus was busy pillaging new lands in the south and searching for gold.
« Last Edit: 04/19/2018 03:46 pm by bad_astra »
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Offline WarpTech

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Re: Woodward's effect
« Reply #1394 on: 04/19/2018 05:07 pm »
...
You contradicted yourself. The QV is Lorentz invariant, therefore there is no "drag" force dependent on velocity like you have with air, or wrt "the road".
...

The problem isn't that there is no drag force with the QV. The problem is that there can be no process at all that lets you know how fast you are going with respect to the QV. This means that a drive reacting against the QV must produce constant acceleration with constant power or otherwise you can measure how fast you are going with respect to the QV. But constant acceleration with constant power is a violation of conservation of energy because energy goes with the square of velocity.

I'm ignoring your comments about gravity because you probably will not give any consideration to my Engineering Model of Quantum Gravity, which is what I am using since it is based in QED, a well tested QFT.

Regarding kinetic energy "Ke", it would be more accurate to say;

ΔKe = (1/2)m*(Δv)2

Where it is the change in velocity "relative to where you started", not relative to the QV that matters. One does not need to measure the velocity relative to the QV to determine that there is not constant acceleration for constant power. All that is required is the integral along the path of travel through space-time. It doesn't matter if the drive is pushing on the QV or if it's just a photon rocket. It is the same thing, because the QV I can interact with is comprised of EM fields and photons.

« Last Edit: 04/19/2018 05:08 pm by WarpTech »

Offline RonM

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Re: Woodward's effect
« Reply #1395 on: 04/19/2018 05:23 pm »
I'm ignoring your comments about gravity because you probably will not give any consideration to my Engineering Model of Quantum Gravity, which is what I am using since it is based in QED, a well tested QFT.

You should preface your comments with "Based on my Engineering Model of Quantum Gravity . . ."

That way people will know to comment on your theory instead of thinking you don't understand accepted physics.

Offline ppnl

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Re: Woodward's effect
« Reply #1396 on: 04/19/2018 06:59 pm »
ΔKe = (1/2)m*(Δv)2

Yes but this is just wrong. The problem is that by referencing only a change in velocity it ignores the fact that kinetic energy, like velocity, is frame dependent. Your equation allows you to violate conservation of energy at will by calculating it in two different frames and adding them together as if they were the same thing.

To see how this works consider an electric car. Lets say it accelerates from 0 to 50mph and to do so it pulls one unit of electric energy from the battery. Now how much energy would be pulled from the battery if you accelerated from 50mph to 100mph? Well according to your equation the change in velocity is still just 50mph so again you just need one unit of energy from the battery. But that's just wrong. If you do the experiment you will find that it takes three units of energy to go from 50mph to 100mph. And you would need five more units to get to 150mph. You need ever increasing power to maintain constant acceleration. Your equation implies you only need constant power for constant acceleration. One unit to get to 50mph, another unit to get to 100mph, another unit to get to 150mph... This is what breaks conservation of energy.

If your space drive only used the change in velocity to calculate energy then you could build up a massive amount of kinetic energy using very little power. This energy is extractable and you will have to explain where it comes from.  You can say zero point energy or energy from a different dimension. You can say invisible blue fairies running on treadmills for all I care. But whatever it is is going to be new to physics.

Offline ppnl

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Re: Woodward's effect
« Reply #1397 on: 04/19/2018 07:26 pm »
I'm ignoring your comments about gravity because you probably will not give any consideration to my Engineering Model of Quantum Gravity, which is what I am using since it is based in QED, a well tested QFT.

You should preface your comments with "Based on my Engineering Model of Quantum Gravity . . ."

That way people will know to comment on your theory instead of thinking you don't understand accepted physics.

Um, actually I entered this discussion arguing that he needed a theory beyond conventional physics against his claim that he did not. I have no interest in discussing whether his theory is correct or not. Experimenters will decide that. My concern is whether it is internally consistent and whether he has a good grasp of conventional physics.

Experimenters have long ago shown that his equation for kinetic energy is just not right. It violates conservation of energy threatening to make his theory inconsistent. And his inability to grasp this violation suggests his grasp of basic conservation laws is lacking. It seems to be an inability to grasp Galilean relativity. 

Offline WarpTech

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Re: Woodward's effect
« Reply #1398 on: 04/19/2018 07:38 pm »
ΔKe = (1/2)m*(Δv)2

Yes but this is just wrong. The problem is that by referencing only a change in velocity it ignores the fact that kinetic energy, like velocity, is frame dependent. Your equation allows you to violate conservation of energy at will by calculating it in two different frames and adding them together as if they were the same thing.

To see how this works consider an electric car. Lets say it accelerates from 0 to 50mph and to do so it pulls one unit of electric energy from the battery. Now how much energy would be pulled from the battery if you accelerated from 50mph to 100mph? Well according to your equation the change in velocity is still just 50mph so again you just need one unit of energy from the battery. But that's just wrong. If you do the experiment you will find that it takes three units of energy to go from 50mph to 100mph. And you would need five more units to get to 150mph. You need ever increasing power to maintain constant acceleration. Your equation implies you only need constant power for constant acceleration. One unit to get to 50mph, another unit to get to 100mph, another unit to get to 150mph... This is what breaks conservation of energy.

If your space drive only used the change in velocity to calculate energy then you could build up a massive amount of kinetic energy using very little power. This energy is extractable and you will have to explain where it comes from.  You can say zero point energy or energy from a different dimension. You can say invisible blue fairies running on treadmills for all I care. But whatever it is is going to be new to physics.

Point taken, but you changed the problem. Nobody said anything about starting and stopping and adding up the pieces. That's not what I was thinking at all! You said we have to measure the speed relative to the QV. We don't. A photon rocket expends fuel and radiates momentum and energy. Calculating conservation of energy is simply adding up the photons in the exhaust, and then calculating what "v" would be; given the total exhausted energy and the remaining rest-mass, it is straightforward to calculate the Ke relativistically.
« Last Edit: 04/19/2018 07:39 pm by WarpTech »

Offline ppnl

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Re: Woodward's effect
« Reply #1399 on: 04/19/2018 08:24 pm »
ΔKe = (1/2)m*(Δv)2

Yes but this is just wrong. The problem is that by referencing only a change in velocity it ignores the fact that kinetic energy, like velocity, is frame dependent. Your equation allows you to violate conservation of energy at will by calculating it in two different frames and adding them together as if they were the same thing.

To see how this works consider an electric car. Lets say it accelerates from 0 to 50mph and to do so it pulls one unit of electric energy from the battery. Now how much energy would be pulled from the battery if you accelerated from 50mph to 100mph? Well according to your equation the change in velocity is still just 50mph so again you just need one unit of energy from the battery. But that's just wrong. If you do the experiment you will find that it takes three units of energy to go from 50mph to 100mph. And you would need five more units to get to 150mph. You need ever increasing power to maintain constant acceleration. Your equation implies you only need constant power for constant acceleration. One unit to get to 50mph, another unit to get to 100mph, another unit to get to 150mph... This is what breaks conservation of energy.

If your space drive only used the change in velocity to calculate energy then you could build up a massive amount of kinetic energy using very little power. This energy is extractable and you will have to explain where it comes from.  You can say zero point energy or energy from a different dimension. You can say invisible blue fairies running on treadmills for all I care. But whatever it is is going to be new to physics.

Point taken, but you changed the problem. Nobody said anything about starting and stopping and adding up the pieces. That's not what I was thinking at all! You said we have to measure the speed relative to the QV. We don't. A photon rocket expends fuel and radiates momentum and energy. Calculating conservation of energy is simply adding up the photons in the exhaust, and then calculating what "v" would be; given the total exhausted energy and the remaining rest-mass, it is straightforward to calculate the Ke relativistically.

You may not have been thinking about starting and stopping but you should have been because that is the clear implication of the equation you posted. It is what makes it so wrong.

Yes there is no problem with a photon rocket. There is no important difference between it and any other rocket. Now if you can point to the terrawatt laser beam shining out the ass end of your drive and show me the terrawatt nuclear power plant powering it then I agree there is no violation of conservation of energy and momentum.  Photon rockets are possible if profoundly impractical and horribly wasteful for any velocity short of a good fraction of the speed of light.

But until you show me the planet melting laser and the massive power supply...

If you are in an electric car with the windows blacked out you can still figure out how fast you are traveling by measuring how much energy it takes to accelerate. The faster you are going the more energy it takes to go faster.

If you are pushing against the quantum vacuum you could try the same trick. If it works then you have established a universal frame of reference. That is bad.

If it does not work then you have violated conservation of energy. That is worse.

Through all this you have not said which you believe. How much power does it take to accelerate at perfect efficiency and does that power depend on velocity like a car? You are on the horns of a dilemma here.

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