Author Topic: Woodward's effect  (Read 287751 times)

Offline dustinthewind

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Re: Woodward's effect
« Reply #680 on: 03/01/2016 10:50 PM »
All I am saying is if a projectile with momentum gets trapped in an osculating system and hits one wall giving up 20% of [its] energy but the impact with the other wall only gives back 10% of its energy that the projectile will end up giving its energy to the composite walls in a way that drains its energy.

You seem to be confusing a non-vector form of energy with a vector form.

If the energy of the photon is transferred to an object, the object gets hotter, not faster.

It gets hot if the light is striking a black body that absorbs light.  In this case there is some momentum transfer also.  This could be paralleled to a ballistic pendulum. 

If the light strikes a reflective surface there is also a transfer of energy and momentum.  On an elastic collision very little energy is transferred as heat, and is evident when you bounce a ball that returns to its previous height, losing very little energy. This is different from a plastic collision.  This is why I worked through the math, all in the same frame, above to show that light that reflects, changing the velocity of an object, appears to be doubly red-shifted.  What I got for a solution was two times the Doppler shift.  This is likely connected to a reflected photon providing twice the boost of an absorbed photon or an emitted photon was my conclusion.  I bring this up because it appears to reduce the energy of the reflected photon and energy is also transferred to the propelled object. 

What I am saying above is at a set velocity of the projectile (the speed of light) The equation 100%=1=energy to wall/(total energy)+energy to projectile/(total energy) This ratio is determined by the mass of the wall system and the mass of the photon.  If the mass of the photon appears heavier to one wall than the other then the photon pushes on one wall harder than the other.  The cavity will always after many bounces of the photon accelerate in the direction the photon appears to be heavier in mass even with momentum conserved during the collision (also draining the energy of the photon).  It doesn't appear to depend on which direction the photon comes from (left or right) the cavity will accelerate after many bounces in a certain direction.  This is a sort of paradox that I am considering.   

It has to do with changing the mass of an object but conserving momentum and then applying a force on it [reflection]which is connected to the Woodward effect.  I then linked a paper where they are considering if the mass of the photon changes inside a dielectric. 
You can convert between vector/scalar energy, but there needs to be a second system in play to balance the force, otherwise you are violating CoM. You have to pay for the conversion by adding negative (in a vector sense) momentum somewhere else.
« Last Edit: 03/01/2016 11:51 PM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #681 on: 03/02/2016 05:16 AM »
Dude, momentum equals mass times velocity. If you change the mass and keep the velocity the same... just do the math.

And as I said above the photons that have mass are virtual photons. They can also time travel and go faster than light. That does not mean you can build a time machine.


Offline Paul451

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Re: Woodward's effect
« Reply #682 on: 03/02/2016 08:29 PM »
All I am saying is if a projectile with momentum gets trapped in an osculating system and hits one wall giving up 20% of [its] energy but the impact with the other wall only gives back 10% of its energy that the projectile will end up giving its energy to the composite walls in a way that drains its energy.
If the energy of the photon is transferred to an object, the object gets hotter, not faster.
It gets hot if the light is striking a black body that absorbs light.  In this case there is some momentum transfer also.

No, any body gets hotter¹ if it absorbs energy from a photon. That is distinct from the momentum. Both must be accounted for. You are switching back and forth between them as if they were the same thing.

¹ Or is converted to some other form of non-vector energy, like an electrical charge. Same thing.

The only way to turn a photon's energy into momentum (more than the existing momentum of the photon, which is what you're trying to achieve), is if the energy also creates negative (vector) momentum in another system.

You have this idea about the photon "escaping with energy/momentum" when it is reflected, and we need to capture that "energy" in some way. But that treats momentum as a scalar property, you are ignoring the vector component. The momentum is in that direction. So if you transfer that vector momentum to the object/wall/cavity/whatever, you can only transferring momentum with a vector in that same direction.

Ie, if you reduce the "energy" the photon escapes with, then you have also reduced the momentum available to the object/wall/cavity in the opposite direction. You haven't increased it.

Eg,

If we treat right-to-left as the positive vector again, left-to-right as the negative.

A photon approaches a stationary object/wall/cavity/whatever from right-to-left with 1 arbitrary unit of momentum.

The total momentum for the system is 0 + 1 = 1 unit to the left.

The photon bounces away perfectly elastically and the object/etc is pushed away at 2 units of momentum.

The total momentum for the system is 2 + (-1) = 1 unit to the left. The total momentum of the system is unchanged.

So we introduce some mechanism that reduces the momentum of the photon after it reflects. The nature of the mechanism is irrelevant, the effect is the same.

The total momentum for the system is X + (-0.9) = 1 unit to the left. 'X', the momentum of the object/etc, must therefore be 1.9 units to the left. Less than a perfect reflection, not more. That's because you've transferred an extra -0.1 units of momentum to the object, not +0.1 units. If you somehow take "energy" from the photon, and hence momentum, you also have to take the same amount of momentum from the object.

(Then, separately, we have to account for any change in scalar energy. So that energy must be converted to another scalar form. Usually heat. So the object has less vector momentum than a perfect reflection, but more scalar energy as heat. In other words, any mechanism that "captures" energy from the photon is the same as an inelastic reflection.)

When you talk about a photon being "red-shifted" and "transferring its energy" to the object, you want to turn that missing 0.1 units of rightward momentum into 0.1 units of leftward momentum without balancing the equation. Ie, you are violating CoM.

You can only add momentum to the object/etc if you increase the momentum of the reflected photon in the opposite direction, ie, not by red-shifting it, but by blue-shifting it. Giving the photon more momentum and more energy, not less. It is functionally the same as absorbing the first photon and then re-emitting a new photon at higher frequency. Again, the mechanism is irrelevant. It is functionally the same as a photon-drive. (A pointlessly convoluted solar powered photon drive with a battery to make up the difference, which is effectively the same as a battery-powered photon drive that doesn't absorb the incoming photon.)

Offline dustinthewind

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Re: Woodward's effect
« Reply #683 on: 03/02/2016 11:22 PM »
All I am saying is if a projectile with momentum gets trapped in an osculating system and hits one wall giving up 20% of [its] energy but the impact with the other wall only gives back 10% of its energy that the projectile will end up giving its energy to the composite walls in a way that drains its energy.
If the energy of the photon is transferred to an object, the object gets hotter, not faster.
It gets hot if the light is striking a black body that absorbs light.  In this case there is some momentum transfer also.

No, any body gets hotter¹ if it absorbs energy from a photon. That is distinct from the momentum. Both must be accounted for. You are switching back and forth between them as if they were the same thing.
...
No, they use highly reflective mirrors to bounce lasers because... they don't get hot and melt.  A perfectly reflective surface would not get hot.  Like a superconductor has no resistance to current and doesn't get hot when current flows. 

A mirror gets hot because it has resistance to current but in a good mirror this is small.  (silver has lower resistance to current than copper and silver coating copper cavities increases the Q compared to copper.  If it has a black surface then it basically converts the current/light to heat but also receives an impulse (1/2 of a solar sail).  The parallel to a black surface is a ballistic pendulum.  The ballistic pendulum catches the ball and converts Error:half the impulse it would have received correction: the energy the ball would have escaped with, had it been reflected is converted, into heat.  The pendulum still receives half of the impulse it would have, had it reflected it.  A ballistic pendulum makes heat and moves. 

A perfect reflector is like dropping a super-ball on the floor and it bounces back to the same height and no energy is lost to heat.  Some of the balls momentum from the impact (very small because of the planets large mass) still transfers to the planet so energy/momentum is exchanged but in the real world there are no perfectly elastic collisions so some of that energy, not exchanged as momentum, will be converted to heat.  However, we can have a very close approximation to a perfectly elastic collision in which the energy lost to heat is negligible compared to the exchange of momentum.  That last phrase I used is exactly what I mean when I say an elastic collision.  http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html
http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

I've been very busy lately so haven't had much time yet.  I suspect I can mathematically and with diagram show that its possible to violate momentum direction in a closed, engineered, system after many collisions so next time I post I'll show this and the design, if my intuition is correct, or prove myself wrong.   
« Last Edit: 03/05/2016 02:31 AM by dustinthewind »

Offline Paul451

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Re: Woodward's effect
« Reply #684 on: 03/03/2016 07:26 AM »
A perfect reflector is like dropping a super-ball on the floor and it bounces back to the same height and no energy is lost to heat.

You don't have to keep trying to explain elastic collisions. The issue isn't that other people don't understand what you mean by elastic, the issue is that you say elastic, then describe an inelastic collision, but think you're still talking about an elastic one.

When you try to take back energy from the photon, you've turned your elastic collision into an inelastic one. That a) reduces the momentum transfer available to the reflective surface, and b) converts some of the photons energy into another non-vector form, such as heat.

No, they use highly reflective mirrors to bounce lasers because... they don't get hot and melt.

All mirrors have some inefficiency, they will absorb a certain amount of light.

It's not just black surfaces which absorb light. (Indeed, every non-reflective surface absorbs most light. They merely re-emit most photons except those in its absorption spectrum. That's what separates a white surface from a mirror.)

I suspect I can mathematically and with diagram show that its possible to violate momentum direction in a closed, engineered, system after many collisions

Since such a revelation would garner you a Nobel Prize in physics, I'd suggest a little modesty would be in order.

Don't you think if it was so simple to create a CoM violation from a simple reflective cavity, someone else would have also shown it long, long ago?

Offline ppnl

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Re: Woodward's effect
« Reply #685 on: 03/03/2016 03:14 PM »


You keep using weasel words like "redirecting momentum" or violating "momentum direction". Momentum is inherently directional and the whole point of momentum conservation is conserving that "momentum direction".

You need to stop with the weasel words and accept that "redirecting momentum" is exactly a violation of conservation of momentum. You have weaseled and ducked this obvious fact for long enough. 

Offline dustinthewind

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Re: Woodward's effect
« Reply #686 on: 03/07/2016 11:16 PM »
...
I suspect I can mathematically and with diagram show that its possible to violate momentum direction in a closed, engineered, system after many collisions

Since such a revelation would garner you a Nobel Prize in physics, I'd suggest a little modesty would be in order.

Don't you think if it was so simple to create a CoM violation from a simple reflective cavity, someone else would have also shown it long, long ago?
I don't really expect to find a violation of momentum or a Nobel Prize but I like to put enthusiasm into what I am hoping to find.  It might be considered being a weasel, but I see it that if we really wan't to accomplish propellant-less  propulsion there has to be some way of pushing on something we don't know how to yet (space time itself) or some way of violating momentum such as its direction.  There aren't too many options that I see yet.  Here is a design I was considering if it could possibly violate momentum by pushing on light itself, red-shifting it.  It also looks allot like the Mach/Woodward effect in which the mass of the dielectric is changing when the light passes through it.  The dielectric undergoes acceleration when it changes mass and also red-shifts the light.  The light then goes around a circuit only to undergo the same process.  I've yet to work out the momentum equations for it to see if anything extraordinary could come out. 

Light changing the mass of a dielectric:
URL: Mass change of dielectric media induced by propagation of electromagnetic waves by C.Z. Tan
« Last Edit: 03/08/2016 02:59 AM by dustinthewind »

Offline dustinthewind

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Re: Woodward's effect
« Reply #687 on: 03/15/2016 11:52 PM »
I thought this was interesting.  I was able to derive the Doppler shift of light in a dielectric that later escapes.  This is like the Doppler shift of a star moving away.  It is assumed that the star has extra mass associated with the energy it contains and later it decreases in mass and gets a velocity kick from the emitted photon.  I get the regular Doppler shift as a result. 

For a reflected photon you get twice the Doppler shift which appears to be expected as for radar guns.  I suspect I can use this later as the light passes through the dielectric changing its mass.  Later the dielectric should undergo acceleration by the swinging arm, changing its velocity.  The result should be further Doppler-shifting of the light.  The change in force of the accelerating dielectric should be related to the dielectrics change in mass and acceleration.  Some energy should have been removed from the light from the extra Doppler shift as the light goes around the cavity.  Some work remains to show that the light should gradually lose wavelength as it travels around its circuit.

Attached below is the math worked out in wxMaxima with plots. 
« Last Edit: 03/15/2016 11:54 PM by dustinthewind »

Offline GeneralRulofDumb

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Re: Woodward's effect
« Reply #688 on: 03/17/2016 08:51 AM »
I don't understand the fixation with wanting to violate any physical law. As I understand it, Woodward's device should work not in spite of, but based on existing laws.
The way I see it, momentum is conserved just fine in said device. Like a car that accelerates and 'pushes' the road/earth in the opposite direction, so too does a MET 'push' all causally connected matter in the universe in the opposite direction.
One practical problem with that device however appears to be that mechanically vibrating a solid mass has serious limits on frequencies.
And although photons can be considered particles as well, they do not have mass, so that might make it unlikely to somehow have their masses transiently 'fluctuate'.
Perhaps one should therefore consider using electrons (which do have mass) as a medium to invoke a kind of Woodward-effect. They can conveniently be packed into a kind of stream, or current.

Offline Paul451

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Re: Woodward's effect
« Reply #689 on: 03/17/2016 04:36 PM »
Like a car that accelerates and 'pushes' the road/earth in the opposite direction, so too does a MET 'push' all causally connected matter in the universe in the opposite direction.

That doesn't solve the CoE issue.

At any level of efficiency (force/power), there will be a velocity above which the device gains more kinetic energy than it expends in input energy. Even if the device "pushes all matter in the universe" in proportion with that input energy, it will still eventually be creating more energy than it consumes or transfers.

A car can't go above its "over-unity" velocity because friction with the road limits how fast it can go. If you've ever tried to find the top-speed of a car (or seen it done on TV/youtube), you'll know that it takes more time to accumulate that last ten MPH before topping out than it did to accelerate the first hundred MPH. The efficiency (force/power, or acceleration/energy in this case) changes with velocity, hence the over-unity velocity changes. The same is true with any conventional system, there's always a limiting effect.

For MET not to violate CoE/CoM, its interaction with "all matter in the universe" must be similarly velocity dependent. (Which in general, violates the core idea of relativity.)

But it's worse, the over-unity velocity for many of the proposed reactionless-drives (MET/EMdrive) is so low, just a few km/s, that even if there was a limiting mechanism, the Earth itself is already moving faster than the device's over-unity velocity relative to the sun, and the sun is moving faster than the cut-off relative to the galaxy, and the galaxy moving faster relative to the average background radiation.

In other words, even if these devices really could work, they couldn't work on Earth, or in our solar system, or in our galaxy.

You can't have "works as described" and "works on Earth" but not "violates CoE".

Online M.E.T.

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Re: Woodward's effect
« Reply #690 on: 03/17/2016 07:48 PM »
But the principle of the Mach Effect Thruster is that it is not the electrical energy of the onboard power source that is converted into acceleration. Instead, the powersource merely enables the creation of the gravinertial flux, which then exchanges momentum with the rest of the mass of the universe.

The problem, it seems to me, is that people cannot accept this instantaneous exchange of momentum with the distant matter of the entire universe, as at face value it appears to be happening much faster than the speed of light. But for this Woodward has invoked Wheeler-Feynman absorber theory, involving waves going both backward and forward in time. Something which I freely admit I don't understand at all, but which has not been refuted and which is based on solid theory, according to some very eminent people.

Hence, if the Wheeler-Feynman absorber aspect of Woodward's theory holds true, the mechanism for the instantaneous exchange of energy with the rest of the Universe can be explained, and hence, indeed, a Mach Effect Thruster also doubles as a generator of local energy, leeched from the distant universe.

Thus eventually you don't even need your original electrical power source on board your spacecraft anymore, as the Mach Effect Thruster generates its own surplus power in addition to accelerating your spacecraft to near lightspeed.

Hence, from a local point of view, it does indeed represent a free energy device, but only because it is sourcing this energy from the closed system that is the entire universe. Thus, the surplus energy that is extracted here, thanks to the Mach Effect Thruster, is taken away from the entire rest of the Universe.

So the net energy gain is zero. The system in which this takes place is just colossal - it is the entire Universe.
« Last Edit: 03/17/2016 07:55 PM by M.E.T. »

Offline tchernik

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Re: Woodward's effect
« Reply #691 on: 03/17/2016 08:20 PM »
The refusal to acknowledge that any real, working MET will generate energy in the local level due to constant acceleration at a constant energy expenditure, is something that baffles me up to today.

They are making a daring assertion already, but seem to coy from asserting the next logical one that follows.

By the way, the Emdrive has this problem too, and the proponents seem to also do their best to dodge the topic.

If they already do one daring assertion, they should go to the end and simply say they are taking energy from some cosmic source. But apparently that's too embarrassing to say out loud.

Offline birchoff

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Re: Woodward's effect
« Reply #692 on: 03/17/2016 08:47 PM »
The refusal to acknowledge that any real, working MET will generate energy in the local level due to constant acceleration at a constant energy expenditure, is something that baffles me up to today.

They are making a daring assertion already, but seem to coy from asserting the next logical one that follows.

By the way, the Emdrive has this problem too, and the proponents seem to also do their best to dodge the topic.

If they already do one daring assertion, they should go to the end and simply say they are taking energy from some cosmic source. But apparently that's too embarrassing to say out loud.

Not sure I know any supporters of MET that fight the argument that MET's could be used as power generators. All the supporters I have read on this forum and others seem to readily accept it. That said, supporters are still waiting on Woodward to improve his thrust figures and get those improved figures replicated. So I dont have a problem with any supporter not shouting from the mountain tops that they have a free energy device to save the world when at most the only thing you could point to is a few microNewtons of force that is way below the over-unity threshold.

Online M.E.T.

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Re: Woodward's effect
« Reply #693 on: 03/17/2016 09:31 PM »
Hi Birchoff

Indeed. I have long followed your posts, and the posts of others both here and on Talk Polywell, with regard to Woodward's work. To me it seems a far more promising opportunity than the EMdrive which has garnered so much attention despite no one really having a proper theory on why it supposedly works.

The Woodward effect in contrast seems to be so much more elegant, and based on solid theoretical work supported by a number of very accomplished and respected people.

I simply cannot understand why someone with billions to spare hasn't invested a few million dollars to give Woodward the proper equipment and manpower he needs to prove it once and for all.

It seems clear to me that if the Woodward effect proves to be real and scalable, it will change the fabric of our society in almost unimaginable ways. Instantly oil becomes obsolete. As do airplanes as we currently know them. Anti-gravity technology effectively becomes real, as soon as 1G of thrust becomes available.

Heck, fusion power becomes obsolete before it is even perfected.

And all of that BEFORE we even talk about the implications for space travel. Travel to Alpha Centauri in 5 years becomes possible, at 1G acceleration.

The point is, with so much at stake, it doesn't make sense that he is peddling along in his homemade lab, with the occasional undergraduate student to help him keep his bare bones device patched up and ready for the next miniscule thrust experimental run.

It is really frustrating.

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #694 on: 03/18/2016 07:22 PM »
Not sure I know any supporters of MET that fight the argument that MET's could be used as power generators. All the supporters I have read on this forum and others seem to readily accept it. That said, supporters are still waiting on Woodward to improve his thrust figures and get those improved figures replicated. So I dont have a problem with any supporter not shouting from the mountain tops that they have a free energy device to save the world when at most the only thing you could point to is a few microNewtons of force that is way below the over-unity threshold.

What do you mean no supporters fight the argument that METs could be used as power generators?  Wasn't the whole point of Woodward's essay, posted back on page 29 of this very thread (here is the essay) an attempt to refute that METs can be used as power generators?  A completely misguided attempt at that, because it is riddled with basic mechanics errors as well as a misunderstanding of how to use integrals to calculate work. 

The whole essay is (supposedly) a refutation of the idea that METs ever achieve local overunity with respect to their supplied electrical power.  If he had stuck with the explanation that METs are a sort of gravinertial transistor which only consumes electrical energy to harvest greater amounts of energy from the very distance mass of the universe, then conservation of energy would be satisfied within the framework of a constant thrust to power ratio.  Unfortunately he didn't.  From that essay, it is clear he doesn't believe METs achieve local overunity.   

So unless I'm missing something, the greatest supporter of all refutes that METs can be used for power generation, and that supporter is Woodward himself.

Offline Paul451

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Re: Woodward's effect
« Reply #695 on: 03/18/2016 09:38 PM »
But the principle of the Mach Effect Thruster is that it is not the electrical energy of the onboard power source that is converted into acceleration. Instead, the powersource merely enables the creation of the gravinertial flux, which then exchanges momentum with the rest of the mass of the universe.

That isn't adequate to remove the CoE issue.

The device produces constant force/acceleration at any specific constant input energy. That is velocity independent. However the kinetic energy available increases exponentially with velocity. The two are disconnected, due to the different effect of velocity, hence only one can be balanced by the "reaction" of the rest of the universe.

If the MET somehow produces its force by magically pushing off the rest of the universe, it's only that force that is balanced, proportional to energy input. The velocity-dependent energy output is not being balanced.

Once the velocity is high enough, over the critical value F/P, then the device will produce more kinetic energy than it consumes; and hence more than its reaction with the larger universe. Ie, that surplus is not being "balanced" by the rest of the universe. It's genuinely created energy, not just a local-portion of a larger equation.

[The only way out is if the drive isn't "pushing off" the rest of the universe, but is being dragged by an anisotropy. But that has an inherent velocity limit (the relative velocity of the anisotropy). So force/power ratio changes with velocity. And hence would produce diurnal and annual variations.]

Offline flux_capacitor

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Re: Woodward's effect
« Reply #696 on: 03/19/2016 06:49 PM »
The latest patent Jim Woodward filed in 2013 has been finally granted and assigned to SSI:

US patent 9287840, James F. Woodward, "Parametric amplification and switched voltage signals for propellantless propulsion", issued 2016-03-15, assigned to Space Studies Institute.


It covers the realistic realizations of Mach effects. Attached below as a PDF.

Offline dustinthewind

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Re: Woodward's effect
« Reply #697 on: 03/19/2016 09:50 PM »
...Here is a design I was considering if it could possibly violate momentum by pushing on light itself, red-shifting it.  It also looks allot like the Mach/Woodward effect in which the mass of the dielectric is changing when the light passes through it.  The dielectric undergoes acceleration when it changes mass and also red-shifts the light.  The light then goes around a circuit only to undergo the same process.  I've yet to work out the momentum equations for it to see if anything extraordinary could come out. 

Light changing the mass of a dielectric:
URL: Mass change of dielectric media induced by propagation of electromagnetic waves by C.Z. Tan
bold: edited at a later time
I am pretty sure this should be as simple as considering the frequency in the dielectric remains constant but the wavelength and speed of the photon changes.  f=n/lambda*c/n n=refractive index and n/lambda=photon mass c/n=velocity of photon in dielectric.  Due to the change in relative velocity of the dielectric on encounter/re-emission of the photon the photon is red-shifted in frequency and loses mass but not velocity in free space.  Therefore the dielectric pulls some of the momentum of the photon out as the photon passes through.  Normally in a cavity the energy exchanged with the cavity is a function of the photons effective mass with respect to the cavity so the photon hitting the front wall of the cavity imparts a fraction of the photons energy to the cavity and upon the photon hitting the back wall the photon regains the same fraction of energy.  The only differences is the dielectric is pulling momentum away from the photon when its going one direction.  As a result the photon loses momentum/energy over time inside the trapped cavity.

I can even get rid of the bottom dielectric by using 3 prisms to form a triangle (or 4 prisms to form a rectangle, I believe prisms can have very low reflective losses).  The return path of the photon on the 2nd leg is longer than the path through the dielectric so if the path is just right the photon always hits the osculating dielectric at just the right time. 

Strangely, if the photon enters in with positive momentum and the cavity is set to accelerate in the negative x direction the cavity can extract the photons momentum, on its leg when it has negative momentum, converting the photons positive momentum into negative momentum.  If what I am seeing is correct this may be a way to violate the direction of momentum. 

A real life application might be to make the light packets outside the cavity and inject it and allow it to lose wavelength/energy inside.  Maybe the dielectric could be osculated with a pezio-electric.  Maybe Gaussian wave-packets could be generated using a Josephson junction array.  http://link.springer.com/article/10.1007%2FBF01009578 or

Josephson Junction Oscillator Arrays
Matthew J. Lewis, Dale Durand, and Andrew D. Smith
TRW Space & Technology Group.
Peter Hadley*
Stanford University


The ability of an osculating dielectric to extract/add energy/momentum from light could possibly be observed by passing light through said dielectric of high index while osculating and observing the red-blue shift of the light that passes through.  The thickness and index of the dielectric being tuned to the wavelength of light being used such that the dielectric can reverse velocity by the time the light departs.
« Last Edit: 03/20/2016 02:27 AM by dustinthewind »

Offline birchoff

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Re: Woodward's effect
« Reply #698 on: 03/20/2016 12:17 AM »
Not sure I know any supporters of MET that fight the argument that MET's could be used as power generators. All the supporters I have read on this forum and others seem to readily accept it. That said, supporters are still waiting on Woodward to improve his thrust figures and get those improved figures replicated. So I dont have a problem with any supporter not shouting from the mountain tops that they have a free energy device to save the world when at most the only thing you could point to is a few microNewtons of force that is way below the over-unity threshold.

What do you mean no supporters fight the argument that METs could be used as power generators?  Wasn't the whole point of Woodward's essay, posted back on page 29 of this very thread (here is the essay) an attempt to refute that METs can be used as power generators?  A completely misguided attempt at that, because it is riddled with basic mechanics errors as well as a misunderstanding of how to use integrals to calculate work. 

The whole essay is (supposedly) a refutation of the idea that METs ever achieve local overunity with respect to their supplied electrical power.  If he had stuck with the explanation that METs are a sort of gravinertial transistor which only consumes electrical energy to harvest greater amounts of energy from the very distance mass of the universe, then conservation of energy would be satisfied within the framework of a constant thrust to power ratio.  Unfortunately he didn't.  From that essay, it is clear he doesn't believe METs achieve local overunity.   

So unless I'm missing something, the greatest supporter of all refutes that METs can be used for power generation, and that supporter is Woodward himself.

I engaged with a length argument on this very thready about that issue. And after reading that paper over multiple times the only thing clear in my mind is that I dont believe that essay is meant to answer the question we want it to answer. Yes I know that sounds absurd. I can even admit it sounds like denial. But its the only logical conclusion I can reach after reading that essay and arguing about it here. That last paragraph in that paper in particular feels pretty weird. how can you spend 3/4 of an essay arguing that you cannot arbitrarily choose a time to restart acceleration. But then at the end argue that is exactly what you would need to do.

My gut tells me that if he actually believes a MET behaves in the manner described in that last paragraph then the only thing that he can be describing is the oscillating nature of the fluctuating mass in a MET device. which is constantly being accelerated and deccelerated as its mass fluctuates.

Unfortunately I have no way of confirming if my gut feeling is accurate as I have no line of communication with Woodward or Fern.


Offline birchoff

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Re: Woodward's effect
« Reply #699 on: 03/20/2016 12:40 AM »
Hi Birchoff

Indeed. I have long followed your posts, and the posts of others both here and on Talk Polywell, with regard to Woodward's work. To me it seems a far more promising opportunity than the EMdrive which has garnered so much attention despite no one really having a proper theory on why it supposedly works.

The Woodward effect in contrast seems to be so much more elegant, and based on solid theoretical work supported by a number of very accomplished and respected people.

I simply cannot understand why someone with billions to spare hasn't invested a few million dollars to give Woodward the proper equipment and manpower he needs to prove it once and for all.

It seems clear to me that if the Woodward effect proves to be real and scalable, it will change the fabric of our society in almost unimaginable ways. Instantly oil becomes obsolete. As do airplanes as we currently know them. Anti-gravity technology effectively becomes real, as soon as 1G of thrust becomes available.

Heck, fusion power becomes obsolete before it is even perfected.

And all of that BEFORE we even talk about the implications for space travel. Travel to Alpha Centauri in 5 years becomes possible, at 1G acceleration.

The point is, with so much at stake, it doesn't make sense that he is peddling along in his homemade lab, with the occasional undergraduate student to help him keep his bare bones device patched up and ready for the next miniscule thrust experimental run.

It is really frustrating.

Some billionaire hasnt contributed money to woodward. Simply because no one believes it is possible. Humans who have acquired access to substantial sums of money dont get their because they easily believe. In addition, there is no real need for Woodwards technology right now. Sure Everyone here on NSF can appreciate the wonders a commercial version of a MET would bring to society.  But for the unwashed masses, if that technology isn't already on the shelf ready to be deployed it is of no use. This plus the high level of disbelief pretty much assures that any new propulsion technology will forever remain poorly funded.

Look at it this way. Everything we talk about doing in space on NSF could be done without MET's. Nuclear Thermal Rocket technology could have accomplished the same thing years ago. It could have provided us a thriving Cis-lunar  space along with probably research posts on Mars and Venus. And Thats with technology we already understand. I know the common refrain is that the fission is dangerous. But do you really believe that if humanity really felt it needed space resources to solve some important problem that we would let our fears of fission stop us from using it? I sure as hell dont.

Anyway for now. we will have to wait to see the slow development of the technology. I am very happy to see Heidi Fern continuing to develop the theoretical underpinnings of the technology. And there are replication results from external labs that are supposedly due some time this year. If those turn out to positive replication results as they have been rumored to be. And progress is made in the modelling and redesign. We could be looking forward to yet another milestone in the development of the technology. Which will cause more researchers with time and resources to invest in replications. Science is a slow trying process, even more so for us watching from the cheap seats dying to see some idea we have adopted come to fruition. But it is the best methodology humanity has for understanding nature's secrets.