Author Topic: Woodward's effect  (Read 284926 times)

Offline JohnFornaro

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Re: Woodward's effect
« Reply #640 on: 02/18/2016 08:28 PM »
Electricity, as energy, is like mass, which is like physical propellant, according to the well known equation of equivalence of mass and energy. 

The "F" in the F=ma equation above, has an energy/mass component.  The idea of 'over unity', broadly speaking, is incorrect in the example of kinetic energy ever exceeding the energy expended in accelerating the spaceship.

Per the oracle:

"In physics, a force is any interaction that, when unopposed, will change the motion of an object. In other words, a force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate. Force can also be described by intuitive concepts such as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newtons and represented by the symbol F."

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

The force expended is said in the PDF to be constant without a time variation, and that cannot be the case.  An MET thruster will have to have a source of energy, gasoline, loosely speaking.  If it should work, a spaceship made under this principle, once started, will eventually run out of gas and stop accelerating.  A rocket is a closed system.

There has been some discussion buried in these several threads, that a rotational force might be detected, albeit slightly above the level of noise in the system.  If that is the case, using an MET thruster, one could make an airplane or a boat or a car or a drilling rig, but not a spaceship.

The physics that explains all this, until proven otherwise, is not "fundamentally different".

Gargoyle's comment holds:

"You can't talk about a constant ratio of force to input power in a time invariant fashion for a closed system if you want to respect conservation of energy over time."

In other words, where, on the spaceship, does the energy to propel the system come from?

This discussion of 'over unity' has not yet been shown to have merit.  Shown, as in floating the device into a conference room, or by a peer reviewed paper which holds up to simple scrutiny on conservation of energy.

As an aside, the idea that electricity could be converted into forward momentum would be useful enough on its own merits, by eliminating the need to carry traditional propellants, even if it should still require an on board source of power.
Sometimes I just flat out don't get it.

Offline ppnl

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Re: Woodward's effect
« Reply #641 on: 02/19/2016 07:13 AM »
You can't talk about a constant ratio of force to input power in a time invariant fashion for a closed system if you want to respect conservation of energy over time.

I just read the "Mach Effect Thrusters (Mets) And 'Over-Unity' Energy Production" paper.

As is known, F=ma, that is, F=dp/dt=Ma=Mdv/dt.

The paper goes on to assume that F is constant, but this cannot be, except within a narrow time frame in a closed system such as a rocket.  In classical rocketry, one carries one's F with them as propellant.  In an MET thruster, one also needs to carry one's F with them, in this case a source of electrical power, which will inevitably run out.

Again, the Force can remain constant, within the limits of the machinery for a certain finite amount of time, but eventually, one runs out of F, that is, electricity.

To my thinking, this means that one never gets to an 'over unity' situation, because one runs out of gas, as it were.

If there were ever to be an MET rocket, it would accelerate to some speed, run out of power, and then it would have a constant velocity, until that velocity was perturbed by another gravitational body.

Hopefully not head on.

It is true that constant force over enough time will violate conservation of energy. But a MET with constant force until the battery runs down will also violate conservation of energy. Different frames of reference will disagree on how much energy the MET rocket generated but all will agree on how much energy the batteries contain. For some frames of reference the energy gained by the MET rocket exceeds the energy content of the batteries and thus violates conservation of energy.

A rocket gets around this with it's exhaust. Different frames of reference will disagree on how much energy the chemical rocket gains but also disagree on how much energy the fuel and exhaust contain. This allows you to balance conservation of energy and conservation of momentum in all reference frames.

The mass equivalent of the energy in the battery is so small as to be useless for conserving momentum and so useless for conserving energy. Nothing is expelled anyway so you can't balance momentum anyway. E=MC^2 is simply irrelevant to this discussion. Electrical energy is in no sense like the mass of a propellant.

In order to obey energy conservation you have to find something that the MET is pushing against. The amount of energy needed to accelerate then depends on the relative motion between the MET and the thing it is pushing against.   

Offline JohnFornaro

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Re: Woodward's effect
« Reply #642 on: 02/19/2016 12:08 PM »
"It is true that constant force over enough time will violate conservation of energy."

Well, glad we agree with reality as is currently known by science.

Again, this discussion of 'over unity' has not yet been shown to have merit.  Shown, as in floating the device into a conference room, or by a peer reviewed paper which holds up to simple scrutiny on conservation of energy.

Got math?
Sometimes I just flat out don't get it.

Offline aceshigh

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Re: Woodward's effect
« Reply #643 on: 02/19/2016 12:16 PM »
Why doesn't a Bussard Ramjet violates conservation of energy? It collects fuel from the interstellar medium and keeps accelerating forever (like in the excellent book Tau Zero)

Isn't any analogy with reactionless drives possible, just that the "fuel" it captures is, lets say, more immaterial?

Offline ppnl

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Re: Woodward's effect
« Reply #644 on: 02/19/2016 04:39 PM »
Why doesn't a Bussard Ramjet violates conservation of energy? It collects fuel from the interstellar medium and keeps accelerating forever (like in the excellent book Tau Zero)

Isn't any analogy with reactionless drives possible, just that the "fuel" it captures is, lets say, more immaterial?

A Bussard Ramjet is reacting against an external medium, the interstellar gas. It isn't violating conservation of energy any more than a car. A car reacts against an external medium, the road. That allows a car to accelerate forever although it will need ever increasing amounts of power. A Bussard Ramjet will also need ever increassing amounts of power but it gets it because the faster it goes the more gas it collects. But at some point the energy it gets from fusion will not equal the energy lost by collecting the gas.

Offline ppnl

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Re: Woodward's effect
« Reply #645 on: 02/19/2016 05:08 PM »
"It is true that constant force over enough time will violate conservation of energy."

Well, glad we agree with reality as is currently known by science.

Again, this discussion of 'over unity' has not yet been shown to have merit.  Shown, as in floating the device into a conference room, or by a peer reviewed paper which holds up to simple scrutiny on conservation of energy.

Got math?

Say you have a 10 ton MET rocket. You turn on the rocket and accelerate to ten miles per second. The energy gain will be on the order of 10 * 10^2 = 1000 units of energy in appropriate units. If your battery was charged with 1000 units of energy then there appears to be no problem.

But hold on. Say I watched the events from a different frame of reference. Say I was moving backwards at 10 miles per second. Because motion is relative I'm allowed to take my velocity as zero and say you have an initial velocity of 10 miles per second. So I see you accelerate from 10 miles per second to 20 miles per second. I see your energy gain as 10 * 20^2 - 10 * 10^2 = 3000 units of energy. Since your battery only contains 1000 units of energy I see this as a violation of conservation of energy.

The basic problem is that kinetic energy is frame dependent while battery power is not.

A chemical rocket does not have this problem since while the kinetic energy is still frame dependent the energy of the fuel energy is also frame dependent. The two balance in a way that allows us to conserve energy in all frames of reference.

For the MET to avoid violating conservation of energy you will need to identify something that it is pushing against. Like the Bussard Ranjet pushing against the interstellar medium, the car pushing against the road, the airplane pushing against the air, the rocket pushing against its fuel...

It is that "pushing against" that allows you to balance both energy and momentum in all frames of reference.   

Offline Paul451

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Re: Woodward's effect
« Reply #646 on: 02/19/2016 08:03 PM »
Adding to ppnl's reply:

Why doesn't a Bussard Ramjet violates conservation of energy? It collects fuel from the interstellar medium and keeps accelerating forever (like in the excellent book Tau Zero)

In collecting the fuel, the Bussard ramjet experiences drag. As velocity increases, eventually the drag equals the thrust and the ship can no longer accelerate unless you increase the power input. Or in other words, over time the force output per unit energy input is not constant.

It's the constant thrust per input energy regardless of velocity that causes the over-unity issue.

The only analogy for a reactionless thruster is a photon drive. However, the thrust produced by the perfect emission of a photon, versus the energy absorbed by the photon, means that the velocity at "unity" (energy in equals energy out) is when the emitter is at exactly the speed of light. So a photon drive only achieves "over-unity" when the emitter is travelling faster than light.

Any reactionless thruster with a thrust-out/energy-in ratio greater than a photon drive is able to achieve over-unity at velocities below light-speed. Hence they must either be impossible, or require a very very deep rewriting of fundamental physics (like all of it).

(In theory, you could have an apparently reactionless thruster which is actually interacting with something like dark matter. Ie, it changes the velocity of the dark matter it interacts with like a jet through air. Such a dark-matter-drive won't achieve over-unity due to limits on its interaction with the dark-matter. But the experimental results that supposedly show anomalous force aren't consistent with what you'd expect from dark matter interaction.

Note: interaction with "quantum vacuum", "virtual particles" or "deep background" doesn't work like that, since there's no inherent or preservable velocity in those systems, therefore no velocity-dependency, therefore no upper limit on energy-out. They still have the over-unity issue.)

Offline dustinthewind

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Re: Woodward's effect
« Reply #647 on: 02/21/2016 07:34 AM »
When you use conservation of momentum and you find the change in velocity of the ship with respect to the change in velocity of the photon upon emmission and then you take the integral with respect velocity then you get the initial energy which is zero plus some added constant = the energy divided between the two masses and some heat.  When you divide everything by that energy constant that was added to the system and find the percentage of energy to the photon it is almost 100% and about 0% to the engine.  So I think I get why photon propulsion is inefficient.

However, now lets us consider the light reflected between two free floating and separate mirrors and suddenly photon propulsion becomes much more efficient after many reflections.  In effect the light is red-shifted till it becomes efficient for propulsion.  At least here the energy is divided equally between the two mirrors. 

I wonder if it is possible to have a single device (closed system) that has two mirrors but one mirror is more effective at harnessing energy from the photon than the other mirror.  Maybe by giving the photon some added mass when striking one mirror? (one mirror being a super-conductor possibly?).  http://physics.stackexchange.com/questions/47791/what-do-massive-photons-have-to-do-with-superconductivity .  (The efficiency of photon propulsion having to do partly with the "effective" mass of the photon?)  It might be like firing off a higher frequency photon that you red-shifted into non-existence?  You still have a photon that leaves (possibly at a long enough wavelength) but most of its energy is transferred to one side of the engine?  This vaguely reminds me of Dr. Rodal reminding us of some ones suggestion to have one end plate as fero-magnetic [EM drive thread]. 

Still thinking on this but wondering if it's a possibility. 
« Last Edit: 02/21/2016 08:11 AM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #648 on: 02/21/2016 06:09 PM »
dustinthewind,

The problem is all the equations governing electromagnetism are time symmetric. That is the ultimate source of conservation of momentum. If you want to violate conservation of momentum you will have to disprove the entirety of electromagnetic theory. If it is wrong then you cannot use it to derive your drive. You will have to find where it gives incorrect predictions and use that to derive the correct theory. Then use the correct theory to invent your drive.

That will be hard because they have looked really really hard and the universe seems to be time symmetric at all relevant energy.

   

Offline dustinthewind

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Re: Woodward's effect
« Reply #649 on: 02/22/2016 01:44 AM »
dustinthewind,

The problem is all the equations governing electromagnetism are time symmetric. That is the ultimate source of conservation of momentum. If you want to violate conservation of momentum you will have to disprove the entirety of electromagnetic theory. If it is wrong then you cannot use it to derive your drive. You will have to find where it gives incorrect predictions and use that to derive the correct theory. Then use the correct theory to invent your drive.

That will be hard because they have looked really really hard and the universe seems to be time symmetric at all relevant energy.

I don't intend to violate either conservation of momentum or energy.  I used both in the math below. 

Imagine a ball created out of one wall, inside the ship, similar to the wall emitting a photon.  Now imagine this ball is perfectly elastic and has an initial velocity.  We are going to do a trick where we modify the mass of the ball when it strikes one of the walls such that the energy exchange to the walls is more efficient. When the ball strikes the other wall it is back to its previous mass where the energy exchange is less efficient.  Eventually the ball will slow down because its energy is being given to the structure as a whole.

The ball is in analogy to a photon so instead of slowing down it is red-shifted.

I attached a gif of the math I did below.  It shows the % of energy exchanged between a wall and the ball depends on the mass of the ball.  You then have to consider independently the exchange between two different walls where the mass of the ball is modified with each bounce.  This is the reason I suggested the super-conductor because I think it suggested it could modify the mass of the photon.  (One wall being a superconductor.)  The idea is supposed to conserve momentum. 

The ball will of course change velocity with each bounce but we then sub the new velocity in and figure the exchange of energy when it bounces again.  Eventually most the energy is given to one wall rather than the other.

I think there was an error in the derivation in the image.  I have fixed the derivation.  The % energy exchanged should be 100% if they are equal in mass. 
« Last Edit: 07/14/2016 11:17 PM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #650 on: 02/22/2016 04:02 AM »
dustinthewind,

The problem is all the equations governing electromagnetism are time symmetric. That is the ultimate source of conservation of momentum. If you want to violate conservation of momentum you will have to disprove the entirety of electromagnetic theory. If it is wrong then you cannot use it to derive your drive. You will have to find where it gives incorrect predictions and use that to derive the correct theory. Then use the correct theory to invent your drive.

That will be hard because they have looked really really hard and the universe seems to be time symmetric at all relevant energy.

I don't intend to violate either conservation of momentum or energy.  I used both in the math below. 

Imagine a ball created out of one wall, inside the ship, similar to the wall emitting a photon.  Now imagine this ball is perfectly elastic and has an initial velocity.  We are going to do a trick where we modify the mass of the ball when it strikes one of the walls such that the energy exchange to the walls is more efficient. When the ball strikes the other wall it is back to its previous mass where the energy exchange is less efficient.  Eventually the ball will slow down because its energy is being given to the structure as a whole.

The ball is in analogy to a photon so instead of slowing down it is red-shifted.

I attached a gif of the math I did below.  It shows the % of energy exchanged between a wall and the ball depends on the mass of the ball.  You then have to consider independently the exchange between two different walls where the mass of the ball is modified with each bounce.  This is the reason I suggested the super-conductor because I think it suggested it could modify the mass of the photon.  (One wall being a superconductor.)  The idea is supposed to conserve momentum. 

The ball will of course change velocity with each bounce but we then sub the new velocity in and figure the exchange of energy when it bounces again.  Eventually most the energy is given to one wall rather than the other.

But that is as simple and obvious a violation of conservation of momentum as you could ever hope for. And it leads to an immediate violation of conservation of energy. If you turn on your engine and accelerate up to some non zero momentum then you have created that momentum out of nothing. The only way to avoid it is if you can point to something that was accelerated in the opposite direction and has negative momentum to cancel out your positive momentum.


Offline Paul451

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Re: Woodward's effect
« Reply #651 on: 02/22/2016 03:24 PM »
Imagine a ball created out of one wall, inside the ship, similar to the wall emitting a photon.  Now imagine this ball is perfectly elastic and has an initial velocity. We are going to do a trick where we modify the mass of the ball when it strikes one of the walls such that the energy exchange to the walls is more efficient. When the ball strikes the other wall it is back to its previous mass where the energy exchange is less efficient.  Eventually the ball will slow down because its energy is being given to the structure as a whole.
The ball is in analogy to a photon so instead of slowing down it is red-shifted.

Where does the initial velocity of the ball come from? Or in the case of photons, where do the photons come from?

Ball:

If the ball is accelerated up to its initial velocity by a mechanism within or connected to the ship, then clearly the ship will move in the opposite direction during the set-up. As the ball bounces off the magic-wall, it will lose momentum precisely to return the ship to its original velocity, no net velocity will be added.

If the balls are accelerated up to their initial velocity by the ship, but are allowed to leak away after they reflect off the magic-wall rather than being retained in the bouncy-chamber, then you've just invented a rocket engine - using balls as propellant. It has the same maths as any rocket engine.

If the balls are accelerated by an outside mechanism, you've merely invented a weird ball-wind sailing ship. However, it has the same limits as any sailing ship.

Photon:

If the photons are emitted within the chamber, then the ship accelerates in the opposite direction. It's just a conventional photon drive, all you've done is add an extra reflector. Ie, the magic-wall does nothing except steer the thrust of the drive.

If the photons are not allowed to leave the chamber, then it's not even a drive. Like the ball-chamber, the final velocity equals the initial velocity, with no net change.

If the photons come from from an outside source, your magic-wall is just a solar-sail.

What you are imagining happening, can't happen. Your maths fails to account for half the situation, how the initial energy imbalance is set-up. Once you include the set-up, the problem collapses into a mundane system, a rocket or a sail.

Offline dustinthewind

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Re: Woodward's effect
« Reply #652 on: 02/23/2016 01:32 AM »
dustinthewind,

The problem is all the equations governing electromagnetism are time symmetric. That is the ultimate source of conservation of momentum. If you want to violate conservation of momentum you will have to disprove the entirety of electromagnetic theory. If it is wrong then you cannot use it to derive your drive. You will have to find where it gives incorrect predictions and use that to derive the correct theory. Then use the correct theory to invent your drive.

That will be hard because they have looked really really hard and the universe seems to be time symmetric at all relevant energy.

I don't intend to violate either conservation of momentum or energy.  I used both in the math below. 

Imagine a ball created out of one wall, inside the ship, similar to the wall emitting a photon.  Now imagine this ball is perfectly elastic and has an initial velocity.  We are going to do a trick where we modify the mass of the ball when it strikes one of the walls such that the energy exchange to the walls is more efficient. When the ball strikes the other wall it is back to its previous mass where the energy exchange is less efficient.  Eventually the ball will slow down because its energy is being given to the structure as a whole.

The ball is in analogy to a photon so instead of slowing down it is red-shifted.

I attached a gif of the math I did below.  It shows the % of energy exchanged between a wall and the ball depends on the mass of the ball.  You then have to consider independently the exchange between two different walls where the mass of the ball is modified with each bounce.  This is the reason I suggested the super-conductor because I think it suggested it could modify the mass of the photon.  (One wall being a superconductor.)  The idea is supposed to conserve momentum. 

The ball will of course change velocity with each bounce but we then sub the new velocity in and figure the exchange of energy when it bounces again.  Eventually most the energy is given to one wall rather than the other.

But that is as simple and obvious a violation of conservation of momentum as you could ever hope for. And it leads to an immediate violation of conservation of energy. If you turn on your engine and accelerate up to some non zero momentum then you have created that momentum out of nothing. The only way to avoid it is if you can point to something that was accelerated in the opposite direction and has negative momentum to cancel out your positive momentum.

A photon is traveling through space and has momentum.  In the frame we choose there is also a cavity with no momentum.  The photon is traveling towards the cavity with no momentum and the cavity window opens and lets in this poor soul of a photon.  The window closes behind and the photon is now trapped.  The photon hits one wall of the cavity and loses 1% of its energy to the cavity (much less for a real photon) and is red-shifted.  The photon rebounds and now hits the back wall transferring its momentum to the back wall slowing the cavity again but re-gaining its energy (blue-shifted).  This cavity-photon system now has the momentum of the single photon. 

Now we introduce the magic cavity (joke, as its not really magic, hopefully), where the cavity modifies the mass of the poor soul of the photon that enters and is trapped, but respects conservation of momentum.  We again open the window and the photon enters and we close the window.  This time the photon strikes the wall that modifies its mass slowing the photon and makes the photon more effective at transferring a % of its energy to the cavity.  This time the photon transfers 2% (exaggerated) of its energy to the cavity and then rebounds, having lose a bit more energy is more red-shifted.  As it travels back to the back wall it rebounds but this time only regains 1% of its energy it lost because of its reduced mass with increased velocity.  The cycle continues and the photon continues to lose energy till at last it comes to rest at the back wall (or just red-shifts into oblivion).  The entire energy has been drained from the photon.  When ~100% of the energy is drained from the photon the momentum is transferred to the cavity but in this case all of the momentum has been transferred to the cavity.  In the first example only maybe 1% of the energy from the photon ever resided in the cavity and only periodically.  The momentum of both systems should be conserved.  I am not quite sure yet about when the photon enters the cavity in the opposite direction as the cavity will gain momentum as this seems to violate momentum.

this parallels to using 2 free floating mirrors to more effectively suck the energy out of photons more efficiently but does it inside of a cavity.  The two ships catching photons and trapping them to both accelerate in opposite directions may also have conservation of momentum problems now that I think of it, considering the initial photon's momentum is now divided in two directions, though unevenly. 

With balls the cavity would fill up but with photons it shouldn't. 

I hope this also addresses Paul's inquiry.  If not, just let me know more specifically what the question was.  I don't think it matters where the photons come from as long as they are trapped inside the cavity.  As far as modifying the mass of a photon well... I was considering a superconductor, a dielectric, maybe a photon between two Casimir force effected plates.  I'll have to look around, and think more about it. 

You can change the mass with out changing the momentum.  If you increase the mass you decrease the velocity.  Changing the mass of the ball may be difficult but changing the effective mass of the photon I am not sure sure that is impossible.  One might wonder why light slows [non-locally] near gravitational masses or in dielectrics for example.
« Last Edit: 02/23/2016 04:11 AM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #653 on: 02/23/2016 02:13 AM »

But you cannot change the momentum or mass of either a photon or a ball that way. By definition if you do you violate conservation of momentum.

Photons are the quanta of the electromagnetic force. In mediating that force they can carry negative momentum in the positive direction, have mass and even travel back in time or travel faster than light. But they are only virtual photons. They don't exist. They are children of the uncertainty principle. They borrow what they need to violate whatever but they must pay it back in a way that leaves no violation. Otherwise quantum mechanics would not be time reversible. They are not quantum observables and you should not think of them as physical things involved in physical events. Instead think of them as a way to visualize a mathematical formalism. That mathematical formalism does not allow violations of conservation of momentum any more than it allows sending messages faster than light or backwards in time. All of those things destructively interfere with themselves and cancel out.

In the end a black box, forget about any mechanism inside, that can accelerate to some velocity without pushing something in the opposite direction is violating conservation of momentum. Quantum mechanics does not allow that. 

Online Stormbringer

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Re: Woodward's effect
« Reply #654 on: 02/23/2016 03:19 AM »
so a photon is the quantum of the EMF which is composed of a magnetic component and an electrical component. so what makes a photon? does an electron "date" a monopole? what?

When antigravity is outlawed only outlaws will have antigravity.

Offline ppnl

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Re: Woodward's effect
« Reply #655 on: 02/23/2016 05:44 AM »
so a photon is the quantum of the EMF which is composed of a magnetic component and an electrical component. so what makes a photon? does an electron "date" a monopole? what?

I don't think the question makes sense. If you have a field with the properties of the electric field then when the strength of that field changes it creates a field with the properties of a magnetic field. A changing magnetic field creates an electric field. That is a property of the electric and magnetic field. If you quantize this field you get a point like particle with all the properties of a photon. It must be mass-less because the range of the electric field is infinite, it must have an odd spin number because the electric field is both attractive and repulsive and so on. It takes all its properties from the field that it is a quantization of. Nothing more needs to be said about it.

Now it remains possible that the photon has other properties like for example a very small mass. But that would be new and very unexpected physics for which there is no empirical evidence. In fact the mass would have to be very very tiny because otherwise it would change the nature of the electric field in observable ways. Its properties are constrained by the field that it quantizes.

 

Offline Paul451

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Re: Woodward's effect
« Reply #656 on: 02/23/2016 06:20 PM »
The photon rebounds and now hits the back wall transferring its momentum to the back wall slowing the cavity again but re-gaining its energy (blue-shifted).

That's the part that doesn't happen.

Offline dustinthewind

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Re: Woodward's effect
« Reply #657 on: 02/24/2016 12:04 AM »
The photon rebounds and now hits the back wall transferring its momentum to the back wall slowing the cavity again but re-gaining its energy (blue-shifted).

That's the part that doesn't happen.
How so?

Offline Paul451

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Re: Woodward's effect
« Reply #658 on: 02/24/2016 02:38 PM »
The photon rebounds and now hits the back wall transferring its momentum to the back wall slowing the cavity again but re-gaining its energy (blue-shifted).
That's the part that doesn't happen.
How so?

The photon can't recover its original energy while imparting less momentum on the second wall than it transferred to the first.

After the photon pushes the front wall and hence the ship, the photon is blue-shifted relative to the now-moving back wall. When it reflects, it will therefore impart more energy than it gave the front wall, relative to the new frame-of-reference of the ship. Correcting for the ship's change of frame, the extra energy imparted on the back wall will be exactly the same energy as the front wall, relative to a stationary observer. Ie no net acceleration.

If the energy transfer process is somehow asymmetrical, and the front wall somehow reflects "harder" than the back wall and so the ship remains moving forwards after the back-wall reflection, then the photon will be remain red-shifted (lower energy) relative to the front-wall after the rear-wall bounce. The photon will always have less energy to give the front wall than it has to give the back, proportional to the excess energy from the first bounce. That red/blue asymmetry between the front and back walls will always be equal and opposite to the asymmetry in the hardness-of-reflectivity of the two walls.

The net effect is always going to be a ship that sits stationary, getting a bit warmer as it converts the stored energy that powers the magic wall into waste heat.

Looking back at your description, I get the impression that you've visualised the red/blue asymmetry occurring in the same direction as the reflection asymmetry. That cannot happen.

So ultimately, an individual incoming photon has a specific net energy vector. It doesn't matter what you do, how you transform that photon, the net energy vector for the whole system (photon+ship) must remain the same.

What you've described is ultimately just an elaborate way for the front wall to absorb the energy of the incoming photon, there's no "pumping" effect from the cavity. And if you absorb the energy of the incoming photon, you simply convert one unit-vector of energy from the photon into one unit-vector of energy of the ship. No matter how you absorb that photon's energy, the output can only ever be one unit of energy in the same vector as the original photon.

(If you reflect the incoming photon (solar sail), you can have a system which is two units forward and one unit aft. So your "red-shift cavity" ship is actually less efficient than a solar sail of the same cross-sectional area.)
« Last Edit: 02/24/2016 02:41 PM by Paul451 »

Offline dustinthewind

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Re: Woodward's effect
« Reply #659 on: 02/24/2016 05:14 PM »
The photon rebounds and now hits the back wall transferring its momentum to the back wall slowing the cavity again but re-gaining its energy (blue-shifted).
That's the part that doesn't happen.
How so?

The photon can't recover its original energy while imparting less momentum on the second wall than it transferred to the first.
The first example is of a cavity that doesn't absorb the photons energy. Only the 2nd example does this.

After the photon pushes the front wall and hence the ship, the photon is blue-shifted relative to the now-moving back wall. When it reflects, it will therefore impart more energy than it gave the front wall, relative to the new frame-of-reference of the ship. Correcting for the ship's change of frame, the extra energy imparted on the back wall will be exactly the same energy as the front wall, relative to a stationary observer. Ie no net acceleration.
Correct on no net acceleration, however, the cavity has moved a bit having both accelerated and then decelerated.  This is still the first example where the photon recovers its original wavelength which is (blue shifted) with respect to its previous red shift but technically it just recovers its original wavelength.  We are dealing with a perfectly elastic photon so I should say we are dealing with a superconductive cavity.  The frame I am considering is a frame that is independent of the cavity or the photon.  Originally, it was stationary with the cavity.  Sort of a lab frame.  The wall that first welcomed the photon is moving away from that frame when re-emitting the photon so I consider it red-shifted as well as being red-shifted having imparted some of its energy into the cavity.

If the energy transfer process is somehow asymmetrical, and the front wall somehow reflects "harder" than the back wall and so the ship remains moving forwards after the back-wall reflection, then the photon will be remain red-shifted (lower energy) relative to the front-wall after the rear-wall bounce. The photon will always have less energy to give the front wall than it has to give the back, proportional to the excess energy from the first bounce. That red/blue asymmetry between the front and back walls will always be equal and opposite to the asymmetry in the hardness-of-reflectivity of the two walls.
Ok this is the 2nd example.  It's correct the photon becomes more and more red-shifted in this case.  By the ballistics the photon being heavier when it strikes first wall then more momentum is transferred to the cavity 2%.  With it also being perfectly elastic and throwing the photon back it receives twice the momentum.  On striking the back wall it never recovers its previous wavelength (energy) w.r.t. the lab frame and remains red-shifted.  Having only recovered 1% its original energy and with reflection, twice this, then it is now short by -4%+2%=-2%, or this is just a rough approximation.  With this reduced energy it continues its previous cycle. 

The net effect is always going to be a ship that sits stationary, getting a bit warmer as it converts the stored energy that powers the magic wall into waste heat.
mmm, maybe I didn't clarify early on that the stipulation was that the photon was perfectly elastic.  A real photon of course will also generate waste heat as a real ball isn't perfectly elastic but I am assuming for a superconductor it should be about 100% elastic.  For being perfectly elastic all the energy goes into the ballistics and is converted into momentum (ideal assumption).

Looking back at your description, I get the impression that you've visualised the red/blue asymmetry occurring in the same direction as the reflection asymmetry. That cannot happen.
I am not sure we actually disagree with the change in wavelength.  I was analyzing from a non-accelerated frame that was independent of the cavity, since the cavity is periodically accelerated.

So ultimately, an individual incoming photon has a specific net energy vector. It doesn't matter what you do, how you transform that photon, the net energy vector for the whole system (photon+ship) must remain the same.
Correct, there is conserved energy.  In the first example I gave, the photon only ever gave the cavity a very small percentage of its momentum/energy, and only periodically.  The 2nd example the cavity absorbs everything.

What you've described is ultimately just an elaborate way for the front wall to absorb the energy of the incoming photon, there's no "pumping" effect from the cavity. And if you absorb the energy of the incoming photon, you simply convert one unit-vector of energy from the photon into one unit-vector of energy of the ship. No matter how you absorb that photon's energy, the output can only ever be one unit of energy in the same vector as the original photon.

(If you reflect the incoming photon (solar sail), you can have a system which is two units forward and one unit aft. So your "red-shift cavity" ship is actually less efficient than a solar sail of the same cross-sectional area.)
The cavity should be more efficient than a solar sail in that a solar sail lets the photon escape with energy.  The photon only ever transfers a small amount of its momentum to the solar sail due to its incredibly minuscule mass/energy (mass being related to energy) with respect to the sail.  This is the main reason photon propulsion is so inefficient.  Now two ships with a mirror each and accelerating away from each other is a much more efficient form of propulsion because they suck the wavelength (energy) out of the light.  This is exactly what the 2nd example I gave does but it doesn't have the range problems inherent of two mirrors moving far apart.  In effect it relies on modifying the "effective-relativistic" mass of a photon and applying force periodically on that photon which I suspect is related to the Woodward effect on this thread. 

The trick is finding some method of manipulating the effective mass of a photon.  Some ideas are superconductors are supposed to make photons heavy?, or maybe dielectrics slow light because momentum conserved they increase lights mass, or modifying the vacuum energy with the Casimir force could possibly make light heavier/lighter.  I'm still looking for that trick.

« Last Edit: 02/24/2016 10:44 PM by dustinthewind »