We know that on the electromagnetic wave propagation path, the direction of electric field at a point changes periodically. If there is a charged object at the point also changes its charge property periodically with the same frequency, then the electric field force direction on the object will not change. Therefore, the object will do directional movement. Based on this, it is possible to design a spacecraft without propellant. And the calculation suggests that the thrust can be much higher than radiation pressure. With the Electromagnetic Momentum Conservation Equation, this paper also proves that in a limited space, momentum can be not conserved although it is still conserved in the whole universe.

Quote from: ZhixianLin on 12/11/2015 08:29 amWe know that on the electromagnetic wave propagation path, the direction of electric field at a point changes periodically. If there is a charged object at the point also changes its charge property periodically with the same frequency, then the electric field force direction on the object will not change. Therefore, the object will do directional movement. Based on this, it is possible to design a spacecraft without propellant. And the calculation suggests that the thrust can be much higher than radiation pressure. With the Electromagnetic Momentum Conservation Equation, this paper also proves that in a limited space, momentum can be not conserved although it is still conserved in the whole universe.Thank you for your contribution There are several issues for discussion on your report, for example, to pick one:If a force were to be produced, then work would be done when the electromagnetic engine moves in space (Work = Force * displacement). If the force is produced without energy radiation (or mass outflow), the work is performed without spending energy. Then, the proposed electromagnetic device would be a perpetual-motion machine.It is curious that proponents of propellant-less electromagnetic space propulsion that don't rely on external fields for propulsion, like electromagnetic tethers, etc., seem to avoid the consequence of their device being a perpetual motion machine. If the device is a perpetual motion machine, it could be used for energy purposes here on Earth, and such discussions (of using their electromagnetic self-accelerators for energy production on Earth instead of for transportation purposes) are avoided.

On the topic of conservation of momentum. I have a problem with what happens when two gases of different temperature mix. If you conserve kinetic energy, then the mixed gas has more potential momentum than both gases had previous to mixing,KE = ρ2/2Mass . For a gas the KE is potential directional momentum as it can be given direction by letting it go through an orifice such as a rocket nozzle. Attaching an example using Helium to avoid some complications if you use gases that have higher adiabatic indexes, not that I don't think it applies to all matter that are in the gaseous state of matter.

Quote from: Rodal on 12/18/2015 03:14 pmQuote from: ZhixianLin on 12/11/2015 08:29 amWe know that on the electromagnetic wave propagation path, the direction of electric field at a point changes periodically. If there is a charged object at the point also changes its charge property periodically with the same frequency, then the electric field force direction on the object will not change. Therefore, the object will do directional movement. Based on this, it is possible to design a spacecraft without propellant. And the calculation suggests that the thrust can be much higher than radiation pressure. With the Electromagnetic Momentum Conservation Equation, this paper also proves that in a limited space, momentum can be not conserved although it is still conserved in the whole universe.Thank you for your contribution There are several issues for discussion on your report, for example, to pick one:If a force were to be produced, then work would be done when the electromagnetic engine moves in space (Work = Force * displacement). If the force is produced without energy radiation (or mass outflow), the work is performed without spending energy. Then, the proposed electromagnetic device would be a perpetual-motion machine.It is curious that proponents of propellant-less electromagnetic space propulsion that don't rely on external fields for propulsion, like electromagnetic tethers, etc., seem to avoid the consequence of their device being a perpetual motion machine. If the device is a perpetual motion machine, it could be used for energy purposes here on Earth, and such discussions (of using their electromagnetic self-accelerators for energy production on Earth instead of for transportation purposes) are avoided.... unless electricity in > work out....I don't think that you are right that an electromagnetic drive is a perpetual motion device. If, for example, the QV receives energy, it is obvious why this is a nonconcern.

that don't rely on external fields for propulsion, like electromagnetic tethers, etc.,

you are not right

Pardon me Dr. Rodal? "I do not think you are right ... [content]" does not constitute me saying "You are not right." There is no more clear way to express an opinion than the preface "I do not think that." I genuinely question why you edited my quote and why you read my post in such a fashion, and the only inference that seems reasonable is that you have no interest in engaging with me.If you think an emdrive as conceived on this forum is a perpetual motion device, please explain to me the fashion in which you could extract energy from the following device, and assume it does not push against an external field. One has an emdrive that weighs 1kg and travels 1m over a table's surface when 1000000 watts is supplied to the antenna. (For reasons of simplicity, let's say that this is a burst of RF impulse, not constantly applied.) I think it becomes clear that a such-described drive does not constitute a perpetual motion schema in any case, until the power efficiency reaches a certain figure (that is, the point at which spent electrical energy exceeds the potential kinetic energy of the drive. Up until this point it is just in theory a momentum transfer.) So if I am wrong you should be able to point out why I am wrong under the confines of this thought experiment. I have a feeling you are about to tell me that there is no known method of converting electrical energy directly to kinetic potential-- which we both full well as adults understand-- which is not the debate at hand.The question is whether such a device that actually does so would constitute by definition a "perpetual energy machine." In this sense I hope I am wrong. No issues there. Feel free to tell me so if it is your opinion that's the case. That's how we have a dialogue right?

Quote from: oliverio on 12/18/2015 11:06 pmPardon me Dr. Rodal? "I do not think you are right ... [content]" does not constitute me saying "You are not right." There is no more clear way to express an opinion than the preface "I do not think that." I genuinely question why you edited my quote and why you read my post in such a fashion, and the only inference that seems reasonable is that you have no interest in engaging with me.If you think an emdrive as conceived on this forum is a perpetual motion device, please explain to me the fashion in which you could extract energy from the following device, and assume it does not push against an external field.One has an emdrive that weighs 1kg and travels 1m over a table's surface when 1000000 watts is supplied to the antenna. (For reasons of simplicity, let's say that this is a burst of RF impulse, not constantly applied.) I think it becomes clear that a such-described drive does not constitute a perpetual motion schema in any case, until the power efficiency reaches a certain figure (that is, the point at which spent electrical energy exceeds the potential kinetic energy of the drive. Up until this point it is just in theory a momentum transfer.) So if I am wrong you should be able to point out why I am wrong under the confines of this thought experiment. I have a feeling you are about to tell me that there is no known method of converting electrical energy directly to kinetic potential-- which we both full well as adults understand-- which is not the debate at hand.The question is whether such a device that actually does so would constitute by definition a "perpetual energy machine." In this sense I hope I am wrong. No issues there. Feel free to tell me so if it is your opinion that's the case. That's how we have a dialogue right?Oliveiro, don't you understand that this thread is about ZhixianLin's proposal ?Have you read ZhixianLin's paper?ZhixianLin's paper is NOT about the EM Drive, nor is my post about the EM Drive.

Pardon me Dr. Rodal? "I do not think you are right ... [content]" does not constitute me saying "You are not right." There is no more clear way to express an opinion than the preface "I do not think that." I genuinely question why you edited my quote and why you read my post in such a fashion, and the only inference that seems reasonable is that you have no interest in engaging with me.If you think an emdrive as conceived on this forum is a perpetual motion device, please explain to me the fashion in which you could extract energy from the following device, and assume it does not push against an external field.One has an emdrive that weighs 1kg and travels 1m over a table's surface when 1000000 watts is supplied to the antenna. (For reasons of simplicity, let's say that this is a burst of RF impulse, not constantly applied.) I think it becomes clear that a such-described drive does not constitute a perpetual motion schema in any case, until the power efficiency reaches a certain figure (that is, the point at which spent electrical energy exceeds the potential kinetic energy of the drive. Up until this point it is just in theory a momentum transfer.) So if I am wrong you should be able to point out why I am wrong under the confines of this thought experiment. I have a feeling you are about to tell me that there is no known method of converting electrical energy directly to kinetic potential-- which we both full well as adults understand-- which is not the debate at hand.The question is whether such a device that actually does so would constitute by definition a "perpetual energy machine." In this sense I hope I am wrong. No issues there. Feel free to tell me so if it is your opinion that's the case. That's how we have a dialogue right?

Quote from: Rodal on 12/18/2015 03:14 pmQuote from: ZhixianLin on 12/11/2015 08:29 amWe know that on the electromagnetic wave propagation path, the direction of electric field at a point changes periodically. If there is a charged object at the point also changes its charge property periodically with the same frequency, then the electric field force direction on the object will not change. Therefore, the object will do directional movement. Based on this, it is possible to design a spacecraft without propellant. And the calculation suggests that the thrust can be much higher than radiation pressure. With the Electromagnetic Momentum Conservation Equation, this paper also proves that in a limited space, momentum can be not conserved although it is still conserved in the whole universe.Thank you for your contribution There are several issues for discussion on your report, for example, to pick one:If a force were to be produced, then work would be done when the electromagnetic engine moves in space (Work = Force * displacement). If the force is produced without energy radiation (or mass outflow), the work is performed without spending energy. Then, the proposed electromagnetic device would be a perpetual-motion machine.It is curious that proponents of propellant-less electromagnetic space propulsion that don't rely on external fields for propulsion, like electromagnetic tethers, etc., seem to avoid the consequence of their device being a perpetual motion machine. If the device is a perpetual motion machine, it could be used for energy purposes here on Earth, and such discussions (of using their electromagnetic self-accelerators for energy production on Earth instead of for transportation purposes) are avoided.The energy has been output when the electromagnetic wave source emits electromagnetic wave. And the metal panel absorb energy from the electromagnetic wave. So it is not a perpetual motion machine, the energy is from the electromagnetic wave source.

Quote from: ZhixianLin on 12/19/2015 12:06 amQuote from: Rodal on 12/18/2015 03:14 pmQuote from: ZhixianLin on 12/11/2015 08:29 amWe know that on the electromagnetic wave propagation path, the direction of electric field at a point changes periodically. If there is a charged object at the point also changes its charge property periodically with the same frequency, then the electric field force direction on the object will not change. Therefore, the object will do directional movement. Based on this, it is possible to design a spacecraft without propellant. And the calculation suggests that the thrust can be much higher than radiation pressure. With the Electromagnetic Momentum Conservation Equation, this paper also proves that in a limited space, momentum can be not conserved although it is still conserved in the whole universe.Thank you for your contribution There are several issues for discussion on your report, for example, to pick one:If a force were to be produced, then work would be done when the electromagnetic engine moves in space (Work = Force * displacement). If the force is produced without energy radiation (or mass outflow), the work is performed without spending energy. Then, the proposed electromagnetic device would be a perpetual-motion machine.It is curious that proponents of propellant-less electromagnetic space propulsion that don't rely on external fields for propulsion, like electromagnetic tethers, etc., seem to avoid the consequence of their device being a perpetual motion machine. If the device is a perpetual motion machine, it could be used for energy purposes here on Earth, and such discussions (of using their electromagnetic self-accelerators for energy production on Earth instead of for transportation purposes) are avoided.The energy has been output when the electromagnetic wave source emits electromagnetic wave. And the metal panel absorb energy from the electromagnetic wave. So it is not a perpetual motion machine, the energy is from the electromagnetic wave source.To be explicit, and make transparent the energy conservation issue with your device, we need you to write down the expression that, according to your theory, relates:1) the resulting force from your drive to 2) the Power Input to your drive(I did not find such equation, or a discussion of energy conservation, in your paper).Thanks.

...You can find the equations in my paper.1) the resulting force from your driveThe equation (2-8) is the force2) the Power Input to your driveThe power input depends on E0 and I0, that is the power of electromagnetic wave source and inductive circuits connected with the metal panel. From the equation (2-8), we can see that if we do not changed E0 and I0, but only reduce the the angular frequency ω of electromagnetic wave, the force can also be increased. That means we do not need to increase the power of electromagnetic wave source and inductive circuits, but only reduce the frequency of electromagnetic wave, the force can also be increased.

Quote from: ZhixianLin on 12/19/2015 02:44 am...You can find the equations in my paper.1) the resulting force from your driveThe equation (2-8) is the force2) the Power Input to your driveThe power input depends on E0 and I0, that is the power of electromagnetic wave source and inductive circuits connected with the metal panel. From the equation (2-8), we can see that if we do not changed E0 and I0, but only reduce the the angular frequency ω of electromagnetic wave, the force can also be increased. That means we do not need to increase the power of electromagnetic wave source and inductive circuits, but only reduce the frequency of electromagnetic wave, the force can also be increased.So am I correct in interpreting your equation as stating, that for A) constant power input, and B) constant angular frequency ω of electromagnetic wave, your theory gives a constant force to the spaceship, and hence that for constant mass of the spaceship (since you propose a propellant-less device that does not ejects any mass), you are proposing that for constant power input, the spaceship will achieve constant acceleration ?

Quote from: Rodal on 12/19/2015 02:50 amQuote from: ZhixianLin on 12/19/2015 02:44 am...You can find the equations in my paper.1) the resulting force from your driveThe equation (2-8) is the force2) the Power Input to your driveThe power input depends on E0 and I0, that is the power of electromagnetic wave source and inductive circuits connected with the metal panel. From the equation (2-8), we can see that if we do not changed E0 and I0, but only reduce the the angular frequency ω of electromagnetic wave, the force can also be increased. That means we do not need to increase the power of electromagnetic wave source and inductive circuits, but only reduce the frequency of electromagnetic wave, the force can also be increased.So am I correct in interpreting your equation as stating, that for A) constant power input, and B) constant angular frequency ω of electromagnetic wave, your theory gives a constant force to the spaceship, and hence that for constant mass of the spaceship (since you propose a propellant-less device that does not ejects any mass), you are proposing that for constant power input, the spaceship will achieve constant acceleration ?Yes, if power and angular frequency ω of electromagnetic wave are both constant, the force should be constant.

Quote from: ZhixianLin on 12/19/2015 03:06 amQuote from: Rodal on 12/19/2015 02:50 amQuote from: ZhixianLin on 12/19/2015 02:44 am...You can find the equations in my paper.1) the resulting force from your driveThe equation (2-8) is the force2) the Power Input to your driveThe power input depends on E0 and I0, that is the power of electromagnetic wave source and inductive circuits connected with the metal panel. From the equation (2-8), we can see that if we do not changed E0 and I0, but only reduce the the angular frequency ω of electromagnetic wave, the force can also be increased. That means we do not need to increase the power of electromagnetic wave source and inductive circuits, but only reduce the frequency of electromagnetic wave, the force can also be increased.So am I correct in interpreting your equation as stating, that for A) constant power input, and B) constant angular frequency ω of electromagnetic wave, your theory gives a constant force to the spaceship, and hence that for constant mass of the spaceship (since you propose a propellant-less device that does not ejects any mass), you are proposing that for constant power input, the spaceship will achieve constant acceleration ?Yes, if power and angular frequency ω of electromagnetic wave are both constant, the force should be constant.The same way a generator produces constant "energy" with constant power input to the rotor, an electromagnetic drive should achieve constant "momentum" per constant power input. This should not even really be conceived of as acceleration, because acceleration is the thing that imparts momentum, not the same thing as momentum.

Quote from: Rodal on 12/19/2015 02:50 amQuote from: ZhixianLin on 12/19/2015 02:44 am...You can find the equations in my paper.1) the resulting force from your driveThe equation (2-8) is the force2) the Power Input to your driveThe power input depends on E0 and I0, that is the power of electromagnetic wave source and inductive circuits connected with the metal panel. From the equation (2-8), we can see that if we do not changed E0 and I0, but only reduce the the angular frequency ω of electromagnetic wave, the force can also be increased. That means we do not need to increase the power of electromagnetic wave source and inductive circuits, but only reduce the frequency of electromagnetic wave, the force can also be increased.So am I correct in interpreting your equation as stating, that for A) constant power input, and B) constant angular frequency ω of electromagnetic wave, your theory gives a constant force to the spaceship, and hence that for constant mass of the spaceship (since you propose a propellant-less device that does not ejects any mass), you are proposing that for constant power input, the spaceship will achieve constant acceleration ?If you're about to argue this creates a over unity, please explain how a photon rocket does not suffer from the same over unity.

If you're about to argue this creates a over unity, please explain how a photon rocket does not suffer from the same over unity.

Quote from: SteveD on 12/19/2015 04:32 amIf you're about to argue this creates a over unity, please explain how a photon rocket does not suffer from the same over unity.Steve,It does, but unity occurs when the velocity is exactly the speed of light for a perfect photon emitter, so overunity can only occur at greater than c. Worse for a less than perfect emitter. So it cannot occur in a real photon drive below the speed of light. (And if your drive is going FTL, you've already overturned physics and now you're just stop showing off.)[In a device with constant Power/Acceleration ratio, overunity occurs above the the cross-over velocity which is v(unity) = Power/Force. This has been shown several times in the EMDrive thread, I believe.Power=∆E/∆t. Force=∆p/∆t. (where p is momentum). Substitute terms, cancel ∆t, and v(unity) = ∆E/∆p. Energy to emit a photon = hc/λ, assuming a perfect emitter. Change in momentum due to the emission of that photon = h/λ. Substitute and cancel the common terms h & λ, and v(unity) = c.Therefore overunity only occurs when v > c. Which is impossible.Neat, huh?]

I took a brief look at your paper.For the device you drew, there is a big problem with the circuit, it is missing the most important element for it to actually work as described. There are some other issues with the force calculation, but if you can't figure out the issue with the circuit, I am not sure there is a point in me going into more detail. (I'll let you know the issue if you ask, but I think it would be more beneficial for you to figure it out yourself.)For the momentum conservation section, every manipulation of an equation, statement, and conclusion is wrong. In order to explain why, I need to have an idea of what your math background includes. Ideally you would already know how to use and calculate the following, as well as understand the physical significance of each of these concepts: vectors, derivatives, integrals, dot product, cross product, divergence, curl, and tensors.Please let me know which of those concepts you are comfortable with so I can write a clear explanation. (Note that I don't intend to try to teach multivariable calculus over an internet forum)

Quote from: meberbs on 12/19/2015 02:16 pmI took a brief look at your paper.For the device you drew, there is a big problem with the circuit, it is missing the most important element for it to actually work as described. There are some other issues with the force calculation, but if you can't figure out the issue with the circuit, I am not sure there is a point in me going into more detail. (I'll let you know the issue if you ask, but I think it would be more beneficial for you to figure it out yourself.)For the momentum conservation section, every manipulation of an equation, statement, and conclusion is wrong. In order to explain why, I need to have an idea of what your math background includes. Ideally you would already know how to use and calculate the following, as well as understand the physical significance of each of these concepts: vectors, derivatives, integrals, dot product, cross product, divergence, curl, and tensors.Please let me know which of those concepts you are comfortable with so I can write a clear explanation. (Note that I don't intend to try to teach multivariable calculus over an internet forum)Honestly, I didn't find any problem of the circuit. So, I ask you what is the problem? "vectors, derivatives, integrals, dot product, cross product, divergence, curl, and tensors" I know all of them precisely, maybe more precise than you.Let me guess the problem. Missing battery?

Quote from: ZhixianLin on 12/19/2015 11:57 pmQuote from: meberbs on 12/19/2015 02:16 pmI took a brief look at your paper.For the device you drew, there is a big problem with the circuit, it is missing the most important element for it to actually work as described. There are some other issues with the force calculation, but if you can't figure out the issue with the circuit, I am not sure there is a point in me going into more detail. (I'll let you know the issue if you ask, but I think it would be more beneficial for you to figure it out yourself.)For the momentum conservation section, every manipulation of an equation, statement, and conclusion is wrong. In order to explain why, I need to have an idea of what your math background includes. Ideally you would already know how to use and calculate the following, as well as understand the physical significance of each of these concepts: vectors, derivatives, integrals, dot product, cross product, divergence, curl, and tensors.Please let me know which of those concepts you are comfortable with so I can write a clear explanation. (Note that I don't intend to try to teach multivariable calculus over an internet forum)Honestly, I didn't find any problem of the circuit. So, I ask you what is the problem? "vectors, derivatives, integrals, dot product, cross product, divergence, curl, and tensors" I know all of them precisely, maybe more precise than you.Let me guess the problem. Missing battery?Yes, the circuit is nothing but passive elements, so it won't do anything. I'd recommend you just add a oscillating voltage source rather than designing your own inverter though.You might not be familiar with this site, but this is a friendly forum. "maybe more precise than you" sounds like a personal attack, especially when you have know way of knowing how much background I have working with those concepts. Posts with personal attacks in them will usually just be deleted on this forum.For your momentum conservation calculations here is a description of where you go wrong. As a note, your definition of the Maxwell Stress Tensor in equation 4-2 is the negative of the definition I have seen elsewhere. This is OK, since you have a corresponding negative sign in equation 4-1. I will be using the opposite sign convention as you in my description though.Your first mistake is assuming that the time derivative of momentum is equivalent to the time derivative of the Poynting vector, just because they are both time derivatives, and therefore concluding that the divergence of T is therefore an extra term. In actuality the total force exerted on collection of charges is the volume integral of equation 4-1. This is equal to the time rate of change of the mechanical momentum inside the volume. The volume integral of the Poynting vector (with appropriate constant factor) is momentum that is stored in the electromagnetic fields themselves. Since there is a negative sign on the Poynting vector, it makes sense to move it to the other side of the equation, so we get that the time rate of change of the mechanical momentum, plus the time rate of change of the momentum in the fields is equal to the volume integral of the divergence of the Maxwell Stress Tensor.The volume integral of the divergence of a tensor is equal to the surface integral of the tensor (specifically the integral of T dot da). In this case, It then appears that the surface integral Maxwell Stress Tensor represents the time rate of change of momentum exiting the volume that is being integrated over. In other words, Maxwell's Stress Tensor is the momentum flux density. If the boundary of the volume is the surface of a conductor, the surface integral can represent the force acting on the surface, or in the case of free space, it represents the momentum carried out of the volume through the fields i.e. the momentum of photons leaving the volume.This therefore does not support your conclusion the F = dp/dt does not hold locally. It just means that you have to remember to account for any momentum leaving the volume you are concerned with. The mechanical equivalent is tossing a ball out of a room through a door, and summing the momentum of all objects in the room before the ball leaves, and just after the ball leaves. If you don't account for the fact that momentum was carried out of the room by the ball, it will look like momentum conservation is broken.In summary, your conclusion that momentum can be locally non-conserved is incorrect, and your idea that you can get more momentum from an EM based propulsion (with no mass outflow) than the momentum of emitted photons is not supported.

Yes, it need a battery. Do you mean if the volume has surrounded all the charges, then it will only photons can leave the volume, no tensor can leave?

Quote from: ZhixianLin on 12/21/2015 12:12 amYes, it need a battery. Do you mean if the volume has surrounded all the charges, then it will only photons can leave the volume, no tensor can leave?I assume that English is not your native language based on your location. The second sentence does not make sense grammatically.To answer based on a guess at what you meant: The volume has to surround all of the relevant charges that you are trying to determine the force on. The surface integral of the Maxwell stress tensor accounts for the momentum transfer by any photons leaving (or entering) the volume.If that doesn't answer your question, you will have to restate it.

Quote from: meberbs on 12/21/2015 02:01 amQuote from: ZhixianLin on 12/21/2015 12:12 amYes, it need a battery. Do you mean if the volume has surrounded all the charges, then it will only photons can leave the volume, no tensor can leave?I assume that English is not your native language based on your location. The second sentence does not make sense grammatically.To answer based on a guess at what you meant: The volume has to surround all of the relevant charges that you are trying to determine the force on. The surface integral of the Maxwell stress tensor accounts for the momentum transfer by any photons leaving (or entering) the volume.If that doesn't answer your question, you will have to restate it.I know volume integral is equal to surface integral, you don't have to say that again. The problem is how do you define "the relevant charges(all the charges)". A volume can surround some charges, you can say they are "the relevant charges". But there maybe more charges out of the volume, to surround them too, you have to enlarge the volume. But after that, there maybe still more charges out of the volume. So you have to enlarge the volume again and again, until the volume is equal to the whole universe. So only when the volume is the whole universe, then you can make sure all "the relevant charges" is surrounded."Do you mean if the volume has surrounded all the charges, then it will only photons can leave the volume, no tensor can leave?" The tensor I mean the momentum flow.

...As a note, your definition of the Maxwell Stress Tensor in equation 4-2 is the negative of the definition I have seen elsewhere. This is OK, since you have a corresponding negative sign in equation 4-1. I will be using the opposite sign convention as you in my description though...

Quote from: meberbs on 12/20/2015 04:56 pm...As a note, your definition of the Maxwell Stress Tensor in equation 4-2 is the negative of the definition I have seen elsewhere. This is OK, since you have a corresponding negative sign in equation 4-1. I will be using the opposite sign convention as you in my description though...A small note on the physical significance of the sign convention for the stress tensor.The usual convention for the sign of Maxwell's stress tensor results (for the tensor components normal to a surface) in a positive stress tensor (normal component) for tension, and a negative stress tensor (normal component) under compression. Lin adopts the opposite convention: a negative stress tensor (normal component) corresponding to tension, and a positive stress tensor (normal component) under compression. As you point out Lin accommodates this different convention by an appropriate sign change in the balancing equation 4-1.

Quote from: ZhixianLin on 12/21/2015 02:31 amQuote from: meberbs on 12/21/2015 02:01 amQuote from: ZhixianLin on 12/21/2015 12:12 amYes, it need a battery. Do you mean if the volume has surrounded all the charges, then it will only photons can leave the volume, no tensor can leave?I assume that English is not your native language based on your location. The second sentence does not make sense grammatically.To answer based on a guess at what you meant: The volume has to surround all of the relevant charges that you are trying to determine the force on. The surface integral of the Maxwell stress tensor accounts for the momentum transfer by any photons leaving (or entering) the volume.If that doesn't answer your question, you will have to restate it.I know volume integral is equal to surface integral, you don't have to say that again. The problem is how do you define "the relevant charges(all the charges)". A volume can surround some charges, you can say they are "the relevant charges". But there maybe more charges out of the volume, to surround them too, you have to enlarge the volume. But after that, there maybe still more charges out of the volume. So you have to enlarge the volume again and again, until the volume is equal to the whole universe. So only when the volume is the whole universe, then you can make sure all "the relevant charges" is surrounded."Do you mean if the volume has surrounded all the charges, then it will only photons can leave the volume, no tensor can leave?" The tensor I mean the momentum flow.The volume only needs to include all of the charges that you are trying to determine the force on. In this case you just need to make sure the spacecraft is fully enclosed in the volume. Then, you would have an equation that reads: The force on these charges is equal to the time derivative of their momentum is equal to the momentum leaving the volume (surface integral of the Maxwell Stress Tensor) minus the rate if change of momentum stored in the fields within the volume (volume integral of the time derivative of the Poynting vector). No need for infinite volumes to make it work. You seem to be differentiating photons from the stress tensor. The stress tensor inherently incorporates the photon momentum.

Quote from: meberbs on 12/21/2015 02:50 pmThe volume only needs to include all of the charges that you are trying to determine the force on. In this case you just need to make sure the spacecraft is fully enclosed in the volume. Then, you would have an equation that reads: The force on these charges is equal to the time derivative of their momentum is equal to the momentum leaving the volume (surface integral of the Maxwell Stress Tensor) minus the rate if change of momentum stored in the fields within the volume (volume integral of the time derivative of the Poynting vector). No need for infinite volumes to make it work. You seem to be differentiating photons from the stress tensor. The stress tensor inherently incorporates the photon momentum.In my physics textbook, it says it need infinite volumes to make it work. And in the equation 4-1, -▽∙T and -∂g/∂t have the same status. f can have more -▽∙T but less -∂g/∂t, or have more -∂g/∂t but less -▽∙T, they are both OK. And as you said, the momentum can leave the volume. But the momentum is not photons(Poynting vector). The photons will take a lot of energy out, that's why photon rocket has very low energy efficiency. But the momentum will not take a lot of energy, so if f has more -▽∙T but less -∂g/∂t, then the rocket can have higher energy efficiency.

The volume only needs to include all of the charges that you are trying to determine the force on. In this case you just need to make sure the spacecraft is fully enclosed in the volume. Then, you would have an equation that reads: The force on these charges is equal to the time derivative of their momentum is equal to the momentum leaving the volume (surface integral of the Maxwell Stress Tensor) minus the rate if change of momentum stored in the fields within the volume (volume integral of the time derivative of the Poynting vector). No need for infinite volumes to make it work. You seem to be differentiating photons from the stress tensor. The stress tensor inherently incorporates the photon momentum.

Quote from: ZhixianLin on 12/22/2015 12:15 amQuote from: meberbs on 12/21/2015 02:50 pmThe volume only needs to include all of the charges that you are trying to determine the force on. In this case you just need to make sure the spacecraft is fully enclosed in the volume. Then, you would have an equation that reads: The force on these charges is equal to the time derivative of their momentum is equal to the momentum leaving the volume (surface integral of the Maxwell Stress Tensor) minus the rate if change of momentum stored in the fields within the volume (volume integral of the time derivative of the Poynting vector). No need for infinite volumes to make it work. You seem to be differentiating photons from the stress tensor. The stress tensor inherently incorporates the photon momentum.In my physics textbook, it says it need infinite volumes to make it work. And in the equation 4-1, -▽∙T and -∂g/∂t have the same status. f can have more -▽∙T but less -∂g/∂t, or have more -∂g/∂t but less -▽∙T, they are both OK. And as you said, the momentum can leave the volume. But the momentum is not photons(Poynting vector). The photons will take a lot of energy out, that's why photon rocket has very low energy efficiency. But the momentum will not take a lot of energy, so if f has more -▽∙T but less -∂g/∂t, then the rocket can have higher energy efficiency.I just described how it works without infinite volumes (using the Griffiths textbook for reference). I think you may be confusing what your textbook says. You are right that there can be a trade off between how much force comes from which term, but this will not affect the result. For example, if I turn on a laser at time zero, and then apply the volume integral of your equation 4-1 to determine the reaction force being experienced by the laser at a later time, I will have a different contribution of the 2 terms depending on what volume I choose.Assuming the power output of the laser is P, and I try to determine the reaction force at time t (Note that the force is constant as a function of time and equal to P / c.).Using a spherical volume centered at the emission point of the laser with radius greater than t * c :-The contribution from the tensor T will be 0, since none of the fields will have propagated to the surface, so the surface integral of T will be 0.-The contribution from the rate of change of the Poynting vector integral will be equal to P/c because the energy stored in the fields is increasing at rate P (since Power is the rate of change of energy and Poynting vector is the energy transport vector). The extra divide by c comes from the constant factors that turn it into the time derivative of momentum.Using a spherical volume centered at the emission point of the laser with radius less than t * c :-The Poynting vector integral is now time independent (ignoring the periodic variation from the oscillating of the fields, which will cancel with the corresponding periodic variation in the Maxwell Stress Tensor) This means this integral will be zero.- The surface integral of the Maxwell Stress Tensor will therefore show that momentum is leaving the volume at a rate of P / c.You can also choose other odd shaped volumes such that part of the beamwidth of the laser leaves the volume, and part of it is still extending into the volume. This would result in a contribution from both terms. Physics doesn't care what volume you use to calculate, the force on the laser is the same.This shows that the momentum flux through leaving the volume as determined by the stress tensor is the photon momentum. (Note that the tensor also factors in momentum transfer from reaction of a charge inside the volume against a charge outside the volume ("virtual photons" in QED), but this is not a relevant factor for designing a deep space "propellantless" thruster, since by definition, you are now expelling propellant in the form of the charge outside the volume which you are reacting against)As a side note, I have been careless with my references to the Poynting vector, as there is a constant factor (involving ε0 and μ0) difference between what you use in equation 4-3 and what is generally called the Poynting vector. I tend to ignore these since they are mostly just unit conversions that translate between energy and momentum in the context of this discussion, and I have been just referring to that equation as the Poynting vector.Also, when I mention things like power being defined as the rate of change of energy, it is partially for the sake of anyone who is reading this thread that doesn't have much physics background. (this thread is approaching 2000 views)

Quote from: meberbs on 12/22/2015 06:06 amI just described how it works without infinite volumes (using the Griffiths textbook for reference). I think you may be confusing what your textbook says. You are right that there can be a trade off between how much force comes from which term, but this will not affect the result. For example, if I turn on a laser at time zero, and then apply the volume integral of your equation 4-1 to determine the reaction force being experienced by the laser at a later time, I will have a different contribution of the 2 terms depending on what volume I choose.Assuming the power output of the laser is P, and I try to determine the reaction force at time t (Note that the force is constant as a function of time and equal to P / c.).Using a spherical volume centered at the emission point of the laser with radius greater than t * c :-The contribution from the tensor T will be 0, since none of the fields will have propagated to the surface, so the surface integral of T will be 0.-The contribution from the rate of change of the Poynting vector integral will be equal to P/c because the energy stored in the fields is increasing at rate P (since Power is the rate of change of energy and Poynting vector is the energy transport vector). The extra divide by c comes from the constant factors that turn it into the time derivative of momentum.Using a spherical volume centered at the emission point of the laser with radius less than t * c :-The Poynting vector integral is now time independent (ignoring the periodic variation from the oscillating of the fields, which will cancel with the corresponding periodic variation in the Maxwell Stress Tensor) This means this integral will be zero.- The surface integral of the Maxwell Stress Tensor will therefore show that momentum is leaving the volume at a rate of P / c.You can also choose other odd shaped volumes such that part of the beamwidth of the laser leaves the volume, and part of it is still extending into the volume. This would result in a contribution from both terms. Physics doesn't care what volume you use to calculate, the force on the laser is the same.This shows that the momentum flux through leaving the volume as determined by the stress tensor is the photon momentum. (Note that the tensor also factors in momentum transfer from reaction of a charge inside the volume against a charge outside the volume ("virtual photons" in QED), but this is not a relevant factor for designing a deep space "propellantless" thruster, since by definition, you are now expelling propellant in the form of the charge outside the volume which you are reacting against)As a side note, I have been careless with my references to the Poynting vector, as there is a constant factor (involving ε0 and μ0) difference between what you use in equation 4-3 and what is generally called the Poynting vector. I tend to ignore these since they are mostly just unit conversions that translate between energy and momentum in the context of this discussion, and I have been just referring to that equation as the Poynting vector.Also, when I mention things like power being defined as the rate of change of energy, it is partially for the sake of anyone who is reading this thread that doesn't have much physics background. (this thread is approaching 2000 views)You are using a laser which is sending photons. I have said that we should not send photons, we should send tensor. Why don't you use my design to explain directly? And do you mean that the value of -▽∙T depends on the value of -∂g/∂t(without f)?

I just described how it works without infinite volumes (using the Griffiths textbook for reference). I think you may be confusing what your textbook says. You are right that there can be a trade off between how much force comes from which term, but this will not affect the result. For example, if I turn on a laser at time zero, and then apply the volume integral of your equation 4-1 to determine the reaction force being experienced by the laser at a later time, I will have a different contribution of the 2 terms depending on what volume I choose.Assuming the power output of the laser is P, and I try to determine the reaction force at time t (Note that the force is constant as a function of time and equal to P / c.).Using a spherical volume centered at the emission point of the laser with radius greater than t * c :-The contribution from the tensor T will be 0, since none of the fields will have propagated to the surface, so the surface integral of T will be 0.-The contribution from the rate of change of the Poynting vector integral will be equal to P/c because the energy stored in the fields is increasing at rate P (since Power is the rate of change of energy and Poynting vector is the energy transport vector). The extra divide by c comes from the constant factors that turn it into the time derivative of momentum.Using a spherical volume centered at the emission point of the laser with radius less than t * c :-The Poynting vector integral is now time independent (ignoring the periodic variation from the oscillating of the fields, which will cancel with the corresponding periodic variation in the Maxwell Stress Tensor) This means this integral will be zero.- The surface integral of the Maxwell Stress Tensor will therefore show that momentum is leaving the volume at a rate of P / c.You can also choose other odd shaped volumes such that part of the beamwidth of the laser leaves the volume, and part of it is still extending into the volume. This would result in a contribution from both terms. Physics doesn't care what volume you use to calculate, the force on the laser is the same.This shows that the momentum flux through leaving the volume as determined by the stress tensor is the photon momentum. (Note that the tensor also factors in momentum transfer from reaction of a charge inside the volume against a charge outside the volume ("virtual photons" in QED), but this is not a relevant factor for designing a deep space "propellantless" thruster, since by definition, you are now expelling propellant in the form of the charge outside the volume which you are reacting against)As a side note, I have been careless with my references to the Poynting vector, as there is a constant factor (involving ε0 and μ0) difference between what you use in equation 4-3 and what is generally called the Poynting vector. I tend to ignore these since they are mostly just unit conversions that translate between energy and momentum in the context of this discussion, and I have been just referring to that equation as the Poynting vector.Also, when I mention things like power being defined as the rate of change of energy, it is partially for the sake of anyone who is reading this thread that doesn't have much physics background. (this thread is approaching 2000 views)

Lin, we need you to further describe what you mean by "sending a tensor". Maxwell's stress tensor components are either tension or compression normal to a surface, or shear parallel to a surface.Hence your statement "sending a tensor" can be literally interpreted as:1) sending a tensile (pressure) or compressive force distribution normal to a surfaceand/or2) sending a shear parallel to a surface(and these force distributions are going to be balanced on the opposite surface of the infinitesimal cube defining the tensor)Do you mean applying an electromagnetic stress tensor component (a force distribution applied to a surface)? If so, how do you propose to apply an electromagnetic force distribution, through space, (via the electromagnetic vector fields E and B), other than by using photons?

Quote from: Rodal on 12/22/2015 06:43 pmLin, we need you to further describe what you mean by "sending a tensor". Maxwell's stress tensor components are either tension or compression normal to a surface, or shear parallel to a surface.Hence your statement "sending a tensor" can be literally interpreted as:1) sending a tensile (pressure) or compressive force distribution normal to a surfaceand/or2) sending a shear parallel to a surface(and these force distributions are going to be balanced on the opposite surface of the infinitesimal cube defining the tensor)Do you mean applying an electromagnetic stress tensor component (a force distribution applied to a surface)? If so, how do you propose to apply an electromagnetic force distribution, through space, (via the electromagnetic vector fields E and B), other than by using photons?"sending a tensor" I mean send the momentum flux."Do you mean applying an electromagnetic stress tensor component". Yes, that's right.I have said that in the equation 4-1, -▽∙T and -∂g/∂t have the same status. So if -∂g/∂t(photons) can be sent, then why not -▽∙T(momentum flux)?In my design, when the electric field of electromagnetic wave generate electric field force on the metal panel, it generate some momentum flux. It is obviously that the momentum flux can not go to the electromagnetic wave source if the source is very far away. So the momentum flux will go into the open space, that is sending momentum flux.

Quote from: ZhixianLin on 12/23/2015 12:32 amQuote from: Rodal on 12/22/2015 06:43 pmLin, we need you to further describe what you mean by "sending a tensor". Maxwell's stress tensor components are either tension or compression normal to a surface, or shear parallel to a surface.Hence your statement "sending a tensor" can be literally interpreted as:1) sending a tensile (pressure) or compressive force distribution normal to a surfaceand/or2) sending a shear parallel to a surface(and these force distributions are going to be balanced on the opposite surface of the infinitesimal cube defining the tensor)Do you mean applying an electromagnetic stress tensor component (a force distribution applied to a surface)? If so, how do you propose to apply an electromagnetic force distribution, through space, (via the electromagnetic vector fields E and B), other than by using photons?"sending a tensor" I mean send the momentum flux."Do you mean applying an electromagnetic stress tensor component". Yes, that's right.I have said that in the equation 4-1, -▽∙T and -∂g/∂t have the same status. So if -∂g/∂t(photons) can be sent, then why not -▽∙T(momentum flux)?In my design, when the electric field of electromagnetic wave generate electric field force on the metal panel, it generate some momentum flux. It is obviously that the momentum flux can not go to the electromagnetic wave source if the source is very far away. So the momentum flux will go into the open space, that is sending momentum flux.Both terms -∂g/∂t and-▽∙T are terms of an equation of dynamic equilibrium. The term ▽∙T arises from static equilibrium (just as static equilibrium in the theory of continuum mechanics of deformable bodies). The term -∂g/∂t arises from the dynamic aspect of electromagnetism.g is related to linear momentum carried by the (macroscopic) electromagnetic fields ( E and B) . At the particle level this linear momentum is carried by the virtual {and/or real} photons associated with the macroscopic E and B fields. ▽∙T physically corresponds to the total {instantaneous} EM field linear momentum per unit time flowing through the surface.For example, for mechanical forces, one has Newton's 2nd law for a rigid body:m*∂v/∂t - F = 0where F is the applied force. So, in this case applying a force results in an acceleration, the equation of equilibrium has two terms that balance each other. Applying a force or an acceleration are equivalent ways to describe the same thing.Similarly for the electromagnetic equation of dyamic equilibrium, one can describe the behavior macroscopically by the electromagnetic fields (E and B). Both of these fields are due, at the particle level, to photons. Both terms -∂g/∂t and-▽∙T are going to arise in the dynamic equation of equilibrium as a result of these fields.

Quote from: Rodal on 12/24/2015 02:41 amQuote from: ZhixianLin on 12/23/2015 12:32 amQuote from: Rodal on 12/22/2015 06:43 pmLin, we need you to further describe what you mean by "sending a tensor". Maxwell's stress tensor components are either tension or compression normal to a surface, or shear parallel to a surface.Hence your statement "sending a tensor" can be literally interpreted as:1) sending a tensile (pressure) or compressive force distribution normal to a surfaceand/or2) sending a shear parallel to a surface(and these force distributions are going to be balanced on the opposite surface of the infinitesimal cube defining the tensor)Do you mean applying an electromagnetic stress tensor component (a force distribution applied to a surface)? If so, how do you propose to apply an electromagnetic force distribution, through space, (via the electromagnetic vector fields E and B), other than by using photons?"sending a tensor" I mean send the momentum flux."Do you mean applying an electromagnetic stress tensor component". Yes, that's right.I have said that in the equation 4-1, -▽∙T and -∂g/∂t have the same status. So if -∂g/∂t(photons) can be sent, then why not -▽∙T(momentum flux)?In my design, when the electric field of electromagnetic wave generate electric field force on the metal panel, it generate some momentum flux. It is obviously that the momentum flux can not go to the electromagnetic wave source if the source is very far away. So the momentum flux will go into the open space, that is sending momentum flux.Both terms -∂g/∂t and-▽∙T are terms of an equation of dynamic equilibrium. The term ▽∙T arises from static equilibrium (just as static equilibrium in the theory of continuum mechanics of deformable bodies). The term -∂g/∂t arises from the dynamic aspect of electromagnetism.g is related to linear momentum carried by the (macroscopic) electromagnetic fields ( E and B) . At the particle level this linear momentum is carried by the virtual {and/or real} photons associated with the macroscopic E and B fields. ▽∙T physically corresponds to the total {instantaneous} EM field linear momentum per unit time flowing through the surface.For example, for mechanical forces, one has Newton's 2nd law for a rigid body:m*∂v/∂t - F = 0where F is the applied force. So, in this case applying a force results in an acceleration, the equation of equilibrium has two terms that balance each other. Applying a force or an acceleration are equivalent ways to describe the same thing.Similarly for the electromagnetic equation of dyamic equilibrium, one can describe the behavior macroscopically by the electromagnetic fields (E and B). Both of these fields are due, at the particle level, to photons. Both terms -∂g/∂t and-▽∙T are going to arise in the dynamic equation of equilibrium as a result of these fields. "Both terms -∂g/∂t and-▽∙T are going to arise", Yes, both terms will arise. In my design, too. But the difference between my design and a photon thruster is that a photon thruster does not send momentum flux out, only send photons out. But my design will send some momentum flux out, it will send some photons out too, but that is side effect.In my design, the electric field force on the metal panel can be considered as external force, that's why my design does not follow momentum conservation law(Newton's version).In Newton's 2nd law, m*∂v/∂t = ∂p/∂t, and p match g, then In Newton's 2nd law there is not term to match -▽∙T. So the Electromagnetic Momentum Conservation Equation does not follow the Newton's 2nd law if -▽∙T is not zero.You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law.

Quote from: ZhixianLin on 12/24/2015 03:38 amQuote from: Rodal on 12/24/2015 02:41 amQuote from: ZhixianLin on 12/23/2015 12:32 amQuote from: Rodal on 12/22/2015 06:43 pmLin, we need you to further describe what you mean by "sending a tensor". Maxwell's stress tensor components are either tension or compression normal to a surface, or shear parallel to a surface.Hence your statement "sending a tensor" can be literally interpreted as:1) sending a tensile (pressure) or compressive force distribution normal to a surfaceand/or2) sending a shear parallel to a surface(and these force distributions are going to be balanced on the opposite surface of the infinitesimal cube defining the tensor)Do you mean applying an electromagnetic stress tensor component (a force distribution applied to a surface)? If so, how do you propose to apply an electromagnetic force distribution, through space, (via the electromagnetic vector fields E and B), other than by using photons?"sending a tensor" I mean send the momentum flux."Do you mean applying an electromagnetic stress tensor component". Yes, that's right.I have said that in the equation 4-1, -▽∙T and -∂g/∂t have the same status. So if -∂g/∂t(photons) can be sent, then why not -▽∙T(momentum flux)?In my design, when the electric field of electromagnetic wave generate electric field force on the metal panel, it generate some momentum flux. It is obviously that the momentum flux can not go to the electromagnetic wave source if the source is very far away. So the momentum flux will go into the open space, that is sending momentum flux.Both terms -∂g/∂t and-▽∙T are terms of an equation of dynamic equilibrium. The term ▽∙T arises from static equilibrium (just as static equilibrium in the theory of continuum mechanics of deformable bodies). The term -∂g/∂t arises from the dynamic aspect of electromagnetism.g is related to linear momentum carried by the (macroscopic) electromagnetic fields ( E and B) . At the particle level this linear momentum is carried by the virtual {and/or real} photons associated with the macroscopic E and B fields. ▽∙T physically corresponds to the total {instantaneous} EM field linear momentum per unit time flowing through the surface.For example, for mechanical forces, one has Newton's 2nd law for a rigid body:m*∂v/∂t - F = 0where F is the applied force. So, in this case applying a force results in an acceleration, the equation of equilibrium has two terms that balance each other. Applying a force or an acceleration are equivalent ways to describe the same thing.Similarly for the electromagnetic equation of dyamic equilibrium, one can describe the behavior macroscopically by the electromagnetic fields (E and B). Both of these fields are due, at the particle level, to photons. Both terms -∂g/∂t and-▽∙T are going to arise in the dynamic equation of equilibrium as a result of these fields. "Both terms -∂g/∂t and-▽∙T are going to arise", Yes, both terms will arise. In my design, too. But the difference between my design and a photon thruster is that a photon thruster does not send momentum flux out, only send photons out. But my design will send some momentum flux out, it will send some photons out too, but that is side effect.In my design, the electric field force on the metal panel can be considered as external force, that's why my design does not follow momentum conservation law(Newton's version).In Newton's 2nd law, m*∂v/∂t = ∂p/∂t, and p match g, then In Newton's 2nd law there is not term to match -▽∙T. So the Electromagnetic Momentum Conservation Equation does not follow the Newton's 2nd law if -▽∙T is not zero.You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law.When you state << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body (*) in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case.Newton's law, in general, for deformable bodies is:▽∙T + rho * b - rho * ∂v/∂t = 0where T is the Cauchy stress (defined with respect to the deformed configuration as per usual sign convention), rho is the mass density and b is the body force. So, there is no big difference between the equation of dynamic equilibrium in Continuum Mechanics for deformable bodies, and the one for Electromagnetism, for the general case. Both must contain the term ▽∙T in order to enforce equilbrium for a deformable body. People working in solid mechanics, mechanical and aerospace engineers, use such an equation containing ▽∙T to solve practical problems dealing with stresses in structures.In the above expression, ∂v/∂t should be interpreted as a convected time derivative. (for small strains, and small displacements as in metal deformations under usual forces for example, it is approximately the time rate). For highly deformable bodies, for example, for large strains of solid bodies or for general deformation of fluids, the convected time rate of the vector v, becomes "the material derivative" ∂vi/∂t + vk ∂vi/∂yk where ∂v/∂t is taken at a constant spatial coordinate y, and I introduce index notation for clarity:▽∙T + rho * bi - rho * ∂vi/∂t - rho * vk ∂vi/∂yk = 0_______Concerning <<You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law>>. It is known, that quantum mechanics, for a huge ensemble of particles (in this case a huge ensemble of {real or virtual} photons) gives the same results at the macro level as classical electromagnetism. In your invention, it looks like the number of particles is so large that there is no need to invoke quantum mechanics (and you are not invoking quantum coherence and decoherence), as the behavior should be fully explainable at the macro level._____________(*) rigid body = an idealization of a material as having an infinite modulus of elasticity, such that it does not deform under application of a finite stress. Bodies having a finite modulus of elasticity will deform under a stress gradient ▽∙T, and hence the term ▽∙T must be included in Newton's law when considering a real material that has a finite modulus of elasticity, in order to satisfy equilibrium.

Quote from: Rodal on 12/24/2015 01:49 pmQuote from: ZhixianLin on 12/24/2015 03:38 am"Both terms -∂g/∂t and-▽∙T are going to arise", Yes, both terms will arise. In my design, too. But the difference between my design and a photon thruster is that a photon thruster does not send momentum flux out, only send photons out. But my design will send some momentum flux out, it will send some photons out too, but that is side effect.In my design, the electric field force on the metal panel can be considered as external force, that's why my design does not follow momentum conservation law(Newton's version).In Newton's 2nd law, m*∂v/∂t = ∂p/∂t, and p match g, then In Newton's 2nd law there is not term to match -▽∙T. So the Electromagnetic Momentum Conservation Equation does not follow the Newton's 2nd law if -▽∙T is not zero.You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law.When you state << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body (*) in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case.Newton's law, in general, for deformable bodies is:▽∙T + rho * b - rho * ∂v/∂t = 0where T is the Cauchy stress (defined with respect to the deformed configuration as per usual sign convention), rho is the mass density and b is the body force. So, there is no big difference between the equation of dynamic equilibrium in Continuum Mechanics for deformable bodies, and the one for Electromagnetism, for the general case. Both must contain the term ▽∙T in order to enforce equilbrium for a deformable body. People working in solid mechanics, mechanical and aerospace engineers, use such an equation containing ▽∙T to solve practical problems dealing with stresses in structures.In the above expression, ∂v/∂t should be interpreted as a convected time derivative. (for small strains, and small displacements as in metal deformations under usual forces for example, it is approximately the time rate). For highly deformable bodies, for example, for large strains of solid bodies or for general deformation of fluids, the convected time rate of the vector v, becomes "the material derivative" ∂vi/∂t + vk ∂vi/∂yk where ∂v/∂t is taken at a constant spatial coordinate y, and I introduce index notation for clarity:▽∙T + rho * bi - rho * ∂vi/∂t - rho * vk ∂vi/∂yk = 0_______Concerning <<You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law>>. It is known, that quantum mechanics, for a huge ensemble of particles (in this case a huge ensemble of {real or virtual} photons) gives the same results at the macro level as classical electromagnetism. In your invention, it looks like the number of particles is so large that there is no need to invoke quantum mechanics (and you are not invoking quantum coherence and decoherence), as the behavior should be fully explainable at the macro level._____________(*) rigid body = an idealization of a material as having an infinite modulus of elasticity, such that it does not deform under application of a finite stress. Bodies having a finite modulus of elasticity will deform under a stress gradient ▽∙T, and hence the term ▽∙T must be included in Newton's law when considering a real material that has a finite modulus of elasticity, in order to satisfy equilibrium.Merry Christmas, Rodal!You are talking about the Continuum Mechanics, then it proved that vacuum is also continuum. It means vacuum is just like the water can also be pushed. We can push vacuum in vacuum with electromagnetic fields.Electric field force has much higher efficiency than radiation pressure in using energy, so my design is not a photon thruster.Just imagine, if we do not use the metal panel but just put a still charged object on the electromagnetic wave propagation path. And we know in half a cycle the electric field force direction of electromagnetic wave will not change, so we can calculate the average electric field force on the object in half a cycle. Because the initial state of the object is still, so the energy of the object will all come from the electromagnetic wave. After you calculated the average electric field force, then you can compare it with radiation pressure. And you will see that electric field force has much higher efficiency than radiation pressure in using the energy electromagnetic wave.

Quote from: ZhixianLin on 12/24/2015 03:38 am"Both terms -∂g/∂t and-▽∙T are going to arise", Yes, both terms will arise. In my design, too. But the difference between my design and a photon thruster is that a photon thruster does not send momentum flux out, only send photons out. But my design will send some momentum flux out, it will send some photons out too, but that is side effect.In my design, the electric field force on the metal panel can be considered as external force, that's why my design does not follow momentum conservation law(Newton's version).In Newton's 2nd law, m*∂v/∂t = ∂p/∂t, and p match g, then In Newton's 2nd law there is not term to match -▽∙T. So the Electromagnetic Momentum Conservation Equation does not follow the Newton's 2nd law if -▽∙T is not zero.You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law.When you state << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body (*) in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case.Newton's law, in general, for deformable bodies is:▽∙T + rho * b - rho * ∂v/∂t = 0where T is the Cauchy stress (defined with respect to the deformed configuration as per usual sign convention), rho is the mass density and b is the body force. So, there is no big difference between the equation of dynamic equilibrium in Continuum Mechanics for deformable bodies, and the one for Electromagnetism, for the general case. Both must contain the term ▽∙T in order to enforce equilbrium for a deformable body. People working in solid mechanics, mechanical and aerospace engineers, use such an equation containing ▽∙T to solve practical problems dealing with stresses in structures.In the above expression, ∂v/∂t should be interpreted as a convected time derivative. (for small strains, and small displacements as in metal deformations under usual forces for example, it is approximately the time rate). For highly deformable bodies, for example, for large strains of solid bodies or for general deformation of fluids, the convected time rate of the vector v, becomes "the material derivative" ∂vi/∂t + vk ∂vi/∂yk where ∂v/∂t is taken at a constant spatial coordinate y, and I introduce index notation for clarity:▽∙T + rho * bi - rho * ∂vi/∂t - rho * vk ∂vi/∂yk = 0_______Concerning <<You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law>>. It is known, that quantum mechanics, for a huge ensemble of particles (in this case a huge ensemble of {real or virtual} photons) gives the same results at the macro level as classical electromagnetism. In your invention, it looks like the number of particles is so large that there is no need to invoke quantum mechanics (and you are not invoking quantum coherence and decoherence), as the behavior should be fully explainable at the macro level._____________(*) rigid body = an idealization of a material as having an infinite modulus of elasticity, such that it does not deform under application of a finite stress. Bodies having a finite modulus of elasticity will deform under a stress gradient ▽∙T, and hence the term ▽∙T must be included in Newton's law when considering a real material that has a finite modulus of elasticity, in order to satisfy equilibrium.

"Both terms -∂g/∂t and-▽∙T are going to arise", Yes, both terms will arise. In my design, too. But the difference between my design and a photon thruster is that a photon thruster does not send momentum flux out, only send photons out. But my design will send some momentum flux out, it will send some photons out too, but that is side effect.In my design, the electric field force on the metal panel can be considered as external force, that's why my design does not follow momentum conservation law(Newton's version).In Newton's 2nd law, m*∂v/∂t = ∂p/∂t, and p match g, then In Newton's 2nd law there is not term to match -▽∙T. So the Electromagnetic Momentum Conservation Equation does not follow the Newton's 2nd law if -▽∙T is not zero.You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law.

Quote from: ZhixianLin on 12/25/2015 12:20 amQuote from: Rodal on 12/24/2015 01:49 pmQuote from: ZhixianLin on 12/24/2015 03:38 am"Both terms -∂g/∂t and-▽∙T are going to arise", Yes, both terms will arise. In my design, too. But the difference between my design and a photon thruster is that a photon thruster does not send momentum flux out, only send photons out. But my design will send some momentum flux out, it will send some photons out too, but that is side effect.In my design, the electric field force on the metal panel can be considered as external force, that's why my design does not follow momentum conservation law(Newton's version).In Newton's 2nd law, m*∂v/∂t = ∂p/∂t, and p match g, then In Newton's 2nd law there is not term to match -▽∙T. So the Electromagnetic Momentum Conservation Equation does not follow the Newton's 2nd law if -▽∙T is not zero.You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law.When you state << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body (*) in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case.Newton's law, in general, for deformable bodies is:▽∙T + rho * b - rho * ∂v/∂t = 0where T is the Cauchy stress (defined with respect to the deformed configuration as per usual sign convention), rho is the mass density and b is the body force. So, there is no big difference between the equation of dynamic equilibrium in Continuum Mechanics for deformable bodies, and the one for Electromagnetism, for the general case. Both must contain the term ▽∙T in order to enforce equilbrium for a deformable body. People working in solid mechanics, mechanical and aerospace engineers, use such an equation containing ▽∙T to solve practical problems dealing with stresses in structures.In the above expression, ∂v/∂t should be interpreted as a convected time derivative. (for small strains, and small displacements as in metal deformations under usual forces for example, it is approximately the time rate). For highly deformable bodies, for example, for large strains of solid bodies or for general deformation of fluids, the convected time rate of the vector v, becomes "the material derivative" ∂vi/∂t + vk ∂vi/∂yk where ∂v/∂t is taken at a constant spatial coordinate y, and I introduce index notation for clarity:▽∙T + rho * bi - rho * ∂vi/∂t - rho * vk ∂vi/∂yk = 0_______Concerning <<You talk about "the particle level". As we know, in quantum mechanics, particles usually does not follow Newton's 2nd law. And we know the smaller the volume is, the higher probability that -▽∙T will not be zero, then the higher probability that particles does not follow Newton's 2nd law. This can explain why in quantum mechanics, particles usually does not follow Newton's 2nd law>>. It is known, that quantum mechanics, for a huge ensemble of particles (in this case a huge ensemble of {real or virtual} photons) gives the same results at the macro level as classical electromagnetism. In your invention, it looks like the number of particles is so large that there is no need to invoke quantum mechanics (and you are not invoking quantum coherence and decoherence), as the behavior should be fully explainable at the macro level._____________(*) rigid body = an idealization of a material as having an infinite modulus of elasticity, such that it does not deform under application of a finite stress. Bodies having a finite modulus of elasticity will deform under a stress gradient ▽∙T, and hence the term ▽∙T must be included in Newton's law when considering a real material that has a finite modulus of elasticity, in order to satisfy equilibrium.Merry Christmas, Rodal!You are talking about the Continuum Mechanics, then it proved that vacuum is also continuum. It means vacuum is just like the water can also be pushed. We can push vacuum in vacuum with electromagnetic fields.Electric field force has much higher efficiency than radiation pressure in using energy, so my design is not a photon thruster.Just imagine, if we do not use the metal panel but just put a still charged object on the electromagnetic wave propagation path. And we know in half a cycle the electric field force direction of electromagnetic wave will not change, so we can calculate the average electric field force on the object in half a cycle. Because the initial state of the object is still, so the energy of the object will all come from the electromagnetic wave. After you calculated the average electric field force, then you can compare it with radiation pressure. And you will see that electric field force has much higher efficiency than radiation pressure in using the energy electromagnetic wave.Merry Christmas, Lin !Yes, you are correct that the Quantum Vacuum can be thought of as a fluid, and actually there are papers from Universities studying the Quantum Vacuum as a fluid, starting with the great Nobel Prize winner Dirac, who called it a "sea". (*)Your proposed drive sounds too good to be true, and I don't believe in Santa Claus (although Santa is supposed to come through my chimney in a few hours ) Therefore there must be something wrong with your concept, because any drive better than a photon rocket, would be too good to be believed.I would start by looking for "hidden momentum" to find something wrong. There must be a field calculation not being included in your analysis that brings us back to our unexciting reality Yes, I would start by looking for "hidden momentum" not being included in your equations that balances the propulsion.__________(*) However, most physicists think that one cannot use the Quantum Vacuum (QV) to do anything useful because the QV is inmutable and not degradable, and because it is supposed to be the lowest state of energy, so it cannot be disturbed and you cannot extract energy from it.

...Yes, it is too good to be true. But I can not find any bugs or errors of my design yet. I wish I am not fooling myself.

Quote from: ZhixianLin on 12/25/2015 02:01 am...Yes, it is too good to be true. But I can not find any bugs or errors of my design yet. I wish I am not fooling myself. Well, while we think of what could be wrong with the concept, what do you think of modifying the part in your paper discussing Newton's law, since << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case. For a deformable (solid or fluid) continuum, the term ▽∙T must be included in Newton's law, as discussed above.

Quote from: Rodal on 12/25/2015 02:42 amQuote from: ZhixianLin on 12/25/2015 02:01 am...Yes, it is too good to be true. But I can not find any bugs or errors of my design yet. I wish I am not fooling myself. Well, while we think of what could be wrong with the concept, what do you think of modifying the part in your paper discussing Newton's law, since << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case. For a deformable (solid or fluid) continuum, the term ▽∙T must be included in Newton's law, as discussed above.Honestly, I am not quite sure about that. But I think change -▽∙T to -∫v▽∙T will be more rigorous.

Quote from: ZhixianLin on 12/25/2015 03:24 amQuote from: Rodal on 12/25/2015 02:42 amQuote from: ZhixianLin on 12/25/2015 02:01 am...Yes, it is too good to be true. But I can not find any bugs or errors of my design yet. I wish I am not fooling myself. Well, while we think of what could be wrong with the concept, what do you think of modifying the part in your paper discussing Newton's law, since << In Newton's 2nd law there is not term to match -▽∙T.>> that's only true for the special case of a rigid body in static equilibrium in the absence of body forces, and only true, because in that case ▽∙T = 0. So, the ▽∙T term is still there, it is only that its value is zero for that special case. For a deformable (solid or fluid) continuum, the term ▽∙T must be included in Newton's law, as discussed above.Honestly, I am not quite sure about that. But I think change -▽∙T to -∫v▽∙T will be more rigorous.There is no point in multiplying the gradient by the velocity vector and integrating.The way I presented the equation is the correct rigorous way:▽∙T + rho * bi - rho * ∂vi/∂t - rho * vk ∂vi/∂yk = 0This equation can be found in multiple rigorous books (most notably the treatises by Truesdell and Toupin and Truesdell and Noll in Handbuch der Physik, and most books in Continuum Mechanics). For easy Internet reference (in case you don't have access to the Handbuch der Physik, please refer to the Wikipedia article on the Cauchy momentum equation for example:https://en.wikipedia.org/wiki/Cauchy_momentum_equationAlso see this article by Brown University:http://www.brown.edu/Departments/Engineering/Courses/En221/Notes/Conservation_Laws/Conservation_Laws.htmor this chapter for Cauchy's equation of motion for fluids:http://www.owlnet.rice.edu/~ceng501/Chap5.pdfAgain, Newton's law as you presented it is only valid for non-deformable materials, having an infinite modulus of elasticity, in other words it is a simplistic generalization that no real material in the whole Universe follows, because all materials and fluids are deformable to some extent. The correct expression for Newton's law must contain ▽∙T, the gradient of the stress tensor.▽∙T, the gradient of the stress tensor, appears in the equlibrium equations for fluids, for deformable solids and for electromagnetism. Hence Newton's law (when properly stated for deformable continuum) is equally applicable in fluid and solid mechanics as well as in electromagnetism.

...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?

Quote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it

Quote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.

Quote from: ZhixianLin on 12/26/2015 05:13 amQuote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.<< there is no non-Continuum Mechanics for electromagnetism>>This statement is a double negative. Double-negatives implies a positive statement: in this case you are stating that since there is not any non-Continuum Mechanics for electromagnetism, that you are admitting the truth: that Maxwell's Electromagnetism is a Continuum theory....3) Einstein showed that there was no aether. He eventually replaced the aether with a continuous gravitational field that permeates the whole Universe. The theory of General Relativity is a CONTINUUM theory as well......

Quote from: ZhixianLin on 12/26/2015 05:13 amQuote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.<< there is no non-Continuum Mechanics for electromagnetism>>This statement is a double negative. Double-negatives implies a positive statement: in this case you are stating that since there is not any non-Continuum Mechanics for electromagnetism, that you are admitting the truth: that Maxwell's Electromagnetism is a Continuum theory.But then, you appear to go back, as you state << So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law>>1) The equations you are using for Electromagnetism 4-1 and 4-2 are Continuum equations2) Maxwell conceived those equations as being contained in a continuous aether (a medium with finite modulus of elasticity, NOT with infinite modulus of elasticity)3) Einstein showed that there was no aether. He eventually replaced the aether with a continuous gravitational field that permeates the whole Universe. The theory of General Relativity is a CONTINUUM theory as well4) The Quantum Vacuum is continuous5) You have to use Cauchy's generalization of Newton's law, that contains the stress gradient, because the Newton's law you are using in your paper is a simplification, that neglects deformation of the continuum. The Newton's law that you are using assumed INFINITE modulus of elasticity. There is no medium in the Universe with an infinite modulus of elasticity. The Newton's law F = ma you are using is a simplification used in elementary classes, that completely neglects the stress gradient. The stress gradient is not zero in general, because all mediums are deformable. You must use the stress gradient in your discussion of Newton's law. When you discuss Newton's law without including the stress gradient you are discussing an unreal medium that has no stress gradient and which is not deformable. Concerning the Quantum Vacuum see Paul Dirac's paper.

Quote from: Rodal on 12/26/2015 12:41 pmQuote from: ZhixianLin on 12/26/2015 05:13 amQuote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.<< there is no non-Continuum Mechanics for electromagnetism>>This statement is a double negative. Double-negatives implies a positive statement: in this case you are stating that since there is not any non-Continuum Mechanics for electromagnetism, that you are admitting the truth: that Maxwell's Electromagnetism is a Continuum theory.But then, you appear to go back, as you state << So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law>>1) The equations you are using for Electromagnetism 4-1 and 4-2 are Continuum equations2) Maxwell conceived those equations as being contained in a continuous aether (a medium with finite modulus of elasticity, NOT with infinite modulus of elasticity)3) Einstein showed that there was no aether. He eventually replaced the aether with a continuous gravitational field that permeates the whole Universe. The theory of General Relativity is a CONTINUUM theory as well4) The Quantum Vacuum is continuous5) You have to use Cauchy's generalization of Newton's law, that contains the stress gradient, because the Newton's law you are using in your paper is a simplification, that neglects deformation of the continuum. The Newton's law that you are using assumed INFINITE modulus of elasticity. There is no medium in the Universe with an infinite modulus of elasticity. The Newton's law F = ma you are using is a simplification used in elementary classes, that completely neglects the stress gradient. The stress gradient is not zero in general, because all mediums are deformable. You must use the stress gradient in your discussion of Newton's law. When you discuss Newton's law without including the stress gradient you are discussing an unreal medium that has no stress gradient and which is not deformable. Concerning the Quantum Vacuum see Paul Dirac's paper.What is the continuum(medium) in vacuum for Newton's law?Newton's law think vacuum is empty, so Newton's law can not use vacuum as continuum. But electromagnetism think vacuum is not empty, so electromagnetism can use vacuum as continuum.<< there is no non-Continuum Mechanics for electromagnetism>>I mean electromagnetism is always Continuum Mechanics theory, because vacuum is every where in the universe(even in water or air).In vacuum, Newton's law has no continuum, but electromagnetism has(the vacuum). That's why in vacuum Newton's law use the simplified version equation, but electromagnetism equation use Continuum Mechanics version. It is obviously that my drive is running in vacuum, you can't ignore that.

Quote from: ZhixianLin on 12/27/2015 12:17 amQuote from: Rodal on 12/26/2015 12:41 pmQuote from: ZhixianLin on 12/26/2015 05:13 amQuote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.<< there is no non-Continuum Mechanics for electromagnetism>>This statement is a double negative. Double-negatives implies a positive statement: in this case you are stating that since there is not any non-Continuum Mechanics for electromagnetism, that you are admitting the truth: that Maxwell's Electromagnetism is a Continuum theory.But then, you appear to go back, as you state << So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law>>1) The equations you are using for Electromagnetism 4-1 and 4-2 are Continuum equations2) Maxwell conceived those equations as being contained in a continuous aether (a medium with finite modulus of elasticity, NOT with infinite modulus of elasticity)3) Einstein showed that there was no aether. He eventually replaced the aether with a continuous gravitational field that permeates the whole Universe. The theory of General Relativity is a CONTINUUM theory as well4) The Quantum Vacuum is continuous5) You have to use Cauchy's generalization of Newton's law, that contains the stress gradient, because the Newton's law you are using in your paper is a simplification, that neglects deformation of the continuum. The Newton's law that you are using assumed INFINITE modulus of elasticity. There is no medium in the Universe with an infinite modulus of elasticity. The Newton's law F = ma you are using is a simplification used in elementary classes, that completely neglects the stress gradient. The stress gradient is not zero in general, because all mediums are deformable. You must use the stress gradient in your discussion of Newton's law. When you discuss Newton's law without including the stress gradient you are discussing an unreal medium that has no stress gradient and which is not deformable. Concerning the Quantum Vacuum see Paul Dirac's paper.What is the continuum(medium) in vacuum for Newton's law?Newton's law think vacuum is empty, so Newton's law can not use vacuum as continuum. But electromagnetism think vacuum is not empty, so electromagnetism can use vacuum as continuum.<< there is no non-Continuum Mechanics for electromagnetism>>I mean electromagnetism is always Continuum Mechanics theory, because vacuum is every where in the universe(even in water or air).In vacuum, Newton's law has no continuum, but electromagnetism has(the vacuum). That's why in vacuum Newton's law use the simplified version equation, but electromagnetism equation use Continuum Mechanics version. It is obviously that my drive is running in vacuum, you can't ignore that.http://arxiv.org/abs/1501.06763https://en.wikipedia.org/wiki/Superfluid_vacuum_theoryhttps://en.wikipedia.org/wiki/Dirac_seahttp://phys.org/news/2011-08-dark-illusion-quantum-vacuum.htmlhttp://resonance.is/news/quantum-weirdness-replaced-by-classical-fluid-dynamics/

Quote from: Rodal on 12/27/2015 12:35 amQuote from: ZhixianLin on 12/27/2015 12:17 amQuote from: Rodal on 12/26/2015 12:41 pmQuote from: ZhixianLin on 12/26/2015 05:13 amQuote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.<< there is no non-Continuum Mechanics for electromagnetism>>This statement is a double negative. Double-negatives implies a positive statement: in this case you are stating that since there is not any non-Continuum Mechanics for electromagnetism, that you are admitting the truth: that Maxwell's Electromagnetism is a Continuum theory.But then, you appear to go back, as you state << So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law>>1) The equations you are using for Electromagnetism 4-1 and 4-2 are Continuum equations2) Maxwell conceived those equations as being contained in a continuous aether (a medium with finite modulus of elasticity, NOT with infinite modulus of elasticity)3) Einstein showed that there was no aether. He eventually replaced the aether with a continuous gravitational field that permeates the whole Universe. The theory of General Relativity is a CONTINUUM theory as well4) The Quantum Vacuum is continuous5) You have to use Cauchy's generalization of Newton's law, that contains the stress gradient, because the Newton's law you are using in your paper is a simplification, that neglects deformation of the continuum. The Newton's law that you are using assumed INFINITE modulus of elasticity. There is no medium in the Universe with an infinite modulus of elasticity. The Newton's law F = ma you are using is a simplification used in elementary classes, that completely neglects the stress gradient. The stress gradient is not zero in general, because all mediums are deformable. You must use the stress gradient in your discussion of Newton's law. When you discuss Newton's law without including the stress gradient you are discussing an unreal medium that has no stress gradient and which is not deformable. Concerning the Quantum Vacuum see Paul Dirac's paper.What is the continuum(medium) in vacuum for Newton's law?Newton's law think vacuum is empty, so Newton's law can not use vacuum as continuum. But electromagnetism think vacuum is not empty, so electromagnetism can use vacuum as continuum.<< there is no non-Continuum Mechanics for electromagnetism>>I mean electromagnetism is always Continuum Mechanics theory, because vacuum is every where in the universe(even in water or air).In vacuum, Newton's law has no continuum, but electromagnetism has(the vacuum). That's why in vacuum Newton's law use the simplified version equation, but electromagnetism equation use Continuum Mechanics version. It is obviously that my drive is running in vacuum, you can't ignore that.http://arxiv.org/abs/1501.06763https://en.wikipedia.org/wiki/Superfluid_vacuum_theoryhttps://en.wikipedia.org/wiki/Dirac_seahttp://phys.org/news/2011-08-dark-illusion-quantum-vacuum.htmlhttp://resonance.is/news/quantum-weirdness-replaced-by-classical-fluid-dynamics/Sorry, Rodal. I tired of explanation. Is the Superfluid Vacuum a Newton's theory? When did Newton say that vacuum is superfluid?

Quote from: ZhixianLin on 12/27/2015 12:48 amQuote from: Rodal on 12/27/2015 12:35 amQuote from: ZhixianLin on 12/27/2015 12:17 amQuote from: Rodal on 12/26/2015 12:41 pmQuote from: ZhixianLin on 12/26/2015 05:13 amQuote from: Rodal on 12/26/2015 04:01 amQuote from: ZhixianLin on 12/26/2015 12:06 am...I think the equation 4-1 also works under non-Continuum Mechanics, so the equation 4-4 should also be non-Continuum Mechanics. I am comparing them all under non-Continuum Mechanics. The comparison is in order to prove that momentum can be not conserved. If I change the equation 4-4 to Continuum Mechanics form, then how can I prove momentum can be not conserved?And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?Equation 4-1 and 4-2 are Continuum equations, because they are electromagnetic (Maxwell) equations for continuum fields (the E and B fields, and the stress tensor T are defined for a continua). Therefore, the generalized form of Newton's law for deformable continuum media should be used instead of the simplified version assuming infinitely rigid non-deformable objects.As to your final question <<And if my design works, then finally we have to acknowledge that momentum can be not conserved. So why don't we just declare that momentum can be not conserved first?>> that is quite a conundrum isn't it? So at the moment I am leaning that your design is too good to work, that there must be "hidden momentum" to cancel it, and we just have to find it In fact, there is no non-Continuum Mechanics for electromagnetism. Because vacuum is every where in our universe, you can't find a place without vacuum. For Newton's Continuum Mechanics, it needs water, air or some other continuum. But in vacuum, there is no continuum for Newton's Continuum Mechanics. So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law.Just because in vacuum electromagnetism must be Continuum Mechanics, but in vacuum there can not be Newton's Continuum Mechanics, and that's why electromagnetism is different with Newton's law.<< there is no non-Continuum Mechanics for electromagnetism>>This statement is a double negative. Double-negatives implies a positive statement: in this case you are stating that since there is not any non-Continuum Mechanics for electromagnetism, that you are admitting the truth: that Maxwell's Electromagnetism is a Continuum theory.But then, you appear to go back, as you state << So how can I use Newton's Continuum Mechanics in vacuum? In vacuum, we should just use Newton's second law>>1) The equations you are using for Electromagnetism 4-1 and 4-2 are Continuum equations2) Maxwell conceived those equations as being contained in a continuous aether (a medium with finite modulus of elasticity, NOT with infinite modulus of elasticity)3) Einstein showed that there was no aether. He eventually replaced the aether with a continuous gravitational field that permeates the whole Universe. The theory of General Relativity is a CONTINUUM theory as well4) The Quantum Vacuum is continuous5) You have to use Cauchy's generalization of Newton's law, that contains the stress gradient, because the Newton's law you are using in your paper is a simplification, that neglects deformation of the continuum. The Newton's law that you are using assumed INFINITE modulus of elasticity. There is no medium in the Universe with an infinite modulus of elasticity. The Newton's law F = ma you are using is a simplification used in elementary classes, that completely neglects the stress gradient. The stress gradient is not zero in general, because all mediums are deformable. You must use the stress gradient in your discussion of Newton's law. When you discuss Newton's law without including the stress gradient you are discussing an unreal medium that has no stress gradient and which is not deformable. Concerning the Quantum Vacuum see Paul Dirac's paper.What is the continuum(medium) in vacuum for Newton's law?Newton's law think vacuum is empty, so Newton's law can not use vacuum as continuum. But electromagnetism think vacuum is not empty, so electromagnetism can use vacuum as continuum.<< there is no non-Continuum Mechanics for electromagnetism>>I mean electromagnetism is always Continuum Mechanics theory, because vacuum is every where in the universe(even in water or air).In vacuum, Newton's law has no continuum, but electromagnetism has(the vacuum). That's why in vacuum Newton's law use the simplified version equation, but electromagnetism equation use Continuum Mechanics version. It is obviously that my drive is running in vacuum, you can't ignore that.http://arxiv.org/abs/1501.06763https://en.wikipedia.org/wiki/Superfluid_vacuum_theoryhttps://en.wikipedia.org/wiki/Dirac_seahttp://phys.org/news/2011-08-dark-illusion-quantum-vacuum.htmlhttp://resonance.is/news/quantum-weirdness-replaced-by-classical-fluid-dynamics/Sorry, Rodal. I tired of explanation. Is the Superfluid Vacuum a Newton's theory? When did Newton say that vacuum is superfluid?There was no concept of the Supefluid vacuum at the time of Newton. When bringing up Newton's law it is better to be done consistently, with today's knowledge and not with Newton's knowledge (Cauchy extended to defomable media Newton's concept). During Maxwell's time (after Newton) the medium for electromagnetism was thought to be the aether, which was conceived as a material medium having a finite modulus of elasticity (it was NOT considered to be infinitely rigid).The quantum vacuum as a fluid was first discussed by Nobel Prize winner Paul Dirac, as far as I know.The most up-to-date theory on the vacuum as a fluid is the Superfluid vacuum theory. Using the Superfluid vacuum theory as a foundation seems better to me than stating <<EWEFFT looks like a violation of Newton's Law, but it does not violate any principle of electromagnetism.>>.In any case, as stated before it seems to me that your drive performance is too good to be believed , and probably there is hidden momentum (not taken into account in the formulation) that would prevent it.We have to find the missing "hidden momentum" that would make this (better than a photon rocket) performance impossible. If we find the missing "hidden momentum" in your drive discussion, this discussion about the proper way to discuss Newton's law would become unnecessary and pointless.

...Hey, Rodal. Are you working in NASA?"We have to find the missing", Who is that "We"?I am comparing electromagnetism with Newton's theory, not comparing electromagnetism with other modern physics theory.The "hidden momentum" should be in vacuum.

Quote from: ZhixianLin on 12/27/2015 02:59 am...Hey, Rodal. Are you working in NASA?"We have to find the missing", Who is that "We"?I am comparing electromagnetism with Newton's theory, not comparing electromagnetism with other modern physics theory.The "hidden momentum" should be in vacuum.By "we", I meant "you and I, and whoever else that reads this thread that is interested in whether your idea is possible".To me the first step is trying to prove your theory wrong, by finding hidden momentum. The immediate next step is to do an experiment and see what mother Nature has to say about it

Quote from: Rodal on 12/27/2015 03:03 amQuote from: ZhixianLin on 12/27/2015 02:59 am...Hey, Rodal. Are you working in NASA?"We have to find the missing", Who is that "We"?I am comparing electromagnetism with Newton's theory, not comparing electromagnetism with other modern physics theory.The "hidden momentum" should be in vacuum.By "we", I meant "you and I, and whoever else that reads this thread that is interested in whether your idea is possible".To me the first step is trying to prove your theory wrong, by finding hidden momentum. The immediate next step is to do an experiment and see what mother Nature has to say about it I am not finding. If you are not in NASA, then where can you find resource to do experiment?

Quote from: ZhixianLin on 12/27/2015 03:20 amQuote from: Rodal on 12/27/2015 03:03 amQuote from: ZhixianLin on 12/27/2015 02:59 am...Hey, Rodal. Are you working in NASA?"We have to find the missing", Who is that "We"?I am comparing electromagnetism with Newton's theory, not comparing electromagnetism with other modern physics theory.The "hidden momentum" should be in vacuum.By "we", I meant "you and I, and whoever else that reads this thread that is interested in whether your idea is possible".To me the first step is trying to prove your theory wrong, by finding hidden momentum. The immediate next step is to do an experiment and see what mother Nature has to say about it I am not finding. If you are not in NASA, then where can you find resource to do experiment?In the USA there is venture capital to fund innovative ideas, but of course venture capitalists are not going to easily depart with their money, they of course will demand some kind of verification. You will be competing with many other inventors that want to get the little amount of venture capital money available (space propulsion is much less attractive to venture capitalists than other fields like biotechnology).The first step is to look for missing "hidden momentum" that nullifies the idea. The next step, or at the same time, is to think of what an experiment would look like: how would you actually go about building such an experiment. Can you think of how to do such an experiment at home, with a minimum budget? For example, see the different Do It Yourself experiments in the EM Drive threads (of course you would need to have experience in this area so as to be SAFE, as we don't want you to get electrocuted !) Or, do you have access to a University laboratory?The process of thinking about how would you actually build your invention with actual physical parts will be helpful in uncovering problems with the invention and show why it may not be possible for it to be better than a photon rocket.

...No, I can't access any laboratory and I am nobody. I don't even know how to do a reliable simulation for my design. Can you do a simulation for it?

Quote from: ZhixianLin on 12/28/2015 12:09 am...No, I can't access any laboratory and I am nobody. I don't even know how to do a reliable simulation for my design. Can you do a simulation for it?You are somebody: you proved this by taking the step to report your idea in this great forum (NSF) and get a number of responses. You just have to follow it up by thinking of how you would go about implementing your idea with actual components: this will:1) Help you find any reasons why your idea cannot be reduced to practice. If you find an error or you find you cannot reduce it to practice, then modify your idea or come up with a new one. 2) Enable you to write a patent (if in the process of reducing your idea to practice you create something that is not already known "to those skilled in the aerospace art").3) Enable you to perhaps conduct your own Do It Yourself experiment (if you can safely do so) or get others to conduct their safe experiments.

...Just want to make sure I am following the discussion correctly.Maxwells equations for Electromagnetism take as an assumption that EM waves/photons must propagate through some continuous "aether" which has a finite modulus of elasticity.Fast forward to modern timesAre we assuming the aether which is assumed in Maxwells work would be considered to be dirac's quantum vaccum sea? If so given all of the valid observations we have for Maxwell's equations does that mean the quantum vaccum sea must also be continuous with a finite modulous of elasticity?

An observer who resides inside such vacuum and is capable of creating or measuring the small fluctuations would observe them as relativistic objects - unless their energy and momentum are sufficiently high to make the Lorentz-breaking corrections detectable. If the energies and momenta are below the excitation threshold then the superfluid background behaves like the ideal fluid, therefore, the Michelson–Morley-type experiments would observe no drag force from such aether

Like a big scale version of putting a negative magnet on a toy car and making a mount for a positive magnet to hang off same car an inch back causing it to move forward but on large scale?

Quote from: birchoff on 01/03/2016 06:46 am...Just want to make sure I am following the discussion correctly.Maxwells equations for Electromagnetism take as an assumption that EM waves/photons must propagate through some continuous "aether" which has a finite modulus of elasticity.Fast forward to modern timesAre we assuming the aether which is assumed in Maxwells work would be considered to be dirac's quantum vaccum sea? If so given all of the valid observations we have for Maxwell's equations does that mean the quantum vaccum sea must also be continuous with a finite modulous of elasticity?...Wikipedia (https://en.wikipedia.org/wiki/Superfluid_vacuum_theory ) puts it this way:...

Still no one has found the technical error. Is it so difficult?

Quote from: ZhixianLin on 01/29/2016 12:10 amStill no one has found the technical error. Is it so difficult?Yes, solving the equations for electromagnetism is a difficult task. Your paper does not thoroughly address the forces involved, account for radiation from the flat plate, or adequately address the form and source of the incident radiation field.If the posts here and in other threads have not convinced you that you cannot generate more thrust than a photon rocket without emitting mass, and you want to pursue your idea further, Dr. Rodal has provided some good suggestions on where to go from here. You will have to make the time investment to follow through on it though, since it is not likely someone here will have the time to do it for you.

So it means that the technical error is not so obviously, even you can not point out it easily. Just use the Newton's law of conservation of momentum is easy, but that can not provide any technical details. And I think if the design works, obviously it will change the world completely. So it is definitely worth to do detailed analysis, even it is difficult.

Quote from: ZhixianLin on 01/30/2016 11:42 pmSo it means that the technical error is not so obviously, even you can not point out it easily. Just use the Newton's law of conservation of momentum is easy, but that can not provide any technical details. And I think if the design works, obviously it will change the world completely. So it is definitely worth to do detailed analysis, even it is difficult.No, it means the error is that your original analysis is incomplete. I easily pointed out areas where it is incomplete.The general case of Maxwell's equations including special relativity is known to conserve momentum with the only special piece being photons which are massless particles that have momentum proportional to their energy. This is enough to be certain that the design won't work as claimed if you complete the analysis. Unless I feel like practicing solving complicated equations, I have no reason to do the math to complete your analysis for you. As I already said, you can take steps to complete your analysis if you think that there is any reason to do so, but I (and others who understand that EM theory conserves momentum) have no reason to do so ourselves.

So you think EmDrive is also impossible?

I have provided a simple version in a previous post, it is not complicated:"Just imagine, if we do not use the metal panel but just put a still charged object on the electromagnetic wave propagation path. And we know in half a cycle the electric field force direction of electromagnetic wave will not change, so we can calculate the average electric field force on the object in half a cycle. Because the initial state of the object is still, so the energy of the object will all come from the electromagnetic wave. After you calculated the average electric field force, then you can compare it with radiation pressure. And you will see that electric field force has much higher efficiency than radiation pressure in using the energy electromagnetic wave."If you wish, you can check the simple version yourself, not for me.

Quote from: ZhixianLin on 01/31/2016 11:55 pmSo you think EmDrive is also impossible?Unless you bring in new physics, the EM drive cannot work. This has been thoroughly covered on the EMdrive thread. (Note that there are some difficulties defining momentum conservation in general relativity, but many would still consider finding a loophole in GR for propellant less thrust new physics). Quote from: ZhixianLin on 01/31/2016 11:55 pmI have provided a simple version in a previous post, it is not complicated:"Just imagine, if we do not use the metal panel but just put a still charged object on the electromagnetic wave propagation path. And we know in half a cycle the electric field force direction of electromagnetic wave will not change, so we can calculate the average electric field force on the object in half a cycle. Because the initial state of the object is still, so the energy of the object will all come from the electromagnetic wave. After you calculated the average electric field force, then you can compare it with radiation pressure. And you will see that electric field force has much higher efficiency than radiation pressure in using the energy electromagnetic wave."If you wish, you can check the simple version yourself, not for me.That example may be simpler than your design, but it is not simple.I found 3 good papers covering various aspects of this problem.This paper calculates the oscillations the occur in the charged particle when a realistic wavepacket passes by. This results in re-radiation due to the accelerations and changes the spectrum of frequencies present. Their main conclusions are correct (changes to radiation frequency and the fact that the magnetic field does have a significant role in the applied force) However there is an issue due to them ignoring some of the (relativistic) forces. Considering the claim that the particle ends up at rest despite some EM energy now travelling in different directions makes it clear that momentum must not have been conserved.This paper simplifies the analysis by using a non-physical EM wave of infinite extent, but this helps them to calculate the full relativistic equations of motion. The final result matches the oscillations described by the other paper, but including the relativistic effects and radiation damping results in the particle accelerating in the direction of the Poyting vector, balancing the momentum issue I pointed out with the previous paper.Note that the momentum related to the oscillations (and in particular the momentum not in the same direction as the Poynting vector of the incident wave) would be balanced by the instantaneous asymmetry in the fields produced and radiated by the charge. These fields would balance the momentum at all times if you bothered to calculate the EM momentum. (I do not intend to, working out all of the math would take more time than I feel like spending on this.)I am just linking this paper for fun, since it shows how quantum mechanics complicates things. I do not believe it uses relativistic quantum mechanics (also called QED - Quantum ElectroDynamics).Also, looking through these papers reminded me of another specific issue with your setup. The electrons travel slower than light, so the balancing charge to the charge on your plate will have to be nearby. Its distance will be less than one half-wavelength of the wave, so the opposite charge will almost certainly be close enough to be affected by the incident wave.

"That example may be simpler than your design, but it is not simple"Now I make the simple version simpler. We assume that the charge is very heavy or it is connected with a very heavy object. And because half a cycle time is a very short time, so the charge will almost not move in half a cycle time. So we can consider the charge is always still in half a cycle time, and then we do not need to consider anything like oscillations, and it will be very simple. It is so simple that we don't even have to calculate it, just need the charge has enough amount of charge, the electric field force on the charge will be higher than radiation pressure.

It is so simple that we don't even have to calculate it, just need the charge has enough amount of charge, the electric field force on the charge will be higher than radiation pressure.

"The electrons travel slower than light, so the balancing charge to the charge on your plate will have to be nearby. Its distance will be less than one half-wavelength of the wave, so the opposite charge will almost certainly be close enough to be affected by the incident wave."Electromagnetic wave can propagate a very long distance, why you assume that the electromagnetic wave source must be nearby?

Quote from: ZhixianLin on 02/02/2016 02:20 am"That example may be simpler than your design, but it is not simple"Now I make the simple version simpler. We assume that the charge is very heavy or it is connected with a very heavy object. And because half a cycle time is a very short time, so the charge will almost not move in half a cycle time. So we can consider the charge is always still in half a cycle time, and then we do not need to consider anything like oscillations, and it will be very simple. It is so simple that we don't even have to calculate it, just need the charge has enough amount of charge, the electric field force on the charge will be higher than radiation pressure.The q/m ratio is a proportionality constant in many of the equations, increasing m and then increasing q will just cancel out. Also, the second paper I cited does the full equations to determine the motion so trying to simplify it is pointless. QuoteIt is so simple that we don't even have to calculate it, just need the charge has enough amount of charge, the electric field force on the charge will be higher than radiation pressure.You seem to have trouble with the following concept: If you make simplifying assumptions, and then get a result that breaks something such as conservation of momentum, it means there was probably a problem with your assumptions.In this case, you are not accounting for the fact that increasing the charge increases the force and therefore acceleration of the charge. This causes effects such as radiation damping/reaction which complicate everything. Hand waving doesn't work, you have to do the calculations, and the second paper I cited shows the general result, you can plug in different values for q and m if you want, but the basic characteristics and direction of the motion will not change.Edit: I almost forgot the simpler issue that an electron is a point particle, so a naive radiation pressure calculation would yield 0 force, since Force = Pressure*Area and Area = 0. This is clearly not the actual case, and a charged particle has an effective area in which it disturbs the incident field. Increasing the charge increases this effective area, so there is no paradox with the force increasing. (This is a simple explanation that describes the general effects, more accurate details are in the papers I cited)Quote from: ZhixianLin on 02/02/2016 02:20 am"The electrons travel slower than light, so the balancing charge to the charge on your plate will have to be nearby. Its distance will be less than one half-wavelength of the wave, so the opposite charge will almost certainly be close enough to be affected by the incident wave."Electromagnetic wave can propagate a very long distance, why you assume that the electromagnetic wave source must be nearby?I am talking about the charge on the plate in your original design, not the source of the EM wave. If there is charge on the plate, and you are forcing oscillations of the charge between positive and negative, then the balancing charge must be nearby. Specifically, the maximum distance away this charge can be is equivalent to half of a wavelength. Since the amount to which you can focus a beam of photons is limited by its wavelength, the opposite charge will also feel force from the incident EM wave.

In the radiation damping force formula, I can't figure out how the q/m can be cancel out.

Anyway, no one has done a experiment to test it. And Dr. Rodal didn't say that it is not worth to do the experiment. I don't believe God is so cruel that will forbid humans have space travel.

Quote from: ZhixianLin on 02/02/2016 11:55 amIn the radiation damping force formula, I can't figure out how the q/m can be cancel out.I said many of the equations, not all of them. The radiation damping force in particular ends up being net directed in the direction of the direction of the Poynting vector of the incident wave, so this would not produce a force in the direction of the electric field.Quote from: ZhixianLin on 02/02/2016 11:55 amAnyway, no one has done a experiment to test it. And Dr. Rodal didn't say that it is not worth to do the experiment. I don't believe God is so cruel that will forbid humans have space travel. I am not going to discourage performing the experiment. Dr. Rodal's advice to you on this was excellent, start by documenting exactly how you would build the device (don't worry about the force measurement part yet) and go from there:Quote from: Rodal on 12/29/2015 01:12 amQuote from: ZhixianLin on 12/28/2015 12:09 am...No, I can't access any laboratory and I am nobody. I don't even know how to do a reliable simulation for my design. Can you do a simulation for it?You are somebody: you proved this by taking the step to report your idea in this great forum (NSF) and get a number of responses. You just have to follow it up by thinking of how you would go about implementing your idea with actual components: this will:1) Help you find any reasons why your idea cannot be reduced to practice. If you find an error or you find you cannot reduce it to practice, then modify your idea or come up with a new one. 2) Enable you to write a patent (if in the process of reducing your idea to practice you create something that is not already known "to those skilled in the aerospace art").3) Enable you to perhaps conduct your own Do It Yourself experiment (if you can safely do so) or get others to conduct their safe experiments.I am very pro-space travel. Right now the existing science for chemical and electric propulsion leaves room for practical travel within the solar system. Our best bets I can see for interstellar travel would be through general relativity (although all useful solutions I have heard of require negative energy densities), or new physics we will discover such as the true nature of dark matter and dark energy.

The radiation damping force direction is the opposite direction of Poynting vector, right? Radiation damping force is not electric field force, but it doesn't mean that there is no electric field force. It seems that it is just because there is electric field force, and it push the charge, and then the charge generate the radiation damping force.

Quote from: ZhixianLin on 02/03/2016 01:10 amThe radiation damping force direction is the opposite direction of Poynting vector, right? Radiation damping force is not electric field force, but it doesn't mean that there is no electric field force. It seems that it is just because there is electric field force, and it push the charge, and then the charge generate the radiation damping force.Read the second paper I had linked in a previous post. They showed that what normally is considered the radiation damping force, in this case ends up having 2 terms, one that accelerates the particle, and another that provides increasing resistance as the particle approaches the speed of light. (This is expected since relativity requires that the particle velocity not exceed light speed.)Unfortunately, electrodynamics does not behave very intuitively. It makes some sense that a charge placed in the middle of a plane wave would be accelerated in the direction of the Poynting vector, but it is not obvious at first glance how this happens when considering the electric and magnetic fields acting on the charge.

Quote from: ZhixianLin on 02/02/2016 02:20 am"The electrons travel slower than light, so the balancing charge to the charge on your plate will have to be nearby. Its distance will be less than one half-wavelength of the wave, so the opposite charge will almost certainly be close enough to be affected by the incident wave."Electromagnetic wave can propagate a very long distance, why you assume that the electromagnetic wave source must be nearby?I am talking about the charge on the plate in your original design, not the source of the EM wave. If there is charge on the plate, and you are forcing oscillations of the charge between positive and negative, then the balancing charge must be nearby. Specifically, the maximum distance away this charge can be is equivalent to half of a wavelength. Since the amount to which you can focus a beam of photons is limited by its wavelength, the opposite charge will also feel force from the incident EM wave.

I just found a paper from AIAA which suggest that propellant less propulsion is possible.Source: http://arc.aiaa.org/doi/abs/10.2514/6.2001-3654The electromagnetic stress-tensor as a possible space drive propulsion conceptABSTRACTThe Heaviside force, by virtue of its nonzero value in vacuum, would appear to offer the germ of a real space-drive. Classical electrodynamic interactions may be viewed as transmitted continuously through deformations in the state of the electromagnetic field, which acts like a stressed elastic medium. Maxwell stresses not only provide a controllable net momentum flow, but also a physical mechanism acting on the fabric of space-time. In response to this unbalanced force it has been asserted that a spacecraft will move off with equal and opposite momentum. The electromagnetic interaction mechanism under consideration suggests the basis for a novel development in electrical machine technology and a possible space-drive for space transportation. The physical basis of this concept is examined in this paper, and experimental investigations are described. Supporting analyses and historical background are included.INTRODUCTIONPerhaps the most intriguing challenge facing twenty-first century space-flight is the novel concept of a "Space-Drive". John W. Campbell and Sir Arthur C. Clarke are usually credited with conceiving this visionary hypothesis. Without proposing any physical mechanism (for which one might cultivate some emerging technology to exploit) they articulated what such an astonishing apparatus will do: a space-drive is a propulsion mechanism that acts directly upon the fabric of free-space. (Actually, the notion of a space-drive was discussed in the engineering literature almost a dozen years earlier by Joseph Slepian. See below.) Remarkably, spacecraft employing such space-drive devices would not have to convey any reaction mass to eject as propellant.

Quote from: meberbs on 02/02/2016 05:17 amQuote from: ZhixianLin on 02/02/2016 02:20 am"The electrons travel slower than light, so the balancing charge to the charge on your plate will have to be nearby. Its distance will be less than one half-wavelength of the wave, so the opposite charge will almost certainly be close enough to be affected by the incident wave."Electromagnetic wave can propagate a very long distance, why you assume that the electromagnetic wave source must be nearby?I am talking about the charge on the plate in your original design, not the source of the EM wave. If there is charge on the plate, and you are forcing oscillations of the charge between positive and negative, then the balancing charge must be nearby. Specifically, the maximum distance away this charge can be is equivalent to half of a wavelength. Since the amount to which you can focus a beam of photons is limited by its wavelength, the opposite charge will also feel force from the incident EM wave.If you think so, then we can move the EM wave source towards right about half of a wavelength, make the plate just touch the edge of EM wave beam.

Quote from: ZhixianLin on 02/04/2016 12:06 amQuote from: meberbs on 02/02/2016 05:17 amQuote from: ZhixianLin on 02/02/2016 02:20 am"The electrons travel slower than light, so the balancing charge to the charge on your plate will have to be nearby. Its distance will be less than one half-wavelength of the wave, so the opposite charge will almost certainly be close enough to be affected by the incident wave."Electromagnetic wave can propagate a very long distance, why you assume that the electromagnetic wave source must be nearby?I am talking about the charge on the plate in your original design, not the source of the EM wave. If there is charge on the plate, and you are forcing oscillations of the charge between positive and negative, then the balancing charge must be nearby. Specifically, the maximum distance away this charge can be is equivalent to half of a wavelength. Since the amount to which you can focus a beam of photons is limited by its wavelength, the opposite charge will also feel force from the incident EM wave.If you think so, then we can move the EM wave source towards right about half of a wavelength, make the plate just touch the edge of EM wave beam.This idea is not new. I had this idea back nearly 20 years ago and wrote about it here; https://www.researchgate.net/publication/228542228_Warp_Drive_propulsion_within_Maxwell's_equationsDustin (@Dustinthewind) also came up with the idea a few years ago, and there are numerous posts in the EmDrive thread between he and I, and many others.David Waite also came up with the idea and made some Youtube videos about it here; And I've read papers about similar ideas going all the way back to the 1960's. It comes around and is re-discovered every few years.What you need to learn is that the Electromagnetic field "IS" photons, and any field momentum "momentum flux" that is carried away is always mediated by photons. So the idea works, because what you end up with is a phased array of antennas that broadcast EM waves (photons) unidirectionally, and this pushes the antennas the other way. It IS a photon rocket and it will not deliver more thrust than a photon rocket.The concept you have where, the momentum flux is being carried away to a distant absorber is known as Absorber Theory. It's a viable model, but even then, what is being emitted are still just photons which carry away the momentum. So in the end, it is still just a photon rocket that would be most efficient when the output transmission is well collimated, like a laser.

It is irrelevant how many years ago someone had an idea which is physics only...To count as priority date for an invention the idea must be reduced to practical device...10 years ago I've built such device and still have the working model and evidence of built date...It has very low thrust to weight ratio and very high power to force ratio...Since then I've conceived several new ideas and built prove of concepts...The latest, Orman Force Drive simulations show that it has at least 100:1 thrust/weight ratio and 98% efficiency... Currently the prove of concept device is undergoing tests...See the Orman Force declaration and the circumstances that led to invalidation of Lorentz force:https://www.youtube.com/watch?v=lhldn0ef138&feature=youtu.be

Quote from: MathewOrman on 01/17/2019 06:58 amIt is irrelevant how many years ago someone had an idea which is physics only...To count as priority date for an invention the idea must be reduced to practical device...10 years ago I've built such device and still have the working model and evidence of built date...It has very low thrust to weight ratio and very high power to force ratio...Since then I've conceived several new ideas and built prove of concepts...The latest, Orman Force Drive simulations show that it has at least 100:1 thrust/weight ratio and 98% efficiency... Currently the prove of concept device is undergoing tests...See the Orman Force declaration and the circumstances that led to invalidation of Lorentz force:https://www.youtube.com/watch?v=lhldn0ef138&feature=youtu.beThe question is not of patentability, but of usefulness. The paper you attached is trivially wrong. The units simply do not even work out in your first equation, and units are also wrong in your second equation in a way that is fundamentally inconsistent with the first equation as well. The rest seems to be you not understanding how EMF works. The magnetic flux through a loop needs to change to generate EMF, which does not happen with linear motion of the loop through a uniform field. If the Lorentz force law did not work, electric motors, generators, and countless other modern devices would not work.I'd explain the proper ways to run such experiments, but you would probably be better off picking up a good textbook on electrodynamics, me trying to tutor someone in it over the web would not be very effective. If you show some willingness to learn, I could find some useful resources to point you to.Also as to the working device with the very high power to force ratio, if you actually are measuring the right thing, and not some experimental artifact the ratio is greater than c=3e8 m/s and it works because the Lorentz force law works just fine, including the consequences that imply that photons carry momentum, which has been measured many times. There are easier and more efficient way to get up to a 1/c force/power ratio (I prefer this direction of ratio since bigger is better in this case) But even that is too low to be useful.

Need to examine it more carefully...

The Orman Force declaration is not published for negotiation but for the record...

And I do no care for hand waving...

I can only consider experimental evidence that invalidates Orman Force...

That includes the scalar form equations of Orman Force for moving and stationary charge in magnetic field...Now, the type of Lorentz equation that works is force on current conducting wire stationary or moving in uniform magnetic field and such effect is called Laplace force which Lorentz claimed as his but it is not...Lorentz never did do any experiments and that is why all math of his fails experimental confirmation, including ether and Lorentz contraction...

Quote from: MathewOrman on 01/17/2019 07:44 amNeed to examine it more carefully...The righthand side of the first equation has units of force / time. The right hand side of the second equation has units of force / distance. Both of them you claim equals a force on the left side. This is literally irreconcilably inconsistent, and has as much meaning as stating 1=0.Quote from: MathewOrman on 01/17/2019 07:44 amThe Orman Force declaration is not published for negotiation but for the record...So you are simply declaring that every single scientist on the planet for the last century is incompetent. Please research the Dunning–Kruger effect. You may want to take back the insult implied by your "declaration."Quote from: MathewOrman on 01/17/2019 07:44 amAnd I do no care for hand waving...Then you should do less of it.Quote from: MathewOrman on 01/17/2019 07:44 amI can only consider experimental evidence that invalidates Orman Force... You mean like the countless experiments that show the Lorentz force works, or the countless practical applications such as the fact that the power plant that generates the power you use every day in fact successfully generates power?Quote from: MathewOrman on 01/17/2019 07:44 amThat includes the scalar form equations of Orman Force for moving and stationary charge in magnetic field...Now, the type of Lorentz equation that works is force on current conducting wire stationary or moving in uniform magnetic field and such effect is called Laplace force which Lorentz claimed as his but it is not...Lorentz never did do any experiments and that is why all math of his fails experimental confirmation, including ether and Lorentz contraction...Historical reasons for force naming conventions are irrelevant. A current is just a bunch of individual charges moving around, so the distinction you are making between charges and currents is meaningless. You unsupported and incorrect assertion that Lorentz's math doesn't match with experiment demonstrates that you have no clue what you are talking about. Lorentz contraction is a documented effect. The aether doesn't exist, as has been experimentally proven, but the aether was believed to be necessary by most scientists until Einstein came up with special relativity, which built upon work by Lorentz. (Simplifying the story a bit)

The right side of equation simply state that Orman Force is proportional to excess charge in coulombs to acceleration of charge and strength of magnetic field...And the unit of force is in Newtons...

As for experimental evidence of Lorentz force on wire segment moving across uniform constant magnetic field with constant velocity there is none for over 100 year, now...

And in my paper there is and example from MIT what they teach now and what fails in my experiment...

Finally, please provide a link to at least one experimental setup which confirms Lorentz law and equation of charge moving in constant velocity in constant and uniform magnetic field...

Quote from: MathewOrman on 01/17/2019 08:37 amThe right side of equation simply state that Orman Force is proportional to excess charge in coulombs to acceleration of charge and strength of magnetic field...And the unit of force is in Newtons...The symbol = does not mean proportional to, it means equal. If you really meant to use a symbol that means proportional, then for your formulas to be useful, you also would have to define what the constant of proportionality is. (Which would necessarily be different between the 2 equations, since the units would be different.)Quote from: MathewOrman on 01/17/2019 08:37 amAs for experimental evidence of Lorentz force on wire segment moving across uniform constant magnetic field with constant velocity there is none for over 100 year, now...Completely false statement that ignores what I have already written.Quote from: MathewOrman on 01/17/2019 08:37 amAnd in my paper there is and example from MIT what they teach now and what fails in my experiment...It doesn't fail experiment, though it appears you failed to understand how to test it.Quote from: MathewOrman on 01/17/2019 08:37 amFinally, please provide a link to at least one experimental setup which confirms Lorentz law and equation of charge moving in constant velocity in constant and uniform magnetic field...I posted a link to a video in the emDrive thread that does exactly this.

I expected that you would provide this very link :-)

I have to disappoint you: In this setup there is no place where electrons move in constant linear velocity...Electrons are accelerated by the electric field generated by anode and after they pass anode aperture they are decelerated or pulled back by the anode

thus the curve liner trajectory of electrons are due to and consistent with Orman Force law and equation...

To confirmed it I used my own setup where I've placed second anode outside the glass of my Teltron 552 and made the beam curve in opposite direction thus proved as invalidating evidence of Lorentz force...

Now, would you please describe a setup which in your opinion is correct and shows Lorentz force and EMF generated by segment of wire moving in constant linear velocity across uniform magnetic field with constant intensity...

Moving from discussion in the emDrive thread:Quote from: MathewOrman on 01/17/2019 08:48 amI expected that you would provide this very link :-)So? it happens to be one of the first ones in a google search.Quote from: MathewOrman on 01/17/2019 08:48 amI have to disappoint you: In this setup there is no place where electrons move in constant linear velocity...Electrons are accelerated by the electric field generated by anode and after they pass anode aperture they are decelerated or pulled back by the anodeExcept the cathode and anode form a dipole, which means the fields generally cancel out quickly when not directly in between. As the electrons leave the gun fields fall off much quicker (~1/R^3 if they were a dipole, though they may be more like a multipole, which would weaken your argument further), while the fields in between are very high. Any change in velocity is relatively small.Quote from: MathewOrman on 01/17/2019 08:48 amthus the curve liner trajectory of electrons are due to and consistent with Orman Force law and equation...With the information you have provided so far, your equations are not even self-consistent. Your excuse of the electrons being pulled back to the anode does not make sense, as that is contradicted by the circular shape seen in the video, exactly as predicted using the Lorentz force law.Quote from: MathewOrman on 01/17/2019 08:48 amTo confirmed it I used my own setup where I've placed second anode outside the glass of my Teltron 552 and made the beam curve in opposite direction thus proved as invalidating evidence of Lorentz force...You are just describing the effect of adding another source of electric fields to the mix. All that proves is that electric fields deflect electrons, which is the "E" portion of the force law: F = q*(E +v x B) where "x" is the cross product and bold represents vector terms.Quote from: MathewOrman on 01/17/2019 08:55 amNow, would you please describe a setup which in your opinion is correct and shows Lorentz force and EMF generated by segment of wire moving in constant linear velocity across uniform magnetic field with constant intensity...The video does a pretty good approximation of that, though the fields are of course not perfectly uniform. You are the one claiming otherwise, so it is up to you to provide a better setup (and to show numerically how any small deviations from the ideal uniform velocity and uniform fields significantly corrupts the result.)

Second anode according to Lorentz should not change curvature direction but in reality with Teltron 552 it does so you dipole theory fails in reality...

You are pointing to unrelated experiment and I was asking for an existing experiment or your own idea of correct one which pertain to wire moving a magnetic field and NOT of an electron gun and Helmholtz coil.

So, I conclude that you have no idea how to setup such experiment and or not able to provide a link to such...

The equation f=q*(E+v cross B) is false also because the is no constant v in accelerated motion...

Quote from: MathewOrman on 01/17/2019 09:43 amSecond anode according to Lorentz should not change curvature direction but in reality with Teltron 552 it does so you dipole theory fails in reality...Maybe you aren't using some of those words with the commonly accepted meaning, but introducing additional electric fields will deflect the electrons in a different direction when the magnet is off. The specifics would require you to provide a diagram of your setup to explain.Quote from: MathewOrman on 01/17/2019 09:43 amYou are pointing to unrelated experiment and I was asking for an existing experiment or your own idea of correct one which pertain to wire moving a magnetic field and NOT of an electron gun and Helmholtz coil.There is nothing unrelated about what I provided, it is a full demonstration of the Lorentz force, with electons moving with approximately uniform velocity through an approximately uniform electric field.Your request about a moving wire doesn't make sense, because it is the motion of charges that matter, and if there is current flowing through the wire, the contribution from the motion of the wire will be small compared to the velocity of the electrons.Quote from: MathewOrman on 01/17/2019 09:43 amSo, I conclude that you have no idea how to setup such experiment and or not able to provide a link to such...As I have said, there are countless examples of the Lorentz force working, besides the facts about motors and electric generators I described before, and you ignored, there are also examples like the LHC. and other particle accelerators.Here is an example where the charges are being moved by the physical motion of a non-magnetic metal plate past a magnet:Quote from: MathewOrman on 01/17/2019 09:43 amThe equation f=q*(E+v cross B) is false also because the is no constant v in accelerated motion...V is the instantaneous velocity in that formula, which is well defined. In the linked video, the direction of the electron's velocity is constantly changing, which therefore changes the direction of the acceleration, resulting in circular motion. (Constant speed (magnitude, not direction) and an acceleration that is always perpendicular to the velocity results in circular motion.)Anyway, it is well past time for you to provide evidence of something other than unsupported statements that you have run experiments that disprove basic electrodynamics despite the fact that basically all of modern technology is based on electrodynamics working perfectly.

Failed again: in the video the setup has nonuniform magnetic field and relative motion between charges and magnetic field is accelerated...

So, again the second setup is unrelated and you have no idea how to construct your own...

I on the other hand have provided 3 examples of experimental setups where Lorentz law fails and Orman Force and low is supported as if acceleration is null Orman Force is null...

Perhaps you should provide an example of the above from academic sources as you claim that there are countless examples...

By the way, what example in the video shows is Faraday's generator where there is physical brushing against conductive metal surface... With contact less electric field probes of modern oscilloscope the EMF is zero...And there are no brush-less Faraday's generators...

Someone else confirms Orman Force law which predicts when acceleration in motion of the charges is zero the EMF in wire will also be zero:https://www.youtube.com/watch?time_continue=5&v=XSWwrvT_c8w

Failed again....

Quote from: MathewOrman on 01/17/2019 10:20 amFailed again: in the video the setup has nonuniform magnetic field and relative motion between charges and magnetic field is accelerated...Again, the Lorentz force equation uses instantaneous velocity, so your made up requirement for constant velocity is meaningless. In any example, the acceleration induced by the magnetic field changes the velocity, so as soon as you start running the experiment, the condition is lost anyway. (Though magnetic fields on their own only change the direction of velocity.)Quote from: MathewOrman on 01/17/2019 10:20 amSo, again the second setup is unrelated and you have no idea how to construct your own...I know how to construct my own, but I have no reason to do so due to the effectively unlimited number of already existing examples. You still haven't explained why all of modern technology designed using Lorentz force still works.Quote from: MathewOrman on 01/17/2019 10:20 amI on the other hand have provided 3 examples of experimental setups where Lorentz law fails and Orman Force and low is supported as if acceleration is null Orman Force is null...You have provided 0 examples where the Lorentz force fails, as the Lorentz force does not predict EMF for constant motion of a loop of wire through a uniform magnetic field. It is not clear from your experimental descriptions exactly what you did, but my guess at the second one is that you were spinning a magnet around its pole, which doesn't change the field. For the final paper you cited, you seem to have failed to notice they talk only of a singular, linear wire, not a loop. To actually make use of the EMF, you would need a loop, and the loop would have to extend outside the uniform magnetic field, otherwise the EMF from opposite sides of the loop would cancel out. Your claims of the Lorentz force not working are all based on your own misunderstandings of the Lorentz force and incorrect (strawman) claims that the Lorentz force only applies to constant velocity.Quote from: MathewOrman on 01/17/2019 10:20 amPerhaps you should provide an example of the above from academic sources as you claim that there are countless examples...The original experiments were done in the 1800s. Straightforward demonstrations of it are classroom exercises, like the ones I have provided you. Anything else is application, and I have already listed basic applications.Quote from: MathewOrman on 01/17/2019 10:20 amBy the way, what example in the video shows is Faraday's generator where there is physical brushing against conductive metal surface... With contact less electric field probes of modern oscilloscope the EMF is zero...And there are no brush-less Faraday's generators...More unsupported assertions. If friction with the brush caused anything (like a static charge from socks on carpet), the results in the video I provided would not have changed sign with the direction of spin of the disc.Again, you have yet to give your so-called force law a form that results in consistent units. You can't solve anything with it without doing that. Also, you have not demonstrated using it for any practical application, doing numerical calculations would quickly result in you running into problems with the inconsistent units you are using.It is trivial that your made up law cannot explain the fact that the motion in the first link I provided is approximately a circle. If there was any significant backwards acceleration as you claim (there isn't) in the experiment that implies the magnetic force is perpendicular to the direction of acceleration to get the loop started. After that though acceleration of the electrons is clearly a constant pointing to the center of the circle, so if that is being maintained by the magnetic field, suddenly your law has to change to force being parallel to the acceleration instead of perpendicular. Also to support the claim I just made that the acceleration is constant towards the center: that is true due to centripetal acceleration of v^2/r. Radius is obviously constant, and if the velocity was not constant, the the component of acceleration towards the center required to maintain constant radius would change non-linearly. This is not supported by the linear equation you wrote up.Also, if there was significant acceleration back towards the exit of the electron gun, you would get a distorted shape, not a circle that ends up moving away from the electron gun at the end, further disproving your claim that the speed of the electrons was anything but constant.That is 4 different issues with your claim from just 1 experiment-units don't work-direction of force relative to acceleration is undefined, and changes with time in this experiment-force claimed by you is not consistent with experiment resulting in a nice circle.-claims that the anode causes acceleration are not supported by the dataQuote from: MathewOrman on 01/17/2019 11:03 amSomeone else confirms Orman Force law which predicts when acceleration in motion of the charges is zero the EMF in wire will also be zero:https://www.youtube.com/watch?time_continue=5&v=XSWwrvT_c8wThe magnet is spinning around what is presumably its pole. This does not produce a changing magnetic flux through the loop of wire, so the net EMF for the whole loop is expected to be 0. The entire setup is backwards and inside out compared to a homopolar generator (which is what he calls his device). For the incorrect setup he built, the result he got is what is expected.

Why no one can understand my design? It is so simple and with complete theoretical support.

Yes, I expected that you would avoid providing an example of your own or from academic sources because you have no clue and there is none from academia...

So, I will let you have the last answer...