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?]

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.

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.