Angle wire resonator as reactionless drive

[goran d]

We have a wire bent in the middle by 90 degrees, fed with electric field. The wire length is much smaller than the wavelength (e.g. 1/10 or 1/5).
The thing is, i did a simulation of an angle wire like this with current source in the middle, using the 4nec2 antenna simulator.
4nec2 uses the nec2 engine, which is widely used. Guess what - the results show negligible current phase shift between the mid-point and the end-points of the wire. Wire was 10cm, frequency 300MHz. Phase shift was less than a degree. This means that the force due to biot-savart law and lorentz force equation will be greater than 0. It will point in the diagonal, to the direction the angle points.
That's because the B-field touching the wire is in Z direction. Each arm gets force perpendicular to it.
The force from the B-field should be greater than expected radiation pressure.
I did a (rough) simulation (the oscillator fed by e-field) and the force was a lot greater than the equivalent radiation.
I used complex linear equation system.
There is also electric force in the same direction, but much smaller.

Although the phases between my simulator and 4nec2 didn't match, this is not surprising, as they were not simulating exactly the same thing.

[dustinthewind]

I am glad you brought this up as I was also contemplating if this was possible.  I was thinking of doing this with a phased antenna array but at a much lower frequency than it was meant to run.  Most of the time the forces will be symmetric so no propulsion but for a fraction of a second when current reverses in one wire the time retarded forces will be non-symmetric.  Your wires are perpendicular and could work with some reduced efficiency, but we could also use two parallel wires with the current only slightly out of phase between the wires.  If we have a wire spacing of 0.0025m or 0.025cm and the speed of light being about 3E8m/s and we are using your frequency of 300MHz or 3E8Hz then c=f*lambda so wavelength is lambda=c/f = 1m.  Normally you would want the wires spaced at 1/4lambda to get projection of radiation for a phased array but you can still get some at lower frequencies if the currents are out of phase.  Normally, for 0.0025m wire spacing we would want a frequency of 3E10Hz or 3*10^10Hz to be a quarter wavelength apart.  Taking the ratio of the wavelengths we get ratio of tau = 3E8/3E10 = 1E-2 seconds so for 0.01 seconds the time retarded force is non-symmetric before returning to being symmetric.  If we are using a sine wave then the current is only a fraction of what it should be = I_max*sin(pi/2*0.01) = 0.015707317311821*I_max but if we use a square pulse that rises rapidly enough then we could get a max current time retarded non-symmetric force interaction for 0.01 seconds which could be used for propulsion.  On the other hand you have to wind so that the static electric effects work with the magnetic other wise you just got another phased array antenna. 

It is interesting to note the forces of the magnetic field in your design you would be observing would be due to electric field tilting which is from other charges approaching a current.  The two forces during the non-symmetric cycle would be 90 degrees out of phase so adding them together you would get a force of 2*F_max*sin(pi/4) = 1.4142 instead of the forces directly adding together.  The time retarded behavior of wires at 90 angles on the other hand I may have to think about as it is not as simple as two parallel wires. 

[oliverio]

I am glad you brought this up as I was also contemplating if this was possible.  I was thinking of doing this with a phased antenna array but at a much lower frequency than it was meant to run.  Most of the time the forces will be symmetric so no propulsion but for a fraction of a second when current reverses in one wire the time retarded forces will be non-symmetric.  Your wires are perpendicular and could work with some reduced efficiency, but we could also use two parallel wires with the current only slightly out of phase between the wires.  If we have a wire spacing of 0.0025m or 0.025cm and the speed of light being about 3E8m/s and we are using your frequency of 300MHz or 3E8Hz then c=f*lambda so wavelength is lambda=c/f = 1m.  Normally you would want the wires spaced at 1/4lambda to get projection of radiation for a phased array but you can still get some at lower frequencies if the currents are out of phase.  Normally, for 0.0025m wire spacing we would want a frequency of 3E10Hz or 3*10^10Hz to be a quarter wavelength apart.  Taking the ratio of the wavelengths we get ratio of tau = 3E8/3E10 = 1E-2 seconds so for 0.01 seconds the time retarded force is non-symmetric before returning to being symmetric.  If we are using a sine wave then the current is only a fraction of what it should be = I_max*sin(pi/2*0.01) = 0.015707317311821*I_max but if we use a square pulse that rises rapidly enough then we could get a max current time retarded non-symmetric force interaction for 0.01 seconds which could be used for propulsion.  On the other hand you have to wind so that the static electric effects work with the magnetic other wise you just got another phased array antenna. 

It is interesting to note the forces of the magnetic field in your design you would be observing would be due to electric field tilting which is from other charges approaching a current.  The two forces during the non-symmetric cycle would be 90 degrees out of phase so adding them together you would get a force of 2*F_max*sin(pi/4) = 1.4142 instead of the forces directly adding together.  The time retarded behavior of wires at 90 angles on the other hand I may have to think about as it is not as simple as two parallel wires.

If you were traveling at 99.99999% c, from one end of a 90 degree bent were toward the vertex, I believe the other wire would be observed as almost parallel.  Yes?  (A parallel line appears to never intersect. The time experienced by an electron traveling down a wire would be "almost forever", thus it would appear as parallel with slight immeasurable convergence over  the timeframe in relativistic terms from the reference frame of the electron.

[dustinthewind]

I am glad you brought this up as I was also contemplating if this was possible.  I was thinking of doing this with a phased antenna array but at a much lower frequency than it was meant to run.  Most of the time the forces will be symmetric so no propulsion but for a fraction of a second when current reverses in one wire the time retarded forces will be non-symmetric.  Your wires are perpendicular and could work with some reduced efficiency, but we could also use two parallel wires with the current only slightly out of phase between the wires.  If we have a wire spacing of 0.0025m or 0.025cm and the speed of light being about 3E8m/s and we are using your frequency of 300MHz or 3E8Hz then c=f*lambda so wavelength is lambda=c/f = 1m.  Normally you would want the wires spaced at 1/4lambda to get projection of radiation for a phased array but you can still get some at lower frequencies if the currents are out of phase.  Normally, for 0.0025m wire spacing we would want a frequency of 3E10Hz or 3*10^10Hz to be a quarter wavelength apart.  Taking the ratio of the wavelengths we get ratio of tau = 3E8/3E10 = 1E-2 seconds so for 0.01 seconds the time retarded force is non-symmetric before returning to being symmetric.  If we are using a sine wave then the current is only a fraction of what it should be = I_max*sin(pi/2*0.01) = 0.015707317311821*I_max but if we use a square pulse that rises rapidly enough then we could get a max current time retarded non-symmetric force interaction for 0.01 seconds which could be used for propulsion.  On the other hand you have to wind so that the static electric effects work with the magnetic other wise you just got another phased array antenna. 

It is interesting to note the forces of the magnetic field in your design you would be observing would be due to electric field tilting which is from other charges approaching a current.  The two forces during the non-symmetric cycle would be 90 degrees out of phase so adding them together you would get a force of 2*F_max*sin(pi/4) = 1.4142 instead of the forces directly adding together.  The time retarded behavior of wires at 90 angles on the other hand I may have to think about as it is not as simple as two parallel wires.

If you were traveling at 99.99999% c, from one end of a 90 degree bent were toward the vertex, I believe the other wire would be observed as almost parallel.  Yes?  (A parallel line appears to never intersect. The time experienced by an electron traveling down a wire would be "almost forever", thus it would appear as parallel with slight immeasurable convergence over  the timeframe in relativistic terms from the reference frame of the electron.

I think I am not understanding what your getting at.  I thought for copper that the electron velocity is very slow.  Like milimeters/second for DC current.  On the other hand maybe your referencing by 99.999%c as the wave speed in copper?  Not sure what that is.  Supper conductors do have electron velocities at high speeds.  There are fewer free electron Cooper pairs to super-conduct so to get the same current their charges have much higher velocity.    I am not really sure this person has a way to really make the current in the wires out of phase other than the delay in information between the wires as the distance grows.  If there is something to it then maybe it would have relation to the EM drive as its walls are tapered at an angle also. 

I guess it just got me thinking about an idea that relates to the phased array antenna and using lower frequencies. 

Edit2: Ok Oliverio, I see what you mean.  So, it would be wave velocity. 

[oliverio]

I am glad you brought this up as I was also contemplating if this was possible.  I was thinking of doing this with a phased antenna array but at a much lower frequency than it was meant to run.  Most of the time the forces will be symmetric so no propulsion but for a fraction of a second when current reverses in one wire the time retarded forces will be non-symmetric.  Your wires are perpendicular and could work with some reduced efficiency, but we could also use two parallel wires with the current only slightly out of phase between the wires.  If we have a wire spacing of 0.0025m or 0.025cm and the speed of light being about 3E8m/s and we are using your frequency of 300MHz or 3E8Hz then c=f*lambda so wavelength is lambda=c/f = 1m.  Normally you would want the wires spaced at 1/4lambda to get projection of radiation for a phased array but you can still get some at lower frequencies if the currents are out of phase.  Normally, for 0.0025m wire spacing we would want a frequency of 3E10Hz or 3*10^10Hz to be a quarter wavelength apart.  Taking the ratio of the wavelengths we get ratio of tau = 3E8/3E10 = 1E-2 seconds so for 0.01 seconds the time retarded force is non-symmetric before returning to being symmetric.  If we are using a sine wave then the current is only a fraction of what it should be = I_max*sin(pi/2*0.01) = 0.015707317311821*I_max but if we use a square pulse that rises rapidly enough then we could get a max current time retarded non-symmetric force interaction for 0.01 seconds which could be used for propulsion.  On the other hand you have to wind so that the static electric effects work with the magnetic other wise you just got another phased array antenna. 

It is interesting to note the forces of the magnetic field in your design you would be observing would be due to electric field tilting which is from other charges approaching a current.  The two forces during the non-symmetric cycle would be 90 degrees out of phase so adding them together you would get a force of 2*F_max*sin(pi/4) = 1.4142 instead of the forces directly adding together.  The time retarded behavior of wires at 90 angles on the other hand I may have to think about as it is not as simple as two parallel wires.

If you were traveling at 99.99999% c, from one end of a 90 degree bent were toward the vertex, I believe the other wire would be observed as almost parallel.  Yes?  (A parallel line appears to never intersect. The time experienced by an electron traveling down a wire would be "almost forever", thus it would appear as parallel with slight immeasurable convergence over  the timeframe in relativistic terms from the reference frame of the electron.

I think I am not understanding what your getting at.  I thought for copper that the electron velocity is very slow.  Like milimeters/second for DC current.  On the other hand maybe your referencing by 99.999%c as the wave speed in copper?  Not sure what that is.  Supper conductors do have electron velocities at high speeds.  There are fewer free electron Cooper pairs to super-conduct so to get the same current their charges have much higher velocity.    I am not really sure this guy has a way to really make the current in the wires out of phase to be honest but I guess it got me thinking about what might be possible with a phased array.

For all intents and purposes, I think maybe an electron moves through copper at c.  It is not actually the case, but electron pressure propagates through an open current at c, though individual electrons do not.

Thus, electron movement that contributes to electromagnetic events in copper could be thought of as moving through the metal at close to c.

[oliverio]

I am glad you brought this up as I was also contemplating if this was possible.  I was thinking of doing this with a phased antenna array but at a much lower frequency than it was meant to run.  Most of the time the forces will be symmetric so no propulsion but for a fraction of a second when current reverses in one wire the time retarded forces will be non-symmetric.  Your wires are perpendicular and could work with some reduced efficiency, but we could also use two parallel wires with the current only slightly out of phase between the wires.  If we have a wire spacing of 0.0025m or 0.025cm and the speed of light being about 3E8m/s and we are using your frequency of 300MHz or 3E8Hz then c=f*lambda so wavelength is lambda=c/f = 1m.  Normally you would want the wires spaced at 1/4lambda to get projection of radiation for a phased array but you can still get some at lower frequencies if the currents are out of phase.  Normally, for 0.0025m wire spacing we would want a frequency of 3E10Hz or 3*10^10Hz to be a quarter wavelength apart.  Taking the ratio of the wavelengths we get ratio of tau = 3E8/3E10 = 1E-2 seconds so for 0.01 seconds the time retarded force is non-symmetric before returning to being symmetric.  If we are using a sine wave then the current is only a fraction of what it should be = I_max*sin(pi/2*0.01) = 0.015707317311821*I_max but if we use a square pulse that rises rapidly enough then we could get a max current time retarded non-symmetric force interaction for 0.01 seconds which could be used for propulsion.  On the other hand you have to wind so that the static electric effects work with the magnetic other wise you just got another phased array antenna. 

It is interesting to note the forces of the magnetic field in your design you would be observing would be due to electric field tilting which is from other charges approaching a current.  The two forces during the non-symmetric cycle would be 90 degrees out of phase so adding them together you would get a force of 2*F_max*sin(pi/4) = 1.4142 instead of the forces directly adding together.  The time retarded behavior of wires at 90 angles on the other hand I may have to think about as it is not as simple as two parallel wires.

If you were traveling at 99.99999% c, from one end of a 90 degree bent were toward the vertex, I believe the other wire would be observed as almost parallel.  Yes?  (A parallel line appears to never intersect. The time experienced by an electron traveling down a wire would be "almost forever", thus it would appear as parallel with slight immeasurable convergence over  the timeframe in relativistic terms from the reference frame of the electron.

I think I am not understanding what your getting at.  I thought for copper that the electron velocity is very slow.  Like milimeters/second for DC current.  On the other hand maybe your referencing by 99.999%c as the wave speed in copper?  Not sure what that is.  Supper conductors do have electron velocities at high speeds.  There are fewer free electron Cooper pairs to super-conduct so to get the same current their charges have much higher velocity.    I am not really sure this person has a way to really make the current in the wires out of phase other than the delay in information between the wires as the distance grows.  If there is something to it then maybe it would have relation to the EM drive as its walls are tapered at an angle also. 

I guess it just got me thinking about an idea that relates to the phased array antenna and using lower frequencies. 

Edit2: Ok Oliverio, I see what you mean.  So, it would be wave velocity.

I think a superconducting field just loses fewer electrons along the way, and thus works faster for anything where bitrate is important. For something like thrust I believe they would only scale the efficiency of electrons in->force out.

[meberbs]

There is a simple answer to any configuration of wires, charges, currents, magnets, etc. that claims "reactionless drive":

All proposed "self-acceleration" will be equivalent to the reaction from emitted photons. (Most will actually be less efficient than a photon thruster due to emitting radiation in more than one direction.)

This is because the derivation of momentum storage in E-M fields assumes conservation of momentum.

For any ideas like this to be productive, they either should come with an experimental demonstration of greater thrust than an equivalent power laser, or a new theory of Electromagnetism that would allow it to work. The new theory of electromagnetism would have to be able to replicate all known results including special relativity, photon energy and momentum, etc. It would then need a specific description of how and under what conditions it would diverge from the classical E-M theory.

Note that if you come up with a design that you calculate to be a nearly ideal photon thruster at microwave frequencies, it means you invented a highly directional antenna, which may be useful for space communications. This would be a useful discovery, but would be better for "advanced concepts" than "new physics"

If you calculate better than a photon thruster using Maxwell's equations, it means you did your math wrong.

[oliverio]

There is a simple answer to any configuration of wires, charges, currents, magnets, etc. that claims "reactionless drive":

All proposed "self-acceleration" will be equivalent to the reaction from emitted photons. (Most will actually be less efficient than a photon thruster due to emitting radiation in more than one direction.)

This is because the derivation of momentum storage in E-M fields assumes conservation of momentum.

For any ideas like this to be productive, they either should come with an experimental demonstration of greater thrust than an equivalent power laser, or a new theory of Electromagnetism that would allow it to work. The new theory of electromagnetism would have to be able to replicate all known results including special relativity, photon energy and momentum, etc. It would then need a specific description of how and under what conditions it would diverge from the classical E-M theory.

Note that if you come up with a design that you calculate to be a nearly ideal photon thruster at microwave frequencies, it means you invented a highly directional antenna, which may be useful for space communications. This would be a useful discovery, but would be better for "advanced concepts" than "new physics"

If you calculate better than a photon thruster using Maxwell's equations, it means you did your math wrong.

Honest question: could it not be possible to make an energy-variant configuration of some manner demonstrate almost exclusively near-field properties and minimal far-field ones?  It seems to me, analytically speaking, that a photon imparts momentum poorly because far-field effects travel far (sorry for the tautology), not... hard?  The near field would seem to be what an em-propellant should seek to vary.

[oliverio]

To expand briefly on the above, and I may be wrong, it is the case that the photons of the near field and the far field are essentially the same but operating in different configuration.  It seems as though an antenna emits near-field photons, but as you scale the power of the antenna, this effect becomes far less noticeable.  As I understand as well, a photon emitted by the near field essentially pushes back on the radiator as it enters the far field. 

So if one could make a directional antenna that only operates in the near field, it should have a greater thrust than the photon rocket, yes?

[meberbs]

To expand briefly on the above, and I may be wrong, it is the case that the photons of the near field and the far field are essentially the same but operating in different configuration.  It seems as though an antenna emits near-field photons, but as you scale the power of the antenna, this effect becomes far less noticeable.  As I understand as well, a photon emitted by the near field essentially pushes back on the radiator as it enters the far field. 

So if one could make a directional antenna that only operates in the near field, it should have a greater thrust than the photon rocket, yes?

The phrase "only operates in the near field" does not make any sense.

Near field is complicated for all but the simplest antennas, since there are various effects that cancel out such as a photon being emitted from one part of the antenna, only to be absorbed by another part.

Far field is easier to work with for most applications and includes ALL of the photons that actually escape from the antenna. If these are not symmetrically distributed, then the device is a directional antenna, and photon thruster.

The portions of the near field that do not make it to the far field all represent internal interactions between portions of the device and they all cancel (equal and opposite reactions). The far field is therefore the only thing necessary to consider when determining the reaction of the device.

The reason photon thrusters are inefficient is due to the Energy momentum relation for all massless particles in special relativity: E = p*c. (E is energy, p is momentum, c is speed of light (a big number))

If you want a better EM drive than that, you need to find some kind of modification to the known physical laws. As I said above, this means either a theory that reduces to the existing theory under nearly all cases (because existing theory woks very well), or an experimental result that can't be explained by existing theory.

[oliverio]

To expand briefly on the above, and I may be wrong, it is the case that the photons of the near field and the far field are essentially the same but operating in different configuration.  It seems as though an antenna emits near-field photons, but as you scale the power of the antenna, this effect becomes far less noticeable.  As I understand as well, a photon emitted by the near field essentially pushes back on the radiator as it enters the far field. 

So if one could make a directional antenna that only operates in the near field, it should have a greater thrust than the photon rocket, yes?

The phrase "only operates in the near field" does not make any sense.

Near field is complicated for all but the simplest antennas, since there are various effects that cancel out such as a photon being emitted from one part of the antenna, only to be absorbed by another part.

Far field is easier to work with for most applications and includes ALL of the photons that actually escape from the antenna. If these are not symmetrically distributed, then the device is a directional antenna, and photon thruster.

The portions of the near field that do not make it to the far field all represent internal interactions between portions of the device and they all cancel (equal and opposite reactions). The far field is therefore the only thing necessary to consider when determining the reaction of the device.

The reason photon thrusters are inefficient is due to the Energy momentum relation for all massless particles in special relativity: E = p*c. (E is energy, p is momentum, c is speed of light (a big number))

If you want a better EM drive than that, you need to find some kind of modification to the known physical laws. As I said above, this means either a theory that reduces to the existing theory under nearly all cases (because existing theory woks very well), or an experimental result that can't be explained by existing theory.

I certainly follow your logic, I think you're a very sensible poster on this forum. For that reason let me request that you humor my inquiry a bit further.

At or before the border between near and far field, which I understand as gradient, force imparted by photons falls off at a less exponential rate than within what is firmly the far-field.

So here is a thought experiment which I take to be nonparadoxical to you, but it is not so to me:

A laser is attached to (perhaps by a rod) and aimed at an ideal photodiode positioned at the edge of the near-field gradient. The momentum of the photons leaving the laser (inside the near field, right?) should be higher than the ones being "caught" by the photodiode. If this were the case, would it accelerate in open space?

[ZhixianLin]

Do you have any pictures of your drive? Any drive base on radiation pressure will not be better than a photon thruster. You can refer to my drive: http://forum.nasaspaceflight.com/index.php?topic=38996.0

[oliverio]

Do you have any pictures of your drive? Any drive base on radiation pressure will not be better than a photon thruster. You can refer to my drive: http://forum.nasaspaceflight.com/index.php?topic=38996.0

I don't have a drive, only a thought experiment.  I'll make it a bit more relatable.

Imagine you have a cylindrical cavity with endplate injection, but one endplate is made from a perfect reflector and the opposite endplate is made from a perfect photodiode (which is returned to the antenna's powersupply) . Assume that the cavity is about as long as the near-far field boundary.  So if the photons on one side (the photodiode) are far-field particles they ought to possess less momentum but equal energy to the photon that left the far-field.  However, as this photon is absorbed by the photodiode, and electricity is generated, the momentum imparted to the photodiode should be less than the momentum of the emitted photon.

Now, side wall pressure may invalidate that-- but now I then ask the following: if the cylinder has tapered walls, is it not the case that from a photon's reference frame, there exists a geometry of frustrum such that relativistic length contraction would make the frustum's walls appear straight to an observer?

[ZhixianLin]

Do you have any pictures of your drive? Any drive base on radiation pressure will not be better than a photon thruster. You can refer to my drive: http://forum.nasaspaceflight.com/index.php?topic=38996.0

I don't have a drive, only a thought experiment.  I'll make it a bit more relatable.

Imagine you have a cylindrical cavity with endplate injection, but one endplate is made from a perfect reflector and the opposite endplate is made from a perfect photodiode (which is returned to the antenna's powersupply) . Assume that the cavity is about as long as the near-far field boundary.  So if the photons on one side (the photodiode) are far-field particles they ought to possess less momentum but equal energy to the photon that left the far-field.  However, as this photon is absorbed by the photodiode, and electricity is generated, the momentum imparted to the photodiode should be less than the momentum of the emitted photon.

Now, side wall pressure may invalidate that-- but now I then ask the following: if the cylinder has tapered walls, is it not the case that from a photon's reference frame, there exists a geometry of frustrum such that relativistic length contraction would make the frustum's walls appear straight to an observer?

My drive does not need relativity. If your design need relativity, then I can not explain.

[oliverio]

Do you have any pictures of your drive? Any drive base on radiation pressure will not be better than a photon thruster. You can refer to my drive: http://forum.nasaspaceflight.com/index.php?topic=38996.0

I don't have a drive, only a thought experiment.  I'll make it a bit more relatable.

Imagine you have a cylindrical cavity with endplate injection, but one endplate is made from a perfect reflector and the opposite endplate is made from a perfect photodiode (which is returned to the antenna's powersupply) . Assume that the cavity is about as long as the near-far field boundary.  So if the photons on one side (the photodiode) are far-field particles they ought to possess less momentum but equal energy to the photon that left the far-field.  However, as this photon is absorbed by the photodiode, and electricity is generated, the momentum imparted to the photodiode should be less than the momentum of the emitted photon.

Now, side wall pressure may invalidate that-- but now I then ask the following: if the cylinder has tapered walls, is it not the case that from a photon's reference frame, there exists a geometry of frustrum such that relativistic length contraction would make the frustum's walls appear straight to an observer?

My drive does not need relativity. If your design need relativity, then I can not explain.

I'm pretty sure that's a weakness, not a strength. Electromagnetic events should be considered relativistic.

[meberbs]

To expand briefly on the above, and I may be wrong, it is the case that the photons of the near field and the far field are essentially the same but operating in different configuration.  It seems as though an antenna emits near-field photons, but as you scale the power of the antenna, this effect becomes far less noticeable.  As I understand as well, a photon emitted by the near field essentially pushes back on the radiator as it enters the far field. 

So if one could make a directional antenna that only operates in the near field, it should have a greater thrust than the photon rocket, yes?

The phrase "only operates in the near field" does not make any sense.

Near field is complicated for all but the simplest antennas, since there are various effects that cancel out such as a photon being emitted from one part of the antenna, only to be absorbed by another part.

Far field is easier to work with for most applications and includes ALL of the photons that actually escape from the antenna. If these are not symmetrically distributed, then the device is a directional antenna, and photon thruster.

The portions of the near field that do not make it to the far field all represent internal interactions between portions of the device and they all cancel (equal and opposite reactions). The far field is therefore the only thing necessary to consider when determining the reaction of the device.

The reason photon thrusters are inefficient is due to the Energy momentum relation for all massless particles in special relativity: E = p*c. (E is energy, p is momentum, c is speed of light (a big number))

If you want a better EM drive than that, you need to find some kind of modification to the known physical laws. As I said above, this means either a theory that reduces to the existing theory under nearly all cases (because existing theory woks very well), or an experimental result that can't be explained by existing theory.

I certainly follow your logic, I think you're a very sensible poster on this forum. For that reason let me request that you humor my inquiry a bit further.

At or before the border between near and far field, which I understand as gradient, force imparted by photons falls off at a less exponential rate than within what is firmly the far-field.

So here is a thought experiment which I take to be nonparadoxical to you, but it is not so to me:

A laser is attached to (perhaps by a rod) and aimed at an ideal photodiode positioned at the edge of the near-field gradient. The momentum of the photons leaving the laser (inside the near field, right?) should be higher than the ones being "caught" by the photodiode. If this were the case, would it accelerate in open space?

There isn't more momentum in the photons in the near vs far field. An object placed within the near field would have additional forces on it due to the additional fields, but this works both ways. Induced currents and charge distributions in the object would interact with the rest of the device, changing the near field, and possibly the far field as well.

For a different example that might help clarify (or maybe this will confuse things more):

Two electric charges placed near each other will have near field interactions due to Coulomb forces. These forces can be described in terms of "virtual photons." Virtual particles are a quantum mechanical representation of forces, and follow different rules than standard particles. Quantum is weird and subtle, so I won't try to explain the details here, but I will say that in some senses virtual particles aren't "real," they are just the way action at a distance forces are represented in QED. What is relevant is that the speed of light delay is equivalent between the two particles, so the momentum will appear to have been exchanged directly between the particles without having been stored in the fields. Even in different reference frames, this will end up being consistent (although magnetic fields will also appear in the reference frames where the particles are not initially at rest. ) Accelerating charges produce real photons that radiate. Real photons actually have momentum stored in their fields, and need to be accounted for in a momentum balance.

The details get very confusing, so when analyzing a device, it is easiest to just use the fact that classical electrodynamics (which inherently includes special relativity) is a self-consistent theory that conserves both energy and momentum. (Using the definitions for energy and momentum stored in the fields). A sufficiently detailed analysis will always show that the net forces on all of the objects in the near field exactly balances with the photon momentum emitted in the far field.

[oliverio]

Thanks for the reply meberbs, this makes sense that the forces experienced in the near-field aren't the same as a classical "baseball gets thrown" analogy for particles.

The transition point does not happen at a specific time and place, though, right?  As I understand, this is because the zone is a product of gradient self-interference. Does this not imply that we could, by exact positioning of a photodiode, control which virtual photons self-interfere before all photons have actually left it, and create nonsymmetric but conserved forces?  (In this context the photodiode is somewhere between the near field and the far field.)

[dustinthewind]

There is a simple answer to any configuration of wires, charges, currents, magnets, etc. that claims "reactionless drive":

All proposed "self-acceleration" will be equivalent to the reaction from emitted photons. (Most will actually be less efficient than a photon thruster due to emitting radiation in more than one direction.)

This is because the derivation of momentum storage in E-M fields assumes conservation of momentum.

For any ideas like this to be productive, they either should come with an experimental demonstration of greater thrust than an equivalent power laser, or a new theory of Electromagnetism that would allow it to work. The new theory of electromagnetism would have to be able to replicate all known results including special relativity, photon energy and momentum, etc. It would then need a specific description of how and under what conditions it would diverge from the classical E-M theory.

Note that if you come up with a design that you calculate to be a nearly ideal photon thruster at microwave frequencies, it means you invented a highly directional antenna, which may be useful for space communications. This would be a useful discovery, but would be better for "advanced concepts" than "new physics"

If you calculate better than a photon thruster using Maxwell's equations, it means you did your math wrong.

I have to apologize for commenting here without fully understanding the original posters exact device.  I guess I was just a bit excited that it sounded similar to something I had been mulling over.  I have to agree with the above comment in that I think if greater than photon propulsion would work then we would not be pushing with light any more and we would need something else we were pushing against.  A prime candidate would be time and space which would be related to gravitational effects.  That suggest some how uniting electromagnetism with Einstein time retarded field solutions of general relativity probably, which at the moment is a bit beyond me.  One video I know of probably get into it a bit.  Here: I will also post an image of one reason I think that a phased array antenna (see image) only give photon propulsion [even though they operate in near field] because the capacitance effect (charge separation) provides opposing propulsion to the inductive magnetic effects.  You have to take into account time delayed information.  How that would relate to the original posters device, and any theory behind it, I could not say. 

I am curious to ask the original poster "goran d" as to what the non-symmetric forces his simulation estimated at a set current.  If they are large enough then it probably wouldn't be hard to test with an actual experiment.  It is possible the simulator may be at fault or the programming/constraints are off (math is wrong?)

[goran d]

This is because the derivation of momentum storage in E-M fields assumes conservation of momentum.

If the derrivation of momentum storage assumes the conservation laws, then you are deriving it from the combination of Maxwell Equation and Conservation Laws.
This is not the same as getting the conservation laws from only Maxwell's equations.

So, the conclusion that Maxwell Equations imply Conservation Laws is false.
You can't first assume something and then prove it on the basis of the assumption.

[meberbs]

Thanks for the reply meberbs, this makes sense that the forces experienced in the near-field aren't the same as a classical "baseball gets thrown" analogy for particles.

The transition point does not happen at a specific time and place, though, right?  As I understand, this is because the zone is a product of gradient self-interference. Does this not imply that we could, by exact positioning of a photodiode, control which virtual photons self-interfere before all photons have actually left it, and create nonsymmetric but conserved forces?  (In this context the photodiode is somewhere between the near field and the far field.)

You are correct, there is no sharp cutoff, just a point where the near field effects become negligible for a given definition of negligible.

The forces will always be balanced, as long as you include the momentum stored in the fields. You can't get away from momentum conservation, and the only way to get EM momentum away from the rest of the device is through photons. You can redirect the photons in specific directions, but not get better ratios of energy to momentum.

This is because the derivation of momentum storage in E-M fields assumes conservation of momentum.

If the derrivation of momentum storage assumes the conservation laws, then you are deriving it from the combination of Maxwell Equation and Conservation Laws.
This is not the same as getting the conservation laws from only Maxwell's equations.

So, the conclusion that Maxwell Equations imply Conservation Laws is false.
You can't first assume something and then prove it on the basis of the assumption.

Conservation laws all derive from Noether's theorem (although most of the conservation laws were being used well before this theorem was proven).

It is not difficult to find situations in electrodynamics that forces do not appear to be equal and opposite when you consider only the momentum changes in the charged particles. Reconciling this with conservation of momentum, requires that momentum be also stored in the fields. When deriving the equation for momentum in the fields, it is therefore already assumed that momentum is conserved.

I am not proving conservation of momentum by assuming it. (that would be the "correct" usage of "begging the question" by the way). I am pointing out that conservation of momentum is embedded in the way that momentum is assigned to the fields. Maxwell's equations plus conservation of momentum yield equations for the momentum stored and transported by EM fields. Lots of very smart people have reviewed that derivation, and there are no flaws in it. These equations when used correctly cannot yield a result that violates conservation of momentum, because they were derived from conservation of momentum. The "proof" of conservation of momentum is Noether's theorem given the appropriate symmetry (plus it is generally taken as a fundamental law anyway based on all the experimental observations ever made).

[ZhixianLin]

Thanks for the reply meberbs, this makes sense that the forces experienced in the near-field aren't the same as a classical "baseball gets thrown" analogy for particles.

The transition point does not happen at a specific time and place, though, right?  As I understand, this is because the zone is a product of gradient self-interference. Does this not imply that we could, by exact positioning of a photodiode, control which virtual photons self-interfere before all photons have actually left it, and create nonsymmetric but conserved forces?  (In this context the photodiode is somewhere between the near field and the far field.)

You are correct, there is no sharp cutoff, just a point where the near field effects become negligible for a given definition of negligible.

The forces will always be balanced, as long as you include the momentum stored in the fields. You can't get away from momentum conservation, and the only way to get EM momentum away from the rest of the device is through photons. You can redirect the photons in specific directions, but not get better ratios of energy to momentum.

This is because the derivation of momentum storage in E-M fields assumes conservation of momentum.

If the derrivation of momentum storage assumes the conservation laws, then you are deriving it from the combination of Maxwell Equation and Conservation Laws.
This is not the same as getting the conservation laws from only Maxwell's equations.

So, the conclusion that Maxwell Equations imply Conservation Laws is false.
You can't first assume something and then prove it on the basis of the assumption.

Conservation laws all derive from Noether's theorem (although most of the conservation laws were being used well before this theorem was proven).

It is not difficult to find situations in electrodynamics that forces do not appear to be equal and opposite when you consider only the momentum changes in the charged particles. Reconciling this with conservation of momentum, requires that momentum be also stored in the fields. When deriving the equation for momentum in the fields, it is therefore already assumed that momentum is conserved.

I am not proving conservation of momentum by assuming it. (that would be the "correct" usage of "begging the question" by the way). I am pointing out that conservation of momentum is embedded in the way that momentum is assigned to the fields. Maxwell's equations plus conservation of momentum yield equations for the momentum stored and transported by EM fields. Lots of very smart people have reviewed that derivation, and there are no flaws in it. These equations when used correctly cannot yield a result that violates conservation of momentum, because they were derived from conservation of momentum. The "proof" of conservation of momentum is Noether's theorem given the appropriate symmetry (plus it is generally taken as a fundamental law anyway based on all the experimental observations ever made).

"the only way to get EM momentum away from the rest of the device is through photons", Is there any proof of that(except the momentum conservation law)?

[oliverio]

Thanks for the reply meberbs, this makes sense that the forces experienced in the near-field aren't the same as a classical "baseball gets thrown" analogy for particles.

The transition point does not happen at a specific time and place, though, right?  As I understand, this is because the zone is a product of gradient self-interference. Does this not imply that we could, by exact positioning of a photodiode, control which virtual photons self-interfere before all photons have actually left it, and create nonsymmetric but conserved forces?  (In this context the photodiode is somewhere between the near field and the far field.)

You are correct, there is no sharp cutoff, just a point where the near field effects become negligible for a given definition of negligible.

The forces will always be balanced, as long as you include the momentum stored in the fields. You can't get away from momentum conservation, and the only way to get EM momentum away from the rest of the device is through photons. You can redirect the photons in specific directions, but not get better ratios of energy to momentum.

 ....

So I interpret your answer like this: "a LASER itself is a device that prevents photons from leaving the near field in all but one specific direction (which is the maximum of linear momentum potential where photons are concerned)."  In a certain very real sense, this is what a LASER is as opposed to a spherical radiator of RF energy.

I'll have to think on that for a moment, because it answers only part of my question.  To address the unanswered portion I feel there is an explanation lacking of the following: when emitting no photons at all, there is more potential momentum in the near-field of an electromagnetic field than the sum total of all photons leaving a laser's focus.  For that reason, analytically, there seems to be a possibility of dispersing this energy in either the creation of photons (as an antenna or laser normally does) or, it would seem reasonable, directly as momentum given the proper case.

If the field loses potential at the same time as the object gaining it, there is no paradox, right?  Especially given that the field is given potential by a storage device in the first place (i.e. a battery).

[goran d]

Here is a little logical table:
A   B   C   P   Q
0   0   0   1   1
0   0   1   1   1
1   0   0   1   1
1   0   1   1   1
0   1   0   1   1
0   1   1   1   0
1   1   0   0   1
1   1   1   1   1

Here P=A and B implies C
Q=B and C implies A
As you can see there is an entry where P is true while Q is false.
In general, you can't reverse an implication
don't forget,the value of false implies false is true

I talking about the statement that if you derive Poynting  vector from electrodynamics and conservation laws, it's therefore imposiible to get results that disobey conservation laws. This statement is false.
In the truth table, A is conservation laws, B is Maxwell equations, C is Poynting vector
There is an entry in which the derivation is true but the conservation laws can't be implied from Poynting vector and Maxwell's Equations.

[Moe Grills]

    Yada yada yada...
 Michael Faraday was nearly two centuries ahead of you.
Don't you remember the illustration in your college physics text books?
The one showing the thumb and two fingers extended, 90 degrees apart?
That illustrated principle shows that if you had a wire orbiting at LEO, orbiting perpendicular to Earth's magnetic lines of force at about 7.8 km/sec, an electric current passing through the wire, the wire would experience a surprisingly strong perpendicular force acting upon it from Earth's magnetic field. The force acting on the wire would, depending on the direction of the current, gradually thrust the wire from LEO to a higher-orbit; even to GEO if allowed.

[meberbs]

I had been busy with the holidays, so I haven't been posting much recently.

"the only way to get EM momentum away from the rest of the device is through photons", Is there any proof of that(except the momentum conservation law)?

Momentum conservation is one of the most fundamental principles in physics. So much of physics is based on it that just about any proof will use it or a result derived from it. The simple proof for the momentum having to leave through photons is that in special relativity E^2 = (p*c)^2 + (m0*c^2)^2. (p is momentum) This means that the momentum transferred by any massless particle has p = E/c. In the case of EM this is a photon
If you want better than that, your only options are to also expel mass or react against an external object (e.g. the Earth through its magnetic field). This is only a better energy per momentum ratio if you don't include the energy stored in the mass though.

Here P=A and B implies C
Q=B and C implies A
As you can see there is an entry where P is true while Q is false.
In general, you can't reverse an implication
don't forget,the value of false implies false is true

I talking about the statement that if you derive Poynting  vector from electrodynamics and conservation laws, it's therefore imposiible to get results that disobey conservation laws. This statement is false.
In the truth table, A is conservation laws, B is Maxwell equations, C is Poynting vector
There is an entry in which the derivation is true but the conservation laws can't be implied from Poynting vector and Maxwell's Equations.

Derived doesn't mean implied. For a consistent theory such as EM with special relativity included, things do go both ways, there is no way to break conservation of momentum within the theory.

So I interpret your answer like this: "a LASER itself is a device that prevents photons from leaving the near field in all but one specific direction (which is the maximum of linear momentum potential where photons are concerned)."  In a certain very real sense, this is what a LASER is as opposed to a spherical radiator of RF energy.

I'll have to think on that for a moment, because it answers only part of my question.  To address the unanswered portion I feel there is an explanation lacking of the following: when emitting no photons at all, there is more potential momentum in the near-field of an electromagnetic field than the sum total of all photons leaving a laser's focus.  For that reason, analytically, there seems to be a possibility of dispersing this energy in either the creation of photons (as an antenna or laser normally does) or, it would seem reasonable, directly as momentum given the proper case.

If the field loses potential at the same time as the object gaining it, there is no paradox, right?  Especially given that the field is given potential by a storage device in the first place (i.e. a battery).

You are getting into some really subtle details, and I am not really sure how to explain better, so I will give an example from a textbook that shows a quasi-static storage of momentum in the fields balanced by a subtle relativistic effect. Examples like this are sometimes called "hidden momentum", since most would miss the balancing momentum on the first pass.

Take 2 conducting cylinders, that are very long, with radius a and b, with b > a. Place the smaller one inside the (hollow) large one, so that you have a model for a coaxial cable. If you connect a battery between them on one end, so the inner conductor is at higher potential. This will cause the conductors to charge up as a capacitor, with charge per unit length λ = V *2*π*ε0 / ln(b/a) on the inner conductor, and opposite on the outer conductor.

(I will use underline to indicate vectors in the equations below, s is unit vector radial outward, φ is circumferential, and z points away from the end with the battery. The cross product s X φ = z defines the right hand rule for this coordinate set.)

Now, adding a resistor to the other end R will create a current I = V/R. The Electric and Magnetic fields for this setup are E = λ / (2 * π * ε0 * s) s (s is radial distance from center of the cable). B = μ0 * I / (2 * π * s) φ This is only valid between the conductors, since there would be no fields inside the center conductor or outside the outer conductor.

I am ignoring the complications near the edges, and this can be looked at as an analysis of a segment of a very long cable, in which case there are no edge effects within the section under consideration.

The Poynting vector S = E X B / μ0  can be integrated over a cross section of the cable to show that The power transported is equal to I * V, as expected.

The volume integral of μ0 * ε0 * S over a volume including a length L of the cable shows a momentum stored in the fields of μ0 * ε0 * L * I * V = L * I * V / c^2 z.

This momentum doesn't initially make sense, since nothing is moving in this system at first glance and the EM fields are all static. There is a theorem that states that "if the center of mass of a localized system is 0, its total momentum must be 0" (This only applies to linear momentum, not angular momentum). In this case the hidden momentum has to do with the flow of current and relativistic momentum. The charges gain energy, and while currents to the left and right are equal, one current has a smaller number of charges moving faster. In relativity, this means the momentum of the current in one direction does not cancel the momentum in the other direction, by a value that exactly matches the momentum I calculated above.

I found this paper: http://www.physics.princeton.edu/~mcdonald/examples/current.pdf which goes into more detail of some related examples. ( I haven't read the whole thing, but it points out some details that my textbook glossed over)

[goran d]

An attempt of actual proof by counter example that is two perpendicular dipole radiators. Force increases with R^-2, radiation tends to a fixed point (when dipole sizes are same). 

[meberbs]

An attempt of actual proof by counter example that is two perpendicular dipole radiators. Force increases with R^-2, radiation tends to a fixed point (when dipole sizes are same). 

A couple of initial issues that I noted:

1. You make several simplifying assumptions as you go. For this kind of example, you want to avoid that as much as possible, since it is an easy loophole. Specifically:
     a. you neglect d, but then at the end you analyze behavior as R goes to 0. This breaks your previous assumption.
     b. You assume non-relativistic velocities, but as the example I posted above shows, you need to account for relativity to get conservation of momentum to work in EM.
     c. As a result of your assumptions, it seems that you are not accounting for the fields using time retarded fields. (I did not review the form of all of your equations exactly, so I might have missed this)
2. You do not actually calculate the amount of radiation, so your conclusion does not have a solid basis. Your claim that the radiation is constant cannot be simply stated. A single dipole resonator has symmetric fields, and therefore net 0 force on it. 2 resonators are a phased array antenna, and can have a net force on their collection because they can result in a directional beam. Their spacing affects the degree of directionality, which affects the net force.
3. In order to disprove conservation of momentum in EM, you need to show that the surface integral of the Maxwell Stress Tensor is not consistent with the time rate of change of mechanical and electrical momentum. See this post for the set of equations you would have to demonstrate as inconsistent.

Number 2 above is the biggest issue with your calculations. Also I recognize that number 3 is an absurdly high hurdle, since as I am looking at my textbook, I see that the stress tensor comes directly out of the force equation in the derivation. To try to reduce the hurdle, you could show that the Maxwell Stress Tensor does not represent the force per unit area acting on a surface. The main difficulty with this is that you would then have to calculate the momentum over all space outside the volume, which will diverge for radiation unless you include a "turn on" time for the system, but it may be difficult to calculate a physical turn on transient. You could instead calculate momentum change in the volume between two surfaces, (a larger volume enclosing the smaller volume) and you would see the rate of change of EM momentum between the surfaces equal to the difference between the integrals of the Maxwell Stress tensor over each surface. (I am assuming all charges are inside the inner surface). Note that the result will depend on the relative positions of the surfaces, since there will be some surfaces where there is an integer number of wavelengths between the surfaces, so there will not necessarily be a periodic variation of the momentum storage between the surfaces.

I don't recommend you bother with any of these calculations unless you want to practice your calculus, since the equations are all consistent, and you will not find an EM momentum violation unless you neglect something.