Angle wire resonator as reactionless drive

[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.