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  1.  
    Here's a general question about Noether and conservation. Here's two ways of expressing it, here in the case of momentum:

    1. Conservation of momentum is a special case of Noether's theorem, and simply says that if the laws of physics do not change from place to place, then it must hold.

    2. Conservation of momentum is a special case of Noether's theorem, and simply says that because the laws of physics do not change when the position coordinates are reflected about the origin, then it must hold.

    Same goes for energy, but now in respect to time.
    "place to place" => "time to time"
    "when the position coordinates are reflected about the origin" => "when time is reversed."


    Are both these descriptions right, or is one better than the other (superset or something)?
    • CommentAuthorjoshs
    • CommentTimeJul 31st 2013
     
    Posted By: Andrew PalfreymanMy impression is that, as I mentioned, he's simply using the radiation reaction from a 2-element Yagi. In space, that would qualify as "a working thruster", just as would a photon rocket. However, if his optimisations are worth anything (and I don't fully understand them, especially the number of bits required for tuning accuracy - can anyone shed light?), then he's actually doing antenna design optimisation.

    I just got off the phone with him, actually. He's an applied mathematician. He's building something right now. I'm still climbing the learning curve on the patent details, and am going to give him a little email lecture on OU, just as I gave Serrano.
    He might think that if he comes up with some huge Q that he can develop enough radiation pressure to do some good. Really colossal Q's aren't very amenable to apertures of any size. I can't see that as holding any promise at all.
    • CommentAuthorjoshs
    • CommentTimeJul 31st 2013
     
    Posted By: Andrew PalfreymanHere's a general question about Noether and conservation. Here's two ways of expressing it, here in the case of momentum:

    1. Conservation of momentum is a special case of Noether's theorem, and simply says that if the laws of physics do not change from place to place, then it must hold.

    2. Conservation of momentum is a special case of Noether's theorem, and simply says that because the laws of physics do not change when the position coordinates are reflected about the origin, then it must hold.

    Same goes for energy, but now in respect to time.
    "place to place" => "time to time"
    "when the position coordinates are reflected about the origin" => "when time is reversed."


    Are both these descriptions right, or is one better than the other (superset or something)?
    I woould not attempt to express Noether in either of those terms. Noether is a very general theorem that has been applied to CoM, and CoE, so the first statement that CoM falls out from Noether due to spatial invariance of Newton's Laws would be an OK statement to me. CoE falls out provided that one cannot go back in time.
    •  
      CommentAuthorAngus
    • CommentTimeJul 31st 2013
     
    Posted By: joshs
    Posted By: Andrew PalfreymanMy impression is that, as I mentioned, he's simply using the radiation reaction from a 2-element Yagi. In space, that would qualify as "a working thruster", just as would a photon rocket. However, if his optimisations are worth anything (and I don't fully understand them, especially the number of bits required for tuning accuracy - can anyone shed light?), then he's actually doing antenna design optimisation.

    I just got off the phone with him, actually. He's an applied mathematician. He's building something right now. I'm still climbing the learning curve on the patent details, and am going to give him a little email lecture on OU, just as I gave Serrano.
    He might think that if he comes up with some huge Q that he can develop enough radiation pressure to do some good. Really colossal Q's aren't very amenable to apertures of any size. I can't see that as holding any promise at all.


    In optical cavities, which is all I know anything about, very high Q corresponds to very highly reflecting mirrors. The cavity power can be large, but correspondingly little gets out.
    •  
      CommentAuthorAngus
    • CommentTimeJul 31st 2013
     
    My limited understanding of Noether's theorem seems to say that if the laws are symmetric in time (i.e. do not vary as time goes on) then energy is conserved.
    • CommentAuthorjoshs
    • CommentTimeJul 31st 2013
     
    Posted By: Angus
    Posted By: joshs
    Posted By: Andrew PalfreymanMy impression is that, as I mentioned, he's simply using the radiation reaction from a 2-element Yagi. In space, that would qualify as "a working thruster", just as would a photon rocket. However, if his optimisations are worth anything (and I don't fully understand them, especially the number of bits required for tuning accuracy - can anyone shed light?), then he's actually doing antenna design optimisation.

    I just got off the phone with him, actually. He's an applied mathematician. He's building something right now. I'm still climbing the learning curve on the patent details, and am going to give him a little email lecture on OU, just as I gave Serrano.
    He might think that if he comes up with some huge Q that he can develop enough radiation pressure to do some good. Really colossal Q's aren't very amenable to apertures of any size. I can't see that as holding any promise at all.


    In optical cavities, which is all I know anything about, very high Q corresponds to very highly reflecting mirrors. The cavity power can be large, but correspondingly little gets out.
    That is the same for RF cavities. Finite conductivity in the walls results in some absorption of each incident wavefront, limiting Q. Any dielectric loss converts power to heat on each pass as well. It is routine to displace the air with dry nitrogen in high power waveguides in order to reduce such losses even though the wave propagation is incident.
    • CommentAuthorjoshs
    • CommentTimeJul 31st 2013
     
    Posted By: AngusMy limited understanding of Noether's theorem seems to say that if the laws are symmetric in time (i.e. do not vary as time goes on) then energy is conserved.
    My understanding of the application of Noether to energy conservation is that the underlying presumption is that time "goes on".
    •  
      CommentAuthorAngus
    • CommentTimeJul 31st 2013
     
    Ergo, on both ways of looking at it, high Q does not have a lot to do with radiation pressure, but has everything to do with line width.
    • CommentAuthorjoshs
    • CommentTimeJul 31st 2013
     
    Posted By: AngusErgo, on both ways of looking at it, high Q does not have a lot to do with radiation pressure, but has everything to do with line width.
    Or for RF types: percentage bandwidth. I don't see any land yacht races using overgrown LASER pointers as the propulsion units. I wonder why.
  2.  
    Posted By: AngusMy limited understanding of Noether's theorem seems to say that if the laws are symmetric in time (i.e. do not vary as time goes on) then energy is conserved.
    You've put both my alternatives into the same sentence, so I assume you feel that one is as good as the other; "symmetric in time" and "do not vary as time goes on" are the two.
    •  
      CommentAuthorAngus
    • CommentTimeJul 31st 2013
     
    Posted By: Andrew Palfreyman
    Posted By: AngusMy limited understanding of Noether's theorem seems to say that if the laws are symmetric in time (i.e. do not vary as time goes on) then energy is conserved.
    You've put both my alternatives into the same sentence, so I assume you feel that one is as good as the other; "symmetric in time" and "do not vary as time goes on" are the two.


    They aren't really alternatives. In the sense used in physics "symmetric" with respect to a variable means unchanging if that variable is changed. Its not limited to the meaning "same on both sides", or "same in both directions" as I'm sure you know. The laws of physics are time symmetric because they are the same now as yesterday and tomorrow. Actually they are not exactly symmetric with respect to the direction of time - the thermodynamics don't work out.
    •  
      CommentAuthorlegendre
    • CommentTimeAug 3rd 2013
     
    @All

    (This is a repost from another thread, just for continuity)

    Posted By: Andrew PalfreymanI didn't state it was exploitation, but merely asked the question. As for The Woodward Effect, the jury remains out.




    It's funny. The first time I sat down and looked at this, it didn't make sense.. in that I couldn't parse the info. The explanation is unnecessarily complex and long-winded. It was actually looking at this graphic (above) that finally switched-on the light bulb.

    I can't believe that there are maths to support this - are there? Because *if* I understand the premise correctly, I can see the reason it won't work - that is, I can see the basic flaw in reasoning that's leading them to deduce there will be a net thrust.

    I'll take this up in Andrew's 'Propellantless Woo' thread. (And here it is)
    •  
      CommentAuthorlegendre
    • CommentTimeAug 3rd 2013 edited
     
    @Andrew (or anyone else with a proper understanding)

    Am I correct that this 'Woodward effect' centers around the change in rest mass of an inductor or capacitor, as it transitions between charged and discharged states? The basic implementation would be an LC tank circuit, with the physical L and C elements situated diametrically opposed, on opposite ends of a linear transducer element?

    The core tenet being that the L/C elements change rest mass - very slightly, per E=MC^2 - as they charge & discharge?

    Making it go is like this: You set the tank circuit oscillating, and power the transducer probably synched to (or actually with) the tank signal. The xducer pushes the L/C apart when the L is +mass (charged) and the C is -mass (discharged), then on the next half-cycle, the xducer pulls in, while the L is -mass and the C is +mass.

    Each half-cycle produces a slight but consistent net directional thrust along the same vector, alternately pulling towards then kicking off of the more massive side - right?

    If this is actually the theoretical basis, it's got a very, very obvious problem. Nah, it can't be.. people smart enough to figure this out just +have to+ know more than I do about this shit!?

    (ETA: Edits.. and more edits.)
  3.  
    No, that would be too mundane. If going far out, it's best to go really far out. It is not an expression of a change in energy being expressed as an inertial quantity, because E/c^2 is normally an extremely tiny quantity. There is no dispute in the mainstream that this is what we find. This is different. It has to be different because the net energy in the LC circuit you describe never changes, but instead just sloshes around. There's no net change in total energy-inertia in what you describe.

    State of the art for solid state energy density is around 100 KJ/Kg, but that's a dm/m of ~1E-13.
    What's the "obvious thing" you want to say?

    Here's a puzzle

    A pair of supercapacitors are placed on diametrically opposite sides of a wheel.
    The wheel is on Earth and has its axle horizontal. Electrical circuitry (ideal
    and so lossless) connects them, so that they charge and discharge cyclically.
    The charging cycle and the wheel cycle are synchronised so that when the
    connection line is horizontal they are at maximum charge difference, with the
    max charge always on the same side (say right) and when vertical they are at the
    same charge. Because dm=dE/c^2, the wheel will accelerate under gravity,
    endlessly.

    Where does the "free" energy come from?
    •  
      CommentAuthorlegendre
    • CommentTimeAug 3rd 2013 edited
     
    Posted By: Andrew PalfreymanNo, that would be too mundane. If going far out, it's best to go really far out. It is not an expression of a change in energy being expressed as an inertial quantity, because E/c^2 is normally an extremely tiny quantity.


    Very, very tiny.

    There is no dispute in the mainstream that this is what we find. This is different. It has to be different because the net energy in the LC circuit you describe never changes, but instead just sloshes around. There's no net change in total energy-inertia in what you describe.


    In pure Newtonian terms, no, there isn't. Even with the tiny effects of E=MC^2, the differences cancel perfectly. That was the point of my now obviously simple-minded take on the problem.

    State of the art for solid state energy density is around 100 KJ/Kg, but that's a dm/m of ~1E-13.
    What's the "obvious thing" you want to say?


    You already have it, and we agree, far as I can tell. It's that there is no differential in the distribution of PE that might be developed between the L & C elements. PE isn't "contained" in within an object, or even a region of space. It exists in the relationships between two masses, or regions, or entities, or what-have-you.

    Simplest example of my last statement, if you raise a 10lb brick from the floor to a tabletop, you didn't just store X joules of work int he brick. The PE, very real, exists in the interplay of mutual gravitational / electrical forces. In this sense, it's wrong to think that either the L or C should change in dynamic mass with respect to the other - any changes should be equal and opposite.
    •  
      CommentAuthorlegendre
    • CommentTimeAug 3rd 2013
     
    You know, I'm so far out of my area that the above might be among the most confused and misinformed things I've ever written.. and I've no idea, at least not yet.

    Ha.

    At least I won't have to wonder for too long. That's what moletrap does.
  4.  
    If you raise an m Kg brick by h metres in a 1 gee gravitational field, you do m*g*h Joules of work and create that same amount of potential energy. The brick still weighs m*g Newtons - there's no extra m*g*h/c^2 Kg inertia term.

    However, if you charge a capacitor with J Joules, it now possesses extra inertia of J/c^2 Kg.

    Both scenarios describe increased potential energy situations - one gravitational and one electrical.
    Isn't it odd that they are not equivalent from the perspective of inertia?
    • CommentAuthorjoshs
    • CommentTimeAug 3rd 2013
     
    Posted By: legendre@Andrew (or anyone else with a proper understanding)

    Am I correct that this 'Woodward effect' centers around the change in rest mass of an inductor or capacitor, as it transitions between charged and discharged states? The basic implementation would be an LC tank circuit, with the physical L and C elements situated diametrically opposed, on opposite ends of a linear transducer element?

    The core tenet being that the L/C elements change rest mass - very slightly, per E=MC^2 - as they charge & discharge?

    Making it go is like this: You set the tank circuit oscillating, and power the transducer probably synched to (or actually with) the tank signal. The xducer pushes the L/C apart when the L is +mass (charged) and the C is -mass (discharged), then on the next half-cycle, the xducer pulls in, while the L is -mass and the C is +mass.

    Each half-cycle produces a slight but consistent net directional thrust along the same vector, alternately pulling towards then kicking off of the more massive side - right?

    If this is actually the theoretical basis, it's got a very, very obvious problem. Nah, it can't be.. people smart enough to figure this out just +have to+ know more than I do about this shit!?

    (ETA: Edits.. and more edits.)
    In that case it is a fantasy built on a small mathematical error.
    •  
      CommentAuthoralsetalokin
    • CommentTimeAug 3rd 2013 edited
     
    Posted By: Andrew PalfreymanIsn't it odd that they are not equivalent from the perspective of inertia?


    No, because GPE is relative to a reference (h=0) that is external to the lifted mass and is _potential energy_. The stored energy in a capacitor is referenced to the capacitor itself (the voltage (charge) difference between the plates) and isn't potential energy, it is real energy, as the equations reveal. The energy on the cap has the same form as KE, not GPE; it is real, not potential, energy, thus it has mass-inertia.
    •  
      CommentAuthorAngus
    • CommentTimeAug 3rd 2013
     
    The relativist term comes in if you let the brick drop, gaining velocity. When it gets to the bottom some of the potential energy has become a relativistic mass component. If you lower it down gradually with a braked winch the entirety of the potential energy is turned into heat in the brake.