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#460 by GoatGuy on 12 Sep, 2014 19:28
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Hello, folks. From a poster's recommendation at NextBigFuture, I have come here to participate in this and other fora, If y'all will have it. Previous posts at NBF (of mine) have been pasted here; so far, I see little love for the position(s) I've taken.
To "short circuit" the discussion a bit, I have determined that a VERY simple pair of equations brings into question whether the EM-Thruster or Q-Drive is a perpetual-motion / over-unity energy generator or not. These:
if [k] is the EM-drive thrust factor, in Newtons-per-Watt (which has been widely used and quoted) then...
V = 1/k ... is the velocity of the Q-device, where the kinetic energy it creates matches the input energy and...
V = 2/k ... is the velocity of a free-floating spacecraft employing the EM-thruster, where the TOTAL energy invested equals the TOTAL imparted kinetic energy of the spacecraft.
These are important equations; I have not so far discerned that there is a contrary-position regarding the over-unity aspect that these equations give users of the EM-drive device(s). Using the "usual physics" equations for energy, velocity, acceleration, time, mass, force, distance (delta-d) and so on, both of these are really straight forward to prove. Even if all you do is the empirical math to show applicability.
In short:
Ek = ½mV² ... kinetic energy as a function of V
W = Fd ... work (energy) as function of force and delta-distance (don't know how to get delta typed in)
V = at ... delta V as function of acceleration and time (assuming constant)
F = ma ... the usual physics, so
a = F/m
And then some obvious (and algebraically correct) substitutions:
W/t = Fd/t
V = d/t (change in distance over time, nothing new here...)
W/t = FV
W/t is "power", or P, so
P = FV = maV (again, nothing unusual)
When F = k·Pin (k is newtons per watt, Pin is watts) then substituting...
P = k·Pin·V ... now just setting the "critical over-unity point" where both P and Pin are the same (P = Pin) then:
P = k·P·V
1 = k·V
V = 1/k by rearrangement
That one was easy, agreed? And correct, classical Newtonian physics. The second case is a bit harder to prove, but still within reach of ordinary algebra:
Ek = ½mV² ... kinetic energy as a function of V, again.
V = at ... again, now substitute
Ek = ½ma²t² and remembering that [F = ma]...
Ek = F²t²/(2m)
Since the spacecraft/thruster will consume Ein = Pin·t (power times time), then setting Ein = Ek (total energy in, to total kinetic energy of free-flying craft), we get:
Ek == Ein
Ek = Pin·t
Pin·t = F²t²/(2m), and solving for t:
t = 2·Pin·m/F², which is the time at which the kinetic energy of the craft matches the total input energy, and
V = at
V = a·( 2·Pin·m/F² ) (and recalling F = k·Pin AND a = F/m, and a = k·Pin/m...)
V = a·( 2·Pin·m/(k²Pin²) )
V = ( k·Pin/m ) · ( 2·Pin·m/(k²Pin²) ), and with algebraic rearrangement
V = ( 2·k·Pin²·m ) / ( m·k²Pin² ), then cancelling things on top and bottom:
V = 2 / k
Where this then is the velocity of the craft considered in its OWN reference frame, as attaining an amount of delta-V, which equates to kinetic energy, that is EQUAL to the amount of electrical power pumped into the thruster. ANY additional velocity then gives the craft, in its own reference frame, MORE kinetic energy than has been invested into it by way of the thruster.
At (V > 1/k), it is a free energy device.
Thanks for the consideration. I hope there will be a reply that answers this fundamental flaw.
GoatGuy -
#461 by Rodal on 12 Sep, 2014 19:36
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Paul March,
OK, Let's try to discuss the experimental set-up. From one of your previous posts, I understand that the inverted pendulum arm is made of this: http://www.amazon.com/Faztek-15QE1515UL-Aluminum-T-Slotted-Extrusion/dp/B008MQA11C .
Going to the company's website: http://www.faztek.net/technical.html, I understand that the "Faztek 15QE1515UL Aluminum 6063-16 T-Slotted Ultra Light Extrusion with Clear Anodize Finish, 1-1/2" Width x 1-1/2" Height" you use for the inverted pendulum arm has the following Moments of Inertia (in bending):
Ix = Iy = 0.1802 in^4
and the following modulus of elasticity:
E = 10^6 psi
I also understand from your prior post that this moment arm is restrained from rotations around the X and Y (horizontal) axes at the bottom support by two Riverhawk C-flex torsion bearings, and << the center of the two C-flex bearing blocks is 2.38" above and below the centerline of the 24.00" long by 1.50" square aluminum pendulum arm. The long end of the pendulum arm is 15.5" from the torque pendulum's center of rotation, which makes the other short-end of the pendulum arm 8.5">>.
What I need is the arm distance between the bottom support and the supported platform with weight. If I interpreted correctly what you wrote, this length "l" is:
l = 15.5" - 2.38"=13.12 inches =0.33325 meters
I needed the elastic spring constant of the pendulum arm (in swinging motion). I understand that the number you provide << The Riverhawk C-flex torsion bearing's spring constant is a nominal 0.007 in-Lb/deg>> must be for torsion around the vertical "Z" axis normal to the horizontal plane, because this spring constant is extremely low.
So, my interpretation is that we therefore have to calculate the (bending) spring constant of the pendulum arm (in swinging motion). Interpreting the two Riverhawk C-flex torsion bearings constraints as a cantilevered support for the pendulum arm, and using the suggested beam deflection formula in the faztek website for a cantilevered beam acted by a force at the free end:
MaximumDeflection = delta = F * (l^3) / (3 E I )
The (bending) spring constant of the pendulum arm (in swinging motion) for rotation around the X and Y horizontal axes is:
k = F/delta = (3 E I ) / (l^3) = 3 * (10^6 lbf/in^2) * (0.1802 in^4) /(( 13.12 in)^3) = 2393.726 lbf/in
k =( 2393.726 lbf/in) *( 4.44822162 N/lbf) / (0.0254 in/m) = 4.192*10^5 N/m
Is my interpretation correct ? -
#462 by Rodal on 12 Sep, 2014 19:52
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@ Paul March,
We also needed the total supported weight (of all the supported equipment, the drive AND the platform on top of the pendulum arm). I understand that this is 25 lbf (although I wonder whether you included the weight of the supporting platform resting on the pendulum arm as well ?)
Taking this as the total supported mass:
M = 25 lb = 25 lb * (0.453592 kg/lb) = 11.3398 kg
We get a ratio of (swinging motion) spring constant to mass of:
k/M =( 4.192*10^5 kg*(m/s^2)/m)/11.3398 kg = 3.6968*10^4 1/(s^2)
and the gravity term is:
g/l = (9.80665 m/(s^2))/(0.33325 m) = 29.4273 1/(s^2)
So, the stiffness-to-mass term is 1256 times greater than the gravity term, and the frequency (in swinging motion around the X and Y horizontal axes) is:
f = (1/(2 Pi))Sqrt[3.6968*10^4 1/(s^2) - 29.4273 1/(s^2)] = 30.59 Hertz
[I presume that this is an upper bound for the frequency because it assumes that the bearings are providing a cantilevered condition with no swinging flexibility around X an Y, and it presumes that 25 lbf is the total weight supported on top of the pendulum arm, including the platform] -
#463 by Rodal on 12 Sep, 2014 20:26
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Hello, folks. From a poster's recommendation at NextBigFuture, I have come here to participate in this ....
Thanks for the consideration.
GoatGuy
@GoatGuy,
What is your physical explanation for the experimental measurements (of at least 6 different drives) at NASA Eagleworks?, including
A) dependence on the Teflon dielectric resonator (without the dielectric resonator they measured zero thrust) and
B) that when they turned the drive (by 180 degrees rotation around the Z vertical axis), with the dielectric resonator to the "left" instead of to the "right", they got a similar numerical thrust result, but now directed to the opposite direction as compared to the previous orientation (showing a thrust measurement vector dependent on physical orientation of the drive) ? -
#464 by frobnicat on 12 Sep, 2014 20:48
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Hello, folks. From a poster's recommendation at NextBigFuture, I have come here to participate in this and other fora, If y'all will have it. Previous posts at NBF (of mine) have been pasted here; so far, I see little love for the position(s) I've taken.
Speaking for myself as a relatively newcomer, welcome onboard.Quote.../...
At that point, it is a free energy device.
Thanks for the consideration. I hope there will be a reply that answers this fundamental flaw.
GoatGuy
Well, I did post similar argument : http://forum.nasaspaceflight.com/index.php?topic=29276.msg1240286#msg1240286
I don't remember counter arguments : I guess no one can seriously object that a reactionless drive with such Thrust/Power ratio is indeed either a free energy device, or pumping energy from somewhere (see below).
But this very serious and immediate practical consequence seems to be less excitingly investigated than (beyond my understanding) theoretical inconsistencies of the aspects of Machian inertia for critics, and the possibilities of delivering grand pianos to Saturn moons for enthusiasts.
A few caveats :
Your calculations are Newtonian mechanics, when approaching c speeds or Thrust/Power ratios below 1/c its no longer valid, and indeed we have the photon rocket as an example of a "reactionless" drive that is not a free energy device. The claimed thrust/power = k ratio is more than 2 orders of magnitude above that, so it's ok to go Newtonian as a good approximation, also it should be acknowledged and stressed that as soon as a purported k factor is even so slightly better (higher) than 1/c of the photon rocket, then there is energy conservation breaking for some inertial reference frame (also maybe not practical to exploit).
Quibbling : "velocity of the craft considered in its OWN reference frame" could be misread as "the frame of its acquired speed now" (absurd as it would be 0 by definition, but yet). Could we say "in the inertial reference frame of its departure" or something like that (English not my native tongue) ?
Ok so we are breaking energy conservation, great : Noether's theorem show this implies reality not to be time invariant. Hard to swallow much much below cosmological timescales.
Ok so energy is conserved but the acquired energy is pumped from vacuum, not free energy but rel cheap energy, great : quoting myself "tap into vacuum zero-point energy, which would no longer be zero-point... My guess is that all this virtual agitation down there is like a thermal bath and nothing useful (in terms of net work) can come out of it (second principle...) but please someone qualified help." In context :
http://forum.nasaspaceflight.com/index.php?topic=29276.msg1243932#msg1243932
Penultimate point : if the k factor somehow decays with acquired speed (relative to start frame of reference) this could mitigate the free energy aspect. But then the mission profiles are not the same... and it's really difficult to see what would make the system "remember" this particular initial reference frame, and it doesn't prevent the drive to be a free energy generator relative to some well chosen arbitrary inertial reference frame.
Last point : the k=0.4 N/kW figure used for the mission profiles implies possible breakeven starting at speeds of 1/k=2.5km/s relative to a fixed frame. That is if you mount the drive on a rotor, and re-inject the energy of the shaft into the drive, you have not only a theoretical but an almost practical free (or rather cheap) energy generator : the rim of fastest wheels for energy storage is around 2km/s. (Take into account the efficiency of the closed loop, if 0.5 then turn at 5km/s tangential speed, mechanically harder to get at but still conceivable without unobtainium). I'd like to see the experiment done on a freely rotating arm, in an otherwise rotationally symmetric setting around the axis.
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#465 by francesco nicoli on 12 Sep, 2014 21:16
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Anyway, great to have GoatGuy on this forum. NBF is an interesting platform for discussion but faaar less professional than here on many subjects; GoatGuy is by far the more informed discussant there and will add lot of quality here as well. Welcome!
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#466 by GoatGuy on 12 Sep, 2014 21:21
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What is your physical explanation for the experimental measurements (of at least 6 different drives) at NASA Eagleworks?, including
A) dependence on the Teflon dielectric resonator (without the dielectric resonator they measured zero thrust) and
B) that when they turned the drive (by 180 degrees rotation around the Z vertical axis), with the dielectric resonator to the "left" instead of to the "right", they got a similar numerical thrust result, but now directed to the opposite direction as compared to the previous orientation (showing a thrust measurement vector dependent on physical orientation of the drive) ?
I'm sorry, how does that address [V > 1/k case, and V > 2/k cases], which was the point of my posting? I am not questioning whether something was measured. What I am questioning is whether such a finding leads immediately to declaring the device a free energy device after some amount of power is invested in its motion in space. That's, all.
GoatGuy
EDIT / postscript: special alert - I'm aware that when 'k' is equal or less than 1/299,792,000 newtons per watt, conservation of energy is preserved, as that particular number is reserved for perfect radiators of electromagnetic energy. They manage not to break the conservation of energy criterion because they would have to "burn up" a lot of matter (say, optimally, matter-antimatter at perfect efficiency!) ... and the amount of speed/thrust then becomes almost exactly what you'd expect for a rocket that just as efficiently exhausted real reaction mass.
GoatGuy -
#467 by Rodal on 12 Sep, 2014 21:23
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Virtual particles of the vacuum on the other hand don't appear to define a frame of reference, though they might define an "inertial reference" (tm) : Casimir effect for instance don't show different behaviour on different inertial frames (Lorentz invariant, no reference of what would be an absolute 0 speed relative to vacuum) while dynamical Casimir effect allows to measure acceleration in absolute terms (can tell an absolute 0 acceleration relative to it)....
We don't really need to go there either:top physicists nowadays do NOT explain the Casimir effect in terms of "negative mass" or "negative energy" but as a van der Waal (charge) force just as in the Lamb shift. Pauli, Feynman, de Witt were known to express their dissatisfaction with the quantum vacuum explanation. Swchinger found it so distasteful that he came up with his own successful explanation. To me the failure (in his own attempt) by Casimir himself to explain the fine structure constant in terms of the quantum vacuum should be telling enough. Ditto for his failure to explain and predict even the sign of the Casimir force for different shapes (instead of plates).
Why don't we discuss the experimental results and physical explanations for them instead of tearing up controversial theories?
ADDED: To me tearing up a controversial theory is like Don Quixote charging against the windmills. Addressing and explaining the experimental results instead is analogous to physicists explaining quantum mechanics experimental results in the early 20th century to come up with the theory of Quantum Mechanics: the most successful theory to explain nature, yet it was thrusted on us by experimental results so strange that caused Bohr to say <<If quantum mechanics hasn't profoundly shocked you, you haven't understood it yet.>> and Feynman to say <<I think I can safely say that nobody understands quantum mechanics.>>
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#468 by Rodal on 12 Sep, 2014 21:31
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I'd like to see some qualified explanations with Feynman diagrams showing how it's impossible to push on virtual particles (unless they are made real at equivalent energy/mass cost).
I think that Matt Strassler (currently a visiting scholar at Harvard University) does an excellent job here:
http://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/
<<The term “virtual particle” is an endlessly confusing and confused subject for the layperson, and even for the non-expert scientist. I have read many books for laypeople (yes, I was a layperson once myself, and I remember, at the age of 16, reading about this stuff) and all of them talk about virtual particles and not one of them has ever made any sense to me. So I am going to try a different approach in explaining it to you.
The best way to approach this concept, I believe, is to forget you ever saw the word “particle” in the term. A virtual particle is not a particle at all. It refers precisely to a disturbance in a field that is not a particle. A particle is a nice, regular ripple in a field, one that can travel smoothly and effortlessly through space, like a clear tone of a bell moving through the air. A “virtual particle”, generally, is a disturbance in a field that will never be found on its own, but instead is something that is caused by the presence of other particles, often of other fields.>>
ADDED (Feynman diagrams):
<<The language physicists use in describing this is the following: “The electron can turn into a virtual photon and a virtual electron, which then turn back into a real electron.” And they draw a Feynman diagram that looks like Figure 4. But what they really mean is what I have just described in the previous paragraph. The Feynman diagram is actually a calculational tool, not a picture of the physical phenomenon; if you want to calculate how big this effect is, you take that diagram , translate it into a mathematical expression according to Feynman’s rules, set to work for a little while with some paper and pen, and soon obtain the answer.>>
<<This is shown in Figure 7, and the corresponding Feynman diagram is shown in Figure 8. This goes on and on, with a ripple in any field disturbing, to a greater or lesser degree, all of the fields with which it directly or even indirectly has an interaction.
Fig. 8: The Feynman diagram needed to calculate the process shown in Figure 7.
So we learn that particles are just not simple objects, and although I often naively describe them as simple ripples in a single field, that’s not exactly true. Only in a world with no forces — with no interactions among particles at all — are particles merely ripples in a single field! Sometimes these complications don’t matter, and we can ignore them. But sometimes these complications are central, so we always have to remember they are there.>> -
#469 by aero on 12 Sep, 2014 21:38
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@GoatGuy, frobnicat,
Your arguments illustrate that the operational principles of the EM drive satisfy Clark's third law - they are, "indistinguishable from magic". Your arguments do not advance our understanding of those principles. Others have taken the risk and dared to postulate principles but I know of none that are accepted. Go ahead, take the risk, postulate physics sufficiently advanced as to cast light on the difference between operation of the EM drive and magic. -
#470 by 93143 on 12 Sep, 2014 21:55
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I hope there will be a reply that answers this fundamental flaw.
A Mach-effect device is supposed to be interacting with the distant universe; that's inherent in the theoretical derivation. (If I understand correctly, the Wheeler-Feynman stuff doesn't have to be assumed to get this result; Sciama's model and Woodward's application of it just have to do with local gravitational potential.)
Basically, your energy balance isn't complete until you've accounted for the device's interaction with the rest of the matter in its Hubble sphere, whatever form that interaction takes. In other words, you're drawing the box too small.
Furthermore, without something to push on, you aren't just violating conservation of energy; you're violating conservation of momentum too. The equations of mechanics don't work that way, and you can see this if you consider the fact that the kinetic "energy" of a single body is frame-dependent - only a momentum-conserving system exhibits Galilean invariance of kinetic energy changes due to interactions. So really, using Newtonian mechanics to treat a Mach-effect thruster in isolation is nonsensical from the start.
...
I'm not sure how the Q-thruster is supposed to conserve energy; it looks like a ZPE device to me, but then I haven't studied it much...
Regarding the EM-Drive, I thought it was pretty well established that Shawyer's explanation implied a violation of conservation of momentum, and that thus if the drive worked it would be for some other reason. Again, though, I haven't studied it much... -
#471 by frobnicat on 12 Sep, 2014 22:57
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@dr Rodal,
Frankly I'm not qualified to argue about details of vacuum, so maybe I should rather shut my mouth about that
But I will definitely read the link to Matt Srassler when time permits, thanks.
Actually my point in this former post of mine (from which I was excerpting) was that, as far as I know (which is not much in quantum physics, a little more in phenomenological astrophysics) vacuum is Lorentz invariant, and all attempts at putting intrinsic frames of reference of space/vacuum or aether back on the table failed experimentally so far (though there are theories...). It's important as it makes inescapable the "free energy" consequence of "pushing on the vacuum" : breaking momentum conservation (or exchanging net momentum with vacuum) is breaking energy conservation (or exchanging energy with vacuum), one can't go without the other, regardless of possible theories beyond very very well verified quantitative predictions of Newtonian mechanics at speeds<<c, please correct me if I'm wrong on that.
If there is indeed favorite intrinsic frames of reference or aether then yes, I see how it would be possible to push on that with much better efficiencies than with reaction propulsion, without breaking energy conservation (or pumping some from vacuum) but then the k factor would decrease as your absolute speed relative to this aether increase, and mission profiles are not the same. Also some "aether wind" effect should be observed in conjunction with earth rotation (orientation of the experiment relative to the stars). Unless the aether is anchored to nearest massive body (back to old aether problems...).
The point is, this free energy consequence is seldom put forward by, let's say, enthusiasts of space drives. As if maybe free (or cheap) momentum were deemed "more acceptable" than free (or cheap) energy that is so much connoted crackpot science. If it works at all for thrust and spaceships then so well, but please proponents, don't put the free energy generators under the rug as it is a consequence. It sounds sarcastic but I'm not, I'm not against the experiments or the experimenters, clearly I'm very sceptic about a "real" (useful) effect here but ready to be convinced when/if results can be reproduced by sceptics. It needs the sceptics to be convinced that it's worth to try reproduce the experiments. But there is no point in trying to hide the "free energy" consequences.
@aero
yes when Michelson and Morley experiment show the breaking of the rule of speed additivity, this is a result indistinguishable from magic. I'm not qualified and gifted enough to advance the principles at work, should the EMthrust prove to be real. Then maybe the k factor would be observed to decrease with speed in such a way not to break energy conservation or pump it from vacuum, but then don't expect the same mission profiles for spaceships. Or maybe the k factor is indeed free from v effects and, in all logic, we have free energy, even real possible free energy devices tomorrow if k approaches 1N/kW as hoped by proponents.
But we are not at the point where the effect is accepted so I feel no urge to fill the theoretical gap.
Again, I'm not against dedicated minds taking the risk, and quite supportive of inquiring the possible effect, if not by sending a million dollar probe at first, at least by providing means to try to reproduce it and investigate it phenomenologically (magnitude dependence on energy densities ? dependence on orientation relative to stars ? dependence on velocity/acceleration when freely moving ? ...) then we can discuss the theories on firmer ground.
But (again) I'm against a "what if it works" scenario that wont go to all the inescapable consequences, when the consequences are on firm ground that couldn't possibly be overtaken by the hypothesis (I don't see how the possibility to push on vacuum could change the observed fact that Ec=1/2mv² as far as practical mundane useful kinetic energy is concerned, how ?) : this is too reminiscent of bad SciFi. Fantasy is another realm but wont take us to the stars. Why wouldn't you want a nice free energy generator to power up your free momentum drive ?
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#472 by frobnicat on 12 Sep, 2014 23:28
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So really, using Newtonian mechanics to treat a Mach-effect thruster in isolation is nonsensical from the start.
Using Newtonian mechanics to treat the local useful consequences of such a thruster in not nonsensical, the fact that the energy balance would be conserved instantly within the cosmological horizon can be satisfactory for theorists, but as far as engineering goes this is indistinguishable from free energy and doesn't prevent to make accurate mission profiles with Newtonian mechanics or making real unlimited energy generators, assuming the effect is real, regardless of theories. Again, if the effect is real, then very well, we have good spaceships, and good energy generators, free energy, for all practical purpose.
Sorry to put forward my take on this GoatGuy, this was addressed to you.
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#473 by Rodal on 13 Sep, 2014 00:21
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....
@frobnicat, yes, I agree, as I posted before there are substantial problems with the proposed "out of the mainstream" explanations. So, what I propose is that instead we discuss
A) my preference at the moment: alternative explanations for the experimental measurements, or if, you prefer,
B) we discuss in detail Dr. White's equations . We could discuss the fact that, in the end, Dr. White does NOT actually use in his calculations the density of the (zero point) quantum vacuum, but instead he uses what he calls a "local quantum vacuum density" that is several orders of magnitude larger than the density of the zero-point quantum vacuum. Those in this forum that defend or battle the "zero point quantum vacuum" explanation may be defending or battling an un-existing explanation .
So, how about discussing Dr. White's justification for the much higher value he uses for the "local density of the quantum vacuum" and what this implies ...? -
#474 by 93143 on 13 Sep, 2014 00:39
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Again, if the effect is real, then very well, we have good spaceships, and good energy generators, free energy, for all practical purpose.
If the operation of the device is independent of its velocity (Woodward's equation sure looks like it is), and the efficiency in N/W gets high enough for practical application in an energy-generating device, yes, you could do that. I have myself maintained in the past that this sort of technology could revolutionize far more than just space travel.
I have a half-formed idea involving the relativistic Doppler effect that I think might explain the apparent velocity independence of the M-E drive principle, in such a way that a "free-energy" device like what you're describing would essentially be a heat engine powered by the Big Bang. But maybe I should shut up about it until I've had time to get into the theory a little more...as I posted before there are substantial problems with the proposed "out of the mainstream" explanations.
Regarding the Wheeler-Feynman idea specifically, I'm not sure what you've stated constitutes an ironclad case against it. For one thing, it seems to me that gravity and the weak force are two different things, and since we don't even have a theory that unifies them it strikes me as premature to claim that time-asymmetry in one necessarily implies time-asymmetry in the other (though I may be misunderstanding something). For another, according to the Wikipedia page (no, I'm not an expert; can you tell?) on W-F absorber theory, there have been calculations that recover the Lamb shift without requiring self-energy.
I'm an engineer, not a physicist, but this sounds like an open question to me. -
#475 by Rodal on 13 Sep, 2014 00:50
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as I posted before there are substantial problems with the proposed "out of the mainstream" explanations.
Regarding the Wheeler-Feynman idea specifically, I'm not sure what you've stated constitutes an ironclad case against it.
Well, then you are stating your own theory, not Dr. Woodward's, because as per the video I posted, Dr. Woodward answered that his theory requires either the Wheeler-Feynman absorber theory (or even more unsupported by experiments, a constrained theory changing the type of differential equations (elliptic instead of hyperbolic !!!!!)).
Also, you seem to be stating your own theory, not Dr. Woodward's, concerning the incompatibility with the Weak Force, because Dr. Woodward's also stated in that video that his interpretation for the inertia effect is a radiation interaction, and we know that the Weak Force interaction is paramount in a radiation interaction.
But a theory of Physics is more than just words we can discuss in this thread, you would have to replace Dr. Woodward's theory with your own set of consistent equations...
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#476 by Rodal on 13 Sep, 2014 01:03
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it strikes me as premature to claim that time-asymmetry in one necessarily implies time-asymmetry in the other (though I may be misunderstanding something).
It is not premature to claim that there is only one time and not two times. There is not one time for gravitation and another time for the Weak Force, there is only one time in Dr. Woodward's theory. When Dr. Woodward explains his inertia effect as a radiation interaction he is using the same time that he uses for gravitation in his General Relativity equations.For another, according to the Wikipedia page (no, I'm not an expert; can you tell?) on W-F absorber theory, there have been calculations that recover the Lamb shift without requiring self-energy.
Yes, as you know anybody (even without being registered in Wikipedia) can anonymously write and change things in Wikipedia. The alternative calculations that recover the Lamb Shift without requiring self-energy are not generally accepted, and have problems of their own. I wrote a lot of the Wikipedia article on the Beta Distribution ( http://en.wikipedia.org/wiki/Beta_distribution ) (you can see my name associated with most graphs and in the Talk Page), I know what a headache is to deal with this issue by Wikipedia managers, who try to do their best
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#477 by 93143 on 13 Sep, 2014 01:05
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Dr. Woodward's also stated in that video that his interpretation for the inertia effect is a radiation interaction, and we know that the Weak Force interaction is paramount in a radiation interaction.
I'm sorry; you've lost me. What has the weak force got to do with the propagation of changes in a gravity field? -
#478 by Rodal on 13 Sep, 2014 01:08
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Dr. Woodward's also stated in that video that his interpretation for the inertia effect is a radiation interaction, and we know that the Weak Force interaction is paramount in a radiation interaction.
I'm sorry; you've lost me. What has the weak force got to do with the propagation of changes in a gravity field?
Have you actually listened to the video I posted of Dr. Woodward? I'm quoting what Dr. Woodward stated. Perhaps you should state that "Dr. Woodward lost you". Dr. Woodward himself says that it is radiation interaction. -
#479 by Rodal on 13 Sep, 2014 01:16
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Dr . Woodward answers the question with the following statement:
it is a radiation reaction interaction, presumably, that involves the Wheeler-Feynman ...
But I will definitely read the link to Matt Srassler when time permits, thanks.