How the Aether Physics Model came to be
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David Thomson - 12 Feb 2005 03:56 GMT I was posting on another list when one of the readers suggested I use my reply as a general introduction to the Aether Physics Model on these newsgroups. So here is how the Aether Physics Model came to be.
I was working on a flat spiral coil and placing a fluorescent tube between the top load and ground wire of the coil. I noticed that the spark going to the topload was a thin purple streamer and the spark going to the ground was a thick, white streamer.
Until then, I was under the impression that electricity changed its outward form because of the coil arrangement, frequency, current, potential, etc. But here I had a single coil, the same shape feeding both ends. The same frequency existed at both ends. But it became evident that the potential was greater on the topload and the current was greater on the ground end. As a result of these differences, the appearance of the spark changed dramatically.
Thus I surmised that the same electrons in a given coil can manifest in two distinctly different forms. I then posted this observation on the pupman list and began searching the literature for an explanation. Finding none, I decided to investigate physics from the ground up to see if something was missed.
It didn't take but a few days before I realized that charge should be distributed in the math, since it is also distributed in nature. Ed Philips suggested I read Great Experiments in Physics so I purchased a copy. In this book Charles Coulomb himself clearly stated that all charge must be distributed in order for the inverse square law to work.
The first thing I noticed in physics was that Coulomb's constant was precisely equal to a combination of light speed, permeability, permittivity, and a conductance constant, which the established literature does not recognize. I then found that Planck's constant times the conductance constant gives a distributed charge. But I was disappointed that the distributed charge it produced did not equal the elementary charge squared. So I worked with the two distributed charges for a few hours before discovering that they were exactly proportional to each other by way of 8pi times the fine structure constant of the electron. And so I had this equation:
e^2 = 8pia e.emax^2
where e is the elementary charge, a is the fine structure of the electron, and e.emax^2 is the derived value of charge equal to:
e.emax^2 = h * Cd
I then applied the same process to the proton and neutron and found fine structures for both. But since the fine structures and angular momentum of the proton and neutron are not defined in modern physics, I really didn't have an empirical body of data to go by there. However, the lack of data was encouraging since it implied that modern physics was incapable of determining these values based on their chosen model of subatomic behavior.
Three weeks went by before it occurred to me that e.emax^2 could be strong charge, which mediates the strong force. I then realized that if it were, then I was looking at the Unified Field Theory. Further research showed me that it has nothing to do with fields, but is rather a Unified Force Theory. I later found that the concept of fields is very ambiguous in the Standard Model and relativity theories, as modern science apparently does not know what a field really is.
When I calculated the relative value of the forces for the different subatomic particles, looking at both the electrostatic force and the new concept of strong force, I found that the relative strengths of the forces matched the empirical data.
Then I started reading older literature on electromagnetism. I was particularly struck by Ed Leedskalnin's description of electricity as a stream of tiny magnets. Although his view incorrectly denies the electrostatic effect of electrons, he does show a convincing argument for electrons acting as tiny magnets. And then it occurred to me that electromagnetism is caused by the strong charge. Further analysis of the mathematical aspects of the two charges revealed that electrostatic charge has a spherical geometry while strong charge (or electromagnetic charge) has a toroidal geometry.
When I applied this knowledge to my coil observations it made perfect sense. When electrons are magnetically moving sideways in the coil, they produce the bright white streamers. When electrons are moving along their z axis, they produce the thin purple streamers. Electrons moving along their z axis are creating pressure on the electrons in front of them, thus potential increases. Electrons moving sideways along the circular y axis and outward on the x axis (it is a polar coordinate system) they move closer to each and form longitudinal waves of electrons.
Hertz was broadcasting from the top load and Tesla was broadcasting from the ground. This is why Tesla's system of signaling was notably different from Hertz. Whereas Hertz's system caused the electrons to transform into photons and radiate permanently away from the transmitter, thus causing power loss, Tesla's system was pumped into the ground, which is a closed system. This allowed Tesla to longitudinally broadcast current through the electrostatic field of the Earth and have that current return to his transmitter. Thus he could build up resonance within the Earth with very little power loss.
But I as I was working on this Unified Force Theory, it occurred to me that the units were in the wrong dimensions of charge, except three of them. The units of inductance, capacitance, and conductance were in the proper distributed charge, while all other units with charge were in single dimension. Investigating the various changes that take place when the dimensions are corrected in the units led me to the development of the Aether Physics Model.
FrediFizzx - 12 Feb 2005 07:57 GMT | I was posting on another list when one of the readers suggested I use my | reply as a general introduction to the Aether Physics Model on these [quoted text clipped - 72 lines] | has a spherical geometry while strong charge (or electromagnetic charge) has | a toroidal geometry. This looks somewhat like what Weisskopf did back in 1939. I think I told you before that you need to read and really understand Weisskopf's paper "On the Self-energy and the Electromagnetic Field of the Electron". You can find it here;
http://fangio.magnet.fsu.edu/~vlad/pr100/100yrs/html/author/fs2au_12004weisskopf.htm
It's a classic, baby!
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
David Thomson - 12 Feb 2005 23:57 GMT Hi Freddi,
> This looks somewhat like what Weisskopf did back in 1939. I think I told > you before that you need to read and really understand Weisskopf's paper [quoted text clipped - 5 lines] > > It's a classic, baby! Yes, there are a few similarities in the intent of the physics, but there is a vast difference in the method. I produce a completely new set of quantum units derived from quantum measurements, show that the modern units are in error, and precisely quantify the relative strengths of the charges pertaining to the electrostatic charge and electromagnetic charge. And that is just the short list.
I can see how this paper has greatly influenced your work, however. I noticed many similarities between Weisskopf's paper and the equations you have posted in the past.
Dave
FrediFizzx - 13 Feb 2005 02:28 GMT | Hi Freddi, | [quoted text clipped - 3 lines] | > the Self-energy and the Electromagnetic Field of the Electron". You can | > find it here; http://fangio.magnet.fsu.edu/~vlad/pr100/100yrs/html/author/fs2au_12004weisskopf.htm
| > It's a classic, baby! | [quoted text clipped - 4 lines] | pertaining to the electrostatic charge and electromagnetic charge. And that | is just the short list. Your e^2 = 8pia e.emax^2 is simply the same as e^2 = alpha*hbar*c in cgs units. Your emax is just +,- sqrt(hbar*c). You have really done nothing new as far as quantum units go. Wheeler proposed back in 1957 that sqrt(hbar*c) might be spacetime charge in his Quantum Geometrodynamics paper. He was trying to make real particles out of spacetime but never got it to work right. I say the "particles" are already there and are what are defining spacetime.
| I can see how this paper has greatly influenced your work, however. I | noticed many similarities between Weisskopf's paper and the equations you | have posted in the past. There is quite a history involving Weisskopf's paper. If you haven't read it, read "The Second Creation: Makers of the Revolution in Twentieth-Century Physics" by Crease and Mann. Fascinating book.
As far as I can see, any spacetime medium theory should be consistent with most all of quantum theory and relativity. Some re-interpretation is definitely necessary in a few places.
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
David Thomson - 16 Feb 2005 03:29 GMT > Your e^2 = 8pia e.emax^2 is simply the same as e^2 = alpha*hbar*c in cgs > units. Your emax is just +,- sqrt(hbar*c). ??? You really need to double check your math. My equation is in MKS units, but it really doesn't matter. The equation has the same value no matter what system of units is used.
e.emax^2 = 1.400 x 10^-37 coul^2 sqrt(hbar*c) = 1.778 x 10^-13 kg^.5 m^1.5 / sec
You're way off.
> Wheeler proposed back in 1957 that > sqrt(hbar*c) might be spacetime charge in his Quantum Geometrodynamics > paper. What's the reference? sqrt(hbar*c) has no meaning. It certainly does not resolve to charge units.
> As far as I can see, any spacetime medium theory should be consistent with > most all of quantum theory and relativity. Some re-interpretation is > definitely necessary in a few places. I know you have put a lot of thought into this. But I'm still convinced of the accuracy and utility of the Aether Physics Model. I don't think you understand it, though, based upon your equations above.
Dave
FrediFizzx - 16 Feb 2005 09:22 GMT | > Your e^2 = 8pia e.emax^2 is simply the same as e^2 = alpha*hbar*c in cgs | > units. Your emax is just +,- sqrt(hbar*c). [quoted text clipped - 7 lines] | | You're way off. Not way off. I said your emax not emax^2. In SI units it is,
4pi*eps0*hbar*c = 3.5 x 10^-36 coul^2
My relativistic quantum charge unit is bigger than yours. And more simple and reasonable. Your expression is simply;
eps0*hbar*c/2 = 1.3996 x 10^-37 coul^2
Who are you trying to kid with e.emax^2? 4pi*eps0*hbar*c reduces to +,- 1 in natural units. A perfect quantum charge unit. Don't you know that in MKS or SI units the Coulomb constant, k_e, is 1/4pi*eps0. Always. But there is no getting around,
e^2 = alpha*hbar*c in cgs units. Or,
e^2 = alpha*4pi*eps0*hbar*c in SI units.
| > Wheeler proposed back in 1957 that | > sqrt(hbar*c) might be spacetime charge in his Quantum Geometrodynamics | > paper. | | What's the reference? sqrt(hbar*c) has no meaning. It certainly does not | resolve to charge units. JA Wheeler, "On the Nature of Quantum Geometrodynamics," Annals Phys. 2, 604-614 (1957).
See above for charge units. Please learn how to convert to and from cgs and other unit systems.
| > As far as I can see, any spacetime medium theory should be consistent with | > most all of quantum theory and relativity. Some re-interpretation is [quoted text clipped - 3 lines] | the accuracy and utility of the Aether Physics Model. I don't think you | understand it, though, based upon your equations above. I told you to send me a copy of the book. I will read it. You might have some ideas that are useful.
FrediFizzx
David Thomson - 18 Feb 2005 16:17 GMT > | > Your e^2 = 8pia e.emax^2 is simply the same as e^2 = alpha*hbar*c in > cgs [quoted text clipped - 13 lines] > > 4pi*eps0*hbar*c = 3.5 x 10^-36 coul^2 To begin with, your above equation is not expressed in SI units. In SI units it would be:
4pi*eps0*hbar*c = 3.5 x 10^-36 s^2 A^2
Second, your above equation is off by a factor of 8pi.
Also, your claim that "Your emax is just +,- sqrt(hbar*c)" is still wrong even for single dimension charge:
In cgs units: e.emax = .199 * sqrt(hbar * c)
> My relativistic quantum charge unit is bigger than yours. And more simple > and reasonable. Your expression is simply; > > eps0*hbar*c/2 = 1.3996 x 10^-37 coul^2 First of all, my equation is for strong charge. It has nothing to do with relativity. The strong charge (e.emax^2) is equal to e^2 / 8pi * a. The equation e^2 = 8pi * a * e.emax^2 is interpreted as meaning that there are two types of charge in each quantum subatomic particle. There is the electrostatic charge and the electromagnetic charge. The electrostatic charge has a spherical angle while the electromagnetic charge has a steradian angle, which is why e.emax^2 must be multiplied by 4pi. The electrostatic charge has 1 spin and the electromagnetic charge has 1/2 spin, which is why the electromagnetic charge is also multiplied by 2. The electrostatic charge is much smaller than the electromagnetic charge, which is why the electromagnetic charge is reduced by the quantity of alpha.
As a result the small, spherical-angled, 1 spin electrostatic charge is equated to the larger, steradial-angled, 1/2 spin electromagnetic charge.
Explain how epsilon, hbar, and half the speed of light equate the elementary charge squared to the relativistic quantum charge in your system.
> Who are you trying to kid with e.emax^2? Nobody. I presented my rationale for e.emax^2, let's here yours.
> 4pi*eps0*hbar*c reduces to +,- 1 in natural units. A perfect quantum > charge unit. Since when does a non-dimensional number equal a perfect quantum charge unit? And who are you trying to kid with the phrase "natural units?" Also, show how 4pi*eps0*hbar*c reduces to +,- 1.
> Don't you know that in > MKS or SI units the Coulomb constant, k_e, is 1/4pi*eps0. Always. But > there is no getting around, I didn't know that. I thought Coulomb's constant in MKS and SI units was written as:
k.c = 8.988 x 10^9 kg m^3 / sec^2 coul^2 k.c = 8.988 x 10^9 kg m^3 / s^4 A^2
As far as I knew, eps0 was a constant of its own. Expressing Coulomb's constant in terms of the permittivity constant in no way represents the expression of Coulomb's constant in either MKS units or SI units.
> e^2 = alpha*hbar*c in cgs units. Or, > > e^2 = alpha*4pi*eps0*hbar*c in SI units. You apparently don't know what a system of units is. You're expressing elementary charge squared in terms of constants, not systems of units. But even still, give us a narrative of what elementary charge squared is in terms of alpha, hbar and c. For that matter, tell us how hbar was derived from empirical data. What real world meaning does hbar have?
> | > Wheeler proposed back in 1957 that > | > sqrt(hbar*c) might be spacetime charge in his Quantum Geometrodynamics [quoted text clipped - 6 lines] > JA Wheeler, "On the Nature of Quantum Geometrodynamics," Annals Phys. 2, > 604-614 (1957). If I can get ahold of the paper, I'll look at it.
> See above for charge units. Please learn how to convert to and from cgs > and > other unit systems. I think you need to learn the difference between system of units and expressing in terms of constants. Also, you need to be able to describe what it is your equations are telling you.
> | > As far as I can see, any spacetime medium theory should be consistent > with [quoted text clipped - 8 lines] > I told you to send me a copy of the book. I will read it. You might have > some ideas that are useful. Send me your mailing address in a private email. I'll send you a copy.
Dave
FrediFizzx - 18 Feb 2005 19:33 GMT | > | > Your e^2 = 8pia e.emax^2 is simply the same as e^2 = alpha*hbar*c in | > cgs [quoted text clipped - 18 lines] | | 4pi*eps0*hbar*c = 3.5 x 10^-36 s^2 A^2 Bzzzt! What is a coulomb in SI units?
| Second, your above equation is off by a factor of 8pi. Yeah, your 8pi that you have in e^2 = 8pia e.emax^2.
| Also, your claim that "Your emax is just +,- sqrt(hbar*c)" is still wrong | even for single dimension charge: | | In cgs units: e.emax = .199 * sqrt(hbar * c) Yes, your 8pi again.
| > My relativistic quantum charge unit is bigger than yours. And more simple | > and reasonable. Your expression is simply; [quoted text clipped - 22 lines] | | Nobody. I presented my rationale for e.emax^2, let's here yours. Do yourself a favor and just use (hbar*c)/8pi in cgs or (4pi*eps0*hbar*c)/8pi in SI. You might have more credibility doing it that way. That means in natural units, your e.emax^2 = 1/8pi ~= 0.0398. Now you need to explain why this number has never been observed in particle experiments. Or if it has, which ones? Or the sqrt(1/8pi) ~= 0.199? Has this number ever been observed? If so, where? References.
Hbar*c is already well known. I don't have to "rationalize" it; just use it. I make the postulate that hbar is due to "vacuum" processes. It is already known that c is a "vacuum" process.
| > 4pi*eps0*hbar*c reduces to +,- 1 in natural units. A perfect quantum | > charge unit. | | Since when does a non-dimensional number equal a perfect quantum charge | unit? And who are you trying to kid with the phrase "natural units?" Also, | show how 4pi*eps0*hbar*c reduces to +,- 1. In natural units, 4pi*eps0 = 1 since eps0 = 1/4pi; hbar*c = 1; take the square root then the solutions are + or - 1. In natural units, charge is dimensionless. e^2 = alpha.
| > Don't you know that in | > MKS or SI units the Coulomb constant, k_e, is 1/4pi*eps0. Always. But [quoted text clipped - 5 lines] | k.c = 8.988 x 10^9 kg m^3 / sec^2 coul^2 | k.c = 8.988 x 10^9 kg m^3 / s^4 A^2 http://www.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/index.html
| As far as I knew, eps0 was a constant of its own. Expressing Coulomb's | constant in terms of the permittivity constant in no way represents the [quoted text clipped - 9 lines] | terms of alpha, hbar and c. For that matter, tell us how hbar was derived | from empirical data. What real world meaning does hbar have? The above is no different from your expression which is,
e^2 = alpha*8pi*eps0*hbar*c/2 = alpha*4pi*eps0*hbar*c
All you have done is split the inverse coulomb constant in two; 8pi*eps0/2, then shoved eps0/2 into your e.emax^2.
| > | > Wheeler proposed back in 1957 that | > | > sqrt(hbar*c) might be spacetime charge in his Quantum Geometrodynamics [quoted text clipped - 8 lines] | | If I can get ahold of the paper, I'll look at it. This was Wheeler's first attempt to make particles from spacetime. My approach is the inverse. I say particles (quantum objects) already exist and are what are defining spacetime by the geometry of their interactions.
| > See above for charge units. Please learn how to convert to and from cgs | > and [quoted text clipped - 3 lines] | expressing in terms of constants. Also, you need to be able to describe | what it is your equations are telling you. I know perfectly well the difference. In units where c = 1, hbar = charge^2. What does that tell you?
| > | > As far as I can see, any spacetime medium theory should be consistent | > with [quoted text clipped - 10 lines] | | Send me your mailing address in a private email. I'll send you a copy. Will do.
FrediFizzx
David Thomson - 19 Feb 2005 01:34 GMT > | To begin with, your above equation is not expressed in SI units. In SI > | units it would be: > | > | 4pi*eps0*hbar*c = 3.5 x 10^-36 s^2 A^2 > > Bzzzt! What is a coulomb in SI units? Bzzzt! There are no coulombs in SI units.
> | Second, your above equation is off by a factor of 8pi. > > Yeah, your 8pi that you have in e^2 = 8pia e.emax^2. Well, the mathematically correct equation is probably the correct equation?
> | Also, your claim that "Your emax is just +,- sqrt(hbar*c)" is still > wrong [quoted text clipped - 3 lines] > > Yes, your 8pi again. And once again, it belongs there.
> | > Who are you trying to kid with e.emax^2? > | [quoted text clipped - 3 lines] > (4pi*eps0*hbar*c)/8pi in SI. You might have more credibility doing it > that way. That doesn't sound like a rationalization to me. That sounds like ignoring the question.
> That means in natural units, your e.emax^2 = 1/8pi ~= 0.0398. What are natural units?
> Now you need to explain why this number has never been observed in > particle experiments. Or if it has, which ones? Or the sqrt(1/8pi) ~= > 0.199? Has this number ever been observed? If so, where? References. You're the one who threw in the sqrt(1/8pi) term. That is up for you to explain. I show that all charge is distributed. There is no square root function. As for the e.emax^2 term, that shows up in all the quantum units when they are seen as quantum units.
> Hbar*c is already well known. I don't have to "rationalize" it; just use > it. I'm sorry. All science has to be verified. Even 100 years from now our children will have to verify each and every constant just to be sure we were correct. Hbar has no real-world significance. It is arbitrarily presumed.
> I make the postulate that hbar is due to "vacuum" processes. > It is already known that c is a "vacuum" process. c can be measured. Who has measured hbar?
> In natural units, 4pi*eps0 = 1 since eps0 = 1/4pi; hbar*c = 1; take the > square root then the solutions are + or - 1. In natural units, charge > is dimensionless. e^2 = alpha. Are you saying that natural units are cgs units? If you are then e^2 does not equal alpha, it equals 2.307 x 10^-19 gm cm^3 / sec^2. If you're not saying that natural units are cgs units, then please define natural units.
> | > Don't you know that in > | > MKS or SI units the Coulomb constant, k_e, is 1/4pi*eps0. Always. But [quoted text clipped - 8 lines] > > http://www.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/index.html And what is your point? Are you agreeing with me?
> | As far as I knew, eps0 was a constant of its own. Expressing Coulomb's > | constant in terms of the permittivity constant in no way represents the [quoted text clipped - 15 lines] > > e^2 = alpha*8pi*eps0*hbar*c/2 = alpha*4pi*eps0*hbar*c In case you missed it, my expression was:
e^2 = 8pi * a * e.emax^2
That's quite different from what you wrote. And its much simpler as well. As I pointed out earlier, every term in my equation has meaning, which I pointed out in detail. My equation is balanced not only mathematically, but also geometrically.
> All you have done is split the inverse coulomb constant in two; > 8pi*eps0/2, then shoved eps0/2 into your e.emax^2. No, that isn't what I did at all. What I did was find the conductance constant of the Aether, which is equal to 2.112 x 10^-4 siemens and multiply it by the Planck constant:
e.emax^2 = Cd * h
> | > JA Wheeler, "On the Nature of Quantum Geometrodynamics," Annals Phys. > 2, [quoted text clipped - 6 lines] > exist and are what are defining spacetime by the geometry of their > interactions. I don't knock you for looking for answers in your own way. If you can show that particles existed before fields, I would be interested in understanding your theory. This is quite the opposite of my theory. My theory shows that force and angular momentum exist first, then force interacts through charge to produce the Aether, which then absorbs dark matter (angular momentum) to produce subatomic particles. The physical Universe has a non-material cause,. at least that is how I see it.
> | > See above for charge units. Please learn how to convert to and from > cgs [quoted text clipped - 6 lines] > I know perfectly well the difference. In units where c = 1, hbar = > charge^2. What does that tell you? To begin with we can quantify h in terms of the mass of the electron, Compton wavelength and speed of light.
h = m.e * w.C * c
In order to convert h into units of c=1, you would divide h by c and get h = m.e * w.C = 2.21 x 10^-37 gm cm. Arbitrarily dividing h by 2pi will give you the hbar equivalent in units of c = 1. The dimensions of gm cm does not in any way give charge^2. What does that tell you?
Dave
FrediFizzx - 19 Feb 2005 03:11 GMT | > | To begin with, your above equation is not expressed in SI units. In SI | > | units it would be: [quoted text clipped - 4 lines] | | Bzzzt! There are no coulombs in SI units. http://physics.nist.gov/cuu/Units/units.html
Scroll down to Table 3 and presto! There seems to be coulombs listed there for electric charge as a SI derived unit.
| > | Second, your above equation is off by a factor of 8pi. | > [quoted text clipped - 26 lines] | | What are natural units? Oh, dear! I am not sure it is worthwhile continuing this discussion since I swear elsewhere in the previous post I said that " In natural units, 4pi*eps0 = 1 since eps0 = 1/4pi; hbar*c = 1". But for the benifit of lurkers, hbar = c = 1 is natural units or Planck units. In gravity, Newton's G is also set equal to one.
| > Now you need to explain why this number has never been observed in | > particle experiments. Or if it has, which ones? Or the sqrt(1/8pi) ~= [quoted text clipped - 4 lines] | function. As for the e.emax^2 term, that shows up in all the quantum units | when they are seen as quantum units. No, you are the one who threw in sqrt(1/8pi) term by taking out an 8pi. I don't use that "term". You do. If you can't show any experimental evidence for it, then it doesn't mean anything physical. Or show experimental evidence supporting 8pi.
| > Hbar*c is already well known. I don't have to "rationalize" it; just use | > it. | | I'm sorry. All science has to be verified. Even 100 years from now our | children will have to verify each and every constant just to be sure we were | correct. Hbar has no real-world significance. It is arbitrarily presumed. http://pdg.lbl.gov/2004/reviews/consrpp.pdf
Now look where the Particle Data Group/NIST lists hbar*c as a conversion constant ~= 197 MeV fm. Do you really think they are just "kidding" about that?
| > I make the postulate that hbar is due to "vacuum" processes. | > It is already known that c is a "vacuum" process. [quoted text clipped - 8 lines] | not equal alpha, it equals 2.307 x 10^-19 gm cm^3 / sec^2. If you're not | saying that natural units are cgs units, then please define natural units. Natural units are absolutely not cgs units. Electric charge is dimensionless in natural units.
| > | > Don't you know that in | > | > MKS or SI units the Coulomb constant, k_e, is 1/4pi*eps0. Always. But [quoted text clipped - 6 lines] | > | k.c = 8.988 x 10^9 kg m^3 / sec^2 coul^2 | > | k.c = 8.988 x 10^9 kg m^3 / s^4 A^2 http://www.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/index.html
| And what is your point? Are you agreeing with me? It was to show you that in SI units, Coulomb's constant, k_e, is 1/(4pi*eps0). And can simply be written that way.
| > | As far as I knew, eps0 was a constant of its own. Expressing Coulomb's | > | constant in terms of the permittivity constant in no way represents the [quoted text clipped - 33 lines] | | e.emax^2 = Cd * h Run it by us again how you found this Cd? But since your e.emax^2 is eps0*h*c/4pi, we get,
Cd = eps0*c/4pi.
And since we like to see a 4pi with eps0, then,
Cd = 4pi*eps0*c/16pi^2, so in natural units,
Cd = 1/16pi^2 ~= 0.0063
Hmm... don't ever recall seeing that number pop up anywhere.
| > | > JA Wheeler, "On the Nature of Quantum Geometrodynamics," Annals Phys. | > 2, [quoted text clipped - 14 lines] | produce subatomic particles. The physical Universe has a non-material | cause,. at least that is how I see it. Well it just seems more philosophically satisfying to me to have particles (quantum objects) that have always been existing. But no one really knows for sure. Yet. Yeah, you seem to be going somewhat down the quantum geometrodynamics path so you ought to read up on that. BTW, Wheeler couldn't get it to work and eventually bailed on it. But others are probably still working on it.
| > | > See above for charge units. Please learn how to convert to and from | > cgs [quoted text clipped - 16 lines] | you the hbar equivalent in units of c = 1. The dimensions of gm cm does not | in any way give charge^2. What does that tell you? You really need to get a grip with this. E = mc^2; set c = 1 then E = m. What is charge^2/length? It is E, so energy times length is charge^2. In units where c = 1, gm*cm is charge^2.
FrediFizzx
David Thomson - 20 Feb 2005 03:06 GMT > | > | To begin with, your above equation is not expressed in SI units. > In SI [quoted text clipped - 10 lines] > Scroll down to Table 3 and presto! There seems to be coulombs listed > there for electric charge as a SI derived unit. Come on, Fredi. Didn't you look at the last column? It's called a coulomb, but it is expressed as I wrote it, s^2 A^2. You wrote coul^2. coul^2 is the MKS expression of charge.
> | What are natural units? > [quoted text clipped - 3 lines] > benifit of lurkers, hbar = c = 1 is natural units or Planck units. In > gravity, Newton's G is also set equal to one. How can you do physics when everything is equal to 1? I agree, it is not worthwhile continuing this discussion.
> No, you are the one who threw in sqrt(1/8pi) term by taking out an 8pi. > I don't use that "term". You do. If you can't show any experimental > evidence for it, then it doesn't mean anything physical. Or show > experimental evidence supporting 8pi. Have you ever seen this expression before?
G=8pi T
This says that the space-time tensor is equal to 8pi times the mass tensor. Of course, I show that General Relativity is good math based on wrong principles. In my equation:
e^2 = 8pi * a * e.emax^2
I previously stated that alpha is the scale factor for the two equivalent spherical charges. So, in effect, the strong charge on the right is scaled down to the same scale as the electrostatic charge on the left. The strong charge, even though scaled down, is still steradiial in angle and has half spin. The 8pi is the constant that makes steradial, half-spin scaled down strong charge to become equal with spherical angle, one-spin electrostatic charge.
In the Aether Physics Model electrostatic charge is shown to be a quality of Aether, which is donated to angular momentum as it resides in an Aether unit. And as I show in my theory, all matter has strong charge. The strong charge acts upon the Aether (space-time) via the charge force law. The 8pi is the geometric angle, spin conversion factor. In effect, the strong charge is acting as a monopole with the electrostatic charge of the Aether and the attraction is what pulls Aether (space-time) toward massive objects. It has nothing to do with gravity at all. It is just by coincidence that mass is directly proportional to the strong charge that Einstein was able to make his equations work with mass instead of as it should have been with the charges.
So, yes, there is experimental evidence for the existence of the 8pi constant.
> | > Hbar*c is already well known. I don't have to "rationalize" it; > just use [quoted text clipped - 12 lines] > constant ~= 197 MeV fm. Do you really think they are just "kidding" > about that? Yes, I do, although not intentionally. I agree that hbar is widely used in modern physics. I didn't question whether it was widely used or not. All I questioned was the real world origin of hbar. There is no physical process or object that hbar is the quantity of.
> | Are you saying that natural units are cgs units? If you are then e^2 > does [quoted text clipped - 4 lines] > Natural units are absolutely not cgs units. Electric charge is > dimensionless in natural units. It seems everything is dimensionless in natural units. So far you have listed e, hbar, and c. Is there anything else that is dimensionless in natural units that I should know about? Do you have a web page that lists all these units and their relationships to each other? I suppose it would be easy to figure out if everything is equal to 1, but I'd like to see how these units are put to work.
> | > | k.c = 8.988 x 10^9 kg m^3 / sec^2 coul^2 > | > | k.c = 8.988 x 10^9 kg m^3 / s^4 A^2 [quoted text clipped - 5 lines] > It was to show you that in SI units, Coulomb's constant, k_e, is > 1/(4pi*eps0). And can simply be written that way. Well, you are wrong. SI units are a very specific system of units and Coulomb's constant is expressed as I wrote it above. eps0 is a universal constant, not a unit of the SI system. eps0 will be expressed differently in each system of units, but it will always have the same relative place when all the other units are converted to the same system. That is, eps0 is always permittivity of Aether (free space, vacuum, etc).
> | > All you have done is split the inverse coulomb constant in two; > | > 8pi*eps0/2, then shoved eps0/2 into your e.emax^2. [quoted text clipped - 16 lines] > > Cd = 1/16pi^2 ~= 0.0063 I'm sorry, I don't see the relevance of you natural units. Arbitrarily choosing c = 1 doesn't do anything for me. In cgs units, Cd is equal to c/16pi^2.
In cgs units, Cd has the following relationship with Coulomb's constant (k.c=1).
k.c = 1 c = c u.0 = 4pi/c^2 e.0 = 1/4pi Cd = c/16pi^2
k.c = c * Cd * u.0 / e.0
or
Cd = k.c * e.0 / c * u.0
Just as Coulomb's constant, the speed of light, permeability and permittivity are simply constants known through physics, so is the conductance constant.
> Hmm... don't ever recall seeing that number pop up anywhere. 16pi^2 shows up all throughout the Aether Physics Model. It is the geometrical constant of the Aether loxodromes, just as 4pi^2 is the geometrical constant of the toroid, 4pi is the geometrical constant of the sphere, and 2pi is the geometrical constant of the circle. Modern physics chooses to ignore the Aether, so naturally they are not aware of either the conductance constant, the Aether, or the geometry associated with 16pi^2.
> Well it just seems more philosophically satisfying to me to have > particles (quantum objects) that have always been existing. But no one > really knows for sure. Yet. Yeah, you seem to be going somewhat down > the quantum geometrodynamics path so you ought to read up on that. BTW, > Wheeler couldn't get it to work and eventually bailed on it. But others > are probably still working on it. I have succeeded in showing an ontology where the Gforce and dark matter are primary to all. The theory is pleasing to the mind as it is all logically and mathematically connected. All that is needed is the willingness to accept that all physical matter has a non-material origin. Because if you refuse to accept this you will go insane trying to figure out an alternative explanation as to why the math works.
> | To begin with we can quantify h in terms of the mass of the electron, > | Compton wavelength and speed of light. [quoted text clipped - 10 lines] > > You really need to get a grip with this. E = mc^2; set c = 1 then E = m. Wrong, you did the math right earlier when you showed that Cd in "natural units" where c = 1 turns out to be 1/16pi^2 from cgs units. In order to do that, both sides of the equation had to be divided by c. You need to divide both sides of E=mc^2 by c in order to keep the equality. And since by your own admission that E is equal to mc^2, then mc^2=mc^2, and in "natural units" m=m.
I just had another insight. Your natural units are not a system of units at all. There are merely transformed units. If you want all your equations in the MKS system of units to have dimensionless current, you would divide all your equations by coul/sec. And no matter what, if you have an equality, you need to divide both sides of the equation by the same term. This is true with E=mc^2, too.
> What is charge^2/length? It is E, so energy times length is > charge^2. In units where c = 1, gm*cm is charge^2. I'm sorry, but charge^2/length in cgs units is gm^2 cm^5 / sec^4. It is nothing that I know of.
But you are correct that in cgs units, energy times length produces the dimensions of charge. But I show that energy times length produces a photon and the photon is equal to 16pi^2 time charge (in cgs units). So it is shown that the photon and charge units (in the APM) are both related. The photon naturally has the geometrical angle of 16pi^2 (double loxodrome) with reference to charge. You'll get a better handle on this when you receive the book.
Dave
FrediFizzx - 20 Feb 2005 06:50 GMT | > | > | To begin with, your above equation is not expressed in SI units. | > In SI [quoted text clipped - 14 lines] | but it is expressed as I wrote it, s^2 A^2. You wrote coul^2. coul^2 is | the MKS expression of charge. You mean coulomb is the MKS expression of electric charge not coul^2. A coulomb is also the SI derived unit for electric charge. Sure, its SI fundamental units are amp seconds but we don't go around saying with have amp seconds of charge. We say we have so many "coulombs" of charge.
| > | What are natural units? | > [quoted text clipped - 6 lines] | How can you do physics when everything is equal to 1? I agree, it is not | worthwhile continuing this discussion. Not everything is equal to 1. Fundamental electric charge, e, is equal to sqrt(alpha) ~= 1/11.706 in natural units.
| > No, you are the one who threw in sqrt(1/8pi) term by taking out an 8pi. | > I don't use that "term". You do. If you can't show any experimental [quoted text clipped - 4 lines] | | G=8pi T You mean G_uv = 8pi*T_uv, sure. Have you ever seen G_uv = T_uv?
| This says that the space-time tensor is equal to 8pi times the mass tensor. Sure, in units where Newton's G_N = 1 and c = 1. G_uv = T_uv where 8pi*G_N = 1 and c = 1.
| Of course, I show that General Relativity is good math based on wrong | principles. In my equation: [quoted text clipped - 20 lines] | make his equations work with mass instead of as it should have been with the | charges. This is not making sense to me right now. I hope you explain it better in your book.
| So, yes, there is experimental evidence for the existence of the 8pi | constant. What is the reference?
| > | > Hbar*c is already well known. I don't have to "rationalize" it; | > just use [quoted text clipped - 17 lines] | questioned was the real world origin of hbar. There is no physical process | or object that hbar is the quantity of. I think you need to study this a little more. ;-) Hbar is the smallest amount of of angular momentum something could have.
| > | Are you saying that natural units are cgs units? If you are then e^2 | > does [quoted text clipped - 11 lines] | be easy to figure out if everything is equal to 1, but I'd like to see how | these units are put to work. Read Griffiths' "Introduction to Elementary Particles". It is done in cgs units. Where ever you see hbar and/or c, just put a 1. None of the physics changes. It is not hard to figure out what it means to set hbar = c = 1. Mass (or energy) becomes the main units. Everything is in powers of those. E = m = w = 1/length were w is angular frequency. Energy density can just be w^4. If you learn this, then some of the reviews on the Particle Data Group website will be easier to understand since alot of particle physics is done in hbar = c = 1 units. It is a pain in the neck until you get used to it.
| > | > | k.c = 8.988 x 10^9 kg m^3 / sec^2 coul^2 | > | > | k.c = 8.988 x 10^9 kg m^3 / s^4 A^2 [quoted text clipped - 12 lines] | when all the other units are converted to the same system. That is, eps0 is | always permittivity of Aether (free space, vacuum, etc). Eps0 is simply 1/4pi in relation to EM. Now I have it as eps0 = Cvac/(2*lambda) with Cvac being vacuum capacitance and we know that C = Q/V. So we sub in sqrt(hbar*c) in cgs units for spacetime Q then we get,
eps0 = sqrt(hbar*c)/(2V*lambda)
It turns out that this relationship is always equal to 1/4pi in cgs units. So I am not arguing anything against eps0. 4pi pops up all the time in physics so it is most likely a good constant.
| > | > All you have done is split the inverse coulomb constant in two; | > | > 8pi*eps0/2, then shoved eps0/2 into your e.emax^2. [quoted text clipped - 20 lines] | choosing c = 1 doesn't do anything for me. In cgs units, Cd is equal to | c/16pi^2. The relavance is that particle physicists like to reduce important quantities to dimensionless numbers. I just don't remember ever seeing that number.
| In cgs units, Cd has the following relationship with Coulomb's constant | (k.c=1). [quoted text clipped - 14 lines] | permittivity are simply constants known through physics, so is the | conductance constant. Yeah, but I still don't see how you derived it. Looks like you are doing an x=x circular argument above. You just state [k.c = c * Cd * u.0 / e.0]. You need to show how you get Cd without starting with it in an equation. And what does mu0/eps0 mean physically?
| > Hmm... don't ever recall seeing that number pop up anywhere. | [quoted text clipped - 18 lines] | refuse to accept this you will go insane trying to figure out an alternative | explanation as to why the math works. I don't need to go insane. Been there for years now according to my wife. ;-) If dark matter is primary to all then where does this "non-material origin" come from? I think I will stick to the idea that the Universe is defined by very tiny quantum objects that have always existed for awhile. More than a google of them. Much more. If not that, then it has to be pure energy. And what the heck is pure energy?
| > | To begin with we can quantify h in terms of the mass of the electron, | > | Compton wavelength and speed of light. [quoted text clipped - 17 lines] | own admission that E is equal to mc^2, then mc^2=mc^2, and in "natural | units" m=m. No. We changed unit systems here. I give up. When c = 1, E = m. Period.
| I just had another insight. Your natural units are not a system of units at | all. There are merely transformed units. If you want all your equations in | the MKS system of units to have dimensionless current, you would divide all | your equations by coul/sec. And no matter what, if you have an equality, | you need to divide both sides of the equation by the same term. This is | true with E=mc^2, too. Other unit systems are possible. But I think you had better get a grip on conversions to and from natural units first.
| > What is charge^2/length? It is E, so energy times length is | > charge^2. In units where c = 1, gm*cm is charge^2. | | I'm sorry, but charge^2/length in cgs units is gm^2 cm^5 / sec^4. It is | nothing that I know of. Bzzzt! Try again. charge^2/length in cgs is gm*cm^2/sec^2 = energy.
| But you are correct that in cgs units, energy times length produces the | dimensions of charge. Charge^2 not charge.
| But I show that energy times length produces a photon | and the photon is equal to 16pi^2 time charge (in cgs units). So it is | shown that the photon and charge units (in the APM) are both related. The | photon naturally has the geometrical angle of 16pi^2 (double loxodrome) with | reference to charge. You'll get a better handle on this when you receive | the book. What of a photon is equal to 16pi^2 times charge in cgs units? A photon has no charge. Spacetime has the charge.
FrediFizzx
David Thomson - 20 Feb 2005 18:05 GMT > | Come on, Fredi. Didn't you look at the last column? It's called a > coulomb, [quoted text clipped - 6 lines] > fundamental units are amp seconds but we don't go around saying with have > amp seconds of charge. We say we have so many "coulombs" of charge. Here is what you said in a previous message:
> Not way off. I said your emax not emax^2. In SI units it is, > 4pi*eps0*hbar*c = 3.5 x 10^-36 coul^2 You mistakenly expressed charge in MKS units and called it SI units.
> | > | What are natural units? > | > [quoted text clipped - 10 lines] > Not everything is equal to 1. Fundamental electric charge, e, is equal to > sqrt(alpha) ~= 1/11.706 in natural units. That's still dimensions of 1. You would be charting new territory if you proposed that values are also dimensions. It's not that it couldn't be, but we would need to see some kind of explanation for it.
> | > No, you are the one who threw in sqrt(1/8pi) term by taking out an > 8pi. [quoted text clipped - 7 lines] > > You mean G_uv = 8pi*T_uv, sure. Have you ever seen G_uv = T_uv? No, I haven't. But I have seen G(m,n) = 8pi T(m,n)
> | This says that the space-time tensor is equal to 8pi times the mass > tensor. [quoted text clipped - 44 lines] > This is not making sense to me right now. I hope you explain it better in > your book. I didn't include an explanation of General Relativity in this book. I've been asked by Jim to include it, though. He actually thought the GR equation was related to our unified charge equation and recently bought me a book called "Gravitation" to study. I need to read some more on this and give our explanation more depth. In the book, I explain the unifed charge equation in detail, however. So if you understand GR, you'll be able to develop the relationship between these two theories fairly easy on your own.
In any case, I have presented to you an independent source that also uses the 8pi constant in his equation.
> | So, yes, there is experimental evidence for the existence of the 8pi > | constant. > > What is the reference? It's in the Gravitation book. If it were convenient for me to get it right now, I would look it up.
> | All I questioned was the real world origin of hbar. There is no > physical process > | or object that hbar is the quantity of. > > I think you need to study this a little more. ;-) Hbar is the smallest > amount of of angular momentum something could have. I have studied angular momentum in depth. h is the smallest amount of angular momentum the electron can have. Neils Bohr divided h by 2pi because he thought electrons were tiny balls in orbit around a nucleus. By dividing by 2pi, Bohr was hoping to obtain the value of angular momentum for a particular point along the path. But, in reality, the angular momentum is the full trace of the electron in one quantum moment. That is why the electron appears as a cloud. It is an error of 20th Century physics to divide the angular momentum by 2pi.
> Read Griffiths' "Introduction to Elementary Particles". It is done in cgs > units. Where ever you see hbar and/or c, just put a 1. None of the [quoted text clipped - 9 lines] > to > it. I'll have to look at it when I come across it. But I'm certain just from the outset that since angular momentum is said to equal to velocity, which is said to equal 1, that it cannot possibly be correct. You're saying that something with a mass dimension is equal to something without a mass dimension. It's like saying that mass does not exist, and comes from nothing at the same time.
> Eps0 is simply 1/4pi in relation to EM. Now I have it as eps0 = > Cvac/(2*lambda) with Cvac being vacuum capacitance and we know that C = > Q/V. Do we really know that C=Q/V? You're assuming that the units C and V are properly expressed in their dimensions of charge. You'll see my strong case for why it should really be C=4pi/V when you read the book. Similarly for inductance, L=V/4piF^2.
> So we sub in sqrt(hbar*c) in cgs units for spacetime Q then we get, > eps0 = sqrt(hbar*c)/(2V*lambda) > > It turns out that this relationship is always equal to 1/4pi in cgs units. > So I am not arguing anything against eps0. 4pi pops up all the time in > physics so it is most likely a good constant. I agree that 4pi pops up all the time. In fact, I give 4pi a semi-dimension status. It would appear that the spherical constant of 4pi arises from the fact that all quantum existence is spherical in geometry or angle.
> | I'm sorry, I don't see the relevance of you natural units. Arbitrarily > | choosing c = 1 doesn't do anything for me. In cgs units, Cd is equal to [quoted text clipped - 4 lines] > that > number. I still need to hear the rationale for reducing important measurements to dimensionless numbers. As for 16pi^2, here's one that will blow you away. The unit of rotating magnetic field in the Aether Physics Model is equal, in quantum measurements, to m.e * w.C^3 * F.q^2 / e.emax^2. Figure that out in cgs units.
m.e = mass of the electron w.C = Compton wavelength F.q = quantum frequency = c / w.C e.emax^2 is the strong charge of the electron
It comes out to 16pi^2.
The unit of rotating magnetic field is also equal to 16pi^2 * k.c in any system of units. And naturally, since k.c is equal to 1 in the cgs system of units, then rotating magnetic field is equal 16pi^2 in cgs units.
> | In cgs units, Cd has the following relationship with Coulomb's constant > | (k.c=1). [quoted text clipped - 20 lines] > You need to show how you get Cd without starting with it in an equation. > And what does mu0/eps0 mean physically? We get Cd the same way we get Coulomb's constant. It's just there in the data. All Aether constants are absolutes with regard to physical matter. Try deriving the speed of light some day, or permeability or permittivity. It can't be done. These constants simply exist. The conductance constant is no different.
> If dark matter is primary to all then where does this "non-material > origin" come from? I think I will stick to the idea that the Universe is > defined by very tiny quantum objects that have always existed for awhile. That's a cop out. It is perfectly okay to simply accept things as they are. It is the refusal to accept things as they are and insist they must be otherwise that made science what it is today.
> More than a google of them. Much more. If not that, then it has to be > pure > energy. And what the heck is pure energy? That's what I'm saying. The physics shows that Gforce and dark matter gave rise to the physical Universe. There is no such thing as pure energy. There is no such thing as converted mass. The Universe is not a complex mix of a myriad of strange particles. The Universe is extremely simple and arises from extremely simple principles. Instead of trying to define Creation in terms of the human mind, we should train the human mind to see Creation as it is.
> | You need to divide > | both sides of E=mc^2 by c in order to keep the equality. And since by [quoted text clipped - 4 lines] > No. We changed unit systems here. I give up. When c = 1, E = m. > Period. Fredi, you're giving in to the dark side. :-) We must keep our faith in the rules of math. If you change one side of the equation, you must make the same change on the other side of the equation. Yes, you can transform the units on one side to c = 1 for entertainment, but before you can finish the calculation both sides must have the same system of units. You can't equate apples and oranges.
> | I just had another insight. Your natural units are not a system of > units at [quoted text clipped - 9 lines] > Other unit systems are possible. But I think you had better get a grip on > conversions to and from natural units first. I have the grip. You can't have MKS units on one side of the equation and cgs units on the other side when you do the calculation. You are arbitrarily converting one side of the equation to dimensionless velocity. This is perfectly fine. But when you want to know how the right side equates to the left side, then you either have to change the left side to dimensionless velocity or revert the right side back to the same units as the left side. You can't stop halfway. It goes like this:
1ft = 12in
You can convert the 12in to dimensionless units by dividing by 12in.
1ft = 1 (dimensionless unit)
But before you can evaluate the equality you either need to divide 1ft by 12in or multiply 1 by 12in.
> | > What is charge^2/length? It is E, so energy times length is > | > charge^2. In units where c = 1, gm*cm is charge^2. [quoted text clipped - 3 lines] > > Bzzzt! Try again. charge^2/length in cgs is gm*cm^2/sec^2 = energy. charge = gm * cm^3 / sec^2 charge^2 = gm^2 * cm^6 / sec^4 charge^2/length = gm^2 * cm^5 / sec^4
Bzzzt!
> | But you are correct that in cgs units, energy times length produces the > | dimensions of charge. > > Charge^2 not charge. Right. In my system of physics charge is always charge squared.
> | But I show that energy times length produces a photon > | and the photon is equal to 16pi^2 time charge (in cgs units). So it is [quoted text clipped - 7 lines] > > What of a photon is equal to 16pi^2 times charge in cgs units? That's what I'm talking about.
> A photon has no charge. A photon has strong charge, but it has no electrostatic charge. Unless otherwise noted, all charge in the Aether Physics Model is strong charge.
> Spacetime has the charge. Space-time (actually Aether) has both strong charge and electrostatic charge.
Dave
FrediFizzx - 21 Feb 2005 19:54 GMT | > | Come on, Fredi. Didn't you look at the last column? It's called a | > coulomb, [quoted text clipped - 32 lines] | proposed that values are also dimensions. It's not that it couldn't be, but | we would need to see some kind of explanation for it. Have you ever read Feynman's "QED"?
| > | > No, you are the one who threw in sqrt(1/8pi) term by taking out an | > 8pi. [quoted text clipped - 77 lines] | It's in the Gravitation book. If it were convenient for me to get it right | now, I would look it up. Please look it up and give the exact title and author(s) of the book.
| > | All I questioned was the real world origin of hbar. There is no | > physical process [quoted text clipped - 11 lines] | electron appears as a cloud. It is an error of 20th Century physics to | divide the angular momentum by 2pi. h is usually associated with linear momentum and hbar with angular momentum.
| > Read Griffiths' "Introduction to Elementary Particles". It is done in cgs | > units. Where ever you see hbar and/or c, just put a 1. None of the [quoted text clipped - 22 lines] | | Do we really know that C=Q/V? C = Q/V is the definition of capacitance. The capacitance of a system is the proportionality constant of charge to voltage.
| You're assuming that the units C and V are | properly expressed in their dimensions of charge. You'll see my strong case | for why it should really be C=4pi/V when you read the book. Similarly for | inductance, L=V/4piF^2. What units are you in here?
| > So we sub in sqrt(hbar*c) in cgs units for spacetime Q then we get, | > eps0 = sqrt(hbar*c)/(2V*lambda) [quoted text clipped - 63 lines] | It can't be done. These constants simply exist. The conductance constant | is no different. If it is "there in the data"; what data? Reference please.
| > If dark matter is primary to all then where does this "non-material | > origin" come from? I think I will stick to the idea that the Universe is [quoted text clipped - 31 lines] | calculation both sides must have the same system of units. You can't equate | apples and oranges. Please learn how to convert from one system of units to another.
| > | I just had another insight. Your natural units are not a system of | > units at [quoted text clipped - 40 lines] | | Bzzzt! Charge in cgs is sqrt(gm * cm^3 / sec^2). Look it up before you embarrass yourself some more.
| > | But you are correct that in cgs units, energy times length produces the | > | dimensions of charge. [quoted text clipped - 26 lines] | Space-time (actually Aether) has both strong charge and electrostatic | charge. Spacetime has *bound* charge of +,- sqrt(hbar*c) in cgs units. All other charge "units" are fractals of this basic charge including electroweak charge and color charge.
FrediFizzx
David Thomson - 24 Feb 2005 02:49 GMT > | > | So, yes, there is experimental evidence for the existence of the 8pi > | > | constant. [quoted text clipped - 6 lines] > > Please look it up and give the exact title and author(s) of the book. Gravitation by Charles W. Misner, Kip S. Thorne, and John Archibald Wheeler
> | I have studied angular momentum in depth. h is the smallest amount of > | angular momentum the electron can have. Neils Bohr divided h by 2pi [quoted text clipped - 10 lines] > h is usually associated with linear momentum and hbar with angular > momentum. Nope, sorry. h doesn't have the dimensions of linear momentum. Dividing by 2pi cannot change a unit from linear momentum to angular momentum.
> | Do we really know that C=Q/V? > > C = Q/V is the definition of capacitance. The capacitance of a system > is the proportionality constant of charge to voltage. Are you saying that if you take an electrometer and potential meter you can calculate from the two readings the capacitance of a capacitor? Try it sometime and see if it works.
> | You're assuming that the units C and V are > | properly expressed in their dimensions of charge. You'll see my strong [quoted text clipped - 4 lines] > > What units are you in here? These are quantum units as you'll see explained in my book. One of the errors I identify in modern physics is that most of the units have the wrong dimensions of charge. All charge is distributed and must be expressed as distributed units in order for dimensional analysis to work properly. The real kicker is that five units in modern physics are already expressed in the correct dimensions of charge, while all the others are not. The five correct dimensions are permeability, permittivity, conductance, inductance, and capacitance.
> | We get Cd the same way we get Coulomb's constant. It's just there in > the [quoted text clipped - 7 lines] > > If it is "there in the data"; what data? Reference please. Any experiment that measures electrostatic force and yields Coulomb's constant. Any experiment that yields the speed of light. Any experiment that yields the permeability and permittivity constants. By analyzing these constants with quantum measurements it can be seen that there are numerous constants associated with the Aether.
I haven't done it yet, but experiments related to the Casimir effect should provide a direct confirmation of the conductance constant of the Aether. It's just a matter of knowing how to look at the data.
Fred > | > | > What is charge^2/length? It is E, so energy times length is
> | > | > charge^2. In units where c = 1, gm*cm is charge^2. Dave > | > | I'm sorry, but charge^2/length in cgs units is gm^2 cm^5 / sec^4. It is
> | > | nothing that I know of. Fred > | > Bzzzt! Try again. charge^2/length in cgs is gm*cm^2/sec^2 = energy.
Dave > | charge = gm * cm^3 / sec^2
> | charge^2 = gm^2 * cm^6 / sec^4 > | charge^2/length = gm^2 * cm^5 / sec^4 [quoted text clipped - 3 lines] > Charge in cgs is sqrt(gm * cm^3 / sec^2). Look it up before you > embarrass yourself some more. As before, I was expressing charge as it should be, with distributed charge dimensions. The cgs system of units helps to show that I am correct in this regard. Charge as you would write it would have fractional exponents.
> | > | But you are correct that in cgs units, energy times length produces > the [quoted text clipped - 3 lines] > | > | Right. In my system of physics charge is always charge squared.
> | > Spacetime has the charge. > | [quoted text clipped - 4 lines] > other charge "units" are fractals of this basic charge including > electroweak charge and color charge. I present a strong case for two distinct manifestations of charge. I'll stand by it.
Dave
Paul Stowe - 24 Feb 2005 03:36 GMT [Snip...]
>> h is usually associated with linear momentum and hbar with angular >> momentum. > > Nope, sorry. h doesn't have the dimensions of linear momentum. Right... But so what?
> Dividing by 2pi cannot change a unit from linear momentum to angular > momentum. Dimension fixation... That is, focusing on ONE of several interpretations of a dimensional pattern. For example, kg/sec. This can be a linear mass flow (like water thru a pipe) or a harmonic oscillation like the classic mass & spring arrangement. In the case of h (Action) look up the classical concept of Action, it ain't angular momentum. It is the interaction parameter in kinetic theory. Two masses (m) particles moving at speed v multiplied by the mean free path L. In other words,
h = 2mvL
Of course, the dimensions are also equivalent to angular momentum but the point is, one can't discern which from JUST the dimensions.
Paul Stowe
vern@bealenet.com - 24 Feb 2005 03:47 GMT > [Snip...] > [quoted text clipped - 23 lines] > > Paul Stowe Sorry to post off topic, but have you heard from Greywolf42; he hasn't posted in a while.
Vern
Paul Stowe - 24 Feb 2005 05:08 GMT [Snip...]
> Sorry to post off topic, but have you heard from Greywolf42; > he hasn't posted in a while. He's been busy of late. He's still around, usually posting on weekends now. But, doesn't have as much free time as he used to.
Paul Stowe
mingstb@sim-ss.com - 27 Feb 2005 20:16 GMT > Sorry to post off topic, but have you heard from Greywolf42; he hasn't > posted in a while. Hi, Vern.
I'm still around. But I have been busy with other aspects of life for the past few weeks. I'll be mostly doing an occasional lurk for a bit, yet. I haven't had much energy for the newsgroups.
greywolf42 ubi dubium ibi libertas
John C. Polasek - 24 Feb 2005 22:58 GMT snipr
>> | I have studied angular momentum in depth. h is the smallest amount of >> | angular momentum the electron can have. Neils Bohr divided h by 2pi [quoted text clipped - 13 lines] >Nope, sorry. h doesn't have the dimensions of linear momentum. Dividing by >2pi cannot change a unit from linear momentum to angular momentum. snip>
The difference between h and hbar is as follows:
E = hf uses frequency in cycles/second. E is energy x time, with time being the period of one cycle.
hbar is 2pi lower and is used with the higher radian frequency radians/second, so E = hbar x omega Omega is a more "useful" angular rate, generally. v = omega x r.
You can't convert linear into angular momentum just using 2pi, one being ML^2T^-1 and one is MLT^-1.. John Polasek
>----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- >http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups >----= East and West-Coast Server Farms - Total Privacy via Encryption =---- FrediFizzx - 25 Feb 2005 07:46 GMT | > | > | So, yes, there is experimental evidence for the existence of the 8pi | > | > | constant. [quoted text clipped - 8 lines] | | Gravitation by Charles W. Misner, Kip S. Thorne, and John Archibald Wheeler OK, now what page is it on that they derive this 8pi constant? Then if you can, scan it and email it to me. I don't have that book.
| > | I have studied angular momentum in depth. h is the smallest amount of | > | angular momentum the electron can have. Neils Bohr divided h by 2pi [quoted text clipped - 13 lines] | Nope, sorry. h doesn't have the dimensions of linear momentum. Dividing by | 2pi cannot change a unit from linear momentum to angular momentum. The key words above is "associated with".
| > | Do we really know that C=Q/V? | > [quoted text clipped - 4 lines] | calculate from the two readings the capacitance of a capacitor? Try it | sometime and see if it works. http://scienceworld.wolfram.com/physics/Capacitance.html
This *is* the definition of capacitance.
| > | You're assuming that the units C and V are | > | properly expressed in their dimensions of charge. You'll see my strong [quoted text clipped - 13 lines] | correct dimensions are permeability, permittivity, conductance, inductance, | and capacitance. Got your book. Give me a couple weeks to go over it.
| > | We get Cd the same way we get Coulomb's constant. It's just there in | > the [quoted text clipped - 13 lines] | constants with quantum measurements it can be seen that there are numerous | constants associated with the Aether. Please give me one reference that says the conductance of the vacuum is what you say it is. We know that the impedance is ~ 377 ohms. If you invert that to get conductance it is ~ 2.65 x 10^-3 siemens. To get your Cd value it has to be 1/(377ohms*4pi). So what's up with this extra 4pi? Why doesn't this show up in experimental "data". If it does, give us the exact reference.
| I haven't done it yet, but experiments related to the Casimir effect should | provide a direct confirmation of the conductance constant of the Aether. [quoted text clipped - 22 lines] | dimensions. The cgs system of units helps to show that I am correct in this | regard. Charge as you would write it would have fractional exponents. You can't go changing what charge is defined as in a certain unit system. If you want to say that charge^2 in cgs is charge in Dave's unit system that is fine. But don't tell me that charge is charge^2 in cgs.
| > | > | But you are correct that in cgs units, energy times length produces | > the [quoted text clipped - 15 lines] | I present a strong case for two distinct manifestations of charge. I'll | stand by it. Well, I don't really think you are saying much new. Everyone knows that charge by itself doesn't mean much. It takes two to tango thus charge^2. That is why alpha = e^2/(hbar*c) in cgs units. But your big thing here is your Cd factor. If you can't show experimental evidence for it, then you just made it up. Which might be OK, if for some reason it is currently undetectable. In natural units, it seems like it would be eps0^2. But I don't quite follow the connection yet to having eps0 squared. It seems to me that eps0 is already a squared quantity as evident in c = 1/sqrt(eps0*mu0).
FrediFizzx
Bilge - 12 Feb 2005 09:58 GMT David Thomson:
>I was posting on another list when one of the readers suggested I use my >reply as a general introduction to the Aether Physics Model on these [quoted text clipped - 8 lines] >form because of the coil arrangement, frequency, current, potential, etc. >But here I had a single coil, the same shape feeding both ends. That's an unwarranted assumption. The electric field near a sharp point can be extremely large compared with a smooth surface. That's why one spends a lot of effort polishing surfaces which have large potential differences between them if one wants to minimize sparking. By contrast, if one want to initiate sparking, one uses electrodes which have sharp points (or at least a sharp point for the negative side).
>The same >frequency existed at both ends. But it became evident that the potential >was greater on the topload and the current was greater on the ground end. >As a result of these differences, the appearance of the spark changed >dramatically. The potentials aren't really relevent. It's the gradient that matters.
[...]
>It didn't take but a few days before I realized that charge should be >distributed in the math, since it is also distributed in nature. Ed Philips >suggested I read Great Experiments in Physics so I purchased a copy. In >this book Charles Coulomb himself clearly stated that all charge must be >distributed in order for the inverse square law to work. Note that in classical E&M the equations involve charge and current densities.
>The first thing I noticed in physics was that Coulomb's constant was >precisely equal to a combination of light speed, permeability, permittivity, >and a conductance constant, which the established literature does not >recognize. Sure it does. Every time someone write F = e^2/r^2 instead of writing F = ke^2/r^2, that person is recognizing the lack of physical content in the conversion constant `k' by omitting it.
[...]
>where e is the elementary charge, a is the fine structure of the electron, >and e.emax^2 is the derived value of charge equal to: > >e.emax^2 = h * Cd So basically, you've taken some man-made conversion constants, based on man-made definitions of units for fundanental quantities, then declared those to be fundamental and inverted the definitions to obtain the original definitions.
>I then applied the same process to the proton and neutron and found fine >structures for both. But since the fine structures and angular momentum of >the proton and neutron are not defined in modern physics, I really didn't >have an empirical body of data to go by there. Once again, you attribute your lack of knowledge to a deficit of information. What you refer to has existed for nearly a half-century. It's called a dalitz plot, which is a plot of Y vs. I_3 for baryon and meson multiplets with the same J = L + S. The study of those structures is called meson and baryon spectroscopy. It simply isn't possible to have seriously looked for this and not discovered it. There's so much data that it would be impossible to sift through anything but compilations of it found in the particle properties data.
>However, the lack of data >was encouraging since it implied that modern physics was incapable of >determining these values based on their chosen model of subatomic behavior. > >Three weeks went by before it occurred to me that e.emax^2 could be strong >charge, which mediates the strong force. Well no, it can't. In the first place, the strong force consists of three types of charge, not two. That's why the charges are labeled R,G,B instead of +/-. Second, the gluons which carry the strong force also carry the charge in pairs, i.e., the gluons are said to be bicolored. The photon which mediates the electromagnetic force, on the other hand, carries no charge.
>I then realized that if it were, >then I was looking at the Unified Field Theory. Further research showed me >that it has nothing to do with fields, but is rather a Unified Force Theory. >I later found that the concept of fields is very ambiguous in the Standard >Model It's not ambiguous at all. The fields are defined very precisely, as ought to be evident from the fact that there are precise equations which specify them. Specifying something and finding an analytical solution is not the same thing.
>and relativity theories, as modern science apparently does not know >what a field really is. Here's a field theory called qed:
_ L = u(p/ - m)u - j.A + F^uv F_uv
Since L is a lagrangian (density), the equations of motion are completely given by the euler-lagrange equations, just as in any other case. Solving the equations of motion is non-trivial, but so is solving the classical three-body problem and a great many other problems that were conceived long before quantum field theory or relativity.
>When I calculated the relative value of the forces for the different >subatomic particles, looking at both the electrostatic force and the new >concept of strong force, I found that the relative strengths of the forces >matched the empirical data. Why are you surprised? You used constants derived from the data you were trying to match.
>Then I started reading older literature on electromagnetism. I was >particularly struck by Ed Leedskalnin's description of electricity as a >stream of tiny magnets. Although his view incorrectly denies the >electrostatic effect of electrons, he does show a convincing argument for >electrons acting as tiny magnets. Of course. The electron is a spin 1/2. It's straight forward to show that an electron interacts via its charge _and_ a magnetic moment by performing a gordon reduction on the electromagnetic current.
>And then it occurred to me that electromagnetism is caused by the >strong charge. No, it's not. The strong force has its own ``color-magnetic'' interaction.
>Further analysis of the mathematical aspects of the two charges revealed >that electrostatic charge has a spherical geometry while strong charge >(or electromagnetic charge) has a toroidal geometry. The strong force is a vector field, just as is the electromagnetic force.
>When I applied this knowledge to my coil observations it made perfect sense. >When electrons are magnetically moving sideways in the coil, they produce >the bright white streamers. When electrons are moving along their z axis, >they produce the thin purple streamers. What you are calling ``streamers'' is nothing more than the light emitted when ionized atoms and molecules recapture electrons. The difference in color is due to the degree of ionization that occurs, which in turn is due to the rate at which the discharge occurs.
David Thomson - 13 Feb 2005 01:05 GMT Hi Bilge,
> That's an unwarranted assumption. The electric field near a sharp point > can be extremely large compared with a smooth surface. That's why one > spends a lot of effort polishing surfaces which have large potential > differences between them if one wants to minimize sparking. By contrast, > if one want to initiate sparking, one uses electrodes which have sharp > points (or at least a sharp point for the negative side). In the particular experiment I mentioned, both the ground and teminal ends of this coil were connected to contacts that were covered with about an eighth of an inch of plastic coating. The contact between these two coated terminals where the two different types of sparks met were the pins of a flourescent tube. In subsequent experiments I could produce the particular type of spark (thin purple or thick white) by tuning for higher potential or higher current.
> >The same > >frequency existed at both ends. But it became evident that the potential [quoted text clipped - 3 lines] > > The potentials aren't really relevent. It's the gradient that matters. Well, that's exactly what I'm saying. The gradient is between the potential and the current.
> >It didn't take but a few days before I realized that charge should be > >distributed in the math, since it is also distributed in nature. Ed [quoted text clipped - 5 lines] > Note that in classical E&M the equations involve charge and current > densities. The charge is still expressed as single dimension charge. Charge is not distributed just because it exists on a surface. In fact, that is my point. Area is distributed as a surface, but charge is a dimension, too. In order for charge to be distributed it must also be expressed as a distribution, i.e. coul^2, not coul.
> >The first thing I noticed in physics was that Coulomb's constant was > >precisely equal to a combination of light speed, permeability, [quoted text clipped - 5 lines] > F = ke^2/r^2, that person is recognizing the lack of physical content > in the conversion constant `k' by omitting it. That's called "pretending", not "recognizing." Nature does not just throw out conversion constants. If a constant exists, it is due to a real phenomenon.
> >where e is the elementary charge, a is the fine structure of the > >electron, [quoted text clipped - 6 lines] > then declared those to be fundamental and inverted the definitions > to obtain the original definitions. They aren't man-made conversion constants. They are nature made absolute constants. Basically, what you and others have done is take a natural phenomenon and sweep it under the rug because you don't understand it.
> >I then applied the same process to the proton and neutron and found fine > >structures for both. But since the fine structures and angular momentum [quoted text clipped - 11 lines] > anything but compilations of it found in the particle properties > data. I stand by my statement, the proton and neutron angular momenta and their fine structure constants are not recognized by modern physics. Saying that such things are implied is not the same thing as writing them out in value and dimensions.
> >However, the lack of data > >was encouraging since it implied that modern physics was incapable of [quoted text clipped - 11 lines] > The photon which mediates the electromagnetic force, on the other hand, > carries no charge. Within the above model you can state whatever you want. I present a completely new physics from the same data. You can talk about colors, gluons, pions, and other man-made constructs, but I'm discussing strong force in terms of charge, as it must be in order to be related to mass and electrostatic charge. You can rattle off the last 50 years of SM physics all day and all night, but I have presented a system of quantifying strong force that also produces a very simple and symmetrical Unified Force Theory. BTW, I'm not saying you can't practice physics with whatever model you want, I'm just saying I have found a new system of physics that works better than yours.
> >I then realized that if it were, > >then I was looking at the Unified Field Theory. Further research showed [quoted text clipped - 9 lines] > which specify them. Specifying something and finding an analytical > solution is not the same thing. Equations that produce numbers do not mean that fields are precisely defined. You may know what the field strength will be at a given location, but you still have not defined what the field is.
> >and relativity theories, as modern science apparently does not know > >what a field really is. [quoted text clipped - 10 lines] > other problems that were conceived long before quantum field > theory or relativity. So tell us, then. What is a field? Is it a particle? Is it space-time? Is it photons? You presented the equation, so what can you tell us about fields as a result? Yes, we can tell what the strength of the field will be in a given location, but tell us what a field is.
> >When I calculated the relative value of the forces for the different > >subatomic particles, looking at both the electrostatic force and the new [quoted text clipped - 4 lines] > Why are you surprised? You used constants derived from the data you > were trying to match. I'm surprised because you, and nobody else thought of it. It has apparently never occurred to you or anybody else that strong force would be due to strong charge, instead of the imaginary pi meson particle. I'm surprised because instead of keeping the strong force carrier as a dimension, you and others try to make it a particle.
Apparently you are not surprised that I can quantify strong force from the same data and constants you use, and yet for the past couple years you ridiculed me for pointing it out. And likely, you will continue to tell me that I'm wrong for showing you the correct interpretation of the known data and constants.
> >Then I started reading older literature on electromagnetism. I was > >particularly struck by Ed Leedskalnin's description of electricity as a [quoted text clipped - 6 lines] > moment by performing a gordon reduction on the electromagnetic > current. There is a more fundamental cause for the electron 1/2 spin, which I derived from simple dimensional analysis and by treating the so-called "conversion constants" as actual phenomena.
> >And then it occurred to me that electromagnetism is caused by the > >strong charge. > > No, it's not. The strong force has its own ``color-magnetic'' > interaction. Think about what you are saying. Do you really know what a "color-magnetic" interaction is? Or is this a mental construct you have decided to accept until a better explanation comes along? I show the mathematically derived geometry of the strong charge as well as the geometry of mass and electrostatic charge. I show a convincing argument that the strong charge of subatomic particles can be magnetically aligned to produce macro magnetic fields.
> >Further analysis of the mathematical aspects of the two charges revealed > >that electrostatic charge has a spherical geometry while strong charge > >(or electromagnetic charge) has a toroidal geometry. > > The strong force is a vector field, just as is the electromagnetic > force. In your chosen system of physics it is. But I show a new system of physics where charge has geometry. Also, think about what you are saying here. You're saying the strong force is a vector field. So you're saying the strong force originates outside of the particles being bound. How can you make that statement without defining the space and time outside the particle?
Also, you are claiming the electrostatic charge is the electromagnetic force. I disagree. The electrostatic charge produces the electrostatic force and the electromagnetic charge produces the strong force. Coulomb's force law only works for electrostatic charge and at a distance at that. As the charges become closer the electromagnetic charge comes into play and the force law now increases the force. The Casimir equation is actually transposed from the strong force equation of the electron, in my physics, anyway.
> >When I applied this knowledge to my coil observations it made perfect > >sense. [quoted text clipped - 7 lines] > difference in color is due to the degree of ionization that occurs, > which in turn is due to the rate at which the discharge occurs. It's "nothing more than...", wow, are you a psychic, too? You haven't seen the experiment or directly observed the charges. How do you come up with all this knowledge without measurements or direct observations?
I'm impressed that you responded to one of my posts without being rude, which is why I took the time to respond. But I'm still not surprised by your unfounded skepticism. I'm not going to allow someone, who has absolutely no knowledge of the parameters of my experiments and won't take the time to examine the physics, to convince me to blow off my work. What is the fun of that? I want someone to crush my ego by providing actual proof that my experiment was flawed and that my physics is flawed. I want my day in court. You're not going to take that from me with your unfounded skepticism. Nobody is.
Dave
Bilge - 13 Feb 2005 23:38 GMT David Thomson:
>Hi Bilge, > [quoted text clipped - 12 lines] >type of spark (thin purple or thick white) by tuning for higher potential or >higher current. What's your point? [...]
>> The potentials aren't really relevent. It's the gradient that matters. > >Well, that's exactly what I'm saying. The gradient is between the potential >and the current. That is a non-sequiter.
[...]
>> Sure it does. Every time someone write F = e^2/r^2 instead of writing >> F = ke^2/r^2, that person is recognizing the lack of physical content [quoted text clipped - 3 lines] >out conversion constants. If a constant exists, it is due to a real >phenomenon. Bzzzt. Thanks for playing. Conversion constants don't exist in nature. Nature didn't invent the kilogram or the meter. People did.
David Thomson - 16 Feb 2005 03:32 GMT > >That's called "pretending", not "recognizing." Nature does not just > >throw [quoted text clipped - 3 lines] > Bzzzt. Thanks for playing. Conversion constants don't exist in > nature. Nature didn't invent the kilogram or the meter. People did. Bzzzt. The kilogram and meter are measurements, not conversion constants.
Nature invented mass and length, the dimensions of which the measurements of kilogram and meter are constructed. Mass and length, like conversion constants, are non-material phenomena that are the foundation of material phenomena.
Dave
Bilge - 16 Feb 2005 05:00 GMT David Thomson:
>> >That's called "pretending", not "recognizing." Nature does not just >> >throw [quoted text clipped - 5 lines] > >Bzzzt. The kilogram and meter are measurements, not conversion constants. \epsilon_0 = 8.85 x 10^-12 (A-s^2)^2/kg-m^3 That is a conversion constant, moron.
geraldkelleher@hotmail.com - 12 Feb 2005 13:41 GMT http://www.bodley.ox.ac.uk/cgi-bin/ilej/image1.pl?item=page&seq=9&size=1&id=bm.1 843.10.x.54.336.x.425
Better not to be either an aetherist or a relativist to enjoy the epiphany.The guys prior to Maxwell were already in a jam,Newton had already rejected /did'nt need an aether for his ballistic agenda and was useless for anything other than reflection,refraction and treatise on colors.
Somehow the guys in the early 20th century cobbled together a fantastic story of fiction using historical characters and original text,after Isaac the Ambiguous rejected aether in 1704,Albert and his merry bunch did it all otver again in 1905 !.
C'mon guys,what is worse than coming up with a self-promoting story like relativity is supporting the bloody thing,surely you can't believe that the aehterists are competition.
"The fictitious matter which is imagined as filling the whole of space is of no use for explaining the phenomena of Nature, since the motions of the planets and comets are better explained without it, by means of gravity; and it has never yet been explained how this matter accounts for gravity. The only thing which matter of this sort could do, would be to interfere with and slow down the motions of those large celestial bodies, and weaken the order of Nature; and in the microscopic pores of bodies, it would put a stop to the vibrations of their parts which their heat and all their active force consists in. Further, since matter of this sort is not only completely useless, but would actually interfere with the operations of Nature, and [314] weaken them, there is no solid reason why we should believe in any such matter at all. Consequently, it is to be utterly rejected."
Optics 1704
Lady Chatterly - 12 Feb 2005 15:09 GMT >Better not to be either an aetherist or a relativist to enjoy the >epiphany.The guys prior to Maxwell were already in a jam,Newton had >already rejected /did'nt need an aether for his ballistic agenda and >was useless for anything other than reflection,refraction and treatise >on colors. You have yet again proven to everyone that you are an utter and complete imbecile.
>Somehow the guys in the early 20th century cobbled together a fantastic >story of fiction using historical characters and original text,after >Isaac the Ambiguous rejected aether in 1704,Albert and his merry bunch >did it all otver again in 1905 !. Climb mountains to see lowlands.
>C'mon guys,what is worse than coming up with a self-promoting story >like relativity is supporting the bloody thing,surely you can't believe >that the aehterists are competition. Why do you ask if worse than coming up with a self promoting story like relativity is supporting the bloody thing is?
>"The fictitious matter which is imagined as filling the whole of space >is of no use for explaining the phenomena of Nature, since the motions [quoted text clipped - 9 lines] >weaken them, there is no solid reason why we should believe in any >such matter at all. Consequently, it is to be utterly rejected." Did you mean by 'truly exist '.
>Optics 1704 You're depriving some village of an idiot.
-- Lady Chatterly
"Speaksing the truth is necessary, even if some might not like it. Gee, do I sound like Lady Chatterly in this sentence?" -- Barbara Schwarz
geraldkelleher@hotmail.com - 12 Feb 2005 18:10 GMT Do you know what it feels like to come back to a relativity forum ?, it feels like stumbling across a village where the inhabitants have'nt been told that their heated discussions belong to a bygone era,sort of like trying to pick an argument with a communist when all the world is a decrepit consumerist.
Three years ago this forum was a hotbed of real insults but now it looks like you guys really need each other and that I find repulsive.
Look at Newton's take on aether,it means nothing and neither does his take on planetary motion via terrestial ballistics.So,pick up where Kepler left off and do things properly this time,you all |
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