New Field Theory: Similarities & Differences

[Antonio Lao]

McQueen,

I thought about terminal velocity in the falling of an object in a gravitational field. If we make the analogy for the electrical conduction, the small drift velocity of the electron can be thought of as its almost nearly suspension in an electric field. That is the electron is almost not falling in an electric field.

 

[Antonio Lao]

If the drift velocity of the conducting electron is zero, it is then suspended in the electric field. Reversing the analogy, if the terminal velocity of a falling object in a gravittional field is made to be zero, then the object can be said to be suspended in midair. Maybe this is the subtle point that makes science of aerodynamics possible for the technology of airplanes.

 

[Antonio Lao]

The Earth is known not to have a net effect of a strong electric field except during storms and lightnings but the Earth has a net distribution of a very small magnetic field. The principle behind the working of a magnetic compass. This is supposed to be what got Einstein started in his scientific career. But he, I think, never resolved the mystery of the compass needle.

I am thinking just maybe if we resolve the mystery of the compass needle, we might be able to suspend an object in midair defying the force of gravity. This is pure speculation for now.

 

[Antonio Lao]

This morning, I noted another analogy from GR. The cosmological constant implies a zero drift velocity for the expansion. The more exact analogy is that of a pencil balanced on its point.

If this drift velocity is not zero and also changes with time, then the expansion can accelerate.

 

[Antonio Lao]

By more analogy from electrical conduction, the nonzero drift speed of the universe is the speed of light in vacuum. In a gravitational field, photon take the place of electron in an electric field.

 

[Antonio Lao]

On further speculation could the dimensionless quantity the ratio of electron's drift speed to the speed of light in vacuum be related to other dimensionless quantity such as the fine structure constant.

 

[Antonio Lao]

If the charge to mass ratio of the electron is made to be dimensionless and numerically equal to

$$1.7 \times 10^{11}$$

this is approximately 1/5 of the ratio of the speed of light in vacuum to the electron drift speed.

 

[McQueen]

Antonio

You have gone a long way with me on this , we have argued about how electrical energy is conducted in an electrical conductor , you have during the course of our argument quoted from many books , but in the end had to admit that a consistent theory offering a valid explanation does not seem to exist. What then of my theory ? For instance the theory I have put forward at http://www.geocities.com/natureoflight/id5.html (I quote the URL for the convenience of anyone who might want to look it up ) states that the size or wave-lengths of photons that can be emitted or absorbed by an electron has to be limited to a size of about 10 ^-6 m. and that by thus limiting the maximum size of a photon that an electron could emit , many of the inconsistencies referred to above are solved. That electrical energy is conveyed by photons rather than by electrons would make sense because as I had stated photons are known to convey energy , this would also account for the speed with which an electrical current is established and for the electromagnetic fields which surround a conductor bearing an electrical current. I had also stated that this theory also offers a more acceptable explanation for the phenomenon of magnetism than the present theory of constantly flipping domains. Yet you have ended our discussion in a rather ambiguous way , have you had any new thoughts on the subject , or any comments on my theory.

 

[Antonio Lao]

It so happen that the wavelength of your photon falls within the visible region of the electromagnetic radiation. I am still trying to verify if there is any useable energy from this region.

As for magnetism, the answer to look for is for the question where are the magnetic monopoles.

 

[McQueen]

Antonio
The wave-length of the "conduction" photon is on the border between the visible region and the radio region , a frequency of 1 x 10 ¹⁴ would lie well below the visible wave-length , any frequencies lower than this would be composite wave-lengths. The wave-length of the conduction photon is about 1000Nm while the visible low infrared ends at about 800Nm.

 

[Antonio Lao]

This might be the wavelength used in fiber optics. There are more information density than other means of information transmission. But fiber optics cannot be used to power machineries such as dynamos and generators or household appliances.

 

[McQueen]

Antonio
I feel that you are greatly mistaken . I will get back with a longer post after a bit of preparation.

 

[Antonio Lao]

I will be waiting patiently for your series of replies. Thanks in advance.

 

[McQueen]

Antonio
The situation is not as simple as you seem to imply , my “conduction” photon cannot simply be dismissed as belonging to the visible spectrum. It doesn’t. Let us explore the ways in which it has been possible to determine the wave-lengths of various electromagnetic spectra such as visible light , X-rays and Gamma rays , and radio waves. The method used to determine the wave-lengths of visible light was by use of diffraction grating , similarly to find the wave lengths of X-rays and Gamma rays diffraction through crystals was used and in fact it was this method first though of by Moseley , which is responsible for much of our knowledge of atoms today , lastly for radio waves we have the resonance of a tuned LC circuit. Now consider the problem of a wave length that lies in between the radio wave and the visible light spectrum. You cannot use a diffraction grating to see what the wave-length is for the simple reason that the said frequency does not lie in the visible range , also you cannot use the resonance of an LC circuit for the reason that such LC circuits do not at present exist , lastly you cannot use crystal diffraction for the reason that the concerned wave lengths are longer than the crystal boundaries. So what grounds do you have , taking the above into consideration , to dismiss my hypotheses of a “conduction” photon out of hand. As far as I can see such conditions do not exist , this means that my theory is still valid until proved not to be so.

 

[Antonio Lao]

It's the energy output from the visible region that i am concerned about. This energy, I don't think can be capable of supplying enough electrical energy to run even an ordinary household light bulb of say 25 watts. The light bulb give off the same light wavelength as the input source so to speak. This is just like a perpetual machine that generate something out of nothing. In technology, the input always seem to be larger than the output which makes the efficiency always less than 100%. The heat engine, the electric power plant, etc. The entropy of the system is always increasing.

 

[McQueen]

Antonio

It's the energy output from the visible region that i am concerned about. This energy, I don't think can be capable of supplying enough electrical energy to run even an ordinary household light bulb of say 25 watts. The light bulb give off the same light wavelength as the input source so to speak. This is just like a perpetual machine that generate something out of nothing.

It is precisely the energy of this wave-length that makes the whole New Field Theory possible , if the energy of this wave-length (10^-6) is calculated it is seen to be about 1.214 eV . Now this is where it gets interesting because this energy is close to the energy value of an electron which is naturally 1 eV , this means that whatever energy you are talking about when dealing with currents would also apply to New Field Theory. So I can't really see how you can say that the energy won't be enough. In fact it is the present theory which can't correlate energy in the inductive field with actual induced energy , this is one of the proofs of my theory.

 

[Antonio Lao]

The mass of one electron is about half million electron volt (.5 MeV). To derive energy from this mass, it has to be multiplied by the square of its velocity. if this is the drift velocity then the energy is small (calculation not done).

 

[McQueen]

Antonio
“Electron : An elementary particle which is a constituent of all atoms . Its mass is approx. 9.1 x 10 ^-31 Kg. Its charge is 1.6 x 10 ^–19 C. “
It can be seen from this that the charge of the conduction photon . 1.8 x 10 ^–19 C. approximates very closely to that of the electron itself and would therefore fit in very well with all present calculations for calculating of current , voltage etc.,

 

[Antonio Lao]

I don't know how you derived the charge of the conduction photon. I have been taught in school that photon whether real or virtual never has any electric charge. If photon has electric charge then it will interact with anything that it comes in contact with. Then there will be no electromagnetic radiation of any kind. The whole universe would become pitch black and nothing can be seen. Total darkness. The tendency of nature is always to neutralize the electric charge to minimize the electric potential. This neutralization of electric charge is never more pronounce than in lightnings during the life of electrical storms. This neutralization is responsible for the formation of atoms hence the creation of matter as we know it.

 

[McQueen]

I don't know how you derived the charge of the conduction photon. I have been taught in school that photon whether real or virtual never has any electric charge. If photon has electric charge then it will interact with anything that it comes in contact with. Then there will be no electromagnetic radiation of any kind. The whole universe would become pitch black and nothing can be seen.

You are not mistaken , the photon is electrically neutral , instead of the term “charge “ I should have used the term “energy” . Since the photon is the agent of mediation between electrons , its energy is translated into charge , so what I had written is basically correct. The energy of the photon is derived from the equation $$hc/&lamda$$.

 

[Antonio Lao]

$$\frac {hc}{\lambda}$$

is very similar to what I working on in finding the total energy of the universe. In this form, the energy is purely kinetic.