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19 years 3 weeks ago #13646 by rodschmidt
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by ig.farne</i>
Why do not consider the atom with a star structure with rays (true, no virtual) going to infinite and not with electrons? in this way the atom is expanding to infinite. <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In quantum mechanics, the electrons DO go to infinity.

The most likely position for the electron is right on top of the nucleus; the probability of it being anywhere declines (exponentially) with distance, but never goes to zero. So each atom is infinitely large.

By the way, the most likely distance between the electron and the nucleus is a certain radius (details vary depending on which chemical element and which electron we're talking abou) on the order of an angstrom. At the same time, the most likely position for the electron is right on top of the nucleus. How can these both be true? Divide space up into a cubic grid. Only one cube contains the nucleus. There are many cubes at the distance r. So even though the cube containing the nucleus has a higher probability of containing the electron than any of the cubes at distance r, the sheer number of cubes at distance r is enough to compensate for the lower probability of finding the electron in any one of those cubes. So it's more probable to find the electron at distance r than at distance zero; at the same time, it's more probable to find the electron in the position at the nucleus than at any other position.

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19 years 3 weeks ago #13653 by rodschmidt
Replied by rodschmidt on topic Reply from Rod Schmidt
Ignazio,

Quantum mechanics says that an electron does not simultaneously have an exact position and an exact momentum. The uncertainties in position times the uncertainty in momentum is (at least) hbar. So you could prepare an electron in a state that has a precise momentum but an uncertain location, or a precise location but an uncertain momentum.

There is more to it than that. The dimensions x,y,z are independent, so that the uncertainty in x-momentum times the uncertainty in the x-dimension is hbar; simultaneously the uncertainty in y-momentum times the uncertainty in the y-dimension is hbar, and the same for z. BUT: these are independent, so you could prepare an electron with a precise x-location and a precise y-momentum.

In addition, the same relationship holds between energy and time. Relativity tells us that time is like a dimension of space -- space and time are the same kind of thing, they are interchangeable -- so we can think of energy as a kind of momentum. So we have four coordinates, and four conserved quantities, and each coordinate is "conjugate to" a conserved quantity. (Incidentally, you could say the same about angular position and angular momentum, or you could just derive that fact from xyz.) It all fits together very neatly.

I phrased it "electrons do not simultaneously have an exact position and an exact momentum." Textbooks don't say it this way, they are more conservative and say they cannot simultaneously be measured. This is because of the history of quantum mechanics and how it was developed.

So if you're telling me that quantum mechanics is wrong on this point, and that electrons really do have a precise location and momentum, you have a HUGE burden to overcome, because this is very well established. It seems that maybe you dislike uncertainty. Then you need to work on your likes and dislikes, instead of trying to change the theory.

Circuit theory -- voltage, current, capacitance, inductance, resistance -- can certainly be applied at atomic scales, as long as you are careful. I cannot say that circuit theory should be applied "INSTEAD of" quantum mechanics, this doesn't make sense to me. The two theories should be consistent. If they are inconsistent then circuit theory needs to be adapted to fit quantum mechanics, because QM describes small things and circuit theory was developed to describe large things.

I looked over your DNTH webpage but I did not see where it explains what your theory is.

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18 years 11 months ago #14274 by Larry Burford
Ignazio,

You might have more luck getting someone here to comment on your ideas if you were to learn enough about the primary theories (*) to which this Website is dedicated to be able to point out some place where our stuff doesn't work but yours does.

IOW, say something that might be interesting *to us*.

Regards,
LB

(*) Meta Model, Le Sagian (pushing) gravity and exploded planet hypothesis

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18 years 8 months ago #17290 by ig.farne
Replied by ig.farne on topic Reply from ignazio farne
According to DNTH theory the gravitational and electrostatic fields are simply the effect of spin of the nucleons, with one spin's frequency for the electrostatic field and another spin's frequency for the gravitational field. The spin's frequency of the gravitational field is an integer multiple of spin's frequency of the electrostatic field. Our experiments of our laboratories are confirming that. I have translated my (good) italian into my poor english. Ignazio

ignazio farne

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18 years 7 months ago #17237 by ig.farne
Replied by ig.farne on topic Reply from ignazio farne
At my Web site: www.dnth.org /Italiano/Ricerche scientifiche/Documenti/Strutture atomiche e molecolari/Atomo stellare dell'Elio-4 (immagine GIF), it is now possible to see the image-CAD of final structure (for the DNTH Theory) of nucleus of the atom of Helium-4. The spin of the nucleons are not shown. Any comment would be appreciated. Thanks.

ignazio farne

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