Strong and weak nuclear forces in Meta Model

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20 years 9 months ago #8212 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Astrodelugeologist</i>
<br />How does the Meta Model account for the strong nuclear force and the weak nuclear force?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">MM has not yet been examined that far into QM. It has addressed the nature of light, various paradoxes, and (most recently) the possible nature of protons, electrons, and neutrons. It obviously still has a long way to go. The problem for me is that I don't know the underlying experiments well enough to distinguish required conclusions from theory-dependent conclusions.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I'm especially interested in what the Meta Model has to say about the strong nuclear force being a direct square force instead of an inverse square force.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Although I'm not intending to give a specific answer to your question, MM has no problem in principle coming up with a force of that type. To get together in a nucleus, two protons must first be crushed together with a force stronger than the force of repulsion between them. But once they make contact, their spongy elysium atmospheres would no longer push each other apart, but would envelop the combined mass and keep it squeezed together. It would then take a strong force to break that combination apart to allow elysium to squeeze between them again and reverse the process. The pressure applied by the external elysium would then be proportional to the total number of protons in a nucleus, and therefore to the surface area of all those protons combined. (It would be like the pressure of Earth's atmosphere against Earth's surface, crushing any object whose internal air pressure was less than its external pressure.)

So this could potentially produce an r^2 force because surface area of combined protons is proportional to r^2, provided that the experiments only compare r-values and forces from one nucleus to the next. But such a force would not go as r^2 within any single nucleus as we tried to split it.

I have no idea what the experiments actually show. If you do, please describe it. But this is typical of the problems MM faces. It is a robust model with lots of room to grow, but I for one don't have enough of an in-depth QM background to take it to its limits. -|Tom|-

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20 years 9 months ago #8216 by rousejohnny
Replied by rousejohnny on topic Reply from Johnny Rouse
Tom,

I am not sure any great force is necessary in order for the protons to combine. If the charge is contained within an elysium "bubble" it is not until the two charges make contact that there would be repulsion, but then it would be too late and the two "bubbles" would combine per your description above.

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20 years 9 months ago #8512 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rousejohnny</i>
<br />If the charge is contained within an elysium "bubble" it is not until the two charges make contact that there would be repulsion, but then it would be too late and the two "bubbles" would combine per your description above.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I was referring to the proton model in the 2004 December 15 Meta Research Bulletin. The charge is not contained within anything else. It <i>is</i> the thick, spongy like elysium atmosphere held in place around a proton by gravitons. The more a spring-like sponge is compressed, the more it tries to push back to restore its shape. -|Tom|-

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20 years 9 months ago #8218 by Astrodelugeologist
Replied by Astrodelugeologist on topic Reply from
Although my knowledge of nuclear physics is somewhat amateurish, I've done some reading on the subject.

I did some research, and it turns out that my statement that the strong nuclear force is a direct square force was inaccurate. The force between the quarks within the nucleons is thought to be directly proportional to some function of <i>r</i> (probably something like <i>r^n</i>, with <i>n</i> being a positive number). If it's a direct square force, then <i>n</i> would be 2, but nothing I've been able to find indicates that the value of n has been determined.

The force between the nucleons themselves (called the "residual strong interaction" in the Standard Model) is determined by the Yukawa potential, <i>U(r)=-U<font size="1">0</font id="size1">*(a/r)*e^(-r/a)</i>. <i>U<font size="1">0</font id="size1"></i> is a constant equal to about 4 picojoules (4x10^-12 J); <i>a</i> is a constant equal to about 2.5 femtometers (2.5x10^-15 m). Basically it's an inverse exponential force with a lot of additional constants. It seems that it is not the "real" strong nuclear force, but only a residual "side effect" (mediated by quark-antiquark pairs called mesons) of the strong nuclear force acting between the quarks (mediated by gluons, the gauge bosons of the strong nuclear force). According to the Standard Model, that is.

So, in summary, the strong nuclear force between the quarks seems to be in a direct relationship to some function of distance, while the residual strong interaction between the nucleons seems to be an inverse exponential force.

Regarding your statement that the repulsion between protons is a result of the springyness of the spongy elysium atmosphere, does this mean that elysium follows some analog of Hooke's Law? Or is it more complicated than that?

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20 years 9 months ago #8324 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Astrodelugeologist</i>
<br />Regarding your statement that the repulsion between protons is a result of the springyness of the spongy elysium atmosphere, does this mean that elysium follows some analog of Hooke's Law? Or is it more complicated than that?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Hooke's law would imply a force linear with distance. I would think the actual force law must be more complicated than that. But the model just has not been developed that far, so I'm speculating.

However, the model such as it is suggests that an important distinction must be made between how the force law changes with distance from a single nucleus, versus how the force law changes when more protons and neutrons are added to a nucleus. -|Tom|-

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