The ISU Dialogues, Part Two -- Numerology
Our cups refilled, we got back to
Bogie's Infinite Spongy Universe:
CM: "We're taking a brief detour to discuss Bogie's assertion that the proton contains 699,955,457,517 EEPs and the electron contains 381,239,356 EEPs."
BH: "Why those magic numbers?"
CM: "We'll see in a few moments. I'll quote the relevant passages from post #20:"
Quote:
Relationship #1 is the surface vs. the volume of the proton. Simple math comparing the formula for the surface of a sphere and the formula for the volume of a sphere with these formulas: Surface Area = 4 pi r2; Volume = 4/3 pi r3.
To determine the specific numbers we need the radius.
For relationship #2, we know the relationship between the mass of the electron and the mass of the proton, both consisting entirely of EEPs, albeit at drastically differnet densities of EEPs; 1:1,836.
So we solve for the radius that gives us the same relationship between the surface and the volume of the proton as we get between the mass of the electron and the mass of the proton. That way, since my idea is that the number of EEPs in the electron at rest is equal to the number EEPs on the surface of the proton, and a proton is a sphere in the ISU, then I have the number of EEPS on the surface of a proton.
I did this using the trial and error method, this website, and an Excel spreadsheet and about five minutes. It turns out that a radius of 5,508 units gives us the right relationship, i.e., 1:1,836.
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CM: "No need for a spreadsheet; algebra and fifteen seconds will suffice."
Quote:
Area/Volume = (4 pi r2)/(4/3 pi r3) = 3/r = 1/1836,
therefore r = 3*1836 = 5508.
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DB: "But the units are wrong! The ratio of the masses, 1:1,836, is a dimensionless number, but the ratio of area to volume has dimensions of inverse length!"
CM: "Exactly correct. And that's why this computation is wrong, as so often happens when amateurs finally start putting numbers on their stories. I do offer a way out, though."
DB: "How?"
CM: "Consider a proton to be made up of M EEPs and an electron to be made up of N EEPs. If the EEPs all have equal mass, and binding energy does not enter in, which I realize is a bit of wishful thinking, then M/N = 1836. Let the proton have radius R and the EEP a radius of r. Then if a proton is put together out of M EEPs with no spaces or overlap (possibly another bit of wishful thinking) then M = (4/3 pi R
3)/(4/3 pi r
3) = (R/r)
3. Now let us imagine trying to put N EEPs on the surface of this proton. The area of a great circle of an EEP would be 4 pi r
2, so N of these squeezed together on the surface, again ignoring gaps and overlaps would be N = (4 pi R
2)/(4 pi r
2) = (R/r)
2. So finally 1836 = M/N = (R/r), that is the radius of the proton must be 1836 times the radius of an EEP if the mass of the EEPs on the surface of the proton is to add up to the mass of the electron.
CM: "Knowing that the radius of the proton is about 1.3x10
-15 m then tells us the radius of an EEP. Of course this is still numerology and there is precious little physical content in it."
BH: "Amen!"
CM: "After this sidebar, I'd like to get back to discussing post #2."
To be continued ...