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Topic: rough-deriving rest-masses of 0.5, 938, 105 MeV #1253 New Physics
#1373 ATOM TOTALITY 5th ed

Replies: 4   Last Post: Feb 24, 2013 5:11 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
rough-deriving rest-masses of 0.5, 938, 105 MeV #1253 New Physics
#1373 ATOM TOTALITY 5th ed

Posted: Feb 23, 2013 4:36 PM

Rough-deriving rest-masses of 0.5, 938, and 105 MeV. Of course those
are electron, proton and muon respectively.

What follows is some rough play calculations to see
if I can come at all close to the numbers. And keep in mind that the
calculations are counting of the ridges and troughs of standing waves
that have looped back around creating a standing wave. A photon has no
rest-mass because the wave has no circles or loops in it but is more
or less a straight line wave. A particle that has rest mass has part
of its wave circling around into forming a circle or surface area or
volume.

I am working with just plain spheres and circles but I need to work
with ellipsoids for more precision.

Now I need all three formulas of circumference, surface area and
volume:
(a) C = pi(2r)
(b) S.A. = 4pi(r^2)
(c) V = 4/3(pi)(r^3)

Now I have not found out the characteristics of the wavelength of the
hydrogen atom electron or proton so I am going to assume with a
hypothetical number,
the number 300 as a unit of measure, purely hypothetical to see if 300
gives me 0.5, 938, and 105 MeV.

For the electron we have 300 x 300 = 90,000. And if the radius is
90,000 then pi(2r) is 540,000 and for this hypothetical 540,000 is
close enough to 0.5 MeV

For the muon we have (300)^3 = 27,000,000. So if the radius is 300
then the volume is 4/3(pi)(r^3) so that 4 x 27,000,000 is 108,000,000
which is close enough to 105 MeV for the muon.

Now for the proton we have surface area but the inverse of the
electron so we have (90,000)^2
= 8,100,000,000 divided by 6 from 2pi of circumference since inverse
we have 1,350,000,000
which is not a good estimate of the proton rest-mass
of 938,000,000 eV.

Now if I was working with ellipsoids for the proton I believe I could
easily get 938,000,000 rather than get 1,350,000,000 with spheres. Now
I do recall the physics literature has the tear drop shape for the
nucleus of radioactive elements. The tear drop shape is an example of
ellipsoids. Now I have to investigate whether the tear drop shape has
some exacting numbers or whether it is just a qualitative description
with no quantitative backing.

The reader must keep in mind that leptons are hyperbolic geometry and
the rest mass is the counting up of ridges and troughs of the standing
wave, whereas the baryons of protons and neutrons are elliptic
geometry and the counting up of ridges and troughs of standing waves.

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Archimedes Plutonium
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whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

Date Subject Author
2/23/13 plutonium.archimedes@gmail.com
2/24/13 bacle
2/24/13 bacle
2/24/13 bacle
2/24/13 bacle