The Red Shift Mechanism

[Lazarus]

Strange said:

Close, but no banana.

 

Weight of H atom: 1.007825

Weight of Neutron: 1.008664916

Delta: 0.000839916

H / Delta: 1199.9116578323

N / Delta: 1200.9116578323

P / Delta: 1199.2585768099

 

Using your less accurate numbers gives a result even further from what you claim.

 

And what does "but a better number is .00083985429 AMU" mean?

It almost sounds like you have made it up to give the right answer...

 

None of the ratios appear to be exact or have any physical meaning.

 

 

Reply:

That is the mass of a combined electron and positron.

 

1200 times that number gives the hydrogen mass

the exact 7 digits of accuracy to the number you

use plus 3 more digits.

 

1201 times the number gives the neutron mass

matching your number to 7 digits of accuracy.

 

The probability of pure coincidence is therefore

1 in 10 million. At least it is unlikely.

 

********

 

Strange said:

Please show the math to support this. And show

that it applies to atoms with large numbers of

protons, such as lead and uranium.

 

 

Reply:

For an electron and 2 protons with the protons a distance

of 1 from the electron the attraction between the electron

and a proton is 1. The proton to proton repulsion is .25.

So 1 > .25

 

For 3 protons the attraction between the electron and a

proton is 1. For the repulsion of a proton from another

proton the calculation is:

Half the p to p distance is d/2 = sqr root of 3 divided by 2.

The repulsive force is = 1/3.

The force from the other 2 protons is 2/3 times cos(30).

Net repulsive force on a proton is 1 / 1.732.

So 1 > 1/1.732

 

The arithmetic for 4 protons is a little work.

 

********

 

The 2D models are easy to construct. Models

can be constucted in 3D but it is necessary

to use the proton consisting of electrons and

positrons to build most of them. Greater

attractive force is available..

 

Here are some heavier atoms.

 

Helium 3 Helium 4

alpha particle

 

p p p

 

e e e

(or)

p p p p p

 

e e

 

p p p

 

 

Z=2 A=3 Z=2 A=4 Z=2 A=4

 

 

 

Sodium 23 Sodium 24

 

p

e p

p p p p e

e e p p p p

p p e e

p e e p p p

p p e e

epe p p p p p p

p p e e e

p e e p p p p

p p e e p

e e p p p p p p

p p p p e e

e p p

p

 

Z=11 A=23 Z=11 A=24

 

 

 

Platinum 198

 

 

P P P P

E E E E

P P P P P P P P P P P P

E E E E E E

P P P P P P

E E E E E E

P P P P P P P P P P P P P P P P

E E E E E E E E

P P P P P P P P P P

E E E E E E E E E E

P P P P P P P P P P P P P P P P P P P P

E E E E E E E E E

P P P P P P P P P

E E E E E E E E E

P P P P P P P P P P P P P P P P P P

E E E E E E E E

P P P P P P P P

E E E E E E E E

P P P P P P P P P P P P P P P P P P

E E E E E E E E E

P P P P P P P P P

E E E E E E E E E

P P P P P P P P P P P P P P P P P P P P

E E E E E E E E E E

P P P P P P P P P P

E E E E E E E E

P P P P P P P P P P P P P P P P

E E E E E E

P P P P P P

E E E E E E

P P P P P P P P P P P P

E E E E

P P P P

 

Z=78 A=198

 

********

 

Strange said:

Here's one: you're wrong.

 

Reply:

If you can't give a reason just invoke Authority.

 

I really do appreciate your comments.

[swansont]

Swansont Said:

 

You're jumping the gun (in many ways). A

semi-classical model with elliptical orbits

would have to be tested first, before you

can apply it and claim it to be responsible

for a redshift. For starters, explain how

you get an S state in hydrogen, which has

zero orbital angular momentum, with this model.

 

Reply:

Let's rotate the electron enough to zero the

rotational momentum. We could name it

spin.

No, that doesn't work. We know the orbital angular momentum of the S state is zero. In addition to (or separate from) the electron spin.

 

********

 

Swansont said:

"Heisenberg Uncertainty Principle"

 

Reply:

I am somewhat in agreement with Bohr's

"Don't tell God what to do" but I agree

with Einstein's contention which is

something like, "The cause and

effect should not be thrown out. The lack

of knowledge or ability to measure is

behind the need for probability."

I take it that this non-response response means you have no justification.

[Lazarus]

Swansont said:

No, that doesn't work. We know the orbital angular momentum of the S state is zero. In addition to (or separate from) the electron spin.

Reply:
I am not sure that applies to this model. If this model

generates valid Red Shift values it may not matter.

 

Swansont said:
I take it that this non-response response means you have no justification.

 

Reply:

There are a great many excellent and very usefull algorithms

that incorporate the Heisenberg concept but being able to

calculate the probability of something doesn't mean that the

real mechanism can't exist.

 

I am still fretting about non radiating electron orbits. That seems

to be a solid argument that has to be answered.

[Strange]

That is the mass of a combined electron and positron.

 

Is that a reply to "And what does "but a better number is .00083985429 AMU" mean?

 

If so, it is wrong. The sum of the electron and positron masses is 0.0010971598 (based on the Particle Data Group figures).

 

 

Here are some heavier atoms.

 

Pretty pictures but without the math, I don't believe it works.

[Lazarus]

 

Is that a reply to "And what does "but a better number is .00083985429 AMU" mean?

 

If so, it is wrong. The sum of the electron and positron masses is 0.0010971598 (based on the Particle Data Group figures).

 

 

Pretty pictures but without the math, I don't believe it works.

 

Strange said:

Is that a reply to "And what does "but a better number is .00083985429 AMU" mean?

 

If so, it is wrong. The sum of the electron and positron masses is 0.0010971598 (based on the Particle Data Group figures).

 

Reply:

One possible explaination is that puting a electron and a positon

together could result is mass loss. That is rather than annihilating

each other.

 

Strange said:

Pretty pictures but without the math, I don't believe it works.

 

Reply:

It is easy to see that the light elements could work. The heavier

elements would not work in this cofiguration. By allowing the

protons consisting of electrons and positons share some of

them a stronger bond can be made. Also, that makes it easier

to constuct 3D models.

[Strange]

One possible explaination is that puting a electron and a positon

together could result is mass loss. That is rather than annihilating

each other.

 

Any evidence for that? Or just another ad-hoc guess to shore up a failed idea?

 

 

It is easy to see that the light elements could work. The heavier

elements would not work in this cofiguration. By allowing the

protons consisting of electrons and positons share some of

them a stronger bond can be made. Also, that makes it easier

to constuct 3D models.

 

Without the math, I don't believe it works.

[swansont]

Swansont said:

 

No, that doesn't work. We know the orbital angular momentum of the S state is zero. In addition to (or separate from) the electron spin.

 

Reply:

I am not sure that applies to this model. If this model

generates valid Red Shift values it may not matter.

It applies to reality; your model has to agree with nature. The ground-state of Hydrogen has no orbital angular momentum. Any model that does not agree is wrong.

[Lazarus]

Strange said:

And what does "but a better number is .00083985429 AMU" mean?

 

 

 

Reply:

The number .00083985428 divides the mass of both the proton and neutron evenly to 7 significant digits implying a "building block" for both of them.

 

Do you have a better explanation for that or do you just prefer the 1 in 10 million chance that it is pure coincidence.

 

Also, it pure coincidence that the arithmetic of an electron orbiting a nucleus can show the speed of galaxies. In addition, it is pure coincidence that the synchronizing of the movement of electrons in the Bohr atom shows that in elliptical orbits the electrons will take turns approaching the nucleus.

 

 

 

*****************************************************

 

 

 

Strange said:

Pretty pictures but without the math, I don't believe it works.

 

 

Reply:

I wrote a couple of programs to show the mathematics involved

just for you. One does the positioning of the electrons and

protons and the other does the force calculations. A picture of

the model of the nucleus is followed by the results of the

calculations. The programs are at the end of the post.

 

 

 

 

 

Net force vector (nfx,nfy) on proton number 4

Atomic number is 3 Proton 4 at x= 1.000 y= 1.732

Plus x is repulsive, minus x is attractive

 

Number x y d f nfx nfy q

1 1.000 0.000 1.732 0.333 -0.000 0.333 1

2 -0.500 0.866 1.732 0.333 0.289 0.500 1

3 -0.500 -0.866 3.000 0.111 0.344 0.596 1

5 -2.000 -0.000 3.464 0.083 0.416 0.638 1

6 1.000 -1.732 3.464 0.083 0.416 0.721 1

7 0.500 0.866 1.000 1.000 -0.084 -0.145-1

8 -1.000 -0.000 2.646 0.143 -0.192 -0.238-1

9 0.500 -0.866 2.646 0.143 -0.219 -0.379-1

 

Attractive force electron to proton -1.000

Total repulsive force on proton -0.437

 

 

 

 

 

 

Net force vector (nfx,nfy) on proton number 7

Atomic number is 6 Proton 7 at x= 1.500 y= 0.866

Plus x is repulsive, minus x is attractive

 

Number x y d f nfx nfy q

1 1.000 0.000 1.000 1.000 0.500 0.866 1

2 0.500 0.866 1.000 1.000 1.500 0.866 1

3 -0.500 0.866 2.000 0.250 1.750 0.866 1

4 -1.000 -0.000 2.646 0.143 1.885 0.913 1

5 -0.500 -0.866 2.646 0.143 1.993 1.006 1

6 0.500 -0.866 2.000 0.250 2.118 1.223 1

8 -0.000 1.732 1.732 0.333 2.407 1.056 1

9 -1.500 0.866 3.000 0.111 2.518 1.056 1

10 -1.500 -0.866 3.464 0.083 2.590 1.098 1

11 0.000 -1.732 3.000 0.111 2.646 1.194 1

12 1.500 -0.866 1.732 0.333 2.646 1.527 1

13 1.000 0.577 0.577 3.000 0.047 0.027-1

14 -0.000 1.155 1.528 0.429 -0.374 0.108-1

15 -1.000 0.577 2.517 0.158 -0.530 0.090-1

16 -1.000 -0.577 2.887 0.120 -0.634 0.030-1

17 0.000 -1.155 2.517 0.158 -0.728 -0.097-1

18 1.000 -0.577 1.527 0.429 -0.869 -0.502-1

 

Attractive force electron to proton -3.000

Total repulsive force on proton -1.003

 

 

 

 

 

 

Net force vector (nfx,nfy) on proton number 93

Atomic number is 92 Proton 93 at x= 1.058 y= 0.036

Plus x is repulsive, minus x is attractive

 

Number x y d f nfx nfy q

1 1.000 0.000 0.068 214.487 181.980 113.525 1

2 0.998 0.068 0.068 214.487 371.283 12.684 1

3 0.991 0.136 0.121 68.835 409.688 -44.442 1

 

91 0.991 -0.136 0.185 29.234 489.730 -41.700 1

92 0.998 -0.068 0.121 68.850 524.158 17.925 1

94 1.053 0.108 0.072 191.408 537.220 -173.037 1

 

274 1.004 -0.173 0.216 21.424 -215.954 166.582-1

275 1.014 -0.104 0.147 46.198 -229.816 122.512-1

276 1.019 -0.035 0.081 151.935 -303.600 -10.304-1

 

Attractive force electron to proton -643.469

Total repulsive force on proton -303.775

 

 

*****************************************************

 

Programs

 

' calcbnuc.bas

defdbl a-z

 

' Calc positions of electrons and protons in a nucleus

' Circular models only

'

' Input Atomic number for different atoms

 

' Formlas

 

' r=1 Radius of inner circle of atom

' x=r*cos(theta)' Polar to orthogonal location of proton or electron

' y=r*sin(theta)

' Trig functions

 

' An example of circular models

 

' beryllium 8

' p p p

' e e

' p p

' e e

' p p p

'

 

 

' Numbered

 

'

' 6 2 5

' 10 9

' 3 1

' 11 12

' 7 4 8

'

 

open "calcnuc.txt" for output as#8

open "junk" for output as#9

 

print"What is the atomic number of the atom?"

input z

z=int(z)

atomno=z

if z<3 then

print" The number can't be less than 3"

input a$

goto c9999' Outta here

end if

 

howmany=atomno' Number of protons in inner circle

pi=3.1416' Pi

twopi=pi*2

r=1' Set radius of inner circle to 1

 

anglebig=twopi/howmany' Angle between protons in inner circle

angsmall=anglebig/2' Angle between every particle

 

 

ri=1' ri is radius of inner circle of protons = 1

 

' Calculate radiuso, radius of outer circle of protons

 

x1=1' Coordinates of proton number 1

y1=0

 

x2=r*cos(theta+anglebig)' Coordinates of proton number 2

y2=r*sin(theta+anglebig)

 

xc=r*cos(theta+angsmall)' Coordinates of point on circle to proton 2

yc=r*sin(theta+angsmall)

 

xm=(x1+x2)/2' Mid point between protons 1 and 2

ym=(y1+y2)/2

 

distp1p2=sqr((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))' Distance p1 to p2

 

distmpo=distp1p2*.866'Dist mid to outer.Equilateral triangle

 

rp=distmpo/3' Radius of particles. Diagonal is 2/3 of angle bisector

 

distcl=sqr((xc-xm)*(xc-xm)+(yc-ym)*(yc-ym))'Distance circle vs line

 

ro=1+distmpo-distcl' Radius of outer circle

' Calculate radiusm, radius of middle circle of electrons

 

rm=ro-rp*2' Radius of middle circle

 

theta=0' Angle to particle. A proton is on x axis

 

print#9,

print#9," Inner protons"

print#9,

 

for i=1 to howmany' Do the inner protons

 

x=ri*cos(theta)' Convert from polar to rectangular cordinates

y=ri*sin(theta)

print#8,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle

print#9,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle

 

theta=theta+anglebig' Increment around circle

 

next i

 

print#9,

print#9," Outer protons"

print#9,

 

 

theta=angsmall

 

for i=1 to howmany' Do the outer protons

 

 

x=ro*cos(theta)' Convert from polar to rectangular cordinates

y=ro*sin(theta)

print#8,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle

print#9,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle

 

theta=theta+anglebig' Increment around circle

 

next i

 

print#9,

print#9," Electron"

print#9,

 

 

theta=angsmall

 

for i=1 to howmany' Do the inner electrons

 

 

x=rm*cos(theta)' Convert from polar to rectangular cordinates

y=rm*sin(theta)

print#8,x;",";y;",";-1;atomno;rp'x, y, sign and radius of particle

print#9,x;",";y;",";-1;atomno;rp'x, y, sign and radius of particle

 

theta=theta+anglebig' Increment around circle

 

next i

 

c9999:' Done

 

 

********************************************************************

 

 

' calcrepu.bas

 

' Calculate the force on a vunerable proton in the nucleus

 

open "calcnuc.txt" for input as#1

open "calcrepu.txt" for output as#8

 

dim xx(500)' The electrons and protons

dim yy(500)

dim charge(500)

dim theta(500)

 

ct=0' Count of particles

 

c12:

if eof(1) goto c19' Read in the electrons and protons

ct=ct+1

input#1,xx(ct),yy(ct),charge(ct),atomno,radius

goto C12

 

' Calculate the net force on a proton

 

c19:

particle=ct/3+1' Choose a proton in the outer circle to test

atomno=ct/3' Atomic number of atom

 

print#8," Net force vector (nfx,nfy) on proton number";

print#8,particle

print#8," Atomic number is";atomno;

print#8," Proton";

print#8,particle;

print#8,"at x=";

print#8,using"##.###";xx(particle);

print#8," y=";

print#8,using"##.###";yy(particle)'

print#8," Plus x is repulsive, minus x is attractive"

print#8,

Print#8,"Number x y d f nfx nfy q"

 

gosub calcnet' Calculate net force vector. + is out, - is in.

 

repulsef=sqr(nfx*nfx+nfy*nfy)' Calculate repulsive force

if nfx<0 then repulsef=-repulsef' Negative if attractive

 

print#8,

print#8," Attractive force electron to proton ";using"#####.###";-maxforce

print#8," Total repulsive force on proton ";using"#####.###";repulsef

 

goto c8' Done

 

'-----------------------------------------------------

 

' calcnet subroutine

 

 

calcnet:

 

x=xx(particle)' Remember which proton to test

y=yy(particle)

maxforce=0' Remember maximum e to p force

 

 

nfx=0' Net force vector for the proton

nfy=0

 

for i=1 to ct' Loop through all the particles

if i=particle goto c25' Skip ourself

 

if charge(i)=1 then

dx=x-xx(i)' Distance vector from a proton

dy=y-yy(i)

end if

 

if charge(i)<>1 then

dx=xx(i)-x' Distance vector from an electron

dy=yy(i)-y

end if

 

d=sqr(dx*dx + dy*dy)' Distance

 

ux=dx/d' Unit vector

uy=dy/d

 

f=1/(d*d)' Force = inverse of distance squared

if charge(i)=-1 then

if maxforce<f then maxforce=f' Remember maximun e to p force

end if

 

fx=ux*f' Unit vector times force

fy=uy*f

 

nfx=nfx+fx' Add to net force

nfy=nfy+fy

print#8,using"###";i;

print#8,using"###.###";xx(i);'Print the coordinates, distance and

print#8,using"###.###";yy(i);' force of this proton

print#8,using"###.###";d;

print#8,using"#####.###";f;

print#8,using"#####.###";nfx;

print#8,using"#####.###";nfy;

print#8,charge(i)

 

c25:

next i

return

 

c8:

c9999:

Sorry, pictures didn't take.

Edited August 13, 2013 by Lazarus

[swansont]

There is no "positioning of electrons" without a valid model that has electron positions, and you haven't presented one. The Bohr model is demonstrably wrong — anything based on its mechanics is likewise wrong. There is no point in going past that to analyze any model when the premise is flawed. Any conclusion that might be right is purely accidental.

 

Further, the Bohr atom doesn't have elliptical orbits! They are circular (one part of its wrongness, in many ways, for many reasons).

 

Any analysis past this point of not having a valid model is a wasted effort.

[Strange]

The number .00083985428 divides the mass of both the proton and neutron evenly to 7 significant digits implying a "building block" for both of them.

 

Except it doesn't. Dividing the neutron and proton masses by this gives 1200.999897256 and 1199.3466947701 respectively.

 

 

Also, it pure coincidence that the arithmetic of an electron orbiting a nucleus can show the speed of galaxies.

 

As you have a fake model and arbitrary numbers, no, not really. You can "prove" anything with numerology.

 

 

I wrote a couple of programs to show the mathematics involved

just for you.

Well, if you can't present the math ... I am not going to try and reverse engineer your code.

Edited August 13, 2013 by Strange

[Lazarus]

Strange said:

Well, if you can't present the math ... I am not going to try and reverse engineer your code

 

The test proton has the greatest x value

and is at coordinates (a,b).

 

q is the charge of a particle. (1 or -1)

 

d is the distance from the test particle.

 

d = sqr((a-x)^2 + (b-y)^2)

 

The force vector on the test particle

from another particle is:

 

q(a-x)i q(b-y)j

F = ------- + -------

d^3 d^3

 

The sum of all the force vectors is:

gi + hj

 

The net force, f is;

 

f = sqr(g^2 + h^2)

 

The attraction or repulsion is dependent on

the sign of g.

Swansont said:

Further, the Bohr atom doesn't have elliptical orbits! They are circular (one part of its wrongness, in many ways, for many reasons).

Any analysis past this point of not having a valid model is a wasted effort.

 

Reply:

 

This model is not the Bohr Model, the Bohr-Summerfield Model, the Bohr-Kramers-Slater Model nor the Rydenberg Model. The biggest similarity is that attempts are being made to describe the actual reality that underlies the probability waves. These people and others made a lot of progress accounting for split lines, fine lines and other relationships. If they had known what we now know that it is possible for nuclei to have rotating magnetic fields, which allows distinct orbits, things might have gone in a different direction. Elliptical orbits were considered long ago.

 

The only substantial objection to this model is your point that an electron on a curved path should radiate. Not only is the momentum changing but the kinetic energy of the electron would vary in the orbit.

 

One possible explanation is that the photon is so much bigger than a atom that an electron could make one or more complete circuits while a photon was coming or going. That could allow the kinetic energy the photon leaving in one part of the orbit to be reacquired on the opposite side. As for the momentum, the photon should leave in opposite directions on opposite sides of the ellipse, which might cause them to cancel each other.

[Mellinia]

The test proton has the greatest x value

and is at coordinates (a,b).

 

what is x?

[Lazarus]

 

what is x?

 

x is the distance from the origin of a 2 dimentional coordiate system.

 

I really don't understand why I am getting so many complaints about

the arithmetic. It is just high school math.

 

I am not complaining, I appreciate any comment.

[Klaynos]

I really don't understand why I am getting so many complaints about

the arithmetic.

It's because of unconventional formatting I , look at any text book they all do equation followed by where x is the position in metres etc...

 

Also the maths is where the predictions are where you can tell if an idea agrees with reality our not.

[swansont]

This model is not the Bohr Model, the Bohr-Summerfield Model, the Bohr-Kramers-Slater Model nor the Rydenberg Model.

You referred to it as an "elliptical Bohr atom". If it's not a Bohr atom, maybe you should stop doing that.

 

 

The biggest similarity is that attempts are being made to describe the actual reality that underlies the probability waves. These people and others made a lot of progress accounting for split lines, fine lines and other relationships. If they had known what we now know that it is possible for nuclei to have rotating magnetic fields, which allows distinct orbits, things might have gone in a different direction. Elliptical orbits were considered long ago.

 

Considered? Sure. People considered phlogiston for heat, too. Were these actually adopted as a working model? Why is that?

 

 

The only substantial objection to this model is your point that an electron on a curved path should radiate. Not only is the momentum changing but the kinetic energy of the electron would vary in the orbit.

That's one objection. Angular momentum is another. The larger issue is that you haven't presented an actual model, with comparison to experiment. There may be many other objections beyond these. It's hard to be specific when all there is is hand-waving.

[Mellinia]

 

x is the distance from the origin of a 2 dimentional coordiate system.

 

I really don't understand why I am getting so many complaints about

the arithmetic. It is just high school math.

 

I am not complaining, I appreciate any comment.

 

Then why is there a maximum x for a test photon?

 

You have not explained your maths with physic concepts.

[swansont]

The net force, f is;

 

f = sqr(g^2 + h^2)

 

The attraction or repulsion is dependent on

the sign of g.

You're squaring g, so how does the sign matter?

[Lazarus]

 

Then why is there a maximum x for a test photon?

 

Reply:

The most vunerable protons for expulsion are at the

outer edge of the nucleus.

 

*****************************************************************************************

You have not explained your maths with physic concepts

 

 

Reply:

Place the center of the nucleus at (0,0) of the coordinate system,

Then rotate the nucleus until one of the protons furthest from the

origin is on the positive x axis. That proton is the test proton.

If the net force vector is positive the proton will be expelled, if it is

negative the proton will be held in place. The calculation of the

force vector between two particles is straight forward. Then

you have to add up all the force vectors to find the net force

and the direction of the force.

 

.

You're squaring g, so how does the sign matter?

 

Reply:

g is the component of the net force vector

along the x axis. If g is positive the proton

will be expelled. if negative ithe proton will

stay in place.

Correction to the reply to Mellinia

 

Reply:

Place the center of the nucleus at (0,0) of the coordinate system,

 

---------------------------------------------------------------------------------------------------------

Wrong

Then rotate the nucleus until one of the protons furthest from the

origin is on the positive x axis. That proton is the test proton.

 

Should read:

Then rotate the nucleus until one of the protons closest to the

origin is on the positive x axis. The proton with the smallest

angle from it in the counter clockwise direction is the test proton.

---------------------------------------------------------------------------------------------------------------

 

If the net force vector on the test proton is positive the proton

will be expelled, if it is negative the proton will be held in place.

The calculation of the force vector between two particles is

straight forward. Then you have to add up all the force vectors

to find the net force and the direction of the force.

[imatfaal]

...

c12pict.JPG

 

 

Carbon - with 12 protons and 6 electrons? Perhaps check your very basic chemistry - very positively charged and very not Carbon

[Lazarus]

 

 

Carbon - with 12 protons and 6 electrons? Perhaps check your very basic chemistry - very positively charged and very not Carbon

That is the nucleus, not the atom.

[swansont]

Surely a nucleus is three-dimensional, and the lowest energy state would not be a flat ring structure.

[imatfaal]

That is the nucleus, not the atom.

 

You are trying to re-write physics and you seriously think carbon has 12, XII, twelve, TWELVE protons!!!

 

http://www.webelements.com/carbon/

 

This is first year school chemistry/physics H, He, Li, Be, B, C...

[swansont]

 

You are trying to re-write physics and you seriously think carbon has 12, XII, twelve, TWELVE protons!!!

 

http://www.webelements.com/carbon/

 

This is first year school chemistry/physics H, He, Li, Be, B, C...

 

The claim appears to be that the 6 neutrons are really 6 protons + 6 electrons, but now we have a problem with e.g. C-14 which would have 14 protons and 8 electrons, and undergoes beta decay. You have the electron emitted from the nucleus, but now have to explain where the antineutrino came from. Worse, you have neutron-deficient nuclei which undergo beta-plus decay.

 

Like most of the "model" we have seen, it looks only at the first layer of the issue and ignore the complex layers below them which mainstream science has hashed out through the years; investigations and experiments which led to similar proposals being discarded when failed to predict or disagreed with observation.

[Lazarus]

Surely a nucleus is three-dimensional, and the lowest energy state would not be a flat ring structure.

 

Agreed. The ring structure just demonstrates that electrons

can bind multiple protons and it is easier to visualize.

 

There is something else that had to be resolved to make more

satisfactory models. Three protons attached to an electron is

insufficient to build good 3D models and 4 protons attached to

an electron is too weak of a bond.

 

The solution is indicated by the loss of nominal mass as we go

up the atomic numbers and the fact that the difference between

the mass of the neutron and the hydrogen atom is evenly

divisable to 7 digits of accuracy into both indicating a building

block which could consist of an electron and a positron.. That

would imply that the joining of an eledtron and a positron could

resust in some mass loss as apposed to annialating each other.

The reduced mass of bigger nuclei can result form sharing the

building blocks resulting in a stronger bond between protons.

 

That allows compact models to be constructed.

 

Another problem that had to be solved. The radius of an electron

used to be considered more than twice the radius of the proton.

Now the electron is thought bo be smaller than that. That helps

to let the protons overlap.

[swansont]

Agreed. The ring structure just demonstrates that electrons

can bind multiple protons and it is easier to visualize.

A real structure would be important, because from that you can calculate if the nucleus has any multipole moments and compare that to experiment. But the bigger issue is how this fits in with decay, as I have mentioned, or the complete contradiction with the Heisenberg Uncertainty Principle I think I brought up before.