# Charge cloud/mass cloud

1. Jan 20, 2009

### granpa

regarding probability clouds:
if there is a charge cloud surrounding a particle then could there also be a mass cloud surrounding it? the charge cloud would rotate producing a magnetic moment and the mass cloud would rotate producing angular momentum. both would spin at the same rate but wouldnt necessarily be the same size or shape.

if the charge could separate from the mass then that could explain how an electron can 'fit' into a proton produing a neutron. since the mass doesnt fall into the proton then there would be no need of a neutrino.

it would also explain how the electron can sometimes appear as a point particle and at other times seem to be spread out over the whole atom.

Last edited: Jan 20, 2009
2. Jan 20, 2009

### ZapperZ

Staff Emeritus
Electron = 1/2 spin
Proton = 1/2 spin
Neutron = 1/2 spin.

Let's see you show me how 1/2 +/- 1/2 = 1/2.

Zz.

3. Jan 20, 2009

### granpa

because the neutron would still be surrounded by the spinning mass cloud of the electron. adding that into the equation balances it.

the charge cloud and the mass cloud would always rotate at the same rate. angular momentum is always conserved. as long as the 2 clouds are the same size and shape then the magnetic moment is also conserved. but if the charge cloud separates from (becomes smaller than) the mass cloud then the magnetic moment would not be conserved.

from wikipedia:
We can find a spin magnetic moment for a particle with charge q, mass m, and spin S by calculating:

where g is a dimentionless number, called the g-factor. This number depends on the particle: it is about g = 2 for the electron, g = 5.586 for the proton, and g = − 3.826 for the neutron

magnetic moments of elementary particles:
electron -9284.764
proton +14.106067
neutron -9.66236

and dont forget about 'orbital magnetic moment' which also has to be factored in.

Last edited: Jan 20, 2009
4. Jan 20, 2009

### ZapperZ

Staff Emeritus
That makes absolutely no sense whatsoever.

If you look at the s-orbital of an electron, it has ZERO ANGULAR MOMENTUM, unless you want to contradict quantum mechanics.

Secondly, the spin angular momentum is a conserved quantity as a whole. It means that it doesn't matter where in the composite object it is! Unless you want to argue that an electrons has two different "spin" values, then you are stuck at one major violation of a conservation law.

And this is all before we even talk about the FACT that there is a quark flavor change when there is a transformation between a proton and a neutron.

Besides, why are you doing this? Is there a flaw in quantum mechanics and elementary particle physics that you are trying to challenge? If you are, please do so in the IR forum.

Zz.

5. Jan 20, 2009

### granpa

the s orbital has zero 'orbital magnetic moment' but still has spin.

yes. I said that angular momentum is always conserved. whats the issue?

6. Jan 20, 2009

### ZapperZ

Staff Emeritus
1/2 +/- 1/2 = 1/2?

This is not an issue to you? Or do you think algebra is out of fashion?

Zz.

7. Jan 20, 2009

### granpa

rather than spin, I would prefer to do the math in terms of angular momentum since that is what is always conserved.

whenever the mass cloud and the charge cloud are the same size and shape then the angular momentum will be proportional to the magnetic moment.

as you can see:

magnetic moment isnt simply equal to 'spin'. its more complicated than that.

8. Jan 20, 2009

### ZapperZ

Staff Emeritus
The "spin" quantum number is one of the most tightly conserved quantity there is! The magnetic moment that you are calculating is using a MEASUREMENT to determine many different parameters, which is less exact!

If you are ignoring the non-conservation of spin that is staring at you in your face, then you are claiming that this intrinsic number isn't important. Is this what you wish to do?

I have a feeling you never did any kind of QM before, because no one would dismiss such a thing as easily. I mean, look at, for example, the spin-orbit coupling where J = L + S, is a significant aspect of any undergraduate studies in QM. This is exactly what I'm doing here where the composite particle will have to be reconcilled with the individual components.

This is bordering on crackpottery.

You STILL haven't indicated why you are doing this and why QM isn't more valid than this. Is quark flavor change rather difficult for you to accept?

Zz.

9. Jan 20, 2009

### Hans de Vries

The wavefunction represents a charge/current density AND a magnetic moment density.
(= magnetization). In fact at every point a magnetization/polarization tensor is defined as.

$$M^{\mu\nu} ~=~ \left(\frac{g}{2}\right)\frac{\hbar e}{2m}\left(\frac{1}{2mc}\right)~\mbox{\Huge \bar{\psi}\sigma^{\mu\nu}\psi} ~=~ \left( \begin{array}{cccc} 0 & ~~~P_x & ~~~P_y & ~~~P_z \\ -P_x & 0 & - iM_z & ~~iM_y \\ -P_y & ~~iM_z & 0 & - iM_x \\ -P_z & - iM_y & ~~iM_x & 0 \end{array} \right)$$

Whereby the polarization is zero in the rest-frame and is the result of the Lorentz
transform of the magnetization in a moving frame.

Global rotation of the charge density corresponds with an orbital angular momentum
and an orbital magnetic moment. Spin and orbital angular momentum are two different
things.

Regards, Hans

10. Jan 20, 2009

### granpa

besides calling me a crackpot and telling me I'm wrong you havent actually contributed anything of worth to this thread. I see no reason to converse with you at all.

if spin/mass is equated with the angular momentum of the particle than it is indeed always conserved. while ordinarily angular momentum and magnetic moment are proportional to each other, I am suggesting the hypothesis that maybe they arent always proportional to each other.

Last edited: Jan 20, 2009
11. Jan 20, 2009

### ZapperZ

Staff Emeritus
Questioning why you ignore spin conservation isn't contributing anything? That's a major violation of conservation law and you somehow ignored it.

I pointed out several major flaws with your "model", including the FACT that we know of a quark flavor change. This somehow doesn't even come into your radar...

BTW, that "S" symbol in the magnetic moment that you so dearly love to quote. What do you think that is? What do you think is "S" for a neutron? It is 1/2, which is exactly what I wrote earlier. Where do you think they get this "1/2", especially when the component of the neutron, based on your model, has s=1/2 and 1/2. How do you get 1/2 out of a projection of TWO spin 1/2?

And you don't think this is a problem?

Zz.

12. Jan 20, 2009

### granpa

as I've already told you, the spin of the electron would still reside in its mass cloud even after its charge cloud had fallen into the proton thereby making a neutron. the spin of the neutron would simply be the spin of the proton. fully conserved.

13. Jan 20, 2009

### Hans de Vries

A proton becomes a neutron via a W- boson which turns an up quark into a down quark.
The W- boson can come from an electron which hits the proton.

It's the reverse of the neutron decay described here:
http://hyperphysics.phy-astr.gsu.edu/HBASE/particles/proton.html#c4

Regards, Hans

14. Jan 21, 2009

### malawi_glenn

granpa, before even attempting to do something new, learn the current physics and find out where the anomalies are. A new model/theory must not be able to explain not only the existing data, but also make new predictions and explain the anomalies which the current theory can't account for. Besides this, there are critiria for physical theories - they should be mathematically rigourus. You are just playing with words and intuitive pictures. I mean, define "charge cloud" .. do you understand from where it comes from fully? Do you master QED and Electroweak theory?

15. Jan 21, 2009

### ZapperZ

Staff Emeritus
This has gone far enough.

Zz.