On Quarks and Neutrons
Posted: Wed Feb 20, 2008 1:26 pm
While looking at the description of quarks in the course, I had a look at the description of a proton:
In quarks view: uud
In the RS2 view: 1-1-(1)
Here, 'u' is the 'up' quark, and 'd' is the 'down' quark. Now if the comparison is valid, then we take a look at the 'nuclear' neutron:
In quarks view: udd
In the RS2 view: 1/2-1/2-(1)
Now, what if the the parallel continued, and the nuclear neutron is actually 1-(1)-(1), or equivalently, (1)-(1)-1, or a cosmic proton? So what if the cosmic proton and the proton exist in nuclear motion, with the one turning into the other, inter-converting, so to speak?
The "charge" which the nucleus is supposed to have exists in the "atomic region" of the the RS2 atom, so the cosmic proton would not be taking up any charged characteristics. And the cosmic proton, when it converts into a material proton, can release the cosmic aspect of one of the dimensions as the cosmic positron: electron. This electron then picks up a charge from the atomic region motion, giving the net effect of increasing the 'nuclear' charge. This is probably the mechanism of beta decay. It stands to reason that the proton may also become a cosmic proton by releasing a positron, but the positron being material, would not get charged as the electron did, and is consequently unobservable.
I did some digging thru the recently updated Reciprocity journal, and this is what I found:
Larson, in RECIPROCITY Vol 17 No. 2, page 7 wrote:
These are just tentative thoughts, please post your comments.
In quarks view: uud
In the RS2 view: 1-1-(1)
Here, 'u' is the 'up' quark, and 'd' is the 'down' quark. Now if the comparison is valid, then we take a look at the 'nuclear' neutron:
In quarks view: udd
In the RS2 view: 1/2-1/2-(1)
Now, what if the the parallel continued, and the nuclear neutron is actually 1-(1)-(1), or equivalently, (1)-(1)-1, or a cosmic proton? So what if the cosmic proton and the proton exist in nuclear motion, with the one turning into the other, inter-converting, so to speak?
The "charge" which the nucleus is supposed to have exists in the "atomic region" of the the RS2 atom, so the cosmic proton would not be taking up any charged characteristics. And the cosmic proton, when it converts into a material proton, can release the cosmic aspect of one of the dimensions as the cosmic positron: electron. This electron then picks up a charge from the atomic region motion, giving the net effect of increasing the 'nuclear' charge. This is probably the mechanism of beta decay. It stands to reason that the proton may also become a cosmic proton by releasing a positron, but the positron being material, would not get charged as the electron did, and is consequently unobservable.
I did some digging thru the recently updated Reciprocity journal, and this is what I found:
Larson, in RECIPROCITY Vol 17 No. 2, page 7 wrote:
What I think is happening in the case of quarks is that the geometry being different, the effects of single and solid rotations are seen more clearly. Since the legacy scientists are trying to account for motion which occurs in three dimensions, but do not yet realize that charge is not effective in the 'nuclear' region, it has given rise to charges of -1/3 and +2/3 to the quarks, which is merely a redistribution for a 1d rotation and a 2d solid rotation.If the proton and the cosmic proton lose their charges, which they may do on, or near, contact, they are essentially equivalent, and probably freely interconvertible.
These are just tentative thoughts, please post your comments.