OK, let's backtrack. Apparently we are not on the same page yet. In the spirit of KISS and simplicity, the internationally-accepted value of the proton's charge radius is 0.8768 fm. Is there a valid reason to use anything else?
If Vt is going to have any relevance as a general constant, it must apply to hydrogen. Most of the visible universe is hydrogen, so why should we worry about higher Z nuclei for a general quantum theory ... unless, of course, it is to accommodate a strange hypothesis, where only higher values work? ... and as a general observation, recent findings suggest the radius value will be going down, not up. http://articles.latimes.com/2010/jul/07/science/la-sci-proton-20100708 -----Original Message----- From: Craig Haynie Subject: RE: [Vo]:Quantum Transitional State > To justify using such a radius, the inventor must resort to finding an > argument whereby the nucleus can appear to be much larger - voila: string > theory and sound vibrations, which arguably push the radius into size range > where the numbers finally work. Thus everything now seems to mesh, when this > fiction is applied ... if that is - you can suspend disbelief and overlook > the details of the derivation. > > Of course, all of this machination is (arguably) done in support of a > particular constant - which can appear as a velocity, but which essentially > has little validity in actual experiment, when you look a bit closer. If > there was some baseline validity in experiment - then the liberties taken > with the proton radius would be a bit easier to swallow. > I have a couple of issues with the way that Vt is derived. As a non-scientist, it uses 2 values of which I know very little: 1) To use Hooke's Law, it calculates the maximum repulsion between protons, but uses the coulomb barrier as the limit to which protons could be compressed against each other. It uses the value 2 * 1.409e-15 for this. Force = Q^2 / ( 4*pi*eo (2*1.409 x 10-15 )^2 ) = 29.053 Newtons 2) But then within the nucleus, as the spring displacement, it uses 2 * 1.36e-15 meters as the distance between the protons. This leads me to a question: If the protons are spaced at 2 * 1.36e-15 meters in the nucleus, then wouldn't the force between them be greater than that calculated at the coulomb barrier at 2 * 1.409e-15? Or would there be any force at all, considering the influence of the strong nuclear force? Without this same force, why would the spring constant be correct? As I understand it, the reasons that 1.36e-15 m was chosen as the radius of the proton inside the nucleus, are from experiments where they've tried to measure proton separation inside the nucleus. One in particular is referenced for this value: http://tinyurl.com/38wso8c So if they've measured the distance between protons in the nucleus at 1.36e-15 m, then I think this is a logical place to build an equation like Vt. However, with Vt placed only at 4 significant digits, all correlations with it and other physical constants could be just numerology. It might just be a coincidence that it could be used in atomic equations in place of Planck's Constant. But statistically, this likelihood grows smaller as more accuracy is discovered for Vt. So, since there's no physical relationship between Planck's Constant and the 1.36e-15m proton spacing in the nucleus, but there IS a predicted relationship between Planck's Constant and Vt, and definitive relationship between Vt and the 1.36e-15m, THEN why not use Planck's Constant to calculate the effective proton radius in the nucleus, and see if it doesn't hold up upon further analysis. I have no idea how hard it would be to find a better value than 1.36e-15, but if it could be done, then we could test the predictive value of Vt, and demonstrate a relationship between Planck's Constant and the spacing of protons in the nucleus. Otherwise, it just might be a coincidence that Vt seems to have a relationship with Planck's Constant. Craig On Fri, 2010-12-03 at 08:22 -0800, Jones Beene wrote: > -----Original Message----- > From: Craig Haynie > > > by back-calculating Vt, we can then use it to predict the effective radii > > of protons in the nucleus, which is the variable that seems to be the least > > certain. > > Craig, as you no doubt have noticed in this exercise, the proton radius has > been a stumbling block for a certain controversial alternative theory. > > There is a known value for the proton radius, but it does NOT work for the > house-of-cards theory, and therefore a fictional radius has been "invented" > which is much larger than the known value. > > To justify using such a radius, the inventor must resort to finding an > argument whereby the nucleus can appear to be much larger - voila: string > theory and sound vibrations, which arguably push the radius into size range > where the numbers finally work. Thus everything now seems to mesh, when this > fiction is applied ... if that is - you can suspend disbelief and overlook > the details of the derivation. > > Of course, all of this machination is (arguably) done in support of a > particular constant - which can appear as a velocity, but which essentially > has little validity in actual experiment, when you look a bit closer. If > there was some baseline validity in experiment - then the liberties taken > with the proton radius would be a bit easier to swallow. > > Jones > > > >
