On Fri, 2010-12-03 at 09:23 -0800, Jones Beene wrote: > 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
Good point, but what they found in that study from 1987, is that the density of all nuclei are the same. So if Vt applies, then the value to use might be 1.36e-15 for all nuclei -- unless you wanted to make an exception for hydrogen. However, it looks to me as if they are calculating the value of 1.36e-15 as the effective proton radius, using Planck's Constant. http://tinyurl.com/345cnr9 If anyone wants to help me read it, scroll down to "Microscopic analysis of nucleus-nucleus elastic scattering at intermediate energies", and open the PDF. Search for 1.36. So this means that Planck's Constant can't be derived from Vt if Vt was derived from Planck's Constant. Craig
