When two like charged participles are cooper paired together, do they still have charge? They may not. Their charge may be delocalized and exist at a location that is far distant from the spin part of them.
If they both had the same charge, how could they stick together? A quasi-neutron… just a thought… Cheers: Axil On Thu, May 24, 2012 at 12:57 AM, Axil Axil <janap...@gmail.com> wrote: > *It isn't clear to me why a cooper pair of protons would be of nuclear > dimensions, nor why they would be able to surmount the Coulomb barrier.* > > Essentially, there exists no Coulomb barrier at the point of charge > concentration if that concentration is dense enough. > > These days, I am interested in concentration of electron charge in a > small volume. This is how the Chin reaction works. Rossi’s reaction is > inferior in my opinion as hard to control. > > In the Chin reaction, this negative electric charge concentration on a > nano tube will induce a large number of positive charge holes of equal by > opposite charge. > > Now See > > http://en.wikipedia.org/wiki/Electric-field_screening > > *Electric-field screening* > > > The main point here is that as long as there are many positive ions > between two positive charges; say a proton and a nucleus, their interaction > is *screened* strongly, simply because these many positive charge > carriers can terminate electric field lines. So a free ion attracts ions of > opposite sign, making a little `counter ion cloud' which neutralizes its > charge, and therefore by Gauss's law, basically eliminates the electric > field. > > > The size of this `cloud' is roughly the screening length yD, the parameter > that determines when the exponential `cuts off' the Coulomb interaction in > U(r). A useful formula for yD is due to Debye, which comes from a certain > relatively-easy-to-solve limiting case of interaction of charges with free > ions present where the sum over j is over *all* the ions, and where j > counts the number of ions. As you can see, as you add more and more positve > charges, because the induced charges enter squared, the screening length > goes down, down, down. > > > > See the function for the Debye-Hückel length > > > > where Zj = Qj/C is the integer charge > number<http://en.wikipedia.org/wiki/Charge_number>that relates the charge on > the j-th > ionic species to the elementary > charge<http://en.wikipedia.org/wiki/Elementary_charge> > . > > > > http://en.wikipedia.org/wiki/Debye_length > > > *Debye length******* > ** > > > This formula is often called the *Debye screening length*, and provides a > good first estimate of the distance beyond which Coulomb interactions can > be essentially ignored, as well as the size of the region near a point > charge where opposite-charge counter ions can be found. > > > > > On Wed, May 23, 2012 at 10:08 PM, <mix...@bigpond.com> wrote: > >> In reply to Axil Axil's message of Tue, 22 May 2012 21:44:13 -0400: >> Hi, >> [snip] >> >The cooper pair of protons speculation >> >> It isn't clear to me why a cooper pair of protons would be of nuclear >> dimensions, nor why they would be able to surmount the Coulomb barrier. >> (They only have a reasonable chance of tunneling through it if they get >> close >> enough). >> >> Regards, >> >> Robin van Spaandonk >> >> http://rvanspaa.freehostia.com/project.html >> >> >
