On Tue, Aug 6, 2013 at 3:21 AM, MarkI-ZeroPoint <zeropo...@charter.net> wrote:
> Under normal/usual circumstances, when the atom enters the narrow Casimir > channel it gives up some of its E to the ZPF, and when it exits, it ‘springs > back’ to normal ground-state -- NO overall gain or loss; COE still intact > and we’re not thinking heretically! But, why does it enter the channel? Possibly due to the electrostatic attraction between the partially exposed proton of the H atom due to its electron being in a highly excited state?
