In reply to Bob Higgins's message of Sun, 10 Jan 2016 10:51:47 -0700:
Hi,
This message will only make sense if viewed with a fixed width font.
[snip]
>What Holmlid proposes is that planar hexagonal Rydberg clusters of deuterium
>can form stacks where the inter-nucleus spacing in the stack can be 2.3 pm.
>The hexagonal Rydberg clusters are essentially planar with an inter-nucleus
>spacing that is bigger than D2 gas. So, in one dimension, along the column of
>the stack, Holmlid claims that the inter-nucleus spacing is 2.3 pm, while in
>the other 2 dimensions the inter-nucleus spacing is 100x bigger. From a
>density standpoint, this would be a set of linear strings. How do you ascribe
>density to something that is a linear string? It would certainly be a tensor.
[snip]
I was going to write:-
What makes me highly skeptical of the claim is that I see no way to get two
deuterons (or protons for that matter), within 2.3 pm of one another while the
electrons are hundreds of pm away.
...when it occurred to me that the columns might interleave, such that the
electrons from one layer came between the nuclei from the layers above and
below. The spacing between layers would then be half of 2.3 pm.
Imagine pushing two parallel "cylinders" into one another until the wall of each
reached the axis of the other, with the layers of each "cylinder" interleaving
with those of the other.)
A1 A2
E N E
E N E
E N E
E N E
E N E
E N E
Each E N E layer is actually a single atom where the two E's represent a single
electron in a circular orbit. N stands for nucleus. A1 is the axis of the first
vertical cylinder. A2 is the axis of the second vertical cylinder.
I wonder if coincidentally(?) the vertical separation distance is the fine
structure constant times the radius??
Regards,
Robin van Spaandonk
http://rvanspaa.freehostia.com/project.html
