Linear chain: rigid sliding vs a kink
A row of n bare kernels with a row of n electrons alongside, offset above them so they sit outside
every nucleus. Two ways for charge to move, and the question is which is cheaper.
Why a chain and not the ring. The ring gave the pinning barrier cleanly — no ends, every
site equivalent — but no electrostatic bias can drive a sea around a closed loop, since a
conservative field has zero circulation. A chain has ends, so a potential drop along it is an
ordinary thing to apply, and a single kink can enter at one end and leave at the other. On a ring
kinks must come in pairs.
What the ring already settled. Rigid sliding of the whole sea costs 0.01186 Ha for six
electrons — 54 meV each, a threshold field of 9×108 V/m, about
0.11 V per lattice site. Ten orders of magnitude too stiff to be conduction. But that path
makes every electron crest its barrier at once, and its cost grows with the number of participants.
The kink. Electrons left of j0 keep their registry; those to the right are advanced
by one full site spacing, over a width w. That leaves one ion unpartnered — a vacancy in the
electron row, carrying one unit of charge. Only the kink region pays, so the cost is
independent of chain length, and one kink crossing the chain transports exactly one electron.
The two numbers wanted. Creation = E(kink) − E(uniform), whether a kink can exist
at all. Migration = the variation of E with j0 as the kink is stepped along, which
is the Peierls–Nabarro barrier and sets the threshold field for kink motion. If migration is far
below the rigid 54 meV per electron, the chain conducts by kinks at fields the rigid mode could
never reach.
URL: ?n=9 ions, &a=2.2 spacing, &dy=1.2 how far
the electron row sits above the ion row, &kink=0 uniform reference or
1 with a kink, &j0=4 kink centre, &w=1.0 kink width in
sites, &phi=0 rigid registry offset, &bs=0.25,
&reset clears.
relaxing…