Moulding the mechanical potential to more closely match the entropic potential
The entropic potential contributed by a single particle
in a tube is
TS = kT log(x).
Two independent particles in a tube contribute
TS = 2 kT log(x).
Two independent particles in a crack whose volume increases as x2 contribute
TS = 2 kT log(x2) = 4 kT log(x).
If such particles push against a constant force (corresponding to a potential U(x)=Fx)
then most of the entropic free energy will be irreversibly lost
because the entropic force exerted at small x is far bigger than the load force F.
In order to extract more work, we need to make the expansion more nearly reversible.
What's required is a mechanical potential U(x) that looks similar to TS(x).
Such potentials are easy to come by in molecular biology.
A pair of charged groups with separation x contribute a potential
U(x)=-A/x.
A pair of dipoles contribute a potential
U(x)=-B/(4x^4).
I think the -A/x potential is plausible, and it's closest to
log(x), so my simulations focus on -A/x; but the 1/x^4 potential
also works fine, and I made a few simulations of it too.
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