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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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Site last modified Tue Jul 5 12:32:01 BST 2005