Metropolis Algorithm State Probability Function Question

[ericjang]

Hello,

I am trying to evaluate the Boltzmann relative probabiltiy as part of the Metropolis criterion for molecular docking.

Suppose I have computed the energy of a new 'pose' and I find out that the energy actually increased by 10kJ. So I evaluate:

prob = e^(-deltaE/(kT))
where k = 1.38065*10^-26 kJ //convert to kJ
and suppose we are at room temperature, 298.15K
prob = -10kJ/(298.15K * 1.38065*10^-26)

and then if some random number R is greater than this probability, we accept the change anyway! otherwise we reject.

So here's my question:
if deltaE is positive, then wouldn't prob always evaluate to 0 unless deltaE was on the scale of 10^-24?

Did I get my constants/units right?

[ericjang]

Oops, turns out they were the wrong units! Here is the correct value I should be using...
0.00831446 kJ/mol/K

Can someone explain to me why there are two different kinds of units? kJ/K vs kJ/mol/K, one incorporates mass...

Anyway, it is still kind of odd that at room temp (298.15 K), even if the energy increase is only 12 kilojoules/mol, the probability of accepting the pose becomes less than 1%. Strange, I wonder if that makes local optima easier to trap.

[deleted-71882]

Hello ericjang,

Units are always a problem it seems. To make things worse, someone always invents a new one. I suppose that's the point: units are somewhat arbitrary. One key to handling things properly is to just make sure every number you have is in the same set of units.

So it really doesn't matter what units E, k, and T are in as long as they are all the same . If k is in kJ/(mol K), then E has to be in kJ/mol.

Your last statement about local minima is certainly true. Problems with local minima have no completely satisfactory solution. Starting an annealing process to find an optimum configuration has to be started at a high-enough T to sample the complete solution space. Even problems of less than a hundred degrees of freedom have trouble with local minima. Solve this problem, and you'll make a real name for yourself!