shirtsgroup / cg_openmm

Tools to build coarse grained models and perform simulations with OpenMM
MIT License
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How to do enthalpy/entropy of configurational states #72

Open mrshirts opened 4 years ago

mrshirts commented 4 years ago

OK there are two difference scenarios:

The average of the energy over all configurational states is = \sum_i W_1 E_1 + \sum W_2 E_2 + \sum W_3 E_3, where the W_i are the weights from MBAR corresponding to each configuration in the overall Hamiltonian, where \sum W_j = exp(-f_j) (up to a constant) for j=1,2,3 are the configurations.

So I THINK the average energy/enthalpy of each state is \sum W_j E_j / \sum W_j, with the difference in enthalpy between two states, say 1 and 2, is equal to:

\sum W_1 E_1 / \sum W_1 - \sum W_2 E_2/ \sum W_2

i.e. the weighted average of the configurations consistent with that state, normalized by the probability of the state.

The the entropy difference (at a single T) should be obtainable by (dH - dG)/T = dS.

I could potentially figure out the error propagation, but calculating it by bootstrap is probably better.

This probably should be checked by finite difference of the free energy for each state, since I'm not entirely certain about this derivation!

mrshirts commented 4 years ago

Or could do a numerical fit to \Delta F, and differentiate to get \Delta S, then compute \Delta H by \Delta F + T \Delta S