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dd331
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The probability distribution for some thermodynamic variable x is given by
P = N e^{-A(x)/KT}
where A(x) is the availability, which can be replaced by Hemlholtz free energy F, Gibb's free energy G, etc depending on the conditions imposed. N is just some normalization constant. A(x) can be expanded in a taylor series about the equilibrium conditions,
A(x) = A(x_{0}) + (x - x_{0})(\frac {\partial A} {\partial x})_{x = x_{0}} + \frac{1} {2} (x - x_{0})^{2} (\frac {\partial^2 A} {\partial x^2})_{x = x_{0}} + ...
The second term is 0 since dA/dx = 0 at equilibrium. If we truncate all the other terms, clearly we see that P will be a Gaussian distribution with mean of x_{0} and standard deviation of
\sqrt {\frac {K T} {(\frac {\partial^2 A} {\partial x^2})_{x = x_{0}}}}
What is the justification for truncating this series? This is justified if (x - x0) is small. But why will it be small for big N?
P = N e^{-A(x)/KT}
where A(x) is the availability, which can be replaced by Hemlholtz free energy F, Gibb's free energy G, etc depending on the conditions imposed. N is just some normalization constant. A(x) can be expanded in a taylor series about the equilibrium conditions,
A(x) = A(x_{0}) + (x - x_{0})(\frac {\partial A} {\partial x})_{x = x_{0}} + \frac{1} {2} (x - x_{0})^{2} (\frac {\partial^2 A} {\partial x^2})_{x = x_{0}} + ...
The second term is 0 since dA/dx = 0 at equilibrium. If we truncate all the other terms, clearly we see that P will be a Gaussian distribution with mean of x_{0} and standard deviation of
\sqrt {\frac {K T} {(\frac {\partial^2 A} {\partial x^2})_{x = x_{0}}}}
What is the justification for truncating this series? This is justified if (x - x0) is small. But why will it be small for big N?
