# Why are probabilities distribution of thermodynamic variables tend to Gaussian?

1. Dec 13, 2009

### dd331

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?

2. Dec 14, 2009

### mathman

I am not familiar with the details of the physics. However such truncation would be based on the assumption |x-x0| is small.