Fluktuationen

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We want to examine the fluctuation around the mean energy by calculating the mean $\left\langle E \right\rangle $ and the variance $\sigma_E$ of the energy. \begin{equation} \sigma _E^2 =\left\langle E^2 \right\rangle - \left\langle E \right\rangle ^2 \tag{1} \end{equation} First, we write $\left\langle E \right\rangle$ and $\left\langle E^2 \right\rangle$ in terms of the partition function \begin{equation} \left\langle E \right\rangle = \sum_i P(i) E_i=\left( \frac{1}{Z} \sum_i e^{-\beta E_i}\right) E_i = - \frac{1}{Z} \frac{\partial (Z)}{\partial \beta } = - \frac{\partial log(Z)}{\partial \beta } \end{equation}

\begin{equation} \left\langle E^2 \right\rangle = \sum_i P(i) E_i^2=\left( \frac{1}{Z} \sum_i e^{-\beta E_i}\right) E_i^2 = \frac{1}{Z} \frac{\partial ^2 (Z)}{\partial \beta ^2} \end{equation} plug into equ. (1) and simplify by recognising, that the expression is just the second derivative of $log(Z)$ with respect to $\beta$ \begin{equation} \sigma _E ^2 = \frac{1}{Z} \frac{\partial ^2 (Z)}{\partial \beta ^2} - \left( \frac{1}{Z} \frac{\partial (Z)}{\partial \beta } \right) ^2 = \frac{\partial ^2 log(Z)}{\partial \beta ^2 } \end{equation} differentiating once gives the energy $- \left\langle E \right\rangle $ and we are left with \begin{equation} - \frac{\partial \left\langle E \right\rangle }{\partial \beta }= - \frac{\partial \left\langle E \right\rangle }{\partial T } \frac{\partial T }{\partial \beta } \end{equation} knowing $ T = 1/(k_B \beta)$ gives $\partial T / \partial \beta = -k_b \cdot T^2$. The heat capacity tells you how much energy is needed to heat the temperature by one unit, so $\partial \left\langle E \right\rangle /\partial T =C_V$ and we finally get \begin{equation} \sigma _E = \sqrt{C_V \cdot k_b} \cdot T \end{equation} The heat capacity is proportional to the number of molecules $C_V=O(N k_B)$ and $ \left\langle E \right\rangle = O(Nk_BT)$. For large numbers $N \rightarrow \infty$ of molecules the ratio \begin{equation} \frac{\sigma_E}{ \left\langle E \right\rangle} \propto \frac{1}{\sqrt{N}} \end{equation} goes to zero, which applies for usually considered closed isothermal systems with typical values $\sigma _E \approx 10^{-10} \left\langle E \right\rangle$. In this case the probability distribution is practically a $\delta$-distribution. For very small molecule numbers, the fluctuations become importatnt.