Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm currently studying for an exam in thermodynamics/classic statistical mechanics and 2 things came up which are confusing me.

First the ergodic hypothesis states that it is equal to take the mean value of one micro state with respect to time or take the mean of the whole ensemble. In our textbook the state of the whole phase space is defined as a probability measure $d\mu (x)=\omega (x)dx$ on $\Gamma_N$ = $\mathbb{R}^6$. The ergodic hypothesis is formally stated as: \begin{align} &d\mu_{E}(x)=\Sigma(E)^{-1} \cdot \delta(H(x) - E)\ dx\\ &\Sigma(E)=\int_\Gamma \delta(H(x)-E)dx \end{align} is the only invariant probability measure under $\phi_t$ on $\Gamma(E)$. Where $\phi_t$ is the hamiltonian flow and $\Gamma(E) = \{x \in \Gamma_N|H(x) = E\}$. Up to this point it's clear to me. But after this they say that also for a 'general state': $\omega(x)dx = \int dE(\delta(H(x)-E)\omega(x) dx)$ we can write the time average as an ensemble average: \begin{align} <f> = \int dE\ W_E \cdot \int_\Gamma f(x) d\mu_E(x)\\ W_E = \int\omega(x)\delta(H(x)-E)dx \end{align} Where f is an arbitrary observable. They go on proving the statement: \begin{align} <f>=\int dx\ f(\phi_t(x))\ \omega(x) = \int dx\ dE\ f(\phi_t(x))\ \delta(H(x)-E)\ \omega(x)\\ =\int dE\ W_E \int dx\ \frac{1}{W_E}\ f(\phi_t(x))\ \delta(H(x)-E)\ \omega(x)=\int dE\ W_E \int f(x) \ d\mu_E \end{align} The last step in this deviation is absolutely not clear to me. If just someone could point out what I'm missing this would be great.

Second I got a problem with the definition of the canonical ensemble. We look at two systems in thermic contact with $V'>>V,\ N'>>N$ where V is volume and N is the number of particles. The state of the complete system "0" is the micro canonical ensemble to the energy E. The Hamiltonian is $H_0(x,x') = H(x) + H'(x')$ where we neglect the interaction Hamiltonian. The state of the small system is then given by: \begin{align} \omega(x) = \frac{1}{\Sigma_0(E)} \int_{\Gamma'}dx' \delta(H(x) + H'(x') - E) \end{align} Why is this the case? Again I would be very glad if someone could shortly give me a hint here why this holds... Thanks in advance. Edit: Maybe I just answered the sec ond question myself: Does this simply hold because the states of the complete system are the same as the states of the small system (because they are in thermal contact) and since we only integrate over $\Gamma'$ those are the states we are looking for?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.