3
$\begingroup$

In Landau&Lifshitz V: Statistical Physics the following derivation of the law of increase of entropy is given. I need help understanding several crucial steps; I'll briefly summarize the notations and definitions employed.


Notation and definition of $S$

I refer to a quantum system with a density operator $\rho$. The entropy being defined is the Gibbs entropy, defined for small subsystems of a big isolated system as: $$S=\overline {-\ln\rho} ,$$ where overline denotes an average over the ensemble. Since for these subsystems the the density operator is the canonical one: $$\ln \rho = \alpha+\beta E,$$ this is equivalent to: $$S = \ln \dfrac{1}{\rho (\overline E)}.$$ Introducing the density $\frac{\text d \Gamma}{\text d E}$ of states with energy $E$, the energy distribution (for a quasi continuous spectrum) is given by $W(E)= \rho (E)\frac{\text d \Gamma}{\text d E}$. The width of this distribution is $$\Delta E = 1/W(\overline E)$$ and correspondingly the number of states within that interval is $$\Delta \Gamma = \frac{\text d \Gamma}{\text d E} \Delta E=\frac {1}{\rho (\overline E)},$$ motivating the definition of $S$ as $\ln \Delta \Gamma$.

Finally, for an isolated system with an energy $E$, the entropy is defined as the sum of its small quasi-independent parts: $S = S_1 + S_2 +\dots+S_n$. Since the entropy for a small subsystem defined above is additive, this definition is well posed (that is, given a partition of the system in sufficiently small subsystems, an ulterior partition gives the same value for the total entropy).


The proof of the second law of thermodynamics.

Consider an isolated system with energy $E$, subdivided in $n$ quasi independent small parts. The joint probability density for the energies $E_1,E_2,\dots ,E_n$, since the isolated system distribution is uniform, is given by: $$\text d w \propto\prod _i \dfrac{\text d \Gamma _i}{\text d E_i}\text d E_i. $$

The functions $\Delta \Gamma _i$ and $\Delta E _i$ (defined above) are functions of the mean value of the energies $\overline E _i$. In the same way $S_i=S_i(\overline E _i)$. If we formally consider $S$ and $\Delta E$ as functions of the actual values $E$, we can rewrite the above equation in the form: $$\text d w \propto e^S \prod _i \dfrac{\text d E_i}{\Delta E _i}.$$ While $e^S$ is a rapidly varying function of the energies $E_i$, $\prod _i \Delta E _i$'s dependence from energies is totally inessential so that this is equivalent to: $$\text d w \propto e^S \prod \text d E_i$$


I stop here since from the last equality follows easily that $S$ is maximized at thermal equilibrium.

I have written in italics the points that I don't understand. For example, the statement that “$\Delta \Gamma _i$ and $\Delta E _i$ are functions of $\overline E _i$”. This is false, they are functionals of the distribution $\rho _i$. I don't understand how should I interpret, for example, $\Delta E (E)$. I don't think that the answer is the simple $\Delta E (E) = 1/W(E)$, since in that case it would be totally false that "it's dependence on $E$ is inessential".

I really can't make sense out of this: the probability density of the expected values $\overline E_i$ is indeed $$\dfrac{\text d w}{\text d E_1 \dots \text d E_n} |_{E_i = \overline E_i} = e^S\prod \dfrac{1}{\Delta E _i},$$ where $S$ and $\Delta E_i$ are calculated in the standard way. However I don't see how this quantities can be defined as functions of $E$ to give meaning to the equations above.

Can someone clarify on this derivation and on the definition of the functions $S(E)$, $\Delta \Gamma (E)$, $\Delta E (E)$??

Thank you very much.

$\endgroup$
  • $\begingroup$ partially related : physics.stackexchange.com/q/169874 $\endgroup$ – pppqqq Mar 12 '15 at 21:53
  • $\begingroup$ What section of the Landau & Lifshitz book is this from? I do not see any connection to 2nd law of thermodynamics, by the way. $\endgroup$ – Ján Lalinský Mar 12 '15 at 22:24
  • $\begingroup$ @JánLalinský this is from chapter I, §7. The connection is established in section §8, the "law of increase of entropy". However this is not really what I'm concerned about and probably I should have chosen a different title for the question. $\endgroup$ – pppqqq Mar 12 '15 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.