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 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 $S(E)$, $\Delta \Gamma (E)$, $\Delta E (E)$?

  • $\begingroup$ partially related : physics.stackexchange.com/q/169874 $\endgroup$
    – pppqqq
    Commented Mar 12, 2015 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. $\endgroup$ Commented Mar 12, 2015 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
    Commented Mar 12, 2015 at 22:31
  • $\begingroup$ Closely related (Liouville theorem VS entropy increase) physics.stackexchange.com/q/761468/226902 $\endgroup$
    – Quillo
    Commented Oct 10, 2023 at 11:42

1 Answer 1


Their approach is a bit tricky because they simultaneously use the microcanonical and canonical ensemble. Both are equivalent in the thermodynamic limit, which is why they consider that the subsystems are macroscopic.

The functions $\Delta\Gamma_i$ and $\Delta E_i$ (defined above) are functions of the mean value of the energies $\overline E_i$

Yes, they are functionals of $\rho_i$. However, $\rho_i$ in turn is given by the canonical distribution and is parametrised by only one number. Explicitly, the natural parameter is $\beta$ (inverse temperature). However, since $\overline E_i$ can be calculated from $\beta$, you can invert the relationship and parametrise it directly with $\overline E_i$.

To illustrate this parametrisation (which does not require the system to be macroscopic), take a $2$ state system labeled $\pm$ system of respective energies $\pm1$. You then have ($w$ is now a pmf, not a pdf): $$ w_\pm = \frac{e^{\mp\beta}}{2\cosh(\beta)} \\ \overline E = \tanh\beta $$ so by inverting the relationship: $$ w_\pm = \frac{1\pm\overline E}{2} $$ so all the expected values of the observables are function of $\overline E$. In particular, you can use the previous formulas: $$ \Delta\Gamma_i = \frac{1}{\rho_i(\overline E_i)} \\ \Delta E_i = \frac{1}{W_i(\overline E_i)} $$

If we formally consider $S_i$ and $\Delta E_i$ as functions of the actual values $E_i$ (you forgot the indices in your question)

Instead of assuming that the subsystems have fluctuating energy, you just assume that their energy is fixed and equal to the mean energy. This allows you to avoid integrating over each subsystem. This is justified once again because the subsystems are macroscopic, mathematically, you are using the law of large numbers.

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

This is the usual Laplace method argument. This allows you to get rid of the factor $\prod_i\Delta E_i$ and absorb it in the overall constant prefactor.

Hope this helps.

  • $\begingroup$ I was hoping to get more detail om that last step. Intuitively I can feel why $e^S$ would vary that much faster than $\prod_i\Delta E_i$ that you could absorb it into the constant, but I would like it to be shown a bit more explicitly (and/or I would like to see other examples of this same kind of 'trick'). L&L has a similar elimination of $\prod \Delta E$ in pg. 80 where they derive the canonical distribution from the microcanonical, but there the $\Delta E$ eliminated is a single factor belonging to the much larger heat bath, the energy of which can be considered practically equal to $E_0$, $\endgroup$ Commented Sep 20, 2023 at 14:55
  • $\begingroup$ the energy of the whole closed system, while in the case in question $\prod \Delta E$ is summed over all subsystems where not one is assumed much bigger than the others. It seems then that they mean for it to apply even to larger deviations from the average energy values, especially since they use this formula to motivate the Law of Increase of Entropy later on. $\endgroup$ Commented Sep 20, 2023 at 15:04

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