Denote by $\rho(p,q)$ (the $p$ and $q$ are being used as shorthand for several degrees of freedom), the phase space probability distribution function, (so $\rho\,\text{d}p\text{d}q$ is the probability the system is in a state within the phase space volume $\text{d}p\text{d}q$ at any instant).

I understand that, by Liouville's theorem, $\rho(p, q)$ is an integral of motion, and therefore its dependence on the $p$ and $q$ can be reframed as a dependence on other integrals of motion, i.e. $$ \rho(p,q) = \rho(C_1(p,q),\dots,C_n(p,q)) $$ where $C_i$ are other integrals of motion.

In the Landau and Lifshitz Statistical Physics book (3rd ed.) pg. 11

We may therefore say that the distribution function, being a function of the mechanical invariants, is itself an integral of the motion. It proves possible to restrict very considerably the number of integrals of the motion on which the distribution function can depend. To do this, we must take into account the fact that the distribution $\rho_{12}$ for a combination of two subsystems is equal to the product of the distribution functions $\rho_1$ and $\rho_2$ of the two subsystems separately: $$ \rho_{12} = \rho_1\rho_2\,. $$ Hence $$ \log{\rho_{12}} = \log{\rho_1} + \log{\rho_2}\,, $$ so that the logarithm of the distribution function is an additive quantity. We therefore reach the conclusion that the logarithm of the distribution function must be not merely an integral of the motion, but an additive inte­gral of the motion.

And the book then goes on to conclude that because $\log{\rho(p,q)}$ is an "additive integral of motion", it can only depend upon other additive integrals of motion, hence it must be a linear combination of the system's energy $E(p,q)$, momentum $\vec{P}(p,q)$, and angular momentum $\vec{M}(p,q)$ because these are the only independent additive integrals of motion as stated in a previous L&L book (and discussed here).

I don't understand how we arrive at the requirement that $\rho(p,q)$ must be dependent upon only "additive integrals of motion" from requiring $\log{\rho_{12}}=\log{\rho_1}+\log{\rho_2}$. (I'm also unsure of the precise mathematical definition of an additive integral of motion, since an integral of motion is specific to a particular system, so I'm not sure how one combines integrals of motions of different systems.)

Why can't we just have a system with two completely isolated subsystems, with distribution functions $\rho_1$ and $\rho_2$, where $\rho_1$ and $\rho_2$ can depend on arbitrary integrals of motion? Or even if they do depend on $E$, $\vec{P}$, and $\vec{M}$, it is in a way where $\log\rho$ is not a linear combination of such.


1 Answer 1


Additive here means extensive. That is, if you have a system composed of two distinct subsystems 1 and 2, an additive quantity $A$ obeys $A_{12} = A_1 + A_2$. Since $\log\rho_{12} = \log\rho_1 + \log\rho_2$, $\log\rho$ is additive by definition.

The only way this works is if $\log\rho$ only depends on other additive properties of the system. For the sake of argument, suppose you had something like $\log\rho = T$. This would contradict the additivity of $\log\rho$ because if you have a system composed of subsystems with temperatures $T_1$, $T_2$, it's not true that the temperature $T_{12}$ of the total system is $T_1 + T_2$. So $\log\rho_{12} \neq \log\rho_1 + \log \rho_2$.

Response to the comment: My understanding of Landau-Lifshitz on this is that the delta function is being added in "by hand" to the distribution function. Take a look at Eq. 4.5 followed by 4.6. I don't think they are saying that if you exponentiate 4.5, you get 4.6. Rather, we choose the total $\rho$ so that restricted to a submanifold of given total energy, we'll have, say, $\rho(E_0) = e^{\alpha + \beta E_0}$ constant. The "restricted to a submanifold of given total energy" has to be added in by hand with the delta function. The point is that for this to make sense, the argument of the delta function has to be an additive quantity.

Admittedly, this is kind of rough. Maybe someone else has a better/more rigorous understanding.

  • $\begingroup$ Thanks for the answer. I kind of understand your temperature example, but on the other hand, what I'm still confused about is this. If we have a system, with two isolated subsystems, independent harmonic oscillator say; subsystem 1 has one particle and a distribution function $\rho_1(E_1)=\delta(E_1-5)$ and subsystem 2 has one particle and distribution function $\rho_2(E_2)=\delta(E_2-10)$. Then the distribution function for the entire system is $\rho(E_1, E_2)=\delta(E_1-5)\delta(E_2-10)$, but $\log\rho$ is not a linear combination of the total energy of the system $E$. $\endgroup$
    – eugenhu
    Commented Sep 20, 2019 at 15:27
  • 1
    $\begingroup$ I've edited with a response, although I'm not sure it rises to the level of an answer. $\endgroup$
    – d_b
    Commented Sep 20, 2019 at 16:03

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