Weinberg's proof of Gibbs' $H$-theorem I'm trying to understand Weinberg's explanation of a general $H$-theorem he attributes to Gibbs (Foundations of Modern Physics, p. 35), but I'm having trouble with the mathematical modeling.
The goal is to show that the negative Gibbs-entropy $\int{P(\alpha)d\alpha\ln{P(\alpha)}}$ doesn't increase in time. In the proof Weinberg assumes that there is a "differential rate $\Gamma(\alpha \rightarrow \beta)$ such that the rate at which a system in state $\alpha$ makes a transition to a state within a range $d\beta$ around state $\beta$ is $\Gamma(\alpha \rightarrow \beta) d\beta$." I thought that $\Gamma(\alpha \rightarrow \cdot)$ is the conditional probability density function of moving to some other state immediately after being in $\alpha$; my problem is that this leads to me to expect that the probability of being at state $\alpha$ at time $t$ should satisfy
$$\frac{dP(\alpha)}{dt} = \int{d\beta P(\beta) \Gamma(\beta \rightarrow \alpha)},$$
in analogy to discrete Markov chains. However, Weinberg writes that
$$\frac{dP(\alpha)}{dt} = \int{d\beta [P(\beta) \Gamma(\beta \rightarrow \alpha) - P(\alpha) \Gamma(\alpha \rightarrow \beta)]},$$
the intuitive explanation being that the probability of being at $\alpha$ decreases if the system transitions away from $\alpha$, hence the additional term.
How can Weinberg's equation be justified? Is this an assumption with physical insight, or a result in probability about certain kinds of stochastic systems? If the latter, what kinds of stochastic systems are these?
Finally, the statement of the theorem seems to contradict results that the Gibbs entropy remains constant due to Liouville's theorem (e.g. here). Surely there is some thing that Weinberg assumes or does differently, but I don't know what.
 A: Weinberg's equation seems easier to justify then the preceding equation in the OP. Indeed, suppose the system can be found in state $\alpha$ with probability $P(\alpha, t)$, in state $\beta$ with probability $P(\beta, t)$ and so on - this statement is redundant, since $\alpha$ and $\beta$ are just indices that run over all states, but I want to distinguish two arbitrarily chosen states. At time $t+\Delta t$ we expect that, if the system were in state $\alpha$, it may transition to state $\beta$ with probability $P(\beta,t+\Delta t|\alpha, t)=\Gamma(\alpha\rightarrow\beta)\Delta t$. On the other hand, if the system were in state $\beta$, it may transition to state $\alpha$ with probability $P(\alpha,t+\Delta t|\beta, t)=\Gamma(\beta\rightarrow\alpha)\Delta t$. Thus, the change of probability to be in state $\alpha$ due to these two types of transitions is
$$
P(\alpha, t+\Delta t)- P(\alpha, t) =
P(\alpha,t+\Delta t|\beta, t)P(\beta, t) - P(\beta,t+\Delta t|\alpha, t)P(\alpha,t)
$$
(same can be written for state \beta, but I remind that $\alpha,\beta$ are just indices.)
If we account for all the states, we write
$$
P(\alpha, t+\Delta t)- P(\alpha, t) =
\sum_{\beta\neq \alpha} P(\alpha,t+\Delta t|\beta, t)P(\beta, t) - 
\sum_{\beta\neq \alpha} P(\beta,t+\Delta t|\alpha, t)P(\alpha,t)
$$
(I write sums rather than integrals in order not to bother with the integration measure.)
More formally it could be cast as Smolukhovsky equation:
$$
P(\alpha, t+\Delta t)= \sum_{\beta\neq \alpha} P(\alpha,t+\Delta t|\beta, t)P(\beta, t) + \left[1
- 
\sum_{\beta\neq \alpha} P(\beta,t+\Delta t|\alpha, t)\right]P(\alpha,t)=
\sum_\beta P(\alpha,t+\Delta t|\beta, t)P(\beta, t),
$$
where
$$
P(\alpha,t+\Delta t|\alpha, t)= 1
- 
\sum_{\beta\neq \alpha} P(\beta,t+\Delta t|\alpha, t)$$
is the probability that the system remains in state $\alpha$, which is the same as the probability that it does not leave from $\alpha$ to any other state. Indeed, this equation is just  a restatement of normalization
$$
1= 
\sum_{\beta} P(\beta,t+\Delta t|\alpha, t),$$
that is, regardless of the state at time $t$, the probability of finding the system in any of its states at $t+\Delta t$ is $1$.
Alternative view:
One could start with the equation in the OP
$$
\frac{dP(\alpha)}{dt}=\int d\beta P(\beta)\Gamma(\beta\rightarrow\alpha)
$$
and integrate it over $\alpha$. Since $\int d\alpha P(\alpha)=1$, we would have
$$
0=\int d\beta P(\beta)\int d\alpha\Gamma(\beta\rightarrow\alpha).
$$
Since the transition rates are defined as positive, it would mean that the $P(\beta)=0$ for any state $\beta$, which violates the normalization.
Remark: The whole exercise can be viewed as shuffling money between multiple bank accounts - the money leaving one account must end in the other accounts, but the total remains the same.
A: Your Weinberg's equation is actually called "the master equation" in the context of non-equilibrium statistical mechanics. Just to complement @RogerVadim's answer, the master equation for Markov chains (discrete time and space) can be rewritten in a way that seems more similar to the equation you are asking about.
$$P(\alpha,k+1)-P(\alpha,k)=\sum_\beta W(\beta\to\alpha)P(\beta,k).$$
Where the $W$'s are transition probabilities that have to be normalized, therefore:
$$W(\alpha\to\alpha)=1-\sum_{\beta\ne \alpha} W(\alpha\to\beta).$$
Inserting this in the first equation:
$$P(\alpha,k+1)-P(\alpha,k)=\sum_{\beta\ne \alpha}[ W(\beta\to\alpha)P(\beta,k)-W(\alpha\to\beta)P(\alpha,k)].$$
If time is continuous, you can rewrite the above equation in terms of transition rates (your $\Gamma$'s):
$$\partial_tP(\alpha,t)=\sum_{\beta\ne \alpha}[ \Gamma(\beta\to\alpha)P(\beta,t)-\Gamma(\alpha\to\beta)P(\alpha,t) ].$$
If states are continuous:
$$\partial_tP(\alpha,t)=\int d\beta[ \Gamma(\beta\to\alpha)P(\beta,t)-\Gamma(\alpha\to\beta)P(\alpha,t) ].$$
