Given an ensemble, the probability density in $6N$-dim phase space is $\rho(x,p,t)$, then the evolution of $\rho(x,p,t)$ is Liouville's equation: $$\frac{\partial}{\partial t}\rho(x,p,t)=-\{\rho(x,p,t),H\}$$
The $\bf definition$ of the equilibrium of a system is $\frac{\partial}{\partial t}\rho(x,p,t)=0$, that is $$\{\rho, H\}=0$$
Because $\frac{\partial}{\partial t} \rho=0 \Leftrightarrow \frac{d}{dt} \int dq dp f(p,q)\rho(p,q,t) =0$ for any function $f(p,q)$, then any observables are constant, which is equilvalent to thermodynamic equilibrium.
If in advance we've found all independent conservative quantities of a time-independent Hamiltonian $H$, for example $H$, $U_i$,$\cdots$. Then $\{\rho, H\}=0$ implies $\rho(x,p,t)$ is conservative quantities. Because we've found all independent conservative quantities of $H$, then $\rho$ must be a function of these quantities, that is $\rho=\rho(H,U_i,\cdots)$.
Therefore, $\frac{\partial}{\partial t}\rho\Leftrightarrow$ ensemble is equilibrium$\Leftrightarrow \rho=\rho(H,U_i,\cdots)$
The $\bf definition$ of microcanonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system which has an $\bf exactly\ specified\ total\ energy$.
Certainly if a time independent Hamiltonian $H$ does not have other conservative quantities, we can get $\rho=\rho(H)$ that is equivalent to "a priori equal probability" in microcanonical ensemble, which states that the probability is equal in the hypersurface of constant energy in phase space. But in general many systems can have other conservative quantities apart from total energy.
My question is
- If the above argumemt is correct, then "a priori equal probability" is certainly wrong for microcanonical ensemble with other conservative quantities. Because the equilibrium of system only requires $\rho=\rho(H,U_i,\cdots)$, that is the probability density can be different for the different constant $U_i$ sheaf in hypersurface of constant energy.
And it's very common that a system can have other conservative quantities, e.g.
$$H=\sum_{n=1}^{N}\frac{\mathbf{p_n}^2}{2m}+\sum_{i<j}V(\mathbf{r}_i-\mathbf{r}_j)$$
This Hamiltonian has conservative energy, momentum and angular momentum.
Therefore in equlibrium the $N$-particle system with different total angular momentum can occur with different probability.
e.g. $\rho_1=\alpha(H)+\beta(L_x)$, $\rho_2=\alpha(H)\beta(p_z)\gamma(L_y)$,$\cdots$
All these $\rho_i$ satisfy the requirement $\frac{\partial}{\partial t}\rho_i=0$, i.e. they are all equlibrium ensemble. So is my argument correct? If wrong, please tell me where's wrong.
2.Certainly you can redefine microcanonical ensemble as an ensemble with same $E,V,N$ and all conservative quantities fixed. In this case, we get the "a priori equal probability". So why we have to posulate "a priori equal probability"? Because we have proven it. Secondly I've checked all commen textbook, they all define the microcanonical ensemble as constant E,V,N. There is no requirement that other conservative quantities must be fixed. See pathria p.30, Kardar p.98, Kerson Huang p.80, Le Bellac p.65.
