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In deducing the zeroth law of thermodynamics in micro-canonical ensemble, there is a frequently-mentioned example. Suppose we put two isolated system, system 1 and 2, in contact and allowing them to exchange heat.

The total energy of the combined system is

$$E = {E_1} + {E_2}$$

The total allowed phase space is

$$\Omega (E) = \int d {E_1}{\Omega _1}({E_1}){\Omega _2}(E - {E_1}) = \int d {E_1}\exp [\frac{{{S_1}({E_1}) + {S_2}(E - {E_1})}}{{{k_B}}}]$$

By taking the maximum of the integrand,

$$\frac{{\partial {S_1}({E_1})}}{{\partial E}} + \frac{{\partial {S_2}(E - {E_1})}}{{\partial E}} = 0$$

we get the equilibrium state ##(E_1^{eq},E - E_1^{eq}) ##, which is the macrostate with exponetially larger number of microstates, and also get the equilibrium condition

$$\frac{{\partial {S_1}({E_1})}}{{\partial {E_1}}} = \frac{{\partial {S_2}(E - {E_1})}}{{\partial {E_2}}}$$

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The equal a priori probability postulate, as descripted by David Tong p4 , is

So my question is, does that mean the microsates corresponds to the equilibrium macrosate ##(E_1^{eq},E - E_1^{eq}) ## possses equal probability or all the microstates with constant ##E ## (the blue line in next figure) possess equal probability?

For me, it looks like every reference I found indicates the second statement. But for a microsate corresponds to ##(E_1^{0},E - E_1^{0}) ## which is not at equilibrium, does it have the same probability as a microstate of ##(E_1^{eq},E - E_1^{eq}) ##?

The total energy of the combined system is

$$E = {E_1} + {E_2}$$

The total allowed phase space is

$$\Omega (E) = \int d {E_1}{\Omega _1}({E_1}){\Omega _2}(E - {E_1}) = \int d {E_1}\exp [\frac{{{S_1}({E_1}) + {S_2}(E - {E_1})}}{{{k_B}}}]$$

By taking the maximum of the integrand,

$$\frac{{\partial {S_1}({E_1})}}{{\partial E}} + \frac{{\partial {S_2}(E - {E_1})}}{{\partial E}} = 0$$

we get the equilibrium state ##(E_1^{eq},E - E_1^{eq}) ##, which is the macrostate with exponetially larger number of microstates, and also get the equilibrium condition

$$\frac{{\partial {S_1}({E_1})}}{{\partial {E_1}}} = \frac{{\partial {S_2}(E - {E_1})}}{{\partial {E_2}}}$$

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

The equal a priori probability postulate, as descripted by David Tong p4 , is

*For an isolated system in*.**equilibrium**, all accessible microstates are equally likelySo my question is, does that mean the microsates corresponds to the equilibrium macrosate ##(E_1^{eq},E - E_1^{eq}) ## possses equal probability or all the microstates with constant ##E ## (the blue line in next figure) possess equal probability?

For me, it looks like every reference I found indicates the second statement. But for a microsate corresponds to ##(E_1^{0},E - E_1^{0}) ## which is not at equilibrium, does it have the same probability as a microstate of ##(E_1^{eq},E - E_1^{eq}) ##?

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