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In order to define temperature in terms of statistics, "Concepts of Thermal Physics" by Blundell states:

Let us assume that the first system can be in any one of $Ω_1(E_1)$ microstates and the second system can be in any one of $Ω_2(E_2)$ microstates. Thus the whole system can be in any one of $Ω_1(E_1)Ω_2(E_2)$ microstates.

as expected from the probabilities.

However, when is considered that when the two systems can exchange energy with each other, it is stated that:

For our problem of two connected systems, the most probable division of energy between the two systems is the one which maximizes $Ω_1(E_1)Ω_2(E_2)$, because this will correspond to the greatest number of possible microstates.

My question is: Given the fact that occurs an exchange of energy between the two systems, I was expected that we will have:

$\Omega_{1+2} \geq \Omega_{1}\Omega_{2}$ as the number of microstates for the whole system

Could anyone explain me what is wrong above in my argument?´

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    $\begingroup$ What is your argument exactly? Why does energy exchange imply that the number of states in the joint system is not the product? $\endgroup$ – kaylimekay Jan 21 at 9:22
  • $\begingroup$ @kaylimekay Exactly $\endgroup$ – JoseAf Jan 21 at 9:24
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Your argument is correct. What is not told explicitly here is that before the systems come in contact you can consider $E_1,E_2$ fixed. After you bring them in contact they are no longer fixed but the quantity $E=E_1+E_2$ is fixed. This leaves only $E_1$ as the free parameter (or $E_2$, just pick one). This means that implicitly $$\Omega_{1+2}(E_1)=\Omega_1(E_1)\Omega_2(E_2)=\Omega_1(E_1)\Omega_2(E-E_1).$$ After the systems have reached thermal equilibrium $E_1$ is again fixed because it has reached the unique value that maximizes $\Omega_{1+2}(E_1)$.

I should add that initially $\Omega_{1+2}=\Omega_1\Omega_2$ but after thermalization $\Omega_{1+2}\geq\Omega_1\Omega_2$.

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    $\begingroup$ So, it works because one of the systems is a heat bath, right? $\endgroup$ – JoseAf Jan 21 at 9:38
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    $\begingroup$ @JoseAf Not exactly. A heat bath is defined a system that is so much larger than your system of interest that removing heat from the heat bath doesn't change the heat bath. In this example from Blundell system 1 and 2 could be equal in size. $\endgroup$ – AccidentalTaylorExpansion Jan 21 at 9:45

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