System 1 has energy $E-\epsilon$ and system 2 has energy $\epsilon$. They are both in thermal contact with each other and system 2 only has one microstate (with energy $\epsilon$). Now they have the same temperature.
Why $P(\epsilon) \propto \Omega (E- \epsilon)\times 1=\Omega (E- \epsilon)$?
In other words, why probability of microstate with energy $\epsilon$ is proportional to number of microstates in system 1 and system 2 altogether? What is the rational behind the above proposition?
Sorry if my question is not clear. This is my first time using this site and my English is not too good.
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Let's start with an example. Let's say system $1$ is a die and can be in one of six states (the top face showing 1-6), $\Omega_1=6$. Then let's say system $2$ is a coin and can be in one of two states (heads or tails), $\Omega_2=2$. Now the combined system will be in one of the combinations \begin{equation} \{1H,1T, 2H, 2T, \cdots 6H, 6T\} \end{equation} where the first number is the die face and the second symbol is heads $T$ or tales $T$. There are 12 possible states, which you can verify by writing out explicitly every entry in the list, or you can calculate since the total number of states of the combined system $\Omega$ is the product of the states of each individual system, $\Omega_{\rm combined} =\Omega_1 \Omega_2=6\times 2 = 12$ (since the state of each system is chosen independently).
Now we run the same logic, but with the values you gave in your question, $\Omega_1 = \Omega(E-\epsilon)$ for system 1 and $\Omega_2=1$ for system 2. Multiplying these together, the number of states in the combined system is $\Omega(E-\epsilon)$.
Finally, the fundamental assumption of thermodynamics is that every microstate is equally likely. Therefore, the probability is proportional to the number of microstates; if there are more microstates with a given energy, then there is more probability for the system to have that energy. As a result, $P(\epsilon) \propto \Omega(E-\epsilon)$.
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$\begingroup$ Yes, I understood the first two paragraphs. But my question is really why probability of microstate with energy $\epsilon$ is proportional to $\Omega (E-\epsilon)$ since I thought probability of microstate with energy $\epsilon$ ought to be proportional to $\Omega(\epsilon)$ since the $P(\epsilon) \propto \frac{\Omega(\epsilon)}{\Sigma \Omega}$ where $\Sigma \Omega$ is the total number of possible microstates in system 1. I don't understand why the probability of microstates with energy $\epsilon$ must be proportional to product $\Omega(E-\epsilon)\Omega(\epsilon)$. Why $\Omega(E-\epsilon)$? $\endgroup$– Nik_28Commented Nov 20, 2021 at 10:22
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$\begingroup$ @NikAhmadFaris The total energy of the combined system is $E$ and is conserved. So if system 2 has energy $\epsilon$, then system 1 must have energy $E-\epsilon$. The total number of microstates of the combined system is the product of the microstates for system 1 and system 2. System 1 has $\Omega_1(E-\epsilon)$ states, and system 2 has $\Omega_2(\epsilon)$ states, so the combined system has $\Omega_1(E-\epsilon)\Omega_2(\epsilon)$ states. You also said that $\Omega_2(\epsilon)=1$, so we can drop that factor, and therefore the combined system has $\Omega_1(E-\epsilon)$ states. $\endgroup$– AndrewCommented Nov 20, 2021 at 13:20
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$\begingroup$ Okay Andrew, $P(\epsilon)=\frac{\Omega_1(E-\epsilon)\Omega_2(\epsilon)}{\Sigma_{i}\Omega_{i}\Sigma_{j}\Omega_{j}}=\frac{\Omega_1(E)\Omega_2(\epsilon)}{\Sigma_{i}\Omega_{i}\Sigma_{j}\Omega_{j}}e^{\frac{\epsilon}{kT}}$. So $P(\epsilon)\propto \Omega_1(E-\epsilon)\Omega_2(\epsilon) \propto e^{\frac{\epsilon}{kT}}$. Where $\Sigma_{i}\Omega_{i}$ is sum of possible microstates in system 1 and $\Sigma_{j}\Omega_{j}$ is sum of possible microstates in system 2. Is this correct? $\endgroup$– Nik_28Commented Nov 20, 2021 at 17:53
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$\begingroup$ @NikAhmadFaris No. $P(\epsilon) = \frac{\Omega_1(E-\epsilon)\Omega_2(\epsilon)}{\int d x \Omega_1(E-x) \Omega_2(x)}$, where $x$ runs overall all possible values of the energy of system 2; note you are summing over multiplicities of the combined system, not over each system independently. It's true that $P(\epsilon) \propto \Omega_1(E-\epsilon) \Omega_2(\epsilon)$. The way you get the Boltzmann factor $e^{-\epsilon/kT}$ (note the key minus sign) is by taking the log of $P$, expanding in $\epsilon/E$ which is assumed to be small, and defining $1/T=\partial S/\partial E$ where $S=k\log \Omega$. $\endgroup$– AndrewCommented Nov 20, 2021 at 20:22
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$\begingroup$ There's a lot of resources that go through the argument in more detail; for example these notes: itp.uni-frankfurt.de/~gros/Vorlesungen/TD/… $\endgroup$– AndrewCommented Nov 20, 2021 at 20:24