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In the book "Concepts in Thermal Physics" the Boltzmann distribution is derived with the following assumptions:

  1. There are two systems, one enormous heat reservoir and one comparatively miniscule system.
  2. The two systems are in thermal equilibrium.
  3. The heat reservoir is so large that any energy the smaller system can remove makes no change to its overall temperature.
  4. The large system has an incredibly large number of possibly microstates.
  5. By contrast the small system is assumed to have 1 microstate for every possible energy.

Therefore the energies of each system are $\epsilon$ for the small system and $(E-\epsilon)$ for the large system, with the total energy being $E$. This allows you to formulate the probability of the small system have energy $\epsilon$ as:

$$P(\epsilon) \propto \Omega(E-\epsilon)\times1$$

Where $\Omega(E-\epsilon)$ is the total number of microstates for the reservoir, and $1$ represents the total number of microstates for the system. This allows you to formulate the Boltzmann distribution:

$$P(\epsilon) \propto e^{-\epsilon/k_BT}$$

My question is, when does assumption 5 apply? Is it allowed because the reservoir is assumed so much larger than the small system? Or would you need to derive the Boltzmann distribution in another way for different systems?

Edit: The answer to this question seems to be given here Derivation of Boltzmann distribution - two questions

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    $\begingroup$ You have not included an additional important assumption: the equilibration between system and thermostat. $\endgroup$
    – GiorgioP
    Sep 24 at 13:48
  • $\begingroup$ Is the above edit an acceptable assumption? $\endgroup$
    – Connor
    Sep 24 at 13:50
  • $\begingroup$ Thermal contact is not enough. Gravitational systems have negative specific heat. A simple way to explain how it may be possible, notwithstanding the positive defined character of the canonical specific heat, is that a gravitational system cannot equilibrate with a thermostat. $\endgroup$
    – GiorgioP
    Sep 24 at 14:00
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    $\begingroup$ Is the question still about assumption 4? $\endgroup$ Sep 24 at 14:05
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    $\begingroup$ @GiorgioP Sorry what you're saying is slightly above me, should I say "The systems are in thermal equilibrium"? $\endgroup$
    – Connor
    Sep 24 at 14:25
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Assumption 4 isn't needed to talk about the Boltzmann distribution. Each state has a probability which is proportional to $\exp(-E/k_B T)$, and if there are $g(E)$ different states with the same energy (degeneracy), then the probability to be at any state of energy E is just proportional to $g(E) \cdot \exp(-E/k_B T)$.

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  • $\begingroup$ Does this mean if you were to include $\Omega(\epsilon)$ as something other than $1$ the derivation would give you what you wrote above? (i.e. $g(E) \cdot exp(-E/k_BT)$) $\endgroup$
    – Connor
    Sep 24 at 13:39
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    $\begingroup$ yes, exactly :) $\endgroup$ Sep 24 at 13:40

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