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The probability of finding a particle with energy $E$ according to Maxwell-Boltzmann distribution is: $$ P(E) =\frac{1}{Z}g(E)e^{\frac{-E}{k_BT}} \qquad eq(1)$$ where g(E) is the degeneracy of the energy level $E$.

However, the deduction of this formula is the following:

-Using the following definition of entropy: $S=k_Bln(\Omega(s))$, where $\Omega$ is the multiplicity of a macrostate $s$.

-Considering also a system in contact with a large reservoir, and two possible states: $s_1$ and $s_2$. Then, the ratio of their probabilities is the ratio of the multiplicities of the reservoir that make the system be in that state.

$$ \frac{P(s_1)}{P(s_2)} = \frac{\Omega_R(s_1)}{\Omega_R(s_2)}=\frac{e^{S_R(s_1)}}{e^{S_R(s_2)}}$$

For example, if $\Omega_R(s_1)=10$ and $\Omega_R(s_2)=5$, then its twice as likely to find the system in $s_1$ than $s_2$.

Then by using thermodynamic:

$$ dS_R = \frac{1}{T}dU_R$$ $$ dU_R = -dE$$

where E is the energy of the system and U the reservoir. Using these equations one can easily get to equation 1.

But, my question is the following:

Why in equation (1) there is the degeneracy term $g(E)$? Isn't that being accounted in the deduction when they count the multiplicity?

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P(s), the probability of being in state s, isn't the same as P(E), the probability of having energy E. Suppose you knew perfectly knew P(s); how would you get P(E)? You'd transform the probability densities, which gets you a factor of dE/ds, which maps onto the multiplicity.

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