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?


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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