On the derivation of the canonical distribution

Question

in deriving the canonical distribution $\rho_c$ for a sistem (labeled by 1) of N identical particles in thermal equilibrium with a heat bath (labeled by 2), at one point I get into the Taylor expansion of the entropy of the reservoir . It is justified with the assumption that the dimensions of the latter are way more big that the ones of the system,

$$ E_2 \gg E_1; \quad N_2 \gg N_1; \quad E=E_1 + E_2 $$

So, doing the expansion, $$ S_2(E_2 = E - E_1) \approx S_2(E_2 = E) -E_1 \Big(\frac{\partial S_2}{\partial E_2}\Big)_{E_2 =E} + (...) =k_B ln[\Gamma(E_2 = E - E_1)]$$

So, as I've already shown that $\rho_c \propto \Gamma_2(E_2 = E - E_1),$ taking the exponential of the expansion I can write $$ \Gamma(E_2 = E - E_1) \approx exp\Big[\frac{S_2(E_2 = E)}{k_B}\Big]\cdot exp\Big[\frac{E_1}{k_B T}\Big]$$ After this, it's said that $S_2$ is evaluated at the fixed energy of the universe, and so it's a constant term, suggesting that it can be eliminated from the calculation. In fact, when it gets to the point of finding $\rho_c$, it is formulated as $$ \rho_c = \frac{1}{Z}e^{-\frac{E_1}{k_B T}}$$ where Z is the partition function, that does not contain any term with $S_2$. So I don't get why I shall drop the term with the entropy of the reservoir. One idea is that I can directly eliminate the term $S_2(E_2 = E)$ in the Taylor expansion, or think of it as a sort of renormalization of the entropy, but I'm not sure these ideas are legitimate.

Answer 1

Here, $\rho_{c}(E_{1})$ is just the probability of the system $1$ being found at the given energy $E_{1}$, and so the constant pre-factor factors out of expression.

$Z$ is just the sum of all the $\Gamma(E_{2})$ as $E_{2}$ is varied over all possible discrete values (or we need an integral if the number of admissible energy values are continuous). Also, $\rho_{c}(E^{*}_{1})=\frac{\Gamma(E^{*}_{2}=E-E^{*}_{1})}{\sum_{E_{2}} \Gamma(E_{2})}$. The $exp[\frac{S_{2}}{k_{B}}]$ factors out of both the numerator and denominator of this expression.