Boltzmann distribution derivation from Lagrange Multipliers

Question

Context

In the derivation of the Boltzmann factor and the canonical partition function based essentially on Lagrange multipliers presented here, the equalities, \begin{align*} p_j &= \frac{1}{Z} e^{\frac{\lambda_2 E_j}{k_B}} \\ Z &= \sum_j e^{\frac{\lambda_2 E_j}{k_B}} \end{align*} are established, where $E_j$ refers to the energy of a microstate, $k_B$ is the Boltzmann constant, and $\lambda_2$ is the remaining undetermined multiplier. Then, using the Gibbs entropy $S = - k_B \sum_j p_j \ln(p_j)$ along with some algebra, \begin{gather*} S = -\lambda_2 U + k_B \ln(Z) \end{gather*} Finally, the article suggests the derivative, \begin{gather*} \frac{\partial S}{\partial U} = - \lambda_2 \end{gather*} without further work. Note that the article also improperly uses a total derivative instead of a partial here as the whole point is to make the identification $\frac{\partial S}{\partial U} = - \lambda_2 = \frac{1}{T}$.

My Question

My question comes from the details of working out the partial derivative. I obtained, \begin{align*} \frac{\partial S}{\partial U} &= - \lambda_2 - U \frac{\partial \lambda_2}{\partial U} + \frac{k_B}{Z} \frac{\partial Z}{\partial U} \\ &= - \lambda_2 - U \frac{\partial \lambda_2}{\partial U} + \frac{1}{Z} \sum_j e^{\frac{\lambda_2 E_j}{k_B}} \left[ E_j \frac{\partial \lambda_2}{\partial U} + \lambda_2 \frac{\partial E_j}{\partial U} \right] \\ &= - \lambda_2 + \frac{1}{Z} \sum_j e^{\frac{\lambda_2 E_j}{k_B}} \lambda_2 \frac{\partial E_j}{\partial U} \end{align*} where we have used the statistical definition of $U$ and the equalities above to cancel one of the derivative terms involving $\frac{\partial \lambda_2}{\partial U}$. It is at this point that I have a question. Clearly, for this to work, we must have $ \frac{\partial E_j}{\partial U} = 0$. But I cannot seem to really justify this step to myself. Is it simply that the energies of the particular microstates do not explicitly depend on the average energy? How would one justify this?

Answer 1

Clearly, for this to work, we must have $ \frac{\partial E_j}{\partial U} = 0$. But I cannot seem to really justify this step to myself. Is it simply that the energies of the particular microstates do not explicitly depend on the average energy? How would one justify this?

This is because when we write $$ \frac{\partial S}{\partial U} $$ with the intention of getting inverse of thermodynamic temperature, we have to assume that all the other thermodynamic state variables that $S$ depends on are constant; otherwise the derivative need not be function of temperature. In thermodynamics, $S$ is function of at least $U,V,N$, so the derivative is actually

$$ \frac{\partial S}{\partial U}\bigg{|}_{V,N} = \frac{1}{T}. $$

Assuming constant $V,N$ in quantum theory implies constant Hamiltonian, and thus constant energy levels $E_j$. Changing $U$ in this partial derivative is thus allowed to happen only via changing the probabilities $\rho_j$'s, not via changing the energy levels $E_j$'s.

Answer 2

Imo, your doubt is related to the assumption that the partial derivative of the energy of a microstate, $(E_j$), with respect to the total energy, $(U$), is zero, i.e., $(\frac{\partial E_j}{\partial U} = 0$).

In the context of statistical mechanics, the assumption $(\frac{\partial E_j}{\partial U} = 0$) is essentially an assumption about the statistical ensemble you are working with. Specifically, it assumes that the energy levels of the microstates are not influenced by the total energy of the system, (U). This assumption is valid for certain types of ensembles, such as the microcanonical ensemble or the canonical ensemble.

The microcanonical ensemble describes an isolated system with a fixed total energy, (U). In this ensemble, the energy of each microstate $(E_j$) is a constant because the total energy is fixed, and the system cannot exchange energy with its surroundings. Therefore, $(\frac{\partial E_j}{\partial U} = 0$) in this ensemble.

The canonical ensemble describes a system in contact with a heat bath at a constant temperature, $(T$). In this ensemble, the energy levels $(E_j$) of the microstates are typically assumed to be fixed properties of the system and not influenced by the total energy $(U$). This assumption is reasonable when dealing with macroscopic systems where the interactions between particles are weak, and the energy levels do not significantly change with variations in the total energy.

So, the assumption $(\frac{\partial E_j}{\partial U} = 0$) is justified in the context of these ensembles. Buuuut, it's important to note that this assumption might not hold in all situations. For systems with strong interactions or where the energy levels are explicitly dependent on the total energy, this assumption may not be valid, and you would need to consider more complex models.