[Physics] When to use the Boltzmann distribution and the chemical potential

statistical mechanicsthermodynamics

How do you know when to use the Boltzmann distribution for a particular problem?

I have many polymers connected together in many different possibilities by connector agents. All are in a solvent. I wrote the partition function of the system and used the Boltzmann distribution, with the chemical potential. However, some problems do not require it. How do you identify the situations that do require the chemical potential?

Best Answer

The distribution you use depends on the ''ensemble'' you are working in. There are three kinds of ensembles:

  1. Microcanonical ensemble (or practical: an isolated system): the number of particles $N_s$ and the energy $E_s$ is fixed. This is ONE SPECIFIC REALISATION of the system. You have only one possible value for the energy and number of particles, so the chance of having $N_s$ particles and a total energy of $E_s$ equals one. The number of elements in the microcanonical ensemble is often denoted as $\Omega_{\text{micro}}(E_s,N_s)$, which is the set of all states containing $E_s$ energy and $N_s$ particles.
  2. Canonical ensemble (or practical: closed system): the number of particles $N_s$ is fixed, but the system can exchange energy! So this ensemble contains all of the possible elements of the microcanonical ensemble with $N_s$ particles. This means that every posible value for the energy is possible. The probability of finding the system in a state with energy $E_s$ is given by the Boltzmann-distribution $$P(E)=\frac{1}{\mathcal{Z}}\exp\left(-\frac{E}{k_BT}\right),$$ where $\mathcal{Z}$ is the sum of states (or partition function). The number of elements in the canonical ensemble is often denoted as:$$\Omega_{\text{can}}(N_s)=\sum\limits_{E_s} \Omega_{\text{micro}}(E_s,N_s)$$
  3. Grand canonical ensemble (or practical: open system): the number of particles and energy can vary freely! Now since the number of particles isn't fixed, our entropy can change, adding or removing particles yields more or less possible microstates and hence another entropy. So we need to add a weight, which we call the chemical potential $\mu$, to this change in number of particles so that the change in entropy: $$\Delta S=\frac{\Delta E}{T}+\frac{p}{T}\Delta V-\frac{\mu}{T} \Delta N$$ is correct according to Boltzmann's formula, which basically tells us that: $$P(E,N)\sim \exp\left(\frac{S}{k_B}\right),$$ since the probability of finding the system is proportional to the number of possible microstates. Upon applying this principle we see that we need to add $-\mu\Delta N$ to the energy in the Boltzmann distribution. A trick to memorize this term (but is wrong since the term is due to entropy and not energy) is to say that a particle ''demands'' an energy of $\mu\Delta N$. The probability of finding the system in a state with energy $E$ and number of particles $N$ is given by: $$P(E,N)=\frac{1}{\Xi}\exp\left(-\frac{E-\mu N}{k_BT}\right),$$ where $\Xi$ is the grand canonical partition function. The number of elements in the grand canonical ensemble is also noted as $\Omega_{\text{Grand}}$.

So long story short, you only need the chemical potential when the number of particles is NOT fixed.

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