In Boltzmann distribution, why is the system at the same temperature as the reservoir?

Consider a boltzmann distribution where the total energy of the reservoir and the system is $$E$$. The energy of the system can be $$\epsilon_i$$ and the energy of the reservoir is $$E-\epsilon_i$$.

Now if the system can take on different energies of $$\epsilon_i$$, why can one say that the system is at equilibirum with the reservoir and has a fixed temperature $$T$$ which is same as that of the reservoir?

At one second the system can have energy $$\epsilon_1$$ and at the next second the system can have energy $$\epsilon_2$$, with good probability of it happening. There is thus a net flow of energy between the system and the reservoir. How can then one say that the system and reservoir is at equilibirum and have the same temperature?

• I don't really understand your question. If you know the energy of the system is $\epsilon_1$ then the energy distribution of the system is a Dirac delta, so there is no question of it being a Boltzmann distribution. You need to consider a statistical averaging of some sort for any non-trivial probability distribution to make sense. – By Symmetry Oct 6 '18 at 17:59
• See the Zeroth law of thermodynamics – Alexander Oct 6 '18 at 18:02
• The small system can be so small that it does not really have a temperature. For example a harmonic oscillator. – Pieter Oct 6 '18 at 18:48
• @coniferous_smellerULPBG-W8ZgjR Just one oscillator, a single particle in one dimension. It is easiest when it is supposed to be quantized, with one level per energy, evenly spaced levels. – Pieter Oct 6 '18 at 19:08
• @TaeNyFan It is difficult to apply the Boltzmann distribution to larger systems, because one does not know how the number of microstates varies with energy. For a single gas molecule, the number of microstates is proportional to the kinetic energy, which leads to the Maxwell-Boltzmann distribution of velocities. But how would one do this for a more complex system? And all this is about equilibrium and isolated systems. – Pieter Oct 7 '18 at 13:08

The reservoir is taken to be large enough to provide the system with a very well-defined probability distribution for its energy $$\epsilon_i$$. For the theorists, this means of course that the reservoir is actually infinitely large.

A system is said to be in thermal equilibrium with the reservoir if its energy is found to obey the expected probability distribution provided by the heat reservoir. This means that one should measure the energy of the system over some longer period of time, since one is considering the probability distribution of the energy, not the value of the energy at any single instant. If the system is not in thermal equilibrium with the bath, the distribution of energies will be very different than expected. For instance, if the system is initially colder than the reservoir, its energy will be found to be smaller than expected until the system has equilibrated.

• Doesn't saying that body A and body B is in thermal equilibrium refers to body A having a definite energy and body B having a definite energy with no net exchange of energy between the two bodies? Why is the definition of thermal equilibrium different for the Boltzmann distribution? – TaeNyFan Oct 7 '18 at 12:11
• The phrase "net exchange of energy" is a bit misleading here. Energy can be exchanged all the time between the reservoir and the system, but in the long run, this exchange will average out. That is the idea of thermal equilibrium. Note that in order to have a well-defined notion of temperature, strictly speaking the heat reservoir should be infinitely large so that the energy of the reservoir does not fluctuate. In contrast, the energy of the system is allowed to fluctuate according to a Boltzmann distribution. – Stijn B. Oct 8 '18 at 8:57

Consider a small system in thermal contact with a large system which has coolness $$\beta$$. For simplicity, the small system has just one state per energy level. It could be a harmonic oscillator with a small quantum energy. The large system has a multiplicity $$\Omega_0$$ when the small system is in its ground state.

When the small system absorbs a quantum of energy, the multiplicity of the large system decreases, but its $$\beta$$ remains the same. It is a heat reservoir, either because it is large or because it is for example melting ice in water. This means that the next quantum will change the multiplicity of the reservoir with the same fractional amount. This results in a negative exponential for multiplicity of the heat reservoir as a function of the amount of energy in the small system. Because the small system has only one state per energy, this means that the multiplicity and the probability of finding the total system in a state where the small system has an energy $$E$$ is a negative exponential of $$E$$.

Mathematically, one can start this reasoning with the definition of the thermodynamic beta: $$\beta = \Omega^{-1}\ \frac{{\rm d}\Omega}{{\rm d}E}.$$

One can rewrite that as a differential equation:

$$\frac{{\rm d}}{{\rm d}E} \Omega = - \beta \Omega,$$ where the minus sign is a consequence of the energy $$E$$ of the small system is taken from the heat reservoir. This has the solution $$\Omega = \Omega_0 e^{-\beta E}.$$ The probability of finding the small system in a state with energy $$E$$ is therefore $$P(E) \propto e^{-\beta E}.$$ This is the Boltzmann factor. This derivation relies on the small oscillator being distinguishable, that is why this gives the classical distribution function.