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I have a basic understanding of thermodynamics, and I came across this derivation of the Boltzmann distribution. I understand all of it except the end and I need some clarification.

At the end, the website claims that the number of particles with energy $E_i$ denoted as $n_i$ is given by: $$n_i=\dfrac{N}{\sum_i e^{-\beta E_i}}e^{-\beta E_i}$$

From what I understand, the Boltzmann Distribution tells me the probability of finding a particle with energy $E_i$, so I simply need $\dfrac{n_i}{N}$?

My last question is that the website abruptly claims that $\beta=1/k_BT$, which apparently comes from applying the zeroth law of Thermodynamics. How can show this is true?

Ayumu Kasugano
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  • See https://physics.stackexchange.com/a/206540/26129 – higgsss Apr 03 '18 at 04:22
  • The point is how the Lagrange multiplier $\beta$ is equal to $ \partial \ln \Omega/ \partial E = 1/kT$, where $\Omega$ is the number of microstates, and $T$ is the thermodynamic temperature. In fact, a theorem about constrained extrema (stated and proved in the linked post) guarantees this. "Working with real systems and applying the zeroth law of thermodynamics" is an extraneous statement. – higgsss Apr 03 '18 at 07:12

2 Answers2

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Taking $$\beta = \frac{1}{k_B T}$$ is a standard notation (it is just redefining the temperature parameter). This is to say that it is a definition of $\beta$ rather than a result from the third law of the thermodynamics. The answer by @Pieter however makes a good point that one could equally define coldness and derive this relation.

Probability of a particle occupying energy state $i$ is the ratio of the number of particles in this state to the total number of particles, when the letter goes to infinity, i.e. $$p_i = \lim_{N\rightarrow +\infty}\frac{n_i}{N}.$$ This is actually not a result, but the definition of probability (in the frequentist sense, which is what almost exclusively used in physics).

Update
To address the comments by @NakshatraGangopadhay : It is important that the probability is defined as a limit for an infinitely big system: both $n_i$ and $N$ become infinitely large, but their ratio remains constant.

If we take a finite system of $N$ particles with the probability $p_i$ of finding each particle in state $i$, the probability of finding $n_i$ particles in this state is given by the binomial distribution (ignoring indistinguishability for simplicity): $$P(n_i)={N\choose n_i}p_i^{n_i}(1-p_i)^{N-n_i},$$ so that the average number of particles in the state is $$\langle n_i\rangle = p_i N.$$ We could also calculate the variance as $$\sigma_{n_i}^2 = Np_i(1-p_i).$$ The relative error in the number of particles is then $$\frac{\sigma_{n_i}}{n_i}\propto \frac{1}{\sqrt{N}}.$$

In the limit $N\rightarrow +\infty$ we have $$ \lim_{N\rightarrow +\infty}\frac{n_i}{N}=p_i,\lim_{N\rightarrow +\infty} \frac{\sigma_{n_i}}{n_i}\rightarrow 0.$$

That is, the fluctuations of the particle number around its average become negligible. This thermodynamic limit is crucial for the precise nature of the thermodynamic laws - although we never truly deal with an infinite system of particles, the errors are negligible when applying the laws of thermodynamics to systems with a number of particles of the order of the Avogadro number, $N_A\approx 10^{23}$.

This reasoning very much underlines the Einstein's claim about thermodynamics being beyond doubt.

Roger V.
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  • I had a confusion with this last part, regarding the definition of probability. This basically says, that if you know $p_i$ and you know $N$, you can easily find out $n_i$. However, doesn't this create a small problem? Like if we know $n_i$, we actually know what particular microstate the entire $N$ particle system is in. Since we know there must be $n_i$ particles in this state, the only microstates available to the entire system is the one where there are $n_i$ particles in the state $i$. All other microstates, where there are more or less particles in the $i$ state, must be impossible – Nakshatra Gangopadhay Nov 21 '21 at 23:38
  • However, that cannot be the case, as we can calculate the probability of any microstate, of the entire system, using the Boltzmann factor and the $N$ particle partition function, and the result comes to be finite. Every microstate has some finite probability of occurring depending on the energy. But if we use the definition of probability to find out the actual number of particles in a state, aren't we just neglecting all the other microstates where there are more or less number of particles in that particular state. – Nakshatra Gangopadhay Nov 21 '21 at 23:41
  • We can easily find the probability of a particle being in any of the energy states. From here, we can find the number of particles in each of these microstates. So, if we know the number of particles in each of these microstates, isn't there only one value of total energy possible for the entire system, while all other values of total energy of the entire $N$ particle system have $0$ probability of occurring. By finding the population of each of these microstates, aren't we somewhat restricting our system ? – Nakshatra Gangopadhay Nov 21 '21 at 23:43
  • Can you please point out the flaw in my reasoning ? – Nakshatra Gangopadhay Nov 21 '21 at 23:44
  • @NakshatraGangopadhay see the update in my answer. – Roger V. Nov 22 '21 at 09:12
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When $\beta$ is the same for systems A and B as well as for B and C, also A and C will be in thermal equilibrium. That is what the zeroth law is about. For historical reasons, the connection with the thermodynamic temperature is $\beta = 1/kT$. But one could have used coldness $\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$ instead. For example at room temperature, the coldness is about 4 % per milli-eV. I wrote about that in an earlier answer.