Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In classic thermodynamics one can derive the Maxwell Boltzmann statistics by solving a Lagrange multipliers equation. In this process a new parameter $\beta$ is introduced to take account of the total energy constraint,

$$L = \beta(E-\sum_i p_i \epsilon_i)+...$$

and it is later connected to $\frac{1}{k_B T}$ by inserting our knowledge about perfect gases. On a side note, about every book around seems to skip the passage with a frustrating "it can be shown that".

My problem is that, given similar Lagrange multipliers equation, I'm not sure whether one can in general relate $E$ and $\beta$: it is evident that $\beta$ has no physical significance by itself, and it's always bounded to some parameter in the constraints that it is multiplying.

For example, say that I have an array of cells that can be either occupied by 0, 1 or 2 dots. I have some constraints, one of which is the density of the dots $\rho$. The Lagrange multipliers equation has the form

$$L = \gamma(\rho-p_1-2p_2)+...$$

where $p_i$ is the probability of having a cell with $i$ dots. Is there some way to eliminate the $\gamma$ parameter in the final explicit formula for $p_1$, and use $\rho$ instead?

share|improve this question
add comment

1 Answer 1

You probably realize that the lagrange multiplier $\gamma$ that you introduced is nothing but the chemical potential right?

Coming back to the Lagrange multiplier $\beta$, you have many ways to convince yourself that it is inversely proportional to the temperature. I will just focus on some of them:

  • the first one consists in devising a simple way to measure experimentally $\beta$. The simplest model you can think of is evidently the ideal gas. You can work out easily that the pressure in this case will be inversely proportional to $\beta$. Now, it is logically very important to remember that the canonical ensemble characterized by the set $(\beta, N, V)$ is model independant in the same way that setting the temperature in usual thermodynamics has nothing to do with the system under study. Therefore, if you find it to be an inverse temperature in one case, it has to be it for every case you can imagine.
  • the second one might actually be easier. The Lagrange multiplier method arises in a more global strategy called a statistical inference taht has been much popularized by E.T. Jaynes in the second half of the last century. The idea is to start from Shannon's entropy and to maximize it with respect to some constraints that characterize the statistical ensemble. If you replace the probability within Shannon's entropy by the canonical one, you get the expression for the entropy in the canonical ensemble. It reads:

    $S(\beta,N,V) = -k_B \sum_{m} \frac{e^{-\beta \epsilon_m}}{Q(\beta, N,V)}\ln \left( \frac{e^{-\beta \epsilon_m}}{Q(\beta, N,V)} \right)$

    where the sum runs over all the microstates $m$ and $Q$ is the canonical partition function.

    Now you can compute what is the value of $\frac{\partial S}{\partial \epsilon}$ and you will find out that it is $k_B \beta$...that is $1/T$ if you trust thermodynamics.

  • There is actually another easy way that consists in starting from the microcanonical ensemble and define the temperature from there. If you then subdivide a system into a big part and a small part, you will find directly the Boltzmann weight with the temperature of the big system for any microstate of the small part. Now, if you separately model this small system with your Lagrange multiplier you will find the same law but with $\beta$ instead of $1/k_B T$. Now, this is true whatever the system and if we moreover assume that the partition function is a monotonic function of $\beta$ (unless we are at a phase transition), we have to conclude that $\beta = 1/k_B T$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.