Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I hear very often among my peers and seniors that just as how $\hbar\rightarrow0$ takes me to classical mechanics from quantum mechanics, $k_B\rightarrow0$ will take me to classical thermodynamics from statistical mechanics.

As nice as this sounds, my gut feeling and intuition tells me this is not the right analogy. I think $N\rightarrow\infty$, the ensemble size of statistical mechanical system, is closer to $\hbar\rightarrow0$ in that respect. Is this correct?

If so, what role does $k_B$ play then?

share|cite|improve this question
up vote 3 down vote accepted

They're two different limits in which two different constants are sent to zero and the resulting limiting theory has different names.

However, both of them are limits for dimensionful constants and the analogy is perfect.

The properly derived $\hbar\to 0$ limit of a quantum mechanical theory is a classical theory – its classical limit – in the very same sense in which the properly derived $k\to 0$ limit of the laws of statistical mechanics produce the laws of thermodynamics.

Now, the limit $k\to 0$ and $N\to \infty$ really means the same thing because the implicit assumption in all these limiting procedures is that the macroscopic quantities known from the everyday life are kept finite – and essentially fixed. This is especially the case for the energy and temperature. The thermal energy of $N$ atoms is something like $$ E=3kTN/2 $$ If both $E,T$ are fixed, it's clear that the $k\to 0$ limit is exactly the same thing as the $N\to\infty$ limit. The thermodynamic limit simply means that we're neglecting all effects in a fixed-energy system that are caused by the finiteness of the number of atoms – or, equivalently, the finiteness (nonzero value) of the contribution of a single atom to the whole (which is proportional to $k$).

One has the freedom to describe the limit in many ways in the quantum mechanical case, too. We may say that the classical limit appears as $\hbar\to 0$. But we may also say that the classical limit emerges when $N_J\to \infty$ where $N_J=J_z/\hbar$, for example. When the angular momentum (or the action $S$) is written as a multiple of $\hbar$, the dimensionless coefficients' condition $N_J\to\infty$ is the same thing as $\hbar\to 0$ because their product is fixed. Classical physics needs to neglect all effects caused by the situation when the overall quantities' describing the system are so small that they're comparable to $\hbar$ (which is the regime where the quantum phenomena become important). Again, we're comparing two things, so saying that one of them is infinitely larger than the other is the same thing as saying that the other one is infinitely smaller.

In both situations, one has many options what $N$ or $N_J$ may exactly be. But in both cases, the limit for the dimensionful constant going to zero, whether it's $k$ or $\hbar$, is equivalent to some dimensionless numbers' (those that measure how much the system is larger relatively to the statistical mechanical or quantum "basic blocks" where the more general theory shows in its full glory) going to infinity.

Because the ratio of probabilities of an entropy-changing process and its time reversal goes like $\exp((S_B-S_A)/k)$, we see that for fixed $S_A,S_B$ in macroscopic units, the ratio becomes strictly infinite. So in thermodynamics, i.e. the thermodynamic limit of the statistical considerations, a decreasing entropy is strictly impossible.

Finally, let me mention that the nonrelativistic limit is analogous to the two limits above, too. We may say that the limit involves $c\to\infty$ which is clearly the same thing as $1/c\to 0$: it plays the same role as $k\to 0$, for example. However, we may also say that the actual velocities are much smaller than $c$ in the limit, so $\beta=v/c\to 0$ or $1/\beta\to \infty$. That's analogous to $N\to\infty$ or $N_J\to\infty$ above.

share|cite|improve this answer

Yes, when people talked about the thermodynamic limit, they are always refer to the limit of large number of particle, or $N \to \infty$.

The role of $k_B$ is to link the microscope quantities and macroscopic thermodynamic quantities. In Newtonian Mechanics, we have already defined the momentum and kinetic energy very clear. On the other hand, Thermodynamic are developed separately with the quantities like volume, pressure and entropy. It is not clear how these two set of quantities are related, or converted, to each other, until people find the Boltzmann constant $k_B$. It might be a bit surprise that only one constant can relate all of them together, but it also indicate that how well when they develop Thermodynamics.

One relation is the average kinetic energy and temperature, which are related by $$\frac{1}{2} m \left\langle v^2\right\rangle = \frac{3}{2} k T$$ Similar between energy and temperature can also be seen in the Boltzmann factor: $$p_i = e^{-\frac{E_i}{k_B T}}$$ Another example is the relation between the entropy and number of state: $$S = k_B \ln(\Omega)$$

These relationship relating the thermodynamic quantities can only make sense when we are talking about the average, in which the system may also contact with the external environment. Large number of particles, or $N\to \infty$, guarantee that the fluctuation of the mean value is of order $\mathcal{O}(1/\sqrt{N})$ because of the central limit theorem. Hence, the thermodynamic quantities can be well defined.

A question here is why we dont use directly, say, number of state instead of entropy. One reason is that the quantity can be impractically large. Another reason is that whenever we talk about with thermodynamic quantity, we always means the average quantity with random fluctuation. It is useful for open system since it is continuously exchanging heat and particle with environment, but the quantities of energy as well as particle is not constant.

share|cite|improve this answer

It all depends on what you mean by statistical mechanics and classical thermodynamics. There are at least three versions of statistical mechanics, (1) a pure quantal version featuring density matricies, (2) a semi-quantal version in which energy levels are quantized but other things are classical, and (3) the pure classical statistical mechanics version developed by Boltzmann and Gibbs and which was the father of both (1) and (2).

I'm not sure what you mean by classical thermodynamics. Or, more exactly, what non-classical thermodynamics might be. In general all the versions of statistical mechanics lead to the results of thermodynamics as long as the temperature isn't too low or too high, the volume of the system is much much greater than its surface area, and the number of particles in the system is large, but not so large that the number density ($n/V$) gets to be too big.

This is the best I can do. If you refine your question folks might be able to come up with a better answer.

share|cite|improve this answer
I knew that there was a quantum-statistical mechanics, and classical-statistical mechanics. But what's the difference between (1) and (2)? I'm intrigued! – QuantumDot Dec 17 '12 at 2:00
In (2) one uses energy levels but in most other things classical mechanics is used. For example one has to divide the partition function by $N!$ to account for identical particles. – Paul J. Gans Dec 26 '12 at 20:06

Your Answer


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.