I have started studying thermodynamics and they defined temperature in a statistical way:

$$ \frac{1}{k_BT}=\frac{d \ln(\Omega)}{dE} $$ where $k_B$ is the Boltzmann constant, $T$ is the temperature, $\Omega$ is the number of micro-states for a given energy and $E$ is the energy.

I don't really see how this corresponds to our general idea of temperature. So if someone can explain how this definition encompasses how we, colloquially, define temperature, I would be grateful

  • 3
    $\begingroup$ Possibly relevant: physics.stackexchange.com/q/65229 $\endgroup$
    – QtizedQ
    Jun 19, 2017 at 20:34
  • $\begingroup$ Your title and question body seem to be asking two slightly different things. Are you asking how to see that this definition of temperature corresponds with a more conventional one, or are you asking why it is "good"? In the former case, which more conventional idea of temperature do you have in mind? In other words, what do you mean by "our general idea of temperature"? Or in the latter case, what makes a definition "good"? $\endgroup$
    – David Z
    Jun 19, 2017 at 20:37
  • 3
    $\begingroup$ More on the definition of temperature. $\endgroup$
    – Qmechanic
    Jun 19, 2017 at 21:27

1 Answer 1


A "good" definition (although the term is vague) must replicate at least some of our major intuitions about the notion that is being defined.

A very fundamental everyday thing that a scientist groping for a notion of temperature might be seeking is that the nett rate of heat flow between two bodies of the same temperature must be nought. When a hot body is in contact with a cold one, the former will heat the latter until they are the same "temperature", whatever that means.

Now consider two bodies, with energy $E_1$ and $E_2$. Let the number of microstates compatible with the macroscopic state of each body be $\Omega_1$ and $\Omega_2$. The total number of microstates of the system as a whole compatible with the system's macrostate is $\Omega_1\,\Omega_2$. What happens when we put the bodies into contact with one another, and assume that they are otherwise isolated?

In my answer here I argue that it is overwhelmingly likely is that any system will wind up in a microstate that is very near to the maximum likelihood microstate, just by undergoing a random walk because, for large ensembles, the set of microstates contains states that look very like the maximum likelihood microstate and almost nothing else (I do some simple calculations with the binomial distribution to show this in my other answer).

So, we conclude that heat is going to flow between the bodies until the system finds its maximum likelihood microstate. Assuming all microstates equally likely, this will be the microstate that maximizes $\Omega_1\,\Omega_2$. At equilibrium, $\Omega_1$ and $\Omega_2$ are functions $\Omega(E)$ of the internal energy of each subsystem. So we write down an equation that maximizes $\Omega(E_1)\,\Omega(E_2)$, subject to the constraint that $E_1+E_2=const$ (assuming the two systems are isolated).

We get, of course:

$$\left.\frac{\mathrm{d}\Omega}{\mathrm{d} E}\right|_{E=E_j}+\lambda=0$$

for $j=1,\,2$ and $\lambda$ our Lagrange multiplier. Well, that's our answer: heat will stop flowing between the bodies when and only when:

$$\left.\frac{\mathrm{d}\Omega}{\mathrm{d} E}\right|_{E=E_1}=\left.\frac{\mathrm{d}\Omega}{\mathrm{d} E}\right|_{E=E_2}=-\lambda$$

so if we define temperature to be some function of $\frac{\partial\Omega}{\partial E}$, we replicate the fundamental property that heat will flow between two bodies of different temperature in contact with one another, whilst no heat flows between two bodies of the same temperature.

Okay, then, what about the initial direction of heat flow? Further careful reasoning like the above then shows that it is the body with greater $\frac{\partial\Omega}{\partial E}$ that gains heat. Therefore, temperature has to be a monotonically decreasing function of $\frac{\partial\Omega}{\partial E}$.

The simplest choice is $T^{-1} \propto \frac{\partial\Omega}{\partial E}$, although, from the reasoning above alone, this choice is far from unique. Ultimately the definition was refined after our knowledge of statistical mechanics grew until we realized that many things could be explained by assuming that the statistical concept of entropy is the same as the classical thermodynamic notion of Carnot and Clausius, as I explain in my answer here. If we do this, then indeed the temperature has to be the reciprocal of $\frac{\partial\Omega}{\partial E}$, if we are to replicate Carnot's definition of temperature in terms of ideal heat engine efficiency. See also some further ideas on temperature and its definition in my recent answer here.


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