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

$\beta$ in statistical mechanics is equal to $\frac{1}{k_BT}$ in in thermodynamics, but I do not understand why $\beta\propto T^{-1}$ instead of, say, $\beta\propto T$?

share|cite|improve this question

Because by convention, we want to write $\beta$ as the coefficient in front of energy $E$ in the exponent $\exp(-\beta E)$. Exponents have to be dimensionless so $\beta$ has to have units of inverse energy. That's why it has to be objects such as $\beta=1/kT$ because $kT$ has units of energy. The latter statement holds because the energy per degree of freedom increases with the temperature. At the end, that's the fundamental answer to your question. The exponential is $\exp(-E/kT)$ because $E\sim kT$ and "hot" means "highly energetic". It's linked to the fact that the temperature has an absolute lower bound, the absolute zero, much like energy is bounded from below.

share|cite|improve this answer

Sometimes a definition is just a definition. The underlying math of statistical mechanics demands that we work with $E/kT$ quite a bit, and historically, we've chosen to work with $\beta \equiv 1/kT$. This can make some mathematical manipulations easier: for instance, $\frac{\partial}{\partial \beta} e^{\beta E} = E e^{\beta E}$ is easier to work with than if we explicitly left in a division (but there is "conservation of misery" here, as my advisor likes to say, where the nice things we get for doing this cost in converting from $\partial/\partial \beta$ to $\partial/\partial T$ down the road).

You're entirely free to rewrite the whole of statistical mechanics in terms of some different parameter that is proportional to $T$. Call it $\tau$ if you like. $\tau = kT$. This can be an illustrative exercise for the inquisitive mind, seeing how it makes some expressions simpler and others more laborious.

share|cite|improve this answer

$\beta$ is the fundamental quantity that appears in the laws of statistical mechanics, and comparison with the laws of thermodynamics then implies that $\beta$ is inversely proportional to the temperature. This can be seen from the first law, which says $dU=TdS-PdV$, while statistical mechanics provides at constant volume the formula $dS/k_B=\beta dU$. As a consequence, $\beta=1/k_BT$.

Therefore a ''temperature'' defined to be proportional to $\beta$ would measure coldness rather than hotness.

It is a historical accident that temperature was designed to measure hotness rather than coldness. The correspondence between statistical mechnaics and thermodynamics would be simpler if it were otherwise, but one cannot change history and the resulting tradition deeply rooted in our culture.

share|cite|improve this answer
All so very true. One advantage of using $\beta$ as hinted at by Neumaier is that the absolute zero conundrum goes away. One can make $\beta$ as small as one wishes without having any problems. But you'd have to make $\beta$ infinite to get to "absolute zero". – Paul J. Gans Nov 2 '12 at 23:39

In statistical thermodynamics, when using the method of Lagrange multipliers, we obtain an expression as

$$-\ln \rho = \alpha + \beta H$$

where $\alpha$ and $\beta$ are the multipliers to be determined. Multiplying by the Boltzmann constant and averaging we obtain the entropy

$$\langle S \rangle = k_\mathrm{B}\alpha + k_\mathrm{B} \beta \langle H \rangle$$

Comparing with the thermodynamic entropy for a closed system at constant composition (Euler expression)

$$S = S_0 + \frac{U}{T}$$

you obtain the value $\beta = 1/k_\mathrm{B}T$.

You could try an alternative method by defining a Lagrange multiplier $\beta' = 1/\beta$,

$$-\ln \rho = \alpha + H / \beta'$$

Repeating the above procedure you would obtain the value $\beta' = k_\mathrm{B}T$, but I do not find any advantage in this.

share|cite|improve this answer

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.