I realized that I had never noticed that $\left[ \frac{\hbar}{k_B T} \right]=$ Time. At $T \approx 300 K$, we have $\frac{\hbar}{k_B T} \approx 10$ fs. What, if anything, does this quantity mean? Does this set the time scale for any processes? Is it used in any calculations?


3 Answers 3


You're doing something wrong: the units of $h$ are energy*time, not energy/time.

With that said, this ratio $k_B~T/h$ is the higher end of the frequency of the characteristic vibrations created by random thermal excitations. These vibrations could be phonons, for example, but also photons, and if you have electronic excitations (in chemical bonds for example) then the frequency of light which excites those states normally is now being emitted/reabsorbed by the substance as part of its thermal activity.

So for example define $\beta = 1/(k_B~T);$ then Planck's law says $$I(\nu,\beta) = \frac{ 2 h}{c^2} ~ \nu^{3} ~ \left(e^{\beta~h~\nu}-1\right)^{-1}$$and to find the peak frequency take a derivative with respect to $\nu$ and set it to zero: $$\frac {2h}{c^2}\left[3\nu^2 \left(e^{\beta~h~\nu}-1\right)^{-1} - \nu^3 ~ \beta ~ h ~ e^{\beta~h~\nu}\left(e^{\beta~h~\nu}-1\right)^{-2}\right] = 0,$$or,$$3 \left(1 - e^{-h ~\beta~\nu}\right) = h~\beta~\nu.$$We can quickly solve this expression by defining $f(x) = 3~(1 - e^{-x})$ and computing $f(f(f(\dots f(1)\dots))),$ which converges on some value $f(x) = x$ for $x = 2.821439372122\dots,$ which is probably some complicated transcendental, but whatever.

Therefore we know that the peak frequency for blackbody radiation is $$\nu \approx 2.8214 / (h \beta) = 2.8214 \frac {k_B T}{h},$$which is an alternate form of Wien's displacement law.

  • $\begingroup$ Woops, I had the fraction flipped upside down (calculated it correctly, however). This is exactly what I was looking for. Thanks! $\endgroup$
    – F. Bardamu
    Aug 17, 2015 at 16:38
  • $\begingroup$ @F.Bardamu well if you flip it then it's just the period of the same. You could probably also say that anything which happens on that time scale is "washed out" by thermal effects or something like that; quantum fluctuations happen with $\Delta E~\Delta t \approx h$, looking for fluctuations with energy above $\Delta E = k_B T$ would require the timescale ~ h / k T. $\endgroup$
    – CR Drost
    Aug 17, 2015 at 16:43

As in a previous answer $\frac{\hbar}{k_B T}$ is the coefficient in front of frequency in Planck’s Law (where I am using $\omega = 2\pi \nu $). $$ I_\omega(\omega, T) = \frac{ 2 \hbar}{c^2} (\frac{\omega}{2\pi})^3(e^\frac{\hbar\omega}{k_B T} - 1)^{-1} $$ $c(\frac{\hbar}{k_B T})^{-1}$ is also the acceleration $a$ that gives the Unruh temperature T. $$ a=c(\frac{\hbar}{k_B T})^{-1} $$ Any object accelerating with a constant acceleration $a$ will be bathed in a Planck spectrum of temperature T of thermal radiation from the vacuum. Likewise, it is expected that a non-accelerating observer will see the accelerating object to be emitting a Planck spectrum of temperature T (called Unruh radiation or called Hawking radiation if $a$ is the acceleration at the Schwarzschild radius of a black hole).

It is interesting that your combination of constants which seem to have so much to do with quantum mechanics, can just be replaced with a simple acceleration that seems to have nothing to do with quantum mechanics.


The quantity $\hbar/k_BT$ comes up in studies of strongly correlated materials. This story is a little complicated but pretty interesting. Empirically, it has been found (1) that many strongly correlated materials have a resistance proportional to temperature, and if one works out the scattering timescale associated with this using something like the Drude formula for conductivity it is always quite close to $\hbar/k_BT$. There are some ideas about why this might be the case (2). The arguments go something like this: a typical excitation above the Fermi surface of your system has energy $k_B T$. From energy-time uncertainty principles, that means that if this excitation is sharply defined it must last for time $\hbar/k_BT$. It is therefore conjectured that this time is a bound on how fast dissipation of an excitation in a system, such as electrical resistance in materials, can possibly happen. Strongly interacting systems, which tend to have very strong scattering and dissipation processes, would have universal scaling because they all saturate this bound. Tentative support for this kind of dissipation bound comes from, of all places, studies of black holes via holographic principles (3).

I should emphasize that this is quite a recent and controversial idea. As you might imagine, many people are very skeptical that black holes can be used to learn about quantum materials! However, if it turns out to be true, there is a very nice answer to your question: 10 fs would be the maximum characteristic speed at which an excitation in a room temperature system can decay and diffuse.


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