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.

I have a question which is inspired by considering the light field coming off an incandescent lightbulb. As a blackbody radiation field, the light is in thermal equilibrium at temperature $T$, which implies that each normal mode has a mean energy given by Planck's law, and a random phase. Thus, if I were to look, microscopically, at the electric field, I would see a fairly complicated random function that can only really be considered constant at timescales $\tau\ll \hbar/k_B T$ (at which the corresponding modes have next to no amplitude and therefore do not affect the electric field's time dependence).

This illustrates a general aspect of thermal and thermodynamic equilibrium: they are only relevant concepts when the systems involved are looked at on timescales far longer than their relevant dynamics.

My question, then, is this: are examples of slowly-varying systems (where by "slowly" I mean on the timescales of seconds, or preferably longer) that can be considered to be in thermal equilibrium on timescales longer than that known?

share|improve this question
    
+1 because I Love Lightbulb Questions Ref: physics.stackexchange.com/questions/28345/… –  Argus May 20 '12 at 18:21
    
This question is a little vague--- do you mean that they are instaneously in thermal equilibirium, but slowly varying parameters? Then the answer is yes. Or do you mean an equilibrium system that is able to slowly oscillate while being in equilibrium with respect to the oscillating variable? Here it must be no, since being in thermal equilibrium means that we must be ignorant of the phase of oscillations. But the answer could also be interpreted as yes, oscillating, but we don't know the phase. Can you clarify what you want? –  Ron Maimon May 23 '12 at 6:42
    
I would like a system where we can observe the dynamics but that can be considered to be in thermal equilibrium when it is observed over a suitably long time. Thus not instantaneous thermal equilibrium with slowly varying parameters, but rather thermal equilibrium when averaged over a long time, with dynamical variables changing slowly enough that we can observe them. –  Emilio Pisanty May 24 '12 at 11:20
    
For the case of brownian motion, we can observe the position so that the particle cannot be considered to be in thermal equilibrium; however, if you consider the motion over a long time (i.e. many "kicks" from the medium) then the mean kinetic energy is indeed $\frac{3}{2}k_B T$. –  Emilio Pisanty May 24 '12 at 11:20

3 Answers 3

Good question! When I started writing this answer I couldn't think of an example, but then I realised that Brownian motion fits the bill, as I'll explain below. So please forgive the somewhat tangential introduction:

To a reasonable degree of approximation, temperature can be thought of as energy per degree of freedom. I say approximation because in quantum systems the number of degrees of freedom can depend on the temperature as well, and then it gets complicated, but let's imagine for a moment that we live in a classical world where that doesn't happen.

The range of temperatures we observe in everyday life ranges up to a few thousands of Kelvins. (You can get much higher temperatures in some experiments, but only for very brief periods of time.) Multiplying by Boltzmann's constant, this corresponds to a maximum of around $10^{-19}$ Joules per degree of freedom in any system that you're likely to be able to sit around and watch for a long period of time.

The rate of motion this corresponds to depends on the mass associated with the degree of freedom. For example, if I surround a 1kg pendulum with air at 300K then both the gas molecules and the pendulum will have an average energy of the order $10^{-21}$ J. For the gas molecules this corresponds to a speed of about $500\,\text{ms}^{-1}$, whereas the pendulum will have an average velocity of about $\sqrt{10^{-21}}\sim 10^{-11}\,\text{ms}^{-1}$ once the system reaches equilibrium. However, the amplitude of its movement will be too small to measure, because the force restoring it to its equilibrium position will easily overwhelm its thermal movement.

However, for a mass of intermediate size, such an amount of energy gives rise to motion that is both slow enough to be perceived by the human eye, and of sufficient magnitude to be seen through a microscope. It was first observed by Brown in the case of pollen grains suspended in water. The grains jiggle around all over the place, but when looked at over long time scales they're in thermal equilibrium - their motion will continue in the same way forever. Unlike a pendulum the grains have no force restoring them to their original position, so they will wonder all over the container they're in.

The time scale can be increased by increasing the mass of the grains. However, as the grains get larger it will be harder and harder to eliminate forces that make them want to do things like sticking to the sides of the container. Still, I guess that with the right setup, and precise enough measurements, you could probably observe Brownian motion taking place on as long a time scale as you want.

share|improve this answer

in the following I'm imagining an expanding container of some ideal gas and the terms macroscopic and microscopic are employed in a real-space sense: if we define a "slowly varying macroscopic system" to be one with changes in the macroscopic parameters, that take place on time scales ($\tau$) much larger than the time scales associated with the dynamics of microscopic degrees of freedom ($h/kT$ for instance), then such a system is in equilibrium when looked upon at time scales between (with a geometrical mean) the two ($t_{e}$). For all this to be consistent, roughly speaking the microscopic degrees of freedom should follow a dynamics which amounts to no macroscopic changes on the scale $t_{e}$. Let's try to see how we can go one step further: if the slow changes in the macroscopic parameters (and I'm having the length of the container in mind) are for instance sinusoidal in time, one may try to see whether the system can be considered to be in equilibrium on some much larger time scale $t'_{e}$. This in not possible (in the container example) since the variations of the macroscopic parameters (i.e. the dimensions of the container) are not negligible at time scales (i.e. $\tau$) much smaller than $t'_{e}$. I'm aware that this is a very incomplete answer to your question, but I hope it leads to further insight, maybe through the discussions that follow.

share|improve this answer

Nathaniel is quite right, Brownian motion is indeed an example of what I'm looking for. Inspired on that, here's another example:

Consider a torsion pendulum immersed in air at temperature $T$. Then the mean potential energy of the pendulum's torsion is, for low angles, $\frac{1}{2}\kappa(\Delta\theta)^2=\frac{1}{2}k_B T$. Sensitive measurements of the pendulum's torsion angle are possible by shining a light beam off a mirror on the pendulum, and for appropriate temperatures and torsional constants the motion can be on the timescales of seconds or minutes.

I know I read this during my statistical mechanics course but I can't find a reference. (If anyone knows of one, please chime in!)

I would like to keep the question open for some time to see what other examples turn up. I would be particularly interested in examples with longer timescales, on the order of days or months, or hopefully even astronomical timescales.

share|improve this answer
    
Brownian motion is not a great example, because it is precisely when we see where the Brownian particle is that we reduce it's phase space volume, and it isn't technically in thermal equilibrium anymore. If you are looking for a system that can be considered as in thermal equilibrium but we see a snapshot that isn't, consider star motion in a globular cluster. –  Ron Maimon May 23 '12 at 6:43

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.