# Friction at zero temperature?

By the fluctuation-dissipation theorem (detailed-balance for Langevin equation), $$\sigma^2 = 2 \gamma k_B T$$ where $\sigma$ is the variance of noise, $\gamma$ is a friction coefficient, $k_B$ is Boltzmann's constant, and $T$ is temperature. So in principle, one can have $\gamma\neq 0$ while $T=0$ and $\sigma=0$.

Is it indeed possible to experimentally achieve a system whose temperature and noise approach zero, but whose friction coefficient $\gamma$ does not approach zero?

• If yes, what would be an example of such a system? What is the physical source of friction for such a system?
• If not, why not? Is there some sort of "quantum" correction to the fluctuation-dissipation theorem that rules out such zero-noise, non-zero friction systems?
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The reason you are getting confused is because the theorem only applies to small forcing where the response is linear. For block on block, this is a force which only causes elastic deformation, not sliding of the block. In this case, the phonons you produce by shaking the block will have no dissipation ideally at zero temperatures. –  Ron Maimon Jul 7 '12 at 21:18
Hi Ron, let me try to say in my own words what I think you said, as an exercise to see if I understood you right. Please pardon any inaccuracies. Am I right to think that your answer is "no, friction necessarily approaches zero at zero temperature" ? And the reason you want to give is purely classical: you want to say that the fluctuation-dissipation theorem is only applicable in the linear, near-equilibrium regime, and the appropriate classical correction to it near zero temperature will already force friction to be zero. –  manoj Jul 9 '12 at 6:06
Partly--- the near equilbrium stuff will have no friction, but if you push objects to slide on top of one another, or make big deformations (like running a big current) there's no constraint, because you're not in the linear regime. –  Ron Maimon Jul 9 '12 at 7:54
Side question. Would any type of friction cause heat therefore friction prevents absolute zero? –  Argus Aug 6 '12 at 21:47

Mechanical friction is a perfectly fine example. The coefficient of friction between two materials does not approach zero at absolute zero.

Electrical resistance (as pointed out by Alexander) is another example. Some materials (superconductors) have zero resistance at absolute zero, but by no means all of them!

I would say that $\gamma \neq 0$ while $T=0$ and $\sigma=0$ is the "default" expectation that occurs most of the time. Things like superconductivity and superfluidity are interesting surprises that go against the normal expectation.

The physical sources of friction at absolute zero are generally the same as the physical sources of friction at other temperatures. For example, electrical resistance can come from electrons bumping into grain boundaries or impurities or defects etc. Mechanical friction comes from phonons (vibrations) that are created as the two materials rub against each other ... same as usual.

If a source of friction is temperature-dependent, it can either increase or decrease as you approach absolute zero.

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electrical resistance is not a good example, as theoretically, when all the magnetic impurities become entangled with each other, and the temperature is ridiculously low, and the voltages are ridiculously small, the resistance in any wire will approach zero at zero temeperature even if it isn't superconducting. This is just very difficult to achieve because of the separation between impurities. –  Ron Maimon Jul 7 '12 at 21:16
@RonMaimon: Are you sure about this? In my interpretation the Kondo effect gives a log-increase of resistivity at low temperatures and then a finite value for $T\rightarrow 0$, but not $R=0$. Additionally you still have non-magnetic impurities where the electrons can scatter even at lowest temperatures. –  Alexander Jul 8 '12 at 6:43
@Alexander: My feeling was that the kondo magnetic impurities at insanely low temperature have to freeze out into singlets by pairing their spins, and electrons which do not have enough energy to break the singlets will not scatter irreversibly off these paired Kondo impurities. You can imagine how ridiculous this temperature is just by thinking about how small the aligning/antialigning force will be between two magnetic impurities, so in practice you are right, but I think in theory you must be wrong, because of the third law--- the true ground state shouldn't have macroscopic entropy. –  Ron Maimon Jul 8 '12 at 6:56
... even with a density of magnetic impurities, so the spins of the impurities must pair up. Similarly, for non-magnetic impurities, the electrons will just fill a Fermi surface which fills the levels in the external potential, and you will see resistance go to zero in theory. This is assuming the electrons are still delocalized at zero temperature, and aren't localized by the impurities. This is a pure theoretical question, in practice, metals will have a constant or growing resistance at low temperature, I agree, but the OP asked a question of pure theory, and I am not sure you are right. –  Ron Maimon Jul 8 '12 at 6:58
Thanks, @Steve. I guess this answers part of my question, that friction can remain non-zero as temperature goes to zero. How about noise? Does the noise actually go to zero for mechanical friction at temperature zero? If yes, why do the phonons not lead to any noise? If not, then $\sigma^2=2\gamma k_B T$ is violated, which is ok, but I would like this confirmed. (Perhaps this is what Ron Maimon was referring to when he said that in the non-linear regime, fluctuation-dissipation does not hold?) –  manoj Jul 9 '12 at 15:28
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Isn't the friction here the mechanical analogy to the resistance in a circuit? At $T=0$ the voltage noise is zero but you still have the finite property 'resistance'.

In more general terms, the dissipation is given by the imaginary part of a generalized susceptibility $\chi$ of your physical system. So as long as your system can dissipate energy you can have non-zero friction at any temperature.

An example is a particle in a liquid under Brownian motion (fluctuation and dissipation for Brownian motion).

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A simple approximation of your question, from the semiclassical point of view, could be this.

Imagine a ball running over a surface. This surface is made up of other little balls: these balls are considered little respect to the one running over them so they create a flat surface. Imagine that these little balls start to moving up an down: the surface won't be flat and we call what is experienced by the big ball in this way "friction".

So let the little balls be independent quantum harmonic oscillators. You find out that the energy of one of these is $E=h\nu (n+1/2)$. For the ground state, $T=0K$ and $n=0$ and $E=h\nu/2$. So it's not zero. So at $T=0K$ there is still a remaining movement of the atoms of the lattice that constitute the surface and so they can cause friction in the sense said before.

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This is a horrible description of friction. The bumpiness of a surface does not automatically cause friction, because gravity is a conservative force. There has to be a mechanism turning large-scale motion into microscopic vibrations (i.e. heat). –  Steve B Jul 7 '12 at 19:03
Sorry, what you say is true also because on average over time the surface is flat. –  edwineveningfall Jul 7 '12 at 20:55
Thanks, @edwineveningfall. In this case, won't the noise also be non-zero? –  manoj Jul 9 '12 at 6:09