Talk:Absolute zero

The article mentions that the "ground state" energy is not zero. I assume that it is refering to the Zero Point Energy (ZPE), which arises in the "particle in a box" problem and elsewhere. While it is true that the ZPE is non-zero, is it considered kinetic energy or some other form of energy? (MS)

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It is kinetic energy. You can solve for the velocity of the particle, and the answer is non-zero. (GJ)

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On a related note, at temperatures near absolute zero, atoms move very slowly. At absolute zero are the atoms completely stationary or is there some zero point vibrational frequency? (MS)

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For a particle in a crystal at low temperatures, simple harmonic motion is a better approximation. Again, there is non-zero velocity and kinetic energy in the ground state, except when the particle is instantaneously motionless at the extremes of the SMH. It was about 20 years ago that i knew a little quantum mechanics, so i can't quote formulae any more. Try an undergraduate quantum mechanics text book (sold mine long ago :-() (GJ)

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Finally, at least for the case of temperature defined in terms of the heat engine (Lord Kelvin's definition of temperature), where temperature is related to the efficiency of a reversible heat engine:

  efficiency = 1 - TH/TC

it is impossible to obtain a temperature below 0 K. If such a temperature were possible, it would be possible to develop a thermodynamic cycle that exhausted heat at that temperature. The result would be an engine with efficiency >100%, which violates the first law of thermodynamics. (MS)

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I think you mean 2nd law. I'm not sure if Kelvin's definition is one of the ones that allow negative temperature. Even if it does, the above reasoning might be faulty. At negative temperatures a lot of other things go negative or just plain weird. I called for real thermodynamicists here because my physics is now only half remembered and i am confused about some of these things. (GJ)

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Efficiency >100% means that you get out more energy than you put in. This violates the 1st law (yes, it also violates the 2nd law). In addition efficiency = 100% violates the 2nd law, except for the special case of reversible processes. Since in practice it is pretty much impossible to implement a reversible process, the 2nd law implies it is impossible to even reach absolute zero.(MS)

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Also, even for population inversion, isn't the ground state still defined as the lower energy of the two states?

--Matt Stoker

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Sometimes, sometimes not. In most practical lasers there are three or four relevant quantum states and the population inversion involves only two of them, not the ground state, and for a four level system not the highest. The negative temperature figure comes about only if you include only the atoms (or whatevers) in the two "inverted" states of the population inversion.

It is also possible to build a two level laser system, with the ground state being the low energy state in population inversion, but for various practical reasons these are a lot less fun.

The definition of temperature usually used for such systems is, derivative of entropy with respect to energy (perhaps multiplied by a constant for the system of units you are using). Adding energy to a system with population inversion actually decreases the entropy, and subtracting energy increases the entropy, hence negative temperature.

BTW for such systems, there are two good approximations to absolute zero. One is with all atoms in the low energy state. The other is with all in the high energy state (perfect population inversion).(GJ)

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I did a little reading from: <a class=encyclopedia href="http://www.google.com/search?q=cache:FJjlpW9h4es:math.ucr.edu/home/baez/physics/neg_temperature+absolute+zero&hl=en">Physics FAQ</a>

One important distinction is that the negative temperature, so defined, is for only one mode (eg. the nuclear spin mode). Thus, the entire system (i.e. all modes) still has a positive temperature. This is why such a negative temperature is possible without violating the 1st & 2nd laws of thermodynamics. It would not be possible to exhaust heat to only the negative temperature mode (other modes would still play a role), so T_C would still be >0 and efficiency < 100%. (MS)

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All those "billionths of a degree absolute" record holders are using this kind of definition of temperature, and they are applying to a specialised system such as "only the atomic nuclei in the material, and only for their temperature calculated with energy states in the magnetic resonance system i am running". Geronimo Jones

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Actually, no, the "billionths of a degree above absolute zero" record holders are defining temperature in terms of the velocity of the atoms (translational mode). Since the cooled particles are gaseous atoms, they have no rotational or vibrational modes. For this case nuclear modes should be negligible. They probably are neglecting electronic modes, since they have no way to measure them. This could be an issue, since they are using the momentum transfer associated with photon absorption to slow the moving atoms, which would result in electronic excitation. However, I am fairly certain the lifetimes of the excited electrons are short enough that on average the electronic energy is quickly redistributed to the translational mode via photon emission. In fact, I am pretty sure it is this energy transfer due to photon emission that limits the minimum possible temperature that is obtained (20nK). (I recently attended a lecture on this by William D. Phillips, who received the 1997 Nobel Prize in Physics for this work. He spent quite some time discussing this energy transfer, since the original theoretical calculations indicated a larger energy for the electronic transition than was actually the case and when the experiment resulted in a lower temperature than was theoretically possible they had to go back and revise the theory).

The application of this particular low temperature work is for atomic clocks, the accuracy of which are improved by using very slow moving atoms. Check the NIST website for details. --Matt Stoker

How can they have a gas at a few billionths of kelivns? Doesn't helium, the hardest element to condense, liquify at 4K? I could understand if they could stretch the condensation point down to 1 K or .1 K by playing with the volume, but even just down to a μK is a difference of 6 orders of magnitude. Especially with a gravitational field pulling the atoms toward the bottom of the tank, I couldn't see them getting anywhere near absolute zero without letting the gas condense.

Also interesting to note is that there are rotational modes available to single atoms. The problem is that the amount of energy necessary to access them is enormous compared to the energy involved in temperature. because of the quantization of angular momentum.--BlackGriffen

Check the NIST website [1] (http://www.nist.gov) to be sure, but I believe the reason it is considered a gas and not a solid is that the sample consists of a very small number of atoms (on the order of ten or less) under high vacuum. In order for condensation to occur the atoms must collide and stick together. Under the experimental conditions collisions are fairly rare, so condensation takes a long time. In other words condensation is kinetically limited. Under some conditions, the scientists were able to observe formation of a Bose-Einstein condensate, so some form of condensation does occur. -- Matt Stoker