Is it possible to achieve a negative absolute temperature?

Question

Is "0 K" or "\(-273°\mathrm{C}\)" really the minimum bound on temperature? Is it possible to achieve a negative temperature, and if so, have the properties of a negative temperature been studied?

Comments

It is just like length - it has the minimal value equal to zero.

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That's not a fair comparison.

If you think about the 0°C it has a meaning because it reflects the fact when water freezes. Negative values of °C are acceptable and are found in the universe.

So think about the question that way, is 0K has to do with some fact in the universe?

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Not grad level but did you check the Wiki page? All is there: http://en.wikipedia.org/wiki/Absolute_zero

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Have a look at this article  :" physicists have created an atomic gas with a sub-absolute-zero temperature for the first time"

To understand this article one should have a background in thermodynamics and how temperature is defined in statistical mechanics.

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Hi mostassi,

PhysicsOverflow is for physics at graduate-level and above. This question is not graduate-level, I am therefore voting to close.

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 @moatassi please turn your answer into a comment.

@dilaton: temperatures $<0$K are definitely graduate level. In undergraduate physics, always $T\ge 0$. 

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@ArnoldNeumaier I see, then please feel free to vote down my closevote from the link in order to vote "leave open". If it does not get two net upvotes, it will do nothing to the question that will stay open. I leave the link to the closevote here for (new) members to learn how community moderation works on PO ...

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I have edited the question to make it a bit more appropriate for PhysicsOverflow. The OP @moatassi may revert it, but the edit doesn't really change the meaning of the question as such.

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@Dilaton @Dimension10

Good questions like this must be redirected to Physics Stack Exchange, since they entertain these questions.

@moatassi Please understand that we have decided to focus on currently active research topics. It is not because the question itself is bad.

Most aspects about your question(except perhaps experimental aspects) have been answered many many decades earlier. For instance I think earliest experiments on gas laws confirmed that -273 C is absolute temperature simply by an extrapolation graphically. I did this experiment in a basic lab in my school and I got reasonable answers.

Answer 1

Objects with temperature 0K are infinitely cold, so it is clear from the definition that they cannot get colder. From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=(k_B T)^{-1}$, where $k_B$ is Boltzmann's constant and $T$ temperature in Kelvin. As $T\to 0$, $\beta\to\infty$, proving the statement. 

This doesn't mean that objects with a temperature $<0$K do not exist. However, these are extremely hot and not extremely cold! Indeed, if you insert a negative number into the formula for $\beta$ one gets a negative value, which means that the object is less cold and hence hotter than any object with arbitrarily large positive temperature! 

On the other hand, ordinary objects cannot have negative temperature; it is a theorem of statistical mechanics that translation invariant systems (i.e., those that can in principle move by arbitrary amounts) must have $T\ge 0$. The only systems that can have negative temperature are those embedded inside a rigid material.

Some references:

D. Montgomery and G. Joyce. Statistical mechanics of “negative temperature” states. Phys. Fluids, 17:1139–1145, 1974. 
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19730013937_1973013937.pdf

E.M. Purcell and R.V. Pound. A nuclear spin system at negative temperature. Phys. Rev., 81:279–280, 1951. 
http://prola.aps.org/abstract/PR/v81/i2/p279_1

Section 73 of Landau and E.M. Lifshits. Statistical Physics: Part 1,

Example 9.2.5 in my online book Classical and Quantum Mechanics via Lie algebras.

Comments

see also http://physics.stackexchange.com/a/21863/7924 where the same is explained in terms of (the  less fundamental) thermodynamics. Conforming to our everyday experience, ''A is hotter than B'' means that ''A will spontaneously lose energy to B when brought in thermal contact with each other''.

Answer 2

Temperature is a statistical property.  A physical system can be in various states of various energies. In a statistical description of a macroscopic system, one uses a probability distribution giving the probability for the system to be in a given state. A macroscopic system is said to be at the inverse temperature $\beta$ if the probability for the system to be in a state of energy $E$ is proportional to $e^{-\beta E}$. In particular, $\beta$ very big ("low positive temperature") means that states of low energy are very likely and states of high energy are very unlikely. When $\beta$ decreases ("usual temperature" increases), this difference between low energy states and high enegry states becomes less sensible and disappear when $\beta=0$ ("infinite usual temperature"): for this value of $\beta$, all the states have the same probability to be found. If $\beta$ becomes negative ("negative usual temperature"), the evolution we have described continues smoothly: states of low energy becomes less likely and states of high energy becomes more likely.

So to say that a system has negative temperature simply means that it is more likely to find the system in a high energy state that in a low energy state, and it is clear from the previous continuous evolution of $\beta$ that such system is hotter than a system with infinite positive temperature.

In most common systems in physics, the energy spectrum (i.e. the set of energies of the various possible states) is bounded below and unbounded above. In such case, the law $e^{-\beta E}$ cannot be normalized to give a probability distribution and so $\beta>0$, i.e. positive temperature, is the only possibility. But negative temperatures are possible for systems with bounded (below and above) energy spectrum.

Comments

Remark: this question has already a correct answer but I wanted to insist on the statistical aspect of the topic.