Proof of existence of lowest temperature $0 K$

Im mathematics there is a concept of infinity meaning that whenever you pick a number and say that it is the smallest/Largest there is a way to further reduce/increase that number by subtracting/adding any other number.

But in physics/chemistry i see that the absolute temperature does not have a negative reading and the lowest temperature is $0 K$.

What is the proof that temperatures below zero cannot exist?

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If you like this question, you may also enjoy reading this question. – Qmechanic May 19 '12 at 12:11

In physics, temperature and other concepts in "thermodynamics" (that was known for centuries from macroscopic analyses of the heat engines and similar systems) is given by a more fundamental theory, the so-called "statistical mechanics". According to statistical mechanics, the thermal phenomena are explained by the motion of the atoms and various states in which the atoms may be found (and the number of these states).

In particular, the probability $p_k$ of a state $k$ (in classical physics, the state is described e.g. by the location and velocity of each particle) is given by $$p_k = C \exp(-E_k/k_BT)$$ where $E_k$ is the energy of the state $k$, $k_B$ is Boltzmann's constant converting kelvins to joules, and $T$ is the absolute temperature in kelvins. The coefficient $C$ is a "normalization factor" that is $k$-independent and chosen so that the sum of $p_k$ over $k$ is equal to one (the total probability).

This form makes it clear that $T\lt 0$ isn't allowed: the exponential would be growing with $E_k$ and because there are infinitely many states with ever larger values of $E_k$ (the kinetic energy may grow arbitrarily high, in particular), the probabilities would be getting larger and their sum would diverge: it couldn't be normalized to one.

Before this statistical explanation involving Boltzmann's constant was known, the temperature was a phenomenological quantity measured by a thermometer. One was actually uncertain about any redefinition $T\to f(T)$ where $f(T)$ is a monotonically increasing function. In principle, one may relabel $T$ so that zero kelvins gets mapped to $-\infty$ in another convention for the temperature, for example; try $T_\text{new convention} = \ln (T)$. However, the ideal gases obeyed $pV = nRT$ so at a fixed pressure, the volume of some gas was proportional to the absolute temperature – the same one as one in statistical mechanics, without any redefinition by a function $f$.

So people knew how to measure the "right absolute temperature" even well before statistical mechanics was understood. The usual thermometers relied on the expansion of liquids etc. which are not ideal gases but they're close enough. For ideal gases, where the absolute temperature is proportional to the volume, the statement that $T\gt 0$ is equivalent to the statement that the volume of the ideal gas cannot be negative. You cool it down and it shrinks but it can't shrink below zero.

Volume is about the "shape" but the underlying reason for the positivity of temperature isn't about locations; it is about the motion. Any physical object with quadratic degrees of freedom will carry $k_BT/2$ of kinetic energy per degree of freedom. Again, because the quadratic kinetic energy of the type $mv_x^2/2$ can't be negative, the absolute temperature can't be negative, either.

In lasers and similar devices, one may formally find negative absolute temperatures when the number of atoms at a higher energy level is greater than the number of atoms at a lower energy level. However, this negative temperature can't be brought to equilibrium with all degrees of freedom in a larger object because the number of high-energy states is always divergent. In lasers, one kind of abuses the fact that that the energy of the "interesting degrees of freedom" is bounded both from below and from above (we only allow two or few levels for each atom).

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Spot on physics +1, but no good history--- the recognition of absolute temperature (no reparametrization freedom) came with Carnot, who defined the absolute temperature as the T in $dS={dQ\over T}$, the identification of this with ideal-gas temperature was instantaneous and I think around the 1820s (maybe 1830s). The recognition of conservation of energy dates to the 1840s, and the statistical interpretation dates to the 1860s-1870s when Boltzmann starts to write. Boltzmann's statistical intepretation is not accepted universally until 1910 or so, and some debate lingers into the 1920s. – Ron Maimon May 19 '12 at 4:28
Slight wrinkle: this assumes energy levels unbounded from above, which is not true for many condensed matter systems. In those cases, one can certainly have negative temperature states. Of course one can wave "it's only a meta-stable state", but when there is a good separation of timescales it doesn't matter. – genneth May 19 '12 at 11:39
Thanks, Ron. Genneth, isn't it what the last paragraph of my text are about? – Luboš Motl May 19 '12 at 18:09
Sorry; somehow I completely missed that the first time round. All good then :-D – genneth May 20 '12 at 16:06

Zero kelvin is the temperature at which there is no thermal motion. Since temperature by definition is the average thermal motion (really kinetic energy) of an ensemble of molecules, then it is a matter of definition that there can not be a lower temperature than that zero K--because there is no such thing as "negative" motion. A good analogy is a batting average. You can't have a negative batting average because there are no "negative" hits. Or, as they told me when I was an intern--you can't get less than zero sleep.

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"temperature by definition is the average thermal motion (really kinetic energy) of an ensemble of molecules" this works only in classical case; Fermi gas has actually very high average kinetic energy even at 0K. – Siyuan Ren May 19 '12 at 2:26
Can you explain? – Richardbernstein May 19 '12 at 3:49
en.wikipedia.org/wiki/Fermi_energy but this is a purely quantum effect. All quantum states of electrons have an associated momentum (which may be zero), but because electrons are fermions the highest occupied state will generally have a non-zero momentum even at absolute zero. – John Rennie May 19 '12 at 6:24
Downvoted because this is just too limited a definition. Temperature is the gradient of internal energy to entropy for equilibrium states, nothing else. – genneth May 19 '12 at 11:38

I'd complete @Luboš Motl's answer with the fact that the relevant quantity in thermodynamics is the coldness $\beta=\frac1{k_BT}$, in ${\mathrm J}^{-1}$ (You can see this answer by Arnold Neumaier for references.) It basicaly comes from the definition of thermodynamic temperature :

For systems where the entropy $S$ is a function of the energy $E$, the temperature $T$ can be defined as $$\frac1T=\frac{dS}{dE}$$

For system at thermal equilibrium, $\beta$ varies from $+\infty$ for very cold systems downto to $0$ of very hot systems. Your question can be rephrased as

What is the proof that there is no coldness above $+\infty$ ?

which seems obvious ...

The next question is about coldness below 0. The short answer is : they exist in non equilibrium system, as discussed in this question. They are logically hotter than systems at equilibrium (with $\beta>0$). Including those systems, $\beta$ can take any real value, with the hotter systems corresponding to the smaller $\beta$. This is of course also true for the temperature $T\propto\frac1\beta$, but the fact that any system with $T<0$ is hotter than any system with $T>0$ becomes less intuitive.

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