# What is temperature?

Recently I read an interesting article about negative temperature. I was puzzled because I thought before that temperature has definite meaning in thermodynamics: it tells about how fast atoms jiggle. Now if temperature can be negative, this means temperature has wider meaning...

I wonder if there is any deep physical meaning of temperature, not mathematical one.

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The negative temperature question is essentially a duplicate of physics.stackexchange.com/q/21851/2451 and links therein. – Qmechanic Nov 10 '12 at 11:50

The thermodynamic definition of temperature is

$$T \equiv \left( \frac{\partial S}{\partial U}\right)^{-1}$$

where $S$ is the thermodynamic entropy of the system and $U$ its internal energy. The thermodynamic concept of temperature $T$ is more general than the kinetic temperature $T_\mathrm{kin}$ (which only measures the average speed of molecules) because molecules do more than just translate in space.

A negative temperature simply means that the entropy of the system decreases $\delta S < 0$ when you add more energy $\delta U > 0$.

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Negative temperatures can only occur in systems where the energy spectrum is bounded above.The systems with Negative-temperature have the opposite characteristics. Adding energy reduces their disorder. But they are not cold in the conventional sense that heat will flow into them from the systems at (+)ve temperatures. In fact, systems with (-)ve absolute temperatures contain more atoms in high-energy states than is possible even at the hottest positive temperatures, so heat should always flow from them to systems above 0K.

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To understand negative temperatures, it is best to consider the inverse temperature $\beta \equiv (k_B T)^{-1}$, which appears in the expression of the partition function $Z\equiv \sum_n e^{-\beta E_n}$. A system 1 is hotter than system 2 (i.e., if brought into thermal contacts, heat will flow from 1 to 2) if $\beta_1 < \beta_2$.

For systems with an energy spectrum that is unbound upwards (as is the case for example for the kinetic energy of particles in a gas), the lowest possible value for $\beta$ is zero, i.e., the hotest possible temperature in a gas is $T=\infty$, in which all states are equally populated, irrespective of their energy.

The situation is different however for a system possessing an energy spectrum which is bound upwards (for example an assembly of spins in a magnetic field). In this case, any value of $\beta$ between $-\infty$ and $+\infty$ is possible. A negative value of $\beta$ corresponds to thermodynamic states in which the microscopic states of low energy are less populated than higher energy states (this the situation of population inversion created in lasers and enabling coherent radiation). So a system with negative temperature $T$ is actually hotter than any system with a positive temperature. The apparent paradox only lies in the fact that the natural quantity for measuring "hotness" is $\beta$ rather than $T$.

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The average kinetic energy of the particles is directly related to the temperature of the object.

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That is only true in systems that actually have a translational degree of freedom. The best example for a system without translational freedom (that also exhibits negative temperature) is that of a number of spin-1/2 particles in a line, trapped so that they are unable to move. If put into a magnetic field, the temperature of this system is related to the number of spins "flipped up" vs. spins "flipped down", and has nothing to do with their (non-existent) kinetic energy. juanrga's answer is the right way to define temperature. – ACuriousMind Jul 30 '14 at 12:29
@ACuriousMind: Actually, every physical system has translational degrees of freedom, the system you mention is just an idealization. No trap is rigid enough to totally eliminate the translational degrees of freedom. Iv4Ps' answer has two other problems though. The average kinetic energy is indeed directly related to the temperature of the object, but typically only if two additional conditions are met. First, the system must be classical (for example, the statement fails for a degenerate Fermi gas), second, the system must be close to a thermodynamic equilibrium. – akhmeteli Apr 10 '15 at 9:44

## protected by Qmechanic♦Apr 10 '15 at 9:35

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