# Temperature of an atom

I read somewhere that the temperature of an atom is not defined. The definition of temperature is only for larger systems. Why is this so?

Thermodynamic functions are strictly defined only for macroscopic systems (systems that have an essentially infinite number of atoms). You can't apply them to individual atoms because that would be confusing large-scale averages with individual microscopic values.

Here's an analogy: the average speed of cars on a stretch of highway might be 55 mph, but it's possible that no individual car is traveling at that exact speed.

Another analogy: the average global temperature is rising over long periods of time, but if I look at the thermometer outside my window, the local temperature is falling over a short period of time... I can't say that my local temperature drop disproves global warming, because weather is a different thing from climate.

Suppose that you knew the speed of a single atom. Now, for a macroscopic ideal gas, the average speed of molecules $\overline{v}$ is $$\overline{v} = \sqrt{\frac{8 R T}{\pi M}}$$ It would be wrong for you to plug in the speed of your single atom and solve for its temperature, because that equation was derived by assuming that you had a macroscopic system (with a certain speed distribution over many molecules); it only applies to average velocities, not individual velocities.

This applies to other thermodynamic functions, too. Let's look at a simple chemical example. For a macroscopic crystal, the Gibbs free energy might be written as $$G = Ng(P,T)$$ where $N$ is the number of atoms and $g$ is the Gibbs free energy per atom. Little $g$ is a function of pressure and temperature.

So can we say that little $g$ is the Gibbs free energy of a single atom? No. It's the average Gibbs free energy per atom for a huge number of atoms. To see the difference, consider a small cluster of atoms from that same crystal, the previous equation will need some correction terms:

$$G = Ng(P,T) + a(P,T)N^{2/3} + b(T)\ln{N} + c(P,T)$$ where the "a" term is a surface free energy and the last couple of terms might come from free energy contributed by things like rotation of the cluster (Source: T. L. Hill, Thermodynamics of Small Systems, Dover, 1962.) You can't say that $g(P,T)$ is the Gibbs free energy of a single atom in the cluster because of those correction terms.

TL;DR: Thermodynamic functions are averages over very large numbers of particles; random fluctuations are tiny compared to the mean values. But with small systems, those fluctuations become quite important, and there are additional effects that cause macroscopic averages to poorly describe microscopic systems.

• Very nice answer. But wouldn't the ergodic hypothesis (which in short states that the ensemble-average is equal to the time-average) imply that thermodynamic functions would also be valid for a microscopic system (e.g. a single particle) if I consider the behavior of the system averaged over a very(!) long period of time? – Philipp Jan 26 '15 at 18:44
• @Phillip - I agree with you - I think maybe it would be more correct to say that the instantaneous temperature of a single atom is undefined, even though the time-averaged temperature may be defined. This is also assuming that there is a suitable reference frame for determining the velocity of a single atom in the first place. Also, "very long" for atoms might not be that long for humans. As a rough guess, I think that at room temp a couple of nanoseconds would probably give a good average. – thomij Jan 26 '15 at 19:25
• @Philipp and thomij, yes, you'll be comparing two averages, not an average and a single value. – Fred Senese Jan 26 '15 at 19:47
• Aah i see. I remember reading the derivation of the average kinetic energy of the atoms of an ideal gas, and it involved quite a few assumptions. Thanks for the great answer sir! – Lexicon Jan 27 '15 at 15:17

While generally temperature in textbooks is connected with the speed of atoms/particles, strictly speaking not the velocity, but the distribution of velocity what counts. If you microscopically observe a bunch of particles, their velocity can be whatever as your frame of observation can be any inertia system. When you want to calculate temperature, you choose reference system which has the same velocity as the center of mass of those particles, and then you calculate the average velocities form that reference. The average will be small, if the distribution of velocities is sharp, and will be large if the distribution is broad. The calculated temperature can be low even for fast particles e.g. in a gas flow, if all the particles moving in the same directions, same velocity!

Let's apply this to one atom: you don't have a distribution of different velocities! Once you adjusted your reference point to the center of mass, ie the particle, it has zero velocity and meaningless to look for a distribution of different velocities.

• There are electrons in the atom... – jinawee Oct 28 '18 at 17:51
• @jinawee Electronic temperature can only be relevant is there are electronically excited states below kT, and a trivial coupling to the kinetics of the atom (generally no). Even so, their behavior is not classical, per def, so we don't discuss them in classical theories. – Greg Oct 29 '18 at 4:45