A show on the weather channel said that as a water molecule ascends in the atmosphere it cools. Does it make sense to talk about the temperature of a single molecule?

  • $\begingroup$ I think it makes sense to talk about the kinetic energy of a molecule, which is where the Kinetic Theory of Ideal Gases comes from: en.wikipedia.org/wiki/Kinetic_theory $\endgroup$ – Greg May 24 '13 at 1:29
  • $\begingroup$ That in turn describes the temperature of a collection of molecules. $\endgroup$ – Greg May 24 '13 at 1:34
  • $\begingroup$ The statistical mechanical definition of temperature is T = (∂E/∂S). Since entropy is directly related to the number of states, I suppose you could define a temperature for a molecule. Not sure how it'd be very useful though. $\endgroup$ – Nick May 24 '13 at 4:09

I think it is a mistake, as often happens in popularizations of science.

A water or any molecule may lose kinetic energy and acquire potential energy, but it is the kinetic energy distribution that gives the temperature of an ensemble of molecules. The shape of the distribution shows that there will always be individual molecules at very high energy, in the ensemble, which they acquire from the random individual collisions.

From the link,$$ f_{\varepsilon} \left(\varepsilon\right)\,\mathrm{d}\varepsilon ~=~\sqrt{\frac{1}{\varepsilon \pi kT}} \, \exp{\left(-\frac{\varepsilon}{kT}\right)}\,\mathrm{d}\varepsilon \,,$$and the shape shows that there always exist tails to high energies. The attribution of temperature labels to individual molecules is wrong.

Maxwell–Boltzmann probability density function, where $a=\sqrt{\frac{kT}{m}}$:

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    $\begingroup$ Sure but if you have one molecule coupled to a thermostat and the system is ergodic then the distribution you are talking about can be thought as being a frequency with which each state is visited over a very long period of time. For once ergodicty is useful in this case $\endgroup$ – gatsu May 24 '13 at 8:24
  • $\begingroup$ The curves in the picture appear to be parameterized by something called $a$, but there is no $a$ in the equation they're presumably meant to illustrate. $\endgroup$ – WillO Apr 4 '17 at 13:32
  • $\begingroup$ @WillO it is in the link alpha=sqrt(kT/m), three different temperatures distributions for a fixed mass. $\endgroup$ – anna v Apr 4 '17 at 18:02
  • $\begingroup$ The equipartition theorem relates temperature to degrees of freedom that appear quadratically in the hamiltonian. So potential energy contributes as well, not just kinetic. For an ideal gas there is only kinetic energy so we get simplified treatments. However, I will admit that the entire subject of relating temperature to energy, and thermal energy vs. internal energy, leaves me somewhat confused. $\endgroup$ – garyp Feb 2 at 18:14
  • $\begingroup$ @garyp I am familiar within this type of equipartiton which has no potential energy there.hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/eqpar.html . In any case everything is about mean and average, even in your link,which needs more than one particle to manifest, imo $\endgroup$ – anna v Feb 2 at 18:56

As the other answers have said, temperature is a collective property and can only be defined when you have an assemblage of particles. However by definition in a molecule you have an assemblage of atoms, and they have relative motions described by the vibrational excitations of the molecule.

So if you have a large enough molecule you can look at the excitations of its vibrational modes and use these to define a temperature. In effect what you're doing is saying that the excitation of the vibrational modes is the same as it would be if the molecule was in equilibrium with some enviroment of the defined temperature.

However I don't think this could usefully be applied to a water molecule. The vibrational excitations of water are of greater than thermal energy at ambient temperatures, and in any case there are only two normal modes. I suppose you could look at the rotation of the molecule, but this would give you only a rough guide to temperature.

  • $\begingroup$ If it is a statistical phaenomena, as described by M-B, there must be a huge number of particles.., what is the argument -theory-to say that a single "big" molecule has a T? $\endgroup$ – santimirandarp Sep 29 '17 at 4:56
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    $\begingroup$ @HernanMiraola: the temperature is related to the energy in the internal degrees of freedom by the equipartition theorem. As long as you have enough internal degrees of freedom to be statistically significant you can assign a temperature just by looking at the rotational and vibrational modes of the molecule. $\endgroup$ – John Rennie Sep 29 '17 at 8:03
  • $\begingroup$ Good, but what is enough? And what happens when it is not enough? $\endgroup$ – santimirandarp Sep 29 '17 at 21:28
  • $\begingroup$ @santimirandarp: Basically all the things that are guaranteed to be exact in the macroscopic limit are only approximations in the microscopic limit. For example, the probability of violations of the second law is nonzero. $\endgroup$ – Ben Crowell Feb 2 at 18:34

Without intending any disrespect, I'm quite surprised that several very knowledgeable people have given a wrong, or at least incomplete, answer to this old question.

For a single molecule that is in complete isolation, it is indeed generally not true (or at least not useful) to assign it a temperature, as others have said. Such a system would be more naturally described in the so-called microcanonical ensemble of thermodynamics, and since it can have a well-defined and conserved energy, the usual role of temperature in determining the probability of occupation of different energy states via a Boltzmann distribution is not relevant. Put simply, temperature is only relevant when there is uncertainty about how much energy a system has, which need not be true when it is isolated*.

However, things are different when you have a molecule in an open system, which can freely exchange energy with its surroundings, as is certainly the case for the specific example the OP has described. In this case, as long as the molecule is in equilibrium or quasi-equilibrium with its surroundings, it does indeed have a well-defined temperature. If there are no other relevant conserved quantities, the quantum state of the molecule is described by a diagonal density matrix in the single-particle energy basis that follows the Boltzmann distribution, $\rho=Z^{-1} e^{-\beta H}$ . Practically speaking, this means that if you know that the molecule is at equilibrium with a given temperature, each time you measure it you can know, probabilistically, what the likelihood is that you will see it with a given energy.

*For completeness I will mention that some people have nevertheless tried to extend the idea of temperature to isolated systems, as the wiki mentions, but this temperature doesn't generally behave in the way you expect from open systems, and it isn't a very useful concept.

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    $\begingroup$ I had missed your answer before writing mine, eventually writing about the same idea, in a looser manner! $\endgroup$ – user154997 Sep 29 '17 at 9:49
  • $\begingroup$ @LucJ.Bourhis actually, it seems to me that they are somewhat different. My answer applies equally if there are no relevant internal degrees of freedom. It is just about a closed vs open system. It is essentially an elaboration on gatsu's comment to Anna's answer ( I don't mention ergodicity explicitly, but it is implied by the statement that the system can thermalize to some equilibrium state). $\endgroup$ – Rococo Sep 29 '17 at 15:42
  • $\begingroup$ Since you consider the Boltzmann distribution, you have an entropy $S=-k\text{Tr}(\rho\log\rho)$, which is the quantum equivalent of the $\sum p\log p$ in my answer. That said, indeed, in my answer, the molecule could be isolated. $\endgroup$ – user154997 Sep 29 '17 at 16:04
  • $\begingroup$ This is not really correct. If the molecule is coupled to its surroundings, then the temperature is a property of the whole system, not of the single molecule. $\endgroup$ – Ben Crowell Feb 2 at 18:33
  • $\begingroup$ @BenCrowell I don't think I agree with this. I would say that if a subsystem (of arbitrary size) is in thermal equilibrium with a heat bath at temperature T, it is appropriate to say that the subsystem is itself at temperature T. $\endgroup$ – Rococo Feb 2 at 21:30

It makes sense if all you know about the molecule is its expected energy. Then you can show that it's energy distribution is the Boltzmann distribution $p(E) = e^{-E/kT}$ for some constant $T$, which is related to the expected energy.

So the question reduces to a philosophical view of probabilities. Does it make sense to assign probabilities to a deterministic system? If you accept probabilities as the reflection of your knowledge of the system rather than something intrinsic then it also makes sense to assign temperature to a single molecule.

  • $\begingroup$ But in this case the expected temperature isn't a property of the molecule, but someone's prior belief about the molecule. So a single molecules still doesn't have a temperature, it has a particular but unknown energy. $\endgroup$ – innisfree Apr 20 '17 at 16:30

Thermodynamics makes sense when you have large numbers of particles. For example, the second law of thermodynamics has an extremely low probability of being violated when you have Avogadro's number's worth of particles. However, if you have a very small number of particles, the second law will frequently be violated.

This comes up in nuclear physics, where we routinely deal with nuclei consisting of 50 or 100 or 200 particles. Yes, we do talk about the temperature of an individual nucleus, and it does make sense.

However, a single water molecule is only 3 atoms, and in a system that size, it's nonsense to talk about temperature. In giant molecules, I can easily imagine that you could have enough atoms to talk about the temperature of an individual molecule.

  • $\begingroup$ This sounds like an application of the "sorites problem"...as you add grains of sand to a collection one by one, when does it become a "heap". Similar issue here? $\endgroup$ – Richardbernstein May 24 '13 at 13:21
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    $\begingroup$ @Richardbernstein: No, thermodynamics is just an approximation that becomes better and better as you have more particles. $\endgroup$ – Ben Crowell May 25 '13 at 19:05

I will explain in simple....

We can measure temperature changes when a Source release energy or extract energy.

We know that every thing want to attain equilibrium. Therefore there should be an equality in temperature in all the atoms or molecules of a system.

The temperature of an atom will be same as that of temperature of surrounding.

But a thermometer is an external source of energy from surrounding. So it provides or extract energy from source.

Now Consider an $Ideal$ condition

Consider a atom placed in space and is away from every radiation from space and any atoms or any other gravity or electromagnetic forces.

Then the temperature will be $zero(Kelvin)$ theoretically and you cannot measure it experimentally !!


As stated in the other answers, in the specific scenario depicted in your question, it makes no sense to talk about the temperature of a molecule. But I can't help to widen the picture because there is a very interesting case here: what about the internal degrees of freedom of the molecule? So far everybody has only considered the motion of the whole molecule, because that was the context you implicitly gave, fair enough.

But what about the vibration of the molecular bonds? The rotations of two neighbour parts of the molecule about a bond? This is not as trivial! For small molecules, the number of associated degrees of freedom are too few to warrant talking about temperature. But it is not so for a large protein. There are so many of them that a statistical approach is possible: one can define an entropy, in the usual $\sum p\log p$ way, where $p$ is the probability for a given configuration. Then, as usual, if we have an entropy we can minimise it with the constraint of a given total energy, and a temperature comes out of that.

After writing this, I realised that @Rococo and @JohnRennie had written an answer on a similar line long before mine!

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    $\begingroup$ I believe John Rennie's answer also talked about this 4 years ago. $\endgroup$ – JMac Sep 29 '17 at 10:31

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