# Thermometers respond only to translational kinetic energy: misconception?

Many sources (example) claim that a thermometer measures (responds to) only translational kinetic energy, linked with the motion of the center of mass of molecules.

For example, the link mentioned above states:

...the temperature as ordinarily measured does not account for molecular rotation and vibration.

This seems incorrect to me. To me, the thermometer will react to any kind of kinetic energy that hits it.

Take for example a real diatomic gas at "normal" pressure and temperature. Those molecules are rotating very quickly. Rotation carries with it kinetic energy. The thermometer has no way of distinguishing between one form of kinetic energy and another.

If I'm wrong, can you point me to a textbook reference to correct my understanding?

Edit: add some background, with references to Feynman and Landau.

From a Feynman lecture:

Towards the end of the lecture, he makes remarks about a diatomic gas.

If I use this lingo:

• two atoms, A and B
• AKE: average kinetic energy
• CM: center-of-mass of the single molecule; add mass of A and B
• internal: refers to interactions between A and B; rotation, vibration

then there are two main results.

3(kT/2) = AKE(A) = AKE(B) = AKE(CM) = AKE(internal)

The AKE of the whole molecule can be broken into parts in two ways:

• AKE(A) + AKE(B)
• AKE(CM) + AKE(internal)

He remarks (with they referring to the atoms in the diatomic molecule):

Although they are held together, when they are spinning and turning in there, when something hits them, exchanging energy with them, the only thing that counts is how fast they are moving. That alone determines how fast they exchange energy in collisions. At the particular instant, the force is not an essential point. Therefore the same principle is right, even when there are forces.

He also remarks:

The entire kinetic energy of the molecule can be expressed either as the sum of the kinetic energies of the separate atoms, or as the sum of the kinetic energy of the CM motion plus the kinetic energy of the internal motions.

Landau and Lifshitz, Statiscal Physics, when discussing the Maxwell-Boltzmann distribution for speeds, makes similar assertions. See the top of page 81.

In summary, the above two references don't seem to be saying that a thermometer reacts only to the CM translational KE of a diatomic molecule.

The fact that 3(kT/2) relates to the CM Average KE doesn't mean that thermometers only "feel" that KE. 3(kT/2) relates to other things as well.

A diatomic molecule is capable of transmitting more energy to a thermometer than a monatomic one. It also is capable of absorbing more energy. The point is that at equilibrium, when the total energy transfer to the thermometer by all gas molecules that slow down due to a collision with the thermometer is balanced by the total energy transfer in the other direction to gas molecules that speed up, the average translational kinetic energy of the gas molecules does not depend on what kind of molecules they are.

That is to say, the thing that is measured by a thermometer (the temperature) corresponds to the average center-of-mass kinetic energy per gas molecule.

Feynman explains this thoroughly.

EDIT, following substantial expansion of question:

The fact that 3(kT/2) relates to the CM Average KE doesn't mean that thermometers only "feel" that KE. 3(kT/2) relates to other things as well.

You can try to wriggle out of the conclusion that the center-of-mass motion of a gas molecule has a direct conceptual connection to the temperature; you may even succeed. But one of the surprising things (to me) about ideal gasses is the extent to which you can ignore all differences between molecules--just treat them all as identical particles (you just have to account for differences in heat capacity). This freedom to ignore the intramolecular details is convenient.

For example, take your diatomic gas: it may be tempting to conclude that you should instead regard the temperature as being more closely conceptually tied to the total kinetic energy of each of the atoms, rather than the CM kinetic energy of the entire molecule. After all, as you point out, both average $$\frac{3}{2}kT$$. It seems like you could just ignore the fact that the atoms are tied together in pairs, and say that the temperature is the average total kinetic energy of the constituent atoms. But if you try to carry this way of thinking (in terms of individual atoms rather than molecules) to the ideal gas law, it doesn't work in its usual form: you'd have one ideal gas law for monatomic gases, another for diatomic gases, etc. This is because pressure is due to the the transfer of momentum during a collision (rather than energy) and that impulse depends entirely on the motion of the center of mass. The fact that you can also understand the temperature in terms of only the center-of-mass motion is kind of surprising. And too useful to walk away from.

• That doesn't seem to answer my question directly. Is the thermometer reacting only to translational kinetic energy, associated with the motion of the center of mass of the molecule, or to all forms of kinetic energy? – John Feb 16 at 16:06
• It reacts to all forms of kinetic energy. That doesn’t matter. The thing that a thermometer measures is average translational kinetic energy per molecule. – Ben51 Feb 16 at 16:09
• Why do you qualify it with "translational"? Is that different from "rotational"? – John Feb 16 at 16:12
• @SebastianRiese Temperature is a derived physical concept not a fundamental one, and whilst the thermometer may not care what kind of physical energy it responds to, it is a completely legitimate and meaningful question what role translational and rotational energies play here. Even more interesting in my opinion is the question to what extent it actually responds to infrared radiation rather than kinetic energies. – Thomas Feb 16 at 21:18
• It is true in this sense: if you tell me the average energy per molecule due to its center-of-mass motion only, I will tell you the temperature. It is not true in the sense that rotational or vibrational motion is somehow not transferable to the thermometer. – Ben51 Feb 16 at 22:42

If we define "temperature" as "a useful number proportional to the mean energy stored in some degree of freedom" (which is not the most fundamental definition of temperature, but it's close in this circumstance), then it's possible for the "translational temperature" and the "spin temperature" of a gas to get out of thermal equilibrium with each other. When this happens, a "thermometer" which couples to the translational degree of freedom will measure the translational temperature, while a thermometer which couples to the spin degree of freedom will measure the spin temperature. In general it takes more cleverness and subterfuge to build a "spin thermometer," which is where the sense of your quote comes from.

A naturally-occurring example of such disequilibrium occurs in diatomic hydrogen, $$\rm H_2$$. The two protons in a hydrogen molecules must obey Fermi-Dirac statistics, so their wavefunction must be antisymmetric under exchange. The rotational wavefunctions with quantum number $$L$$ have symmetry $$(-1)^L$$ under exchange — that is, the $$s$$-wave and $$d$$-wave are symmetric, the $$p$$-wave and $$f$$-wave are antisymmetric, and so on. To have the total wavefunction antisymmetric, the states with even $$L$$ must have the two protons in a spin singlet, while the states with odd $$L$$ must have the two protons in a spin triplet. The energies of the rotational states are

$$E_L = \frac{15\rm\,meV}{2}L(L+1)$$

Here the energy of the $$L=1$$ state, a few milli-eV, is consistent with the moment of inertia for a rotor whose mass and size are set by the hydrogen molecule's mass and bond length. There isn't any important correction to the molecular energy due to the proton-proton spin interaction because the proton's magnetic moment is very small. If the hydrogen gas is in thermal equilibrium, the probability of finding a molecule in a state with angular momentum $$L$$ is proportional to

$$P(L) \propto (2S+1)(2L+1) e^{-E_L/kT}$$

This is just the ordinary Boltzmann factor $$e^{-E_\text{state}/kT}$$ and the degeneracy factors for the spin and orbital angular momentum quantum numbers: there are $$2S+1=3$$ ways to orient a spin triplet, but only $$2S+1=1$$ way to orient a spin singlet. At room temperature ($$kT=25\rm\,meV$$), there are nontrivial populations in states with many $$L$$, so hydrogen's heat capacity is $$\frac52R$$. Near its triple point ($$T \approx 30\,\mathrm K = 2.5\,\mathrm{meV}/k$$) the rotational degrees of freedom are "frozen out" as the Boltzmann factor $$e^{-E_L/kT}$$ becomes smaller and smaller for all the states with $$L>0$$. The heat capacity of cold hydrogen gas approaches $$\frac32R$$ as the rotational states become less accessible.

At least, that's what you'd think until you starting making liquid hydrogen.

It turns out that there isn't any internal mechanism for an isolated molecule of $$\rm H_2$$ to convert from a spin triplet to a spin singlet (or vice-versa), but collisions between molecules allow spin exchange pretty effectively. So hydrogen which has been at room temperature for a long time will have a thermal-equilibrium distribution of $$L$$. The requirement that protons obey Fermi-Dirac statistics, and the spin-degeneracy factor $$(2S+1)$$ attached to the Boltzmann factor, conspire to mean that three-quarters of the population will be spin-triplet molecules with $$L$$ odd. When you cool hydrogen vapor down below its triple point, you actually don't bring all of the molecules into the $$L=0$$ ground state. Most of your hydrogen has odd $$L$$ and gets stuck in the $$L=1$$ state, waiting to collide with another $$L=1$$ molecule to be able to release its rotational energy. The even-$$L$$ and odd-$$L$$ molecules actually behave like two miscible but distinct gases, called "parahydrogen" ($$L$$ even) and "orthohydrogen" ($$L$$ odd). They have slightly different heat capacities, because the first excitation in parahydrogen $$L=0\to2$$ has different energy and degeneracy than the first excitation in orthohydrogen $$L=1\to3$$. They also have slightly different densities and other thermal properties.

The rate of ortho-to-para conversion in cold hydrogen (where the $$L=1$$ state should be "frozen out") is proportional to the square of the orthohydrogen density: the orthos have to find each other. When you liquify cold hydrogen vapor its total density increases by a thousandfold, and the rate of orthohydrogen downconversion increases by a factor of a million. Each $$L=1\to0$$ downconversion releases 15 meV, which happens to be approximately the same the latent heat of vaporization. So what happens when you liquify hydrogen gas from room temperature is

1. Your room-temperature "normal hydrogen" (75% ortho, 25% para) falls below its boiling temperature and condenses into a liquid.
2. The orthohydrogens in the dense liquid undergo collisions with each other more rapidly than they did in the sparse gas. They find each other, have spin-exchange collisions, and convert to parahydrogen.
3. The conversion to parahydrogen releases so much heat that all of your hydrogen boils again!
4. If you have kept your parahydrogen vapor in your cryostat, rather than venting it and replacing it with more normal hydrogen from your bottle outside, you can re-liquify the parahydrogen and it will stay liquid this time.

Any "thermometer" that you attach to this system is going to tell you that hydrogen's triple point is about 33 K and its boiling point at ordinary pressure is about 20 K; ordinary thermometers are not sensitive to the extra heat stored in orthohydrogen's orbital angular momentum degree of freedom. You cannot tell the difference between cold $$L=0$$ hydrogen gas and cold $$L=1$$ hydrogen gas using only a thermometer; this statement directly addresses your question.

The problem with the statement you quote, alluded to in the very first sentence of this long answer, is that thermometers don't measure internal energy. A thermometer measures temperature, which is is a system's "willingness to give up its internal energy," to paraphrase Schroeder's excellent intro-thermal textbook. In many systems the temperature and the internal energy are proportional to each other. But it's quite possible for a system with non-interacting degrees of freedom to function at a different temperature for each type of interaction. The most common examples are "spin temperatures," but there are others as well.

• To me, this seems to be tangential to the original question, which is intended on a much simpler level than this. The essence is whether or not this quote in the original question is true: "the temperature as ordinarily measured does not account for molecular rotation and vibration". That's all. – John Feb 16 at 21:42
• I think this is exactly what the question is asking about: an example of a system where molecular rotation may be present or absent but can't be detected with an ordinary thermometer. Here the "rotational heat" is enough to boil a liquid sample, but it's very hard to observe in gas calorimetry and thermometry. – rob Feb 16 at 21:53
• So is the quote I referred to above true or false? The spirit of the question is one of banal physics, not exotic physics. – John Feb 16 at 21:56
• I don't want to say "true/false" to a sentence fragment containing a negated verb and the word "ordinarily." In the "banal" case, the translational, rotational, and vibrational degrees of freedom are all in thermal equilibrium, so the question is moot. But here I've described a material (hydrogen) whose bulk properties (liquid vs. cold vapor) depend on its translational temperature much more strongly than on its rotational temperature. The "two-gas" language for ortho-para conversion lets us recover the toolset of thermal equilibrium; a non-equilibrium "temperature" is a tricky animal. ... – rob Feb 16 at 22:40
• ... But I will say that when I read your quote in your question, I started to think "what is a system where rotational energy is thermodynamically important, but can't be detected by a thermometer?" Orthohydrogen is the first situation that came to mind, and the simplest. And that's the sense in which your source (in context) is correct, while your intuition in this question ("the thermometer will react to any kind of kinetic energy that hits it") omits many important edge cases. – rob Feb 16 at 22:40