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As far as I can tell temperature seems to be defined as something like average kinetic energy per molecule, but not quite. It looks like it measures something proportional to this average kinetic energy, where the coefficient of this proportion depends on the number of independent degrees of freedom by the equipartition principle.

If I put a thermometer in some substance and let it reach equilibrium to measure the temperature, then I would naively expect that the average kinetic energy per molecule in the thermometer would equal the average kinetic energy per molecule in the substance. But this means that the temperature would actually be different.

So, I conclude my naive understanding is wrong and that the average kinetic energy per degree of freedom must reach equilibrium, and if one object has more degrees of freedom it will end up with more energy. I can maybe convince myself of this intuitively by imagining a di-atomic gas and a monoatomic gas interacting.

But then since the thermometer measures something that does reach equilibrium, I suppose it must be directly measuring the average kinetic energy per degree of freedom. Is that accurate?

And if it is, how does it do that? A mercury thermometer measures temperature by thermal expansion. This seems to suggest that thermal expansion is governed by a fixed set of degrees of freedom. I would guess that it would be the translational kinetic energy (more specifically, since the mercury expands in only one dimension, I would say the translational kinetic energy in that direction). Is this an accurate description of the function of a thermometer?

Sam Jaques
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    I think the main misconception here is that temperature is kinetic energy per degree of freedom. That's only true for an ideal gas. A thermometer can't and doesn't need to know what temperature is microscopically for whatever it's measuring; it only needs to give a reading in proportion to its own temperature. – Javier Jul 24 '20 at 19:52
  • @Javier not only for an ideal gas, but for any classical system of particles, see: Is mean kinetic energy related to temperature of a system of interacting classical particles? – Ruslan Jul 25 '20 at 05:55
  • @Ruslan Doesn't the equipartition theorem apply only for quadratic terms in the Hamiltonian? – Javier Jul 25 '20 at 14:39
  • @Javier well, kinetic energy is quadratic in momentum. Not in relativistic regime, of course, but the classical Newtonian case works fine. – Ruslan Jul 25 '20 at 14:42
  • @Ruslan Good point, but still temperature is not kinetic energy per degree of freedom. You could say that the "kinetic contribution" to temperature is that, but there's also terms coming from the potential. – Javier Jul 25 '20 at 15:48
  • @Javier not really, see the derivation in the answer to the question I've linked above. – Ruslan Jul 25 '20 at 15:52
  • So when two systems are in thermal equilibrium, what exactly is in equilibrium between them? Whatever that is, it should be (proportional to) temperature, since temperature is equal at thermal equilibrium. – Sam Jaques Jul 27 '20 at 09:08
  • @Ruslan: When you say "mean kinetic energy", that excludes rotational kinetic energy and other general intra-particle wiggling, right? If so: kinetic energy is precisely 3 degrees of freedom, and equipartition seems to imply that energy/degree of freedom is constant. So measuring mean kinetic energy and measuring energy/degree of freedom are equivalent at equilibrium. – Sam Jaques Jul 27 '20 at 10:41
  • @SamJaques when I say "mean kinetic energy", it means 3 degrees of freedom per atom. Note that the atoms don't rotate (in the classical regime), nor do they wiggle. Such internal motions (applicable to molecules, not atoms!) are taken into account by the general motions of the atoms. The fact that potential energy between the atoms keeps some of them bound to each other to form molecules doesn't change anything. – Ruslan Jul 27 '20 at 10:48
  • If mean kinetic energy per atom reaches equilibrium, then wouldn't the heat capacity be identical for all substances, since they are all just atoms? A purely classical theory of gases predicts a different heat capacity for diatomic vs. monoatomic gases, right? – Sam Jaques Jul 27 '20 at 10:53
  • Heat capacity will differ, because addition of heat also affects the total potential energy, not only kinetic. – Ruslan Jul 27 '20 at 11:24
  • So temperature is proportional to mean kinetic energy per atom, which ends up being equal in thermal equilibrium, but this is also in equilibrium with the potential energy. – Sam Jaques Jul 27 '20 at 11:30
  • What is equilibrated with the potential energy? Temperature? It doesn't have anything to do with potential energy. – Ruslan Jul 27 '20 at 12:19
  • I mean that if I put a diatomic gas at a low temperature in contact with a monoatomic gas at high temperature, the monoatomic gas transfers heat, and some of it ends up as potential energy in the diatomic gas. At equilibrium, there's no net energy flow between the gases nor between potential and kinetic energy. Does that mean (on average) that the increase in kinetic energy of the diatomic gas equals its increase in potential energy? – Sam Jaques Jul 27 '20 at 14:19
  • No, it means that the energy transferred from the monatomic gas to the diatomic one is distributed between kinetic and potential energy of the diatomic gas. The mean kinetic energies of both gases become the same at equilibrium. Extra energy taken from the monatomic gas to be stored as the potential energy of the diatomic gas is the manifestation of the larger heat capacity of the latter. – Ruslan Jul 27 '20 at 16:09

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As far as I can tell temperature seems to be defined as something like average kinetic energy per molecule

No. A glass of water with some ice cubes, in approximately thermal equilibrium in a refrigerator, has a temperature of $O^{\circ}C$. Both the ice and the liquid water have that temperature. But the average thermal energy of the molecules of liquid water is much greater.

The points of phase changes are the best examples that energy delivered to a system (what increases the system internal energy) is not always followed by temperature increase.

Of course temperature could be redefined to match the internal energy of a system, but it will be totally different from our intutive concept of temperature.

The thermodinamical concept of temperature as $$T = \frac{\partial E}{\partial S}$$ where $S$ is the entropy, explains better that relation.

When there is an energy input to a system and there is no phase change, the temperature increases, and that derivative is crescent.

During the phase change, internal energy keeps increasing, but the derivative (temperature) is constant. The increase of energy is linearly proportional to the increase of entropy in this case.

Claudio Saspinski
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  • Can entropy be defined without involving temperature, though? – Sam Jaques Jul 27 '20 at 09:05
  • At phase transition, particles are constantly shifting between phases. At any given moment, the particles in one phase presumably have a different kinetic(thermal?) energy. But that's not surprising, since a substance not in a phase transition has particles of different energy; we only consider the average. If I took a gas and "froze" it in time and picked a section of particles with lower kinetic energy than the rest, I could point to that section and say "the average thermal energy of these molecules is lower, but it is at the same temperature as the rest" but it isn't.... – Sam Jaques Jul 27 '20 at 10:54
  • at the same temperature, because of the process I used to select those particles. Ice vs. water seems like the same phenomenon: if I stop time and look at the sections that are currently ice vs. currently water, there will be a different average, but in a time-averaged equilibrium sense, there are no ice/water sections since all the particles are freely transitioning between the two states. – Sam Jaques Jul 27 '20 at 10:56
  • But in the case of ice and water in thermal equilibrium, different devices as a mercury thermometer and a thermocouple will show the same temperature for each phase. It is more than only the same average temperature. – Claudio Saspinski Jul 27 '20 at 13:40
  • What happens on a molecular level when the same thermometer contacts ice vs. water at 0 C? What process explains that they will both cause the same level of thermal expansion/contraction in the mercury? – Sam Jaques Jul 27 '20 at 14:20
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Maybe it will help to read on the history of thermometers, and how temperature readings developed:

Temperature is a numerical representation of hot or cold compared against baselines, typically the point at which water freezes and boils.

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The concept of measuring temperature is fairly new. The thermoscope — essentially a thermometer without a scale — was the precursor to the modern thermometer. There were several inventors working on thermoscopes around 1593,

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Ferdinand II, the Grand Duke of Tuscany, followed in 1654, inventing the first enclosed thermometer, using alcohol as a liquid. But it still lacked a standardized scale and was not very accurate.

Around the same time, German physicist Daniel Gabriel Fahrenheit met Olaus Roemer, a Danish astronomer, who developed an alcohol-based thermometer using wine. He marked two points on his thermometer — 60 to mark the temperature of boiling water and 7.5 as the point where ice melted.

So until the statistical mechanical model of an ideal gas, there was no connection of temperature with particles and individual kinetic energy.

The connection with kinetic energy helps to intuitively understand the concept of thermal equilibrium, and this link may help., but the connection happened long after the invention of thermometers.

You state:

I suppose it must be directly measuring the average kinetic energy per degree of freedom. Is that accurate?

I do not think so. The kinetic theory allows us, once we know the temperature and the material, using the kinetic theory to calculate the average kinetic energy. Othewise it measures the temperature on the scale of freezin and boiling water.

anna v
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  • But "measures the temperature on the scale of freezing and boiling water" already assumes a consistent, coherent property called "temperature". Experimentally, such a property seems to exist (which I suppose is why thermometers were invented before statistical mechanics), but I'm wondering what this property of "temperature" corresponds to in a statistical mechanics model. – Sam Jaques Jul 27 '20 at 09:15
  • in a quantum statistical mechanics temperature is the black body radiation curve where the energy per unit volume is dependent on temperature http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html . the black body curve uniquely defines temperature . – anna v Jul 27 '20 at 10:10
  • Sure, but (a) classical thermodynamics seemed to have a reasonable definition of temperature without needing quantum mechanics; (b) this doesn't explain a lot of the everyday phenomena associated with temperature. As in: why would a parameter of black-body radiation cause proteins to denature? Why would it cause phase transitions in states of matter? Why would physical contact between objects cause them to reach equilibrium in this parameter much faster than otherwise? Why do some objects show a "bias" (i.e., heat capacity) in how this parameter reaches equilibrium? – Sam Jaques Jul 27 '20 at 10:28
  • @SamJaques Because temperature is directly connected to energy per volume according to black body formula,. Also qm is necessary to understand the blackbody spectrum, themodynamics is in inadequate . In ideal gases temperature is connected to average kinetic energy, but that is not enough for all the other questions – anna v Jul 27 '20 at 10:45