Is it valid to assign a temperature to individual particles within kinetic theory and then claim that the temperature of the gas is simply the average of the temperature of the molecules?

In other words, can we say that the temperature of each molecule is $T=mv^2/3k_b$ where $v$ is the speed of the molecule, and then the temperature of the body is the mean of the temperature of the atoms or molecules comprising that body?


10 Answers 10


The first thing that needs to be said in this discussion is the fundamental connection between statistical physics and thermodynamics. Statistical physics describes microstates and thermodynamics describes macrostates. They are connected by so called thermodynamic limit—limit of infinite size and infinite particle number.

Microscopic parameters describing particles, like kinetic energy, are not automatically equivalent to macroscopic parameters like internal energy. They are found to be equivalent, which is a nontrivial result. The connection between micro and macro world appears only after doing the thermodynamic limit.

It is also important that many quantities are not the features of a single particle, but of the system. For instance, there is no entropy assigned to a single particle like the energy or momentum is, but there is entropy of the system containing one particle, if you specify the states that can be there.

Temperature is most commonly defined in the macroscopic realm as a quantity that lets you compare the state of two different systems that are interacting only by heat transfer. It turns out that this quantity is: $$T :=\left(\frac{\partial U}{\partial S}\right)_{V,N}.$$ Stretching this definition to the microscopic realm means treating the above equation as definition and putting to it the entropy before the thermodynamic limit. It might cause some trouble—like negative temperatures (that are in fact considered for systems that undergo saturation). So you might talk about a temperature of a system containing one particle but:

  1. it will be not a temperature of this particle, but of the system containing it,
  2. it is a very different temperature than you would expect.


1) Temperature for one particle system

Let us first consider how it is usually done in two particle system ($N=2$), before the problem trivializes in the one particle case. Consider number of microstates given two particles in a box, that can have discrete energy states separated by a constant energy portion of $\epsilon$. We divide the box into $n$ virtual compartments.

Counting the microstates. When the total energy in the system is $1\epsilon$, this energy can be either on first particle, or on the second. Particles are the same, so two situations are exactly the same. This is only one possibility. At the same time, particles can fill compartments in $W=n^2$ ways, so there are $n^2$ possibilities.

For the total energy of $2\epsilon$ we will have 2 possibilities - either energy is distributed equally over both particles, or it is all on one of them. The number of total microstates is $W=2n^2$.

You can see that the number of microstates changes with energy. To calculate what the temperature of this system is, you need to express the number of microstates in terms of energy. This is simple combinatorics but it's not the part of the question. Then entropy is: $S=k_b \log(W)$.

Then you have the expression to calculate

$$\frac{1}{T}=\left(\frac{\partial S(U)}{\partial U}\right)_{V,N},$$ and this will be the expression for temperature of this system.

Let us consider number of microstates for a single particle in a box. As before, we will assume volume divided into compartments.

At energy $1\epsilon$ number of microstates is $n$. At energy $2\epsilon$ number of microstates is $n$, at $3\epsilon$ it's the same.

The number of microstates does not change when adding energy to the system. This means $\frac{1}{T}=\left(\frac{\partial S(U)}{\partial U}\right)_{V,N}$ is zero. Since it's not a limit, it means that a temperature defined this way is not infinite, it simply doesn't exist.

2) As for temperature dependence on the observer, the entropy is not dependent on the change of the observer, so the temperature isn't either, if it were then I would start wondering if the definition is right.

  • $\begingroup$ So according to your last point #1, you are stating that it is not valid to assign individual particles a temperature. Correct? $\endgroup$ – looksquirrel101 Feb 17 '20 at 1:38
  • $\begingroup$ Temperature inherits this property from entropy, and entropy is the measure of the number of possible microstates. It is a feature of system. That correspond well with the main reason we even talk about temperature - to compare systems in different states. Unless the particle has any internal structure and we wish to consider it a system, there is no meaning to the notion of entropy of a particle. $\endgroup$ – Licho Feb 17 '20 at 1:52
  • $\begingroup$ Does this answer your question? $\endgroup$ – Licho Feb 18 '20 at 8:22
  • $\begingroup$ Not really. These are concepts that I am aware of. I'm looking for something more quantitative. Perhaps you could elaborate on the temperature of a system containing one particle and how it would be the same for two different inertial observers. $\endgroup$ – looksquirrel101 Feb 18 '20 at 12:01
  • $\begingroup$ +1. I noticed you committed to Materials Stack Exchange, did you notice we are launched now? materials.stackexchange.com Since you already have a physics account you can get signed in automatically if you click. $\endgroup$ – user1271772 May 16 '20 at 16:22

Is it valid to assign a temperature to individual particles within kinetic theory and then claim that the temperature of the gas is simply the average of the temperature of the molecules?

I don't believe so. Temperature is a macroscopic property of a system. We don't normally talk about the temperature of a single particle.

Thanks for your response. I understand that this is not normally done. But I am asking if there is a logical flaw with such an interpretation.

I wouldn't say there is a logical "flaw" per se. It's just that temperature is defined as a macroscopic property of an object that reflects the collective behavior (in this case average translational kinetic energy) of the multiple microscopic particles that make up an object. But consider the following single particle example at the "macroscopic" level.

I have a ball which I throw and give it translational kinetic energy with respect to the ground. The ball is now my "particle". Assuming I throw it in a vacuum (no air friction) what temperature would I assign to the ball based on the velocity I gave it? The temperature I measure on the ball is only due to the collective microscopic kinetic energies internal to the ball. The balls "internal" kinetic energy. In the absence of air friction, the external kinetic energy of the ball, which is due to the velocity of its center of mass with respect to an external (to the ball) frame of reference, has no influence on temperature that I measure on the ball.

Hope this helps.

  • $\begingroup$ Thanks for your response. I understand that this is not normally done. But I am asking if there is a logical flaw with such an interpretation. $\endgroup$ – looksquirrel101 Feb 15 '20 at 12:51
  • $\begingroup$ @LOLKlimateKatastropheKooks I have responded in an update to my answer. Hope it helps. $\endgroup$ – Bob D Feb 15 '20 at 14:44
  • $\begingroup$ I don’t see how that modification is an improvement. The ball has its own internal degrees of freedom, so that is adding a complication that was not intended in the original question. The question then applies to the ball itself. Can we claim that the temperature of the ball is just the average of the temperatures of the individual particles that make up the ball? My thought is that the answer is no, as you seem to believe as well, but I am asking if such a description is logically absurd. $\endgroup$ – looksquirrel101 Feb 15 '20 at 15:27
  • $\begingroup$ @LOLKlimateKatastropheKooks Well, you are free to take it or leave it. But that's all I have to say on the matter. $\endgroup$ – Bob D Feb 15 '20 at 15:46

Temperature is a valid concept for any system in contact with a thermal bath. As such, you can take any subset of your system that is much smaller than the system, and consider it to be in contact with the remainder of the system as its thermal bath. Since the entropy and temperature are related by

$$\frac{1}{T}=-\left.\frac{\partial S}{\partial E}\right|_N,$$

Then by doing the "marble and matchstick" calculation (e.g. Callen, Thermodynamics, Ch.15) and using

$$S = \log\Big(\,{\rm number\,of\,microstates}\Big),$$

it is easy to show that the "temperature" of the subsystem is just

$$ T = \frac{E_{\rm subsystem}}{N_{\rm subsystem}}$$

In particular, you are allowed to take a subsystem consisting of only one particle, in which case its temperature is just its energy, as you are suggesting. However, the concept of temperature might not be very useful in this case.


Is it valid to assign a temperature to individual particles within kinetic theory and then claim that the temperature of the gas is simply the average of the temperature of the molecules?

Rather than take a yes/no position on the question. As "food for thought", I would like to frame this question in the context of whether it is valid to assign a temperature to an individual isolated particle, or to assign a temperature to an individual non isolated particle of a collection of particles.


Temperature, like pressure, is considered to be an intensive property of a system. By intensive, we mean independent of mass. On the surface, this property would seem to justify that an individual particle can be assigned a temperature representative of the collection. After all, a single particle is simply a subsystem having a mass of 1/M where M is the total mass of all the identical particles of the system. Let's see if this works.

Let's say we partitioned in half a large thermodynamically isolated room containing a monatomic ideal gas (e.g., Helium) at room temperature $T$, by a rigid perfectly insulated partition and measured the temperature in each half. We would be fairly confident that the temperatures in each half would be the same, and equal to the temperature originally measured in the whole room. We could even partition each half in half again, and still be confident that all the temperatures will be the same. But can we continue to do this all the way until we are left with a single particle having a temperature $T$?

No. Because our initial confidence was based on the Maxwell-Boltzmann distribution of speeds and kinetic energies for a large collection of particles. As we continue to decrease the size of the volume being isolated, the average kinetic energy of the volume at the instant it is isolated, and thus its temperature, potentially moves further and further away from that of the original collection. When we reach the last particle, its speed and therefore kinetic energy will be constant. Its "temperature" may bear no resemblance to that of the original collection. Every time we repeat this experiment we wind up with a different "temperature" for the last particle.

May we conclude from the above that temperature is not simply an intensive property, but a macroscopic intensive property, and that the assignment of a temperature to an individual particle isolated from a collection of particles, makes no sense?


If assigning a temperature to an individual isolated particle as representative of the temperature of a collection of particles does not make sense, how about assigning a temperature to a single non-isolated particle within the collection?

Returning to the large isolated room of helium gas, we know that at any instant in time the speeds and thus kinetic energies of the individual particles vary according to the Maxwell Boltzmann distribution. On the other hand, since the individual particles are constantly colliding and exchanging kinetic energy with one another, the speeds of individual particles are also continuously changing in time.

If we were able to follow the speed history of an individual particle over a long period of time, and took the average of its speed over that period, what would the average kinetic energy of that particle be? We know that the average kinetic energy of the collection of particles at any given instant in time is constant with a given value. Would it not also be the case that the kinetic energy of any individual particle, selected at random, averaged over a long period of time will be the same as the average of the collection of particles at any given instant in time? Intuitively it would seem so.

In this example it appears we can assign a temperature to a single particle based on its kinetic energy averaged over a long period of time as being the same as the temperature of a collection of particles having the same average kinetic energy at a given instant in time. But we would be assigning the particle a temperature based on its behavior in a collection of particles.

May we conclude from the above that even if we say the assignment of a temperature to an individual particle is valid, it is also inexorably linked to the macroscopic behavior of a collection of particles, and because of that the temperature assigned to the individual particle has to be in the context of the macroscopic behavior of the collection?

Hope this helps.

  • $\begingroup$ So then two different inertial observers moving at different velocities would disagree on the temperature of the particle. Correct? $\endgroup$ – looksquirrel101 Feb 17 '20 at 0:44
  • $\begingroup$ @LOLKlimateKatastropheKooks would two different inertial observers disagree on the temperature of the book on your table? $\endgroup$ – Bob D Feb 17 '20 at 1:22
  • $\begingroup$ They shouldn’t. Do you believe they should? $\endgroup$ – looksquirrel101 Feb 17 '20 at 1:31
  • $\begingroup$ @LOLKlimateKatastropheKooks Of course they shouldn't disagree. That was the point of my comment responding to yours. The internal kinetic energy of a container of gas particles is that associated with the random translational motions of those particles within the container, whether its a billion particles or a single particle. $\endgroup$ – Bob D Feb 17 '20 at 14:18
  • $\begingroup$ The uniform rectilinear motion of the container (inertial motion) has no effect on the internal energy. That is the external kinetic energy of the container of particles, that is, the kinetic energy of the system with respect to an external frame of reference. That kinetic energy is inertial frame dependent. I have updated my answer to clarify it is in reference to random translational motion. $\endgroup$ – Bob D Feb 17 '20 at 14:18

Is it valid to assign a temperature to individual particles within kinetic theory

No, it isn't. According to kinetic theory of gases, temperature of gasses is mapped to average kinetic energy of molecule (which maps to average speed of molecule): $$ T = \frac{2}{3}\,k_B^{-1}\,\overline{E}_k $$ Single molecule speed or kinetic energy shows nothing about gas temperature, thus it's meaningless to define "molecule temperature".


If we would like to somehow define own molecule temperature no matter what - it would be related to atoms (which composes molecule) vibrational energy. This kind of temperature is called "Vibrational temperature". And is defined in thermodynamics as : $$ \theta _{vib}={\frac {h\nu_{vib} }{k_{B}}} $$

Typical vibration frequencies of atoms in a molecule ranges $[10^{13}; 10^{14}] \,\text{Hz}$

This gives for typical $\text{O}_2$ molecule a $2256\,K$ vibrational temperature. Btw, same vibrational temperature equation can be applied in principle to electromagnetic radiation quanta, finding-out "own temperature" of photon. This time substituting electromagnetic wave frequency. However, physical meaning of "vibrational temperature" of photon would be highly questionable.


Strictly, temperature is a property of an ensemble, not a single particle, so one can only with qualifiers speak of the temperature of a single particle such as a molecule in a gas.

When assigning a temperature to a single particle, the right way to do it is to say the temperature is a property of the motion after averaging over the trajectory, not a property at each moment. Therefore, whereas the velocity and speed of the particle changes repeatedly by collisions, its temperature does not because temperature always was an average property after averaging over all the collisions etc. Therefore, when we understand the temperature this way, one finds that in thermal equilibrium all the molecules of a gas have the same temperature as one another.

In the case of laser cooling of single atoms, there is only a single particle in the system. It can happen (and usually does happen) that when illuminated by lasers the momentum of the particle undergoes diffusive heating combined with frictional cooling, so it is not a constant. The atom's kinetic energy fluctuates up and down. In this case it can so happen that the probability distribution of the kinetic energy $\epsilon$ takes the form $P(\epsilon) \propto \exp(-\epsilon/A)$ for some constant $A$. By comparing this to the Boltzmann factor, one may then say that the distribution is 'thermal' and the atom has a 'temperature' equal to $A/k_{\rm B}$. Strictly speaking however this is not a case of thermal equilibrium, which is why I put the word temperature in inverted commas. The laser field here is not in a thermal state, but it so happens that the net result of its interactions with the atom puts the atom in a thermal state of motion.

  • $\begingroup$ "whereas the velocity and speed of the particle changes repeatedly by collisions" If the particle collisions are ideal elastic, and the system is isolated (no black body radiation emitted from the system) the speed of the particle should be constant, no? $\endgroup$ – Bob D Feb 18 '20 at 17:59
  • $\begingroup$ @AndrewSteane - Thank you. This is how I have always understood it. Both with respect to temperature being a property of the ensemble, and that the averaging of the motion over the trajectory of a single particle will yield the same temperature for all particles in equilibrium. $\endgroup$ – looksquirrel101 Feb 18 '20 at 18:44

Well, I did a search engine search with the following words:

magneto-optical trap single atom

By the looks of it: there are multiple publications about experiments that involve bringing down the density of atoms in the trap to single atom observations. I had a quick look at one publication, and I noticed a remark about the 'thermal velocity of the atom'.

Just glancing at the text excerpts in the search results overview I do see the word 'cooling' used. So it looks to me that for people conducting single atom experiments it is common to still use the expression 'cooling' when referring to reducing the velocity of those single atoms.

(Of course, for a proper assessment one should go through a sizable number of publications.)

  • $\begingroup$ This does not address the question. $\endgroup$ – looksquirrel101 Feb 15 '20 at 12:48
  • $\begingroup$ Cleonis: Agreed. The use of phrases like "LASER cooling" in ion traps is pretty common, and confusing when the idea is carried over to statistical mechanics. Plus we have terms like "thermal neutrons" that mean slow or low energy. If someone is reading material that is basically overview of a the topics, where will they look to learn the difference? $\endgroup$ – C. Towne Springer Feb 17 '20 at 1:25
  • $\begingroup$ I have also seen the "temperature" of a single particle referred to by astrophysicists, by which they are indicating the kinetic energy of the particle (relative to Earth) when it arrives in the atmosphere. It may not be thermodynamically "correct" but it is not invalid. $\endgroup$ – Guy Inchbald Feb 20 '20 at 19:43

This is a tricky question because of the phrasing "is it valid to assign a temperature to individual particles". The answer relies on the difference between Thermodynamics and Statistical Mechanics.

Thermodynamics is the study of macroscopic systems. The properties of thermodynamics are therefore macroscopic in nature, starting with fundamental properties such as total internal energy ($U$), entropy ($S$), volume ($V$), and some concept of quantity (i.e., how much "stuff" there is), often denoted by $N$. As a discipline, thermodynamics is a little hard to define; my grad school textbook defined it as "the study of the restrictions on the possible properties of matter that follow from the symmetry properties of the fundamental laws of physics". That may be a little broad, but the author was trying to get at two properties of thermodynamics: (1) it applies generally to all macroscopic systems, regardless of the actual constituents of the system, and (2) unlike other domains (electrodynamics, classical mechanics, etc.) it does not predict specific numerical values for observable quantities; rather, it sets limits (inequalities) on various processes and it establishes relationships among macroscopic properties that may not at first glance seem related. [1]

Statistical mechanics, on the other hand, is the bridge between thermodynamics and the other domains (electrodynamics, classical mechanics, quantum mechanics) that do offer specific predictions about individual particles. It does so by treating a macroscopic system as an ensemble comprised of a very large number of microscopic elements, and uses statistical techniques to derive macroscopic properties.

So given these definitions, the short (but unsatisfying) answer is no: it is not valid to assign a temperature to an individual particle. The reason is that temperature is a thermodynamic concept; it is squarely in the domain of thermodynamics, not statistical mechanics. Temperature is defined by the relationship between total internal energy and entropy: $$ T = \frac{\partial U}{\partial S},$$ with $V$ and $N$ held constant. In other words, temperature is a macroscopic property of bulk matter, and (as is true in all of thermodynamics) there is no concept of individually interacting particles on a microscopic level.

I suspect what you're really looking for, however, is the statistical mechanics treatment. In other words, what you're really interested in is how (thermodynamic) temperature is related to microscopic properties such as the kinetic energy of individual molecules.

The answer to that question starts with how you model the molecules. There are different options with varying levels of complexity. The simplest is to consider a gas and treat the molecules as microscopic sphere-like particles, each of which travels at a constant velocity in a random direction and undergoes random elastic collisions with other particles and with the walls of a container. This is the kinetic theory of gases. Under these assumptions, all the energy in the system ($U$) is in the ½$m v^2$ kinetic energy of the molecules, which is $\frac{3}{2}kT$ on average. This is what you were referring when you asked whether we can say that "the temperature of each molecule is $T=mv^2/3k_b$".

I have two comments

  1. That result relates temperature to the average velocity of the particles. Each particle will actually have a different velocity $v$, which you can think of as drawn from a probability distribution given by the Maxwell Boltzman distribution: $$f(v) ~\mathrm{d}^3v = \left(\frac{m}{2 \pi kT}\right)^{3/2}\, e^{-mv^2/2kT} ~\mathrm{d}^3v$$ However, there's still nothing about this that prevents you from assigning a unique "temperature" to an individual molecule based on its particular velocity. But this leads to the second point:

  2. The result holds only for the rather idealized situation of how we're treating molecules. Things change when you start to treat molecules a little more realistically, and also when you start to consider liquids and solids as well as gases. Rotational and vibrational energy modes start to become important, for example, and in some situations they become dominant.

So even in the statistical mechanics sense, you can't universally assign the "temperature" of a molecule to be $T=mv^2/3k_b$. You can do something similar, which is to find the relation between the thermodynamic temperature and the energy of a molecule (or, more generally a "microstate") for any system in thermodynamic equilibrium. This is done in an average way via the equipartition theorem, and on an individual (probabilistic) level using partition functions. However, this will depend on the system, and there's not much value in using the word "temperature" to describe a property of an individual molecule. On the microscopic level, it's better to stick with well-understood terms such as energy, and reserve the word "temperature" for macroscopic systems.

[1] Callen, Herbert (1985). Thermodynamics and an Introduction to Thermostatistics (2nd ed.). New York: John Wiley & Sons. ISBN 0-471-86256-8.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Feb 20 '20 at 18:06

Physics is neither religion nor mathenatics. It is also good to see different concepts from different angles and not to dismiss ideas too early. Eventually the results from the ideas have to agree with experiment (and survive Occams razor).

It really depends where you try to go from that statement. If eg

A) you then claim that you can take each particle, sort them by their temperature and then violate 2nd law of Thermodynamics: then no , it is not valid to think about it like this B) you try to model the inelastic collision of two particle as a exchange of temperature, which should lead in that model to equalizing temperature, aka velocity: the no, this is alson wrong C) if you just say it and dont do anything with it: then it is not really wrong, but it wont survive Occam eventually D) if you use it as a stepping stone to understand the dynamics of mesoscopic systems better and eventually the new results agree with experiment: then yes, its valid, if this assumption still is central to the new theory

Similar example is the introduction of negative Temperature. It is wrong in Thermodynamics, but a good way to think about the statistical physics of system with bound Hamiltonian and special initial conditions (e.g. spin chains).

Also I believe there are some publications on single particle thermodynamics using the quantum statistical operator.


It depends on what you want to do with your definition. In science definitions are shorthand for phenomena/properties that capture some information that can be used to built up to explain more complicated phenomena/properties.

We define temperature for a macroscopic quantity to label some numerical quantity that equalises when two objects that are allowed to exchange energy. And using this defined quantity of temperature, among other things, we can build up a formalism to study heat and how it can be converted to work. Namely, thermodynamics.

Is it valid to assign a temperature to individual particles within kinetic theory and then claim that the temperature of the gas is simply the average of the temperature of the molecules?

Consider two particles of equal mass but one is is at rest and the other is moving. They are enclosed in a container that elastically scatters them. Now if we define temperature as the kinetic energy then the two particles will keep exchanging velocities which in your case is the same as the temperature. That means the two objects will never have the same temperature.

But you might say, oh but the temperature is defined only for macroscopic quantities and you took only two. Well consider as many particles as you like their velocities will never be equal. Which means they won’t have the same temperature under your definition. What will be equal however, is the average kinetic energy of a sub collection of the particles.

Thus is we want temperature to have the property of being equal between objects in thermal equilibrium, then we can’t use your definition of temperature.


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