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What conditions do a bunch of atoms need to satisfy to have a temperature?

Suppose that we have a beam of helium atoms travelling in a common straight line, equally spaced with the same velocity. If we place a thermometer in the path of this beam, the atoms will impinge on it and heat it up, to register a temperature. However, if we move the thermometer along with the beam, the indicated temperature can be reduced, ultimately to zero.

Now add a second collimated beam, exactly opposite to the first one, on a very close parallel path. In this case, whichever way we move the thermometer, a positive temperature will register, and it seems reasonable to assign the minimal such reading to be the temperature of this "beam dipole". Is this the simplest possible, or least-entropic, system that can be said to have a temperature?

Added in response to replies: I am troubled by the need for entropy to be maximized as a condition for temperature to be meaningful. The above scenario has the unrealistic quality of an idealized micro-experiment (as Maxwell's demon might carry out), which puts it out of range of classical statistical mechanics. So let's consider a macro-scale (even industrial-scale) alternative, at the risk of introducing features that may distract from the essential idea of defining temperature.

Suppose that we have a gas-phase chemical reactor. The aim is to combine the primary reactants and cool the product as quickly as possible so as to minimize secondary reactions among the products. This is done by injecting two symmetrically opposed, narrow, very-high-speed jets of cooled gaseous reactants under a bath of inert liquid coolant. The reaction rates at any given pressure, temperature, and concentration of reactants and products are well known. (For convenience, we may assume that the reaction is neither exothermic nor endothermic, but the change in entropy favours the reaction at the designed parameter settings.) When the reaction rate is measured, there is an implied temperature profile in the reaction chamber, assuming a classical chemical-thermodynamic--fluid-dynamic model of the process (which would surely be very complicated, but whose details need not concern us!).

New question: Is this implied temperature profile fictitious?

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Great question! I could write a great answer if I had time, but it would be an awful lot of work. The short short version is that a system really needs to be in equilibrium to have a temperature, and the two beams travelling in opposite directions are not in equilibrium, since you could certainly extract work from such a system. But that's an over-simplification, because there are lots of systems that have temperatures without being in equilibrium. (E.g.: a human being.) Disentangling that in a formal way is possible, but as I said, a lot of work. –  Nathaniel Jul 19 at 11:05
    
@Nathaniel Well now I'm just really hoping you'll do that work. Any chance you'll do it at some point ? –  ticster Jul 19 at 11:19
    
@ticster maybe, but it's very dependent on me finding time - I'd say don't hold your breath! –  Nathaniel Jul 19 at 13:52

2 Answers 2

No. You wouldn't say that pair of beams has a temperature. Temperature is defined by the zeroth law of thermodynamics, which states that if $A$ is in thermal equilibrium with $B$ and $B$ is in thermal equilibrium with $C$ then $A$ is in thermal equilibrium with $C$ and $A$, $B$ and $C$ are said to have the same temperature.

Temperature is fundamentally a property objects being in thermal equilibrium and it is very difficult to talk about a beam being in thermal equilibrium with anything. This is because a beam of particles has a very low entropy, all the particles have moving in more or less the same direction with more or less the same energy. Almost any interaction of the beam with another object will cause it to scatter and you will no longer have a nice clean beam. Even your thermometer is not at thermal equilibrium with the two beams. The beam is being destroyed as it collides with the thermometer. The thermometer is in a steady state, losing energy as heat at the same rate it gains energy from the beams, but this steady state is not in thermodynamic equilibrium. So the beam does not have a temperature.

If you want the simplest system with a temperature then I would say, depending on your taste for what is simple, either a particle in a box where the walls of the box are in contact with a large thermal reservoir at a fixed temperature or a large number of particles in a box with walls which do not allow any energy in or out. In the first case the particle reaches equilibrium with the walls of the box, so that if you where to measure the energy of the particle at different times (there are some complications here to do with the ergodic hypothesis ) the distribution of results would be a Boltzmann distribution. In the second case the particles would exchange energy with each other to achieve equilibrium so that measurements of the energies of particles would, again, have a Boltzmann distribution.

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In your second example (which I prefer, since it requires only a reflecting surface and not a massive thermal reservoir), the minimal condition seems to be that the particles are dense enough to bounce off each other at a significant rate; otherwise they could maintain a non-Boltzmann energy distribution, given perfectly elastic collisions with the walls. –  John Bentin Jul 19 at 11:23
    
I would say the ultimate criteria is that there should, on average, be no net transfer of energy between particles. This suggests that relevant criteria would be the size of fluctuation you are willing to tolerate and the ratio of the timescale between microscopic interactions, the collision time $\tau$, and the timescale of your measurements $T$. But I think the difference between that and what you said is pretty abstract. –  By Symmetry Jul 19 at 11:38
    
Regarding the thermometer "losing energy as heat at the same rate it gains energy from the beams": Suppose we surround the thermometer with a synthetic thermal blanket maintained at the (time-lagged) temperature of the thermometer (with two small holes to admit and reflect the beam). Would that make the mean energy of the reflected beam particles equal to that of the incoming particles? In the everyday scenario of taking the temperature of a gas, the velocity of each impinging gas molecule changes when it hits the thermometer, but this is not an issue for measuring the temperature of the gas. –  John Bentin Jul 24 at 13:20
    
The problem is that a beam cannot interact with the thermometer without changing its macrostate, so it cannot be in equilibrium with the thermometer, or anything else, so it can't have a temperature. It doesn't matter that a system in a gas changes its energy when it interacts with a thermometer, provided that if it interacts many times there is no transfer of energy on average and not change in the average state of the system. This is true for particles in a gas, which are already random, but is not true for a beam, in which the particles go from a very particular state to a more random one. –  By Symmetry Jul 24 at 14:03

For a system of atoms to have a meaningful temperature, I would say there has to be uncertainty in their state, and there must be a way (at least hypothetically) for the system to exchange energy with some other system.

The first example you give of the two beams is an idealized state with no uncertainty, so assigning a temperature would be impossible. You could imagine however, a beam where the spacing and velocities of the particles varied, and the particles exchange energy with their neighbours. This would have a well-defined temperature, though it wouldn't be easy to measure.

The second scenario you describe is a non-equilibrium reaction, which is also difficult to assign a temperature to. As an example of why, here's a quote:

Imagine, for instance, a system composed of matter and electromagnetic radiation. A thermometer with perfectly reflecting walls will be unaware of radiation and it will indicate the temperature of the gas. In contrast, a thermometer with perfectly black walls will be sensitive to radiation and will indicate a temperature related both to radiation and matter. In equilibrium, gas and radiation have the same temperature, and both thermometers will indicate the same reading. Out of equilibrium, matter and radiation may have different temperatures and different thermometers will yield different readings.

It might still be possible to assign a meaningful temperature to such a mixture, but that's out of my expertise.

For your specific example, it might be possible to assign a temperature profile in the chemical reaction, because a small volume could be in quasi-equilibrium with its surroundings.

Source: http://www.physik.uni-augsburg.de/theo1/hanggi/Casas.pdf This is the paper I pulled the example from. It also seems like a good paper in general about the topic of temperature in non-equilibrium systems.

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(+1) The paper you cite looks really relevant and interesting. –  John Bentin Jul 25 at 9:23

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