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When we measure the temperature of a gas we typically integrate the molecular collisions and wind up with an 'average' temperature due to the sensor comprising a relatively large thermal mass. And measurement of temperature requires heat transfer through that mass.

But if we shrink the thermal mass of the temperature sensor to smaller and smaller sizes we may notice that the temperature measure gets much 'noisier'. But I suspect it's not all electrical noise, but rather the ability of the sensor to start seeing the actual temperature fluctuations in the gas.

So to my primary question - can we predict by theory the distribution of actual temperature of a gas considering unbounded bandwidth? I'm assuming an ideal gas.

And if so, is there a temperature measuring technology that can come close to measuring the theoretical distribution?

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What do you mean by temperature? Let's say you have your gas hooked up to a very tiny thermometer. It has a pointer that jiggles and bumps around. What you'd find is that the pointer fluctuations are mostly related to the mechanical evolution of the thermometer itself, not the system under consideration.

But that's beside the point. Temperature is not just a number on a machine, but rather a concept that we aspire to define and approximate better and better. So really, what is the essence of the concept called temperature?

What we want most from temperature is the notion of equilibrium. Most importantly, two systems at the same temperature should be unaffected when brought into thermal contact. Now, from microscopic mechanics we realize that systems will certainly exchange energy when brought into contact, and so it would seem that it's impossible for the systems to not affect each other. From this it would seem that temperature is fundamentally a flawed concept and cannot apply to microscopic systems.

Statistical mechanics solves this conundrum: what if nobody precisely knew the states of the systems beforehand, that they were randomized? What if system A and system B have states that are so uncertain that when they are brought into contact, their probability distributions are unchanged? For all purposes, there would be no effect by bringing them into contact. Nobody would notice the difference.

In this way two systems may come into contact and certainly affect each other, and certainly exchange energy, yet still be said to be in equilibrium. This allows the concept of temperature to apply to any system, provided the uncertainties have a certain form. However we must then abandon the concept that temperature is a property of a system. Temperature is unlike energy or momentum, in that if you fully describe the state of a system then it ceases to have a well defined temperature. It is only once you begin to speak of average behaviours, of uncertain systems, that the notion of temperature becomes valuable.

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  • $\begingroup$ The concept of temperature is not restricted to equilibrium states. Moreover, instantaneous temperature, like energy or momentum, is a property of a single system. The concept of average temperature is a property of the ensemble, of the whole collection of systems. Precisely the difference between the actual temperature of a system and the temperature of the ensemble is what we call the fluctuation: $\delta T \equiv \tilde{T} - T$. $\endgroup$
    – juanrga
    Aug 5, 2016 at 18:25
  • $\begingroup$ also see physics.stackexchange.com/questions/175833/… $\endgroup$
    – Yrogirg
    Jan 28, 2020 at 2:56
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The instantaneous temperature of a gas system is a fluctuating quantity given by

$$\tilde{T} \equiv \frac{2\tilde{K}}{k_\mathrm{B} N_\mathrm{df}}$$

here $ \tilde{K}$ is the fluctuating kinetic energy and $N_\mathrm{df}$ the number of degrees of freedom. For the usual case $ N_\mathrm{df}=3N$ with $N$ the number of particles, and averaging both sides of the definition we obtain the average temperature.

$$T = \frac{2K}{k_\mathrm{B} 3N}$$

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Depends what sort of precision you're looking for and the scales of the volumes you are considering. Assuming you mean volumes on the scale of your hand or larger, yes. You can do some really neat things with lasers.

However, at some point (getting smaller and smaller) the heisenberg uncertainty principle will start to get in the way.

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  • $\begingroup$ Although people invoke the HUP for all sorts of things, this doesn't mean it's justified. What has the HUP to do with this question, which seems to be fully classical? $\endgroup$
    – ACuriousMind ♦
    Mar 12, 2015 at 23:54

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