From Wikipedia entry on Kinetic Theory
The temperature of an ideal monatomic gas is a measure of the average kinetic energy of its atoms.
Now if I remove all the particles from the box shown below will the temperature be zero?
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asked Dec 19, 2010 at 13:06
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1$\begingroup$ Why do you ruin such an potentially interesting question with such an asinine condition ? $\endgroup$– TROLLHUNTERCommented Dec 19, 2010 at 21:00
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3$\begingroup$ -1. Maybe if this question was asking about vacuum fluctuations or something concrete, it would make sense. As it stands now it is just silly. Not wrong, or misleading, just silly. $\endgroup$– user346Commented Dec 19, 2010 at 22:15
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3$\begingroup$ @space_cadet: you are not being fair. If this question is reformulated a little it will become a perfectly good question asking for what happens when you stick a thermometer into a completely empty box. So it is directly related to the problem of temperature measurement (which is non-trivial) and the true nature of vacuum. $\endgroup$– MarekCommented Dec 19, 2010 at 23:19
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6 Answers
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There's no temperature.
If we use the following definition "temperature is the average kinetic energy of the particles". Then no particles - no temperature. As the first sight this answer doesn't seem to be good enough, but if you want to calculate "average spin" or "average charge" those parameters will have no sense if there's no particles to calculate data on.
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Even if you remove the particles, there will be a thermal radiation coming from the borders. This way a thermometer placed inside will eventually show the temperature of borders.
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$\begingroup$ Correct, but not answering the question (and a thermostat does not show any temperature ;-) $\endgroup$– SklivvzCommented Dec 19, 2010 at 14:02
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$\begingroup$ @Sklivvz damn phone typing correction; I'll edit this. But I don't agree that this doesn't answer question; it is explicitly mentioned that this is about vacuum in a box. $\endgroup$– user68Commented Dec 19, 2010 at 14:11
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$\begingroup$ @mbq, I believe a thermometer will show something similar to the temperature of the borders even if we have a non zero, but very small, number of particles - for which the temperature might be different (in the Boltzmann sense), so a thermometer can't be used to determine the temperature in this case. I think. $\endgroup$– SklivvzCommented Dec 19, 2010 at 14:21
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1$\begingroup$ Agreed with Sklivvz: Assuming the walls are at a uniform temperature, we will see a plank distribution of EM energy. This distribution defines the temperature of the radiation field. Unless this wall temperature is absolute zero, you can't have a perfect vacuum, because there would still be photons. There would also be a few gas particles, as the vapour pressure of any material will be nonzero at any finite temperature, and evaporation and deposition of gas particles would reach an equilibrium. $\endgroup$ Commented Dec 19, 2010 at 15:54
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1$\begingroup$ @mbq: I see, you interpreted the question differently. I've understood the question as a conceptual one. It doesn't talk about precise physics of the box, quantum fluctuations of vacuum or any other real physics; just classical point particles (or elastic balls). That's why I think your answer misses the point. $\endgroup$– MarekCommented Dec 19, 2010 at 19:54
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The notion of temperature doesn't make any sense in complete vacuum (meaning the absence of all objects). It only makes sense as a description of how much some objects wiggle around.
To discuss the thermometer problem, one first needs to know what it means to measure temperature. "You just insert thermometer wait a little and you're done", I can hear people say. Well, not quite. What happens microscopically?
In the simplest case, if you want to measure temperature of a sample you need to attach thermometer to it. Molecules of these objects will interact and eventually will come into thermal equilibrium. Thermometer then has some calibration that tells you that so-and-so temperature corresponds to so-and-so much wiggling of its molecules. Well, it should be obvious that for this to work, concept of thermal equilibrium is essential. But you won't get thermal equilibrium if there are very few molecules of the sample. In particular zero.
Also note that the contact of surfaces is not the only way to attain thermal equilibrium. Any heat transfer process will do and that means any interaction. So you can try to measure the temperature e.g. by electromagnetic radiation. Well, if you insert such a thermometer into completely empty box then depending on the box's properties the electromagnetic radiation will either leave completely and thermometer will show zero or the box would trap radiation and the box would no longer be empty (it would contain photons). In any case, what you are measuring now is not temperature of the vacuum but rather the EM transmission properties of the box.
To summarize: problem of measurement is not a trivial one and it has actually lead physicists to great many discoveries. Noting that you can't simultaneously measure position and momentum gave rise to quantum mechanics. Noting that couplings of our elementary particle theories depend on the energy you input into measurement gave rise to renormalization and better understanding of quantum field theories as a whole. So it's always important to think about what are you actually measuring microscopically.
Now, let me talk about some related things for a bit.
Consider that box full of gas again. As you'd lower temperature of the walls the molecules would transfer their energy to the walls and become slower. Now, you can imagine that by doing this for a long time you'll eventually reach zero temperature and all the movement will stop.
In reality, this is not possible because you'd need infinite time to reach that temperature. And even if you had that time, you have to take uncertainty principle (you can't know the position of an object absolutely precisely) into account. Actually, cooling is a big field of physics in itself and entails various extremely sophisticated techniques that are very close to 0K.
Also note that in reality there is no such thing as vacuum (again in the above sense) because of quantum fluctuations.
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$\begingroup$ If I place a thermometer inside the box and manage to get half of the particles out of the box will temperature the thermometer show decrease in temperature? $\endgroup$ Commented Dec 19, 2010 at 13:28
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1$\begingroup$ Although there's nothing wrong with your answer, I believe that radiation is not "supported" by the OP's definition of temperature (i.e. the classical one). The correct answer to their question, should be simply "undefined" (since the average kinetic energy of zero particles is undefined). $\endgroup$– SklivvzCommented Dec 19, 2010 at 14:01
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$\begingroup$ @Sklivvz: yes, I stated that temperature doesn't make sense for zero particles already in the first paragraph. The rest is just to address the comment above yours. $\endgroup$– MarekCommented Dec 19, 2010 at 14:13
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1$\begingroup$ I dont like this answer, cause its completely offtopic rambling and doesnt answer the OP's question $\endgroup$ Commented Dec 19, 2010 at 20:49
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1$\begingroup$ @kalle: can you be more explicit about what's wrong with my answer? Or is this just a spite down-vote? $\endgroup$– MarekCommented Dec 19, 2010 at 20:52
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According to the above definiton the answer is not defined mathematically, the average kinetic energy is
$\displaystyle 1/N \sum_{i=1}^N m_i*v_i^2/2$
If we plug in N=0 we see that it blows up
Now if we had some other definiton or the gas was not quantized maybe we could take the limit as N->0
Edit: This answers your question if you define a thermometer as something that measures temperature, and you have defined temperature as above.
answered Dec 19, 2010 at 13:19
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$\begingroup$ :-) that is why the question! If you put a thermometer inside vacuum what reading will it show? $\endgroup$ Commented Dec 19, 2010 at 13:21
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1$\begingroup$ @Charmer: if it was a perfectly isolated thermometer then it would stay at whatever temperature you had put it in the box. In reality it would radiate as a black-body radiator and eventually lose its energy until it would come into equilibrium with the radiation in the vacuum. If there were none then it would eventually show zero. But it would not measure the temperature of the "zero number of particles". I think I will update my answer to say what it actually means to measure temperature. $\endgroup$– MarekCommented Dec 19, 2010 at 13:29
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$\begingroup$ Since you are taking the average KE, using a limit won't help :-) $\endgroup$– SklivvzCommented Dec 19, 2010 at 14:02
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2$\begingroup$ @TheMachineCharmer: a true vacuum would likely not stay so for very long if you inserted a thermometer into it--it would peel (a very small, but nonzero number of) atoms off of the surface of the thermometer, and would become filled with photons from the blackbody radiation of the thermometer. $\endgroup$ Commented Dec 19, 2010 at 15:14
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1$\begingroup$ @kalle: please mind your language ;-) Also try to be a little bit less arrogant as saying that this is the only correct answer. It's a definition of a kinetic energy in classical physics. In general it has nothing to do with temperature (e.g. it won't work for temperature of a photon gas) and you didn't elaborate on the connection between energy and temperature at all. The only correct answer (if there were such a thing) would need to talk about averaging problem in general (which just doesn't work for zero particles). But not averaging over kinetic energy in particular ;-) $\endgroup$– MarekCommented Dec 19, 2010 at 22:33
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You cannot remove all the particles from the box. Even if you remove all atoms, still there vill be photons inside, which carry their own kinetic energy. The photons are generated by the box walls and reach with time thermostatic equilibrium with the walls, so the temperature inside the box will be the same as the temperature of the box walls.
Any body placed in such box will be gradually reaching the same themperature as that of the box, through radiation, even if there are no atoms inside.
Only if the box walls are at absolute zero there will be no photons inside.
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1$\begingroup$ You are right, but you are repeating mbq's point. $\endgroup$ Commented Mar 24, 2012 at 3:32
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Vacuum does not contain radiation nor other electromagnetic fields.
Stationary and constant velocity vacuums do not have any temperature. Accelerated vacuum has a temperature - the Unruh temperature:
$${\displaystyle T={\frac {\hbar a}{2\pi ck_{\mathrm {B} }}}}$$
It is very weak. In the acceleration of Earth's gravity it is only $4×10^{−20}$ K.
