Exploring Temperature: Understanding Particle Vibration and Absolute Zero

[vanhees71]

Well vacuum in physics usually means there's nothing, but radiation is something. A vacuum in this strict sense is impossible to prepare thanks to the 3rd Law of thermodynamics.

 

[russ_watters]

With your restrictive understanding of vacuum there would not even be a speed of light in vaccum.

From the first paragraph of the section:

The strictest criterion to define a vacuum is a region of space and time where all the components of the stress–energy tensor are zero. This means that this region is devoid of energy and momentum, and by consequence, it must be empty of particles and other physical fields (such as electromagnetism) that contain energy and momentum.

The first few sections of the article are too loose/colloquial for my taste. It's true that vacuums on Earth are useful even when containing radiation/fields, but they waited too long before giving the complete definition.

Also, this isn't a necessary logical knot, a bit like "alone in an empty room", the speed of light in a vacuum is like saying the space is empty except for the light. Or, the vacuum is what you have before you send the light through.

And not for nothing, but "the speed of light" is an unnecessarily restrictive term, coined because light was the first thing discovered that that speed applied to. "The speed of light" applies even if there is no light involved in whatever the scenario is being examined.

And maybe more to the point, you started this with the suggestion that there can be radiation in a vacuum, which, true or not, is an incorrect response to what you were responding to in post #31: radiation cannot store heat. It carries heat, but you can't, for example, heat up the already released photons in a laser beam.

 

[DrStupid]

@russ_watters many thanks aganin for clarification. I stand corrected.

And maybe more to the point, you started this with the suggestion that there can be radiation in a vacuum, which, true or not, is an incorrect response to what you were responding to in post #31: radiation cannot store heat.

Yes, of course heat cannot be stored but ransferred only. But I didn't want to be nitpicking. I think it is clear what PeterDonis actually means by "store heat".

 

[PeterDonis]

With your restrictive understanding of vacuum there would not even be a speed of light in vaccum.

Sure there is; it just isn't called the "speed of light" because that name has incorrect implications. It's called the "invariant speed". It's a property of spacetime, not light.

 

[PeterDonis]

radiation cannot store heat. It carries heat, but you can't, for example, heat up the already released photons in a laser beam.

Sure you can: just interact them with something. Radiation carries energy, has temperature, and stores heat. To change the heat content of anything, it has to interact with something: you can do that with radiation just as you can with air (though the specific interaction will be different).

The difference with radiation is that, since it has no conserved particle numbers or charges (no baryons, no leptons, no electric charge), the interactions it undergoes can be thought of as destroying the old radiation and creating new radiation. But I don't think that justifies saying that radiation cannot store heat.

 

[russ_watters]

Sure you can: just interact them with something. Radiation carries energy, has temperature, and stores heat. To change the heat content of anything, it has to interact with something: you can do that with radiation just as you can with air (though the specific interaction will be different).

The difference with radiation is that, since it has no conserved particle numbers or charges (no baryons, no leptons, no electric charge), the interactions it undergoes can be thought of as destroying the old radiation and creating new radiation. But I don't think that justifies saying that radiation cannot store heat.

I'm trying hard not to quibble with this usage of the word "heat" as I have been criticized in the past for being a bit loose with it (and in turn I think it is a poorly defined word) and I recognized when I said it that the word "carry" may be cumbersome, but the second paragraph is indeed what I meant. I'll give examples:

-If you have a metal container filled with air and you put a blowtorch to the outside, you heat the air inside.
-If you have a metal container that is a perfect mirrored surface (for certain wavelengths) and a certain amount of radiation bouncing around inside, and you put a blowtorch to it, you can add additional radiation without affecting the already existing radiation in the container.
-If the metal container of air has a movable wall, you can apply mechanical work to generate additional thermal energy in the air. That's not adding heat to the air, that's doing work on the air, per the definition of heat.
-The metal container with the movable wall can also apply mechanical work to the radiation.

And maybe a more direct response to/example of what I was responding to:

A classical vacuum can't be brought into thermal equilibrium with anything either; there has to be something present that can store heat, in which case it is no longer vacuum.

Here's what I was envisioning: The sun radiates toward Earth, not in thermal equilibrium. Any arbitrary volume of space between them has radiation traveling from the sun to the Earth, carrying thermal energy (heat). At any snapshot in time, this volume will contain a certain amount of thermal energy in transit (heat). I'm not sure, so maybe I'm missing a thermal interaction that can be done to increase the amount of heat being carried to Earth while it is in transit.

Contrast that with, say, water flowing from one place to another, which whether in a container/conduit or not can be heated. IMO, this difference is relevant even if the terminology is imprecise.

 

[PeterDonis]

-If you have a metal container filled with air and you put a blowtorch to the outside, you heat the air inside.

Yes.

-If you have a metal container that is a perfect mirrored surface (for certain wavelengths) and a certain amount of radiation bouncing around inside, and you put a blowtorch to it, you can add additional radiation without affecting the already existing radiation in the container.

No, you can't, because there is no way to distinguish "the already existing radiation" from "additional radiation" that you added. The radiation isn't going to have a well-defined photon number anyway since it's not going to be in a Fock state; it will basically be black-body radiation. But the radiation will have a well-defined temperature, and that temperature will increase when you apply the blowtorch. So it certainly seems appropriate to say that you heated the radiation inside.

maybe I'm missing a thermal interaction that can be done to increase the amount of heat being carried to Earth while it is in transit.

There isn't any obvious one in the solar system as it is, but the same would be true for, e.g., a coronal emission from the sun, which is basically a blob of plasma, i.e., a blob of baryons and leptons, not radiation. You would have to put something in the path that could interact with the plasma thermally, but you could do that with the radiation too.

IMO, this difference is relevant

I'm not seeing any relevant difference as far as heat is concerned.

 

[russ_watters]

No, you can't, because there is no way to distinguish "the already existing radiation" from "additional radiation" that you added. The radiation isn't going to have a well-defined photon number anyway since it's not going to be in a Fock state; it will basically be black-body radiation. But the radiation will have a well-defined temperature, and that temperature will increase when you apply the blowtorch.

That surprises me. If I have a green laser shining into this container and then apply a blowtorch causing the container to glow, won't a camera inside record a smooth black-body curve except for an extra-bright green section? Doesn't superposition apply?

There isn't any obvious one in the solar system as it is, but the same would be true for, e.g., a coronal emission from the sun, which is basically a blob of plasma, i.e., a blob of baryons and leptons, not radiation. You would have to put something in the path that could interact with the plasma thermally, but you could do that with the radiation too.

I would think a hot blob of plasma is going to radiate and cool down on its way toward Earth. But yes, you could also have it flow through a large heat exchanger on its way to Earth and cool down via conduction.

I'm not seeing any relevant difference as far as heat is concerned.

Fair enough. I'll try one more way to express what I see:
There are, as far as I can tell, four ways to add thermal energy to air in a tank: Conduction, convection, radiation and mechanical work.

From our discussion it would seem there are two ways to add thermal energy to a tank full of nothing but radiated photons: additional radiation and mechanical work.

Is this difference, 2 vs 4, and the methods themselves "relevant"? I guess I don't think it is something I want to quibble about. I'll call them "different" and leave it to others to decide if they think they are "relevant".

 

[vanhees71]

Yes.
No, you can't, because there is no way to distinguish "the already existing radiation" from "additional radiation" that you added. The radiation isn't going to have a well-defined photon number anyway since it's not going to be in a Fock state; it will basically be black-body radiation. But the radiation will have a well-defined temperature, and that temperature will increase when you apply the blowtorch. So it certainly seems appropriate to say that you heated the radiation inside.

Just to add a bit more: If you have a container and heat the walls, the electrons contained in the walls are rattling more and more, and this random motion leads to em. radiation. Keeping the walls at a certain constant temperature the corresponding radiation inside will also come to thermal equilibrium. This process is easily understood in a kinetic-theory way: There are photons created and absorbed all the time until everything comes to thermal equilibrium, where the absorption and creation rate becomes equal. Fortunately we don't need to solve this kinetic problem to know the equilibrium state of the radiation, but that's given by the maximum-entropy principle (equilibrium is the state of maximal entropy at the given constraints; in our discussed case it's the given temperature of the walls which equivalently means that the mean energy density of the radiation is fixed).

For simplicity let's make the quantization volume a cube of length $L$ and take it as a subvolume within the very big total volume of the cavity. Then we can impose periodic boundary conditions, which makes the calculation a bit simpler. This means a complete basis of the electromagnetic field are bosonic Fock states with the single-particle states chosen as momentum-helicity eigenstates. The momenta $\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}$ and the helicities take the values $\lambda=\pm 1$ (I use natural units with $\hbar=c=k_{\text{B}}=1$). The Hamiltonian is given by
$$\hat{H}=\sum_{\vec{p},\lambda} E_{\vec{p}} \hat{N}(\vec{p},\lambda),$$
Where $\hat{N}(\vec{p},\lambda)$ are the number operators for photons in the wave mode given by $\vec{p}$ and $\lambda$. The energies are $E(\vec{p})=|\vec{p}|$ since photons are massless.

The partition sum thus is given by (with $\beta=1/T$ the inverse temperature)
$$Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
For the following it's more convenient to first calculate a somewhat more general quantity in order to be able to calculate various interesting expectation values, namely
\begin{equation*}
\begin{split}
Z[\alpha]&=\mathrm{Tr} \exp(-\beta \hat{H} + \sum_{\vec{p},\lambda} \alpha(\vec{p},\lambda) \hat{N}(\vec{p},\lambda) = \mathrm{Tr} \exp(\sum_{\vec{p},\lambda}[-\beta E(\vec{p})+\alpha(\vec{p},\lambda)] \hat{N}(\vec{p},\lambda)] \\
&=\prod_{\vec{p},\lambda} \sum_{N(\vec{p},\lambda)=0}^{\infty} \exp[(-\beta E(\vec{p})+\alpha(\vec{p},\lambda))N(\vec{p},\lambda)] \\
&=\prod_{\vec{p},\lambda} \frac{1}{1-\exp(-\beta E(\vec{p})+\alpha(\vec{p},\lambda))}.
\end{split}
\end{equation*}
Now it's easy to calculate the mean number of photons in each wave mode. Given that the Statistical operator is
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H})=\frac{1}{Z} \left [\frac{\partial Z[\alpha]}{\partial \alpha(\vec{p},\lambda)} \right]_{\alpha(\vec{p},\lambda)=0} = \left [ \frac{\partial}{\partial \alpha(\vec{p},\lambda)} \ln Z[\alpha] \right]_{\alpha(\vec{p},\lambda)=0}.$$
Which leads after some algebra to
$$n(\vec{p},\lambda)=\langle N(\vec{p},\lambda) \rangle=\frac{1}{\exp(\beta E(\vec{p}))-1},$$
i.e., the Planck black-body distribution as expected.

This doesn't depend on $\lambda$, i.e., the radiation is completely unpolarized (which is also expected from the equlibrium state being the state of maximal disorder).

BTW: A perfectly reflecting wall won't lead to thermal equilibrium at all since no em. radiation (or photons) can be absorbed. It's perfectly isolating and thus you can't "heat up the radiation", and you'd just have the one "green" cavity mode excited all the time. On the other hand, if the box is not a perfect mirror inside (and that's the case of course for any real-world cavity, although you can indeed make quite perfectly reflecting cavities nowadays) the initially excited radiation will get partially absorbed and the cavity walls heat up and at the end you have Planck radiation (the better the walls are reflecting the longer this will take) with the temperature determined by the total energy within the cavity and the walls.

 

[PeterDonis]

If I have a green laser shining into this container

Do you mean, a green laser originally was shining into the container, which "captured" some of its light, and then the container was sealed? Otherwise the container would have to be open, which I thought it wasn't supposed to be.

If this is what you mean, then I think you could, in principle, tell that the green laser light was there. But first, there is a technicality to dispose of: as @vanhees71 points out, if the walls are perfectly reflecting mirrors, they won't absorb any radiation, and therefore they won't emit any radiation either, which means there's no possibility of heat exchange between the radiation and the walls. If that's the case, then of course you can't heat the radiation by applying a blowtorch to the outside of the tank: but that's because of a special (and impossible in practice) property of the walls, not the radiation.

If, OTOH, we allow the walls to absorb and emit at least some radiation, then, at any nonzero temperature, they will do so, and so even before we apply the blowtorch, there will be some black-body radiation inside the tank as well as the green laser light that got captured. We could tell that the green laser light was there because the intensity at that frequency would be higher than the normal black-body radiation intensity at whatever the radiation temperature was inside the tank. Applying the blowtorch to the outside would raise the radiation temperature inside the tank, but you would still be able to tell that the green laser light was there because there would still be a spike in intensity at that frequency. Over time, however, the green laser light would gradually be absorbed by the walls and re-emitted as black body radiation, so the spike would gradually go away.

But this still doesn't show a fundamental difference between radiation and an ordinary gas, because I could, in principle, do the same thing with an ordinary gas: I could prepare a bundle of gas molecules all carefully set up to have exactly the same kinetic energy, and inject them into a tank that already contained some ordinary air, and this would mean that the kinetic energy distribution inside the tank would not be an exact Maxwell distribution corresponding to the temperature in the tank: it would have a "spike" at whatever kinetic energy I prepared the molecules I injected to have. Of course this "spike" would go away faster (I think) than the "spike" due to green laser light in the above example would, because the injected air molecules would interact with both the existing air molecules and the walls much more strongly than radiation interacts with anything.

I would think a hot blob of plasma is going to radiate and cool down on its way toward Earth.

It will, but in practice, as I understand it, a typical coronal emission doesn't lose very much heat between the Sun and the Earth, because radiation into vacuum is a very poor way of losing heat.

Also, a blob of radiation will cool down somewhat as it travels because it will unavoidably spread--it can't be perfectly collimated.

There are, as far as I can tell, four ways to add thermal energy to air in a tank: Conduction, convection, radiation and mechanical work

You can't add thermal energy to a sealed tank of air by convection: how is air going to flow in? But the other three seem fine to me.

From our discussion it would seem there are two ways to add thermal energy to a tank full of nothing but radiated photons: additional radiation and mechanical work.

I agree that conduction will, at least, be much, much less effective with a tank full of radiation, because, as I said above, ordinary gas molecules interact much more strongly than radiation does. So I would summarize the difference you describe as "radiation interacts much, much more weakly with anything than ordinary gas molecules do".