Is Temperature Truly Absolute Zero in a Vacuum with No Particles?

[DLeuPel]

The understanding that I have of temperature is that it is defined as the vibration of particles. Now, does this mean that in a vacuum where there are no particles the temperature is the absolute 0 ?

 

[anorlunda]

You refer to the thermodynamic definition of temperature Thermodynamics deals with the study of large numbers of particles. But there are other definitions.

The Cosmic Microwave Background (CMB) also has a temperature (roughly 3 degrees K, -270 C, -452 F). It permeates the universe, even when there are no atoms around. Heat energy flows from warmer to colder. So if an object at absolute zero was set adrift in space, it would absorb energy from the CMB and warm up.

 

[Vanadium 50]

Now, does this mean that in a vacuum where there are no particles

How would you measure such a thing?

 

[sophiecentaur]

. It permeates the universe, even when there are no atoms around.

Is that true? Wasn't the CMB originated in the BB itself, where all the matter in the Universe was present when the radiation was formed? I think the 3K figure actually describes the temperature of a body that would be radiating with a spectrum of the CMB.
I think the OP is suffering from (as we all were) having been told half a story and half a definition, in his/her youth. Rather than 'vibrations of particles' there is a much better definition of temperature in most cases which is 'The average Kinetic Energy of particles'. It actually has Units.
There are a number of inconsistencies or confusions of how Temperature is manifest. The 'Temperature of the Sun's Corona is measured as several million K, which really doesn't fit in with the simple model of an 'atmosphere' around a body with surface temperature of 6000K.

 

[DLeuPel]

Is that true? Wasn't the CMB originated in the BB itself, where all the matter in the Universe was present when the radiation was formed? I think the 3K figure actually describes the temperature of a body that would be radiating with a spectrum of the CMB.
I think the OP is suffering from (as we all were) having been told half a story and half a definition, in his/her youth. Rather than 'vibrations of particles' there is a much better definition of temperature in most cases which is 'The average Kinetic Energy of particles'. It actually has Units.
There are a number of inconsistencies or confusions of how Temperature is manifest. The 'Temperature of the Sun's Corona is measured as several million K, which really doesn't fit in with the simple model of an 'atmosphere' around a body with surface temperature of 6000K.

So, how does the heat from the Sun propagate trough vacuum if there are no particles in the vacuum?

 

[jbriggs444]

So, how does the heat from the Sun propagate trough vacuum if there are no particles in the vacuum?

As light. That light has a distribution of frequencies [at least once you settle on a frame of reference to measure it from] and that distribution has a temperature which follows from the average kinetic energy per photon, $E=h \nu$

 

[anorlunda]

Is that true? Wasn't the CMB originated in the BB itself, where all the matter in the Universe was present when the radiation was formed? I think the 3K figure actually describes the temperature of a body that would be radiating with a spectrum of the CMB.

Yes it's true. So are all the other things in your paragraph true. There is no inconsistency among them.

But if you think the BB was at a point, that's wrong. We have Insughts articles on that.

Edit: BTW the CMB is just light like solar radiation is light.

 

[Paul Colby]

How would you measure such a thing?

I get stuck on "thing". In the absence of all particles and radiation (an impossibility AFAICT) is one left with more than one possible state? I think the usual assumption is vacuum is a unique state. This would seem to imply it has no temperature.

 

[anorlunda]

I get stuck on "thing". In the absence of all particles and radiation (an impossibility AFAICT) is one left with more than one possible state? I think the usual assumption is vacuum is a unique state. This would seem to imply it has no temperature.

But there is no such place. If there was, I guess temperature would be undefined.

I don't think vacuum as "a unique state" works. The word state must describe something, not nothing.

But please, lest us not descend into philosophy.

 

[sophiecentaur]

But if you think the BB was at a point, that's wrong.

Sorry - sloppy language there. But the 'stuff' was all around the same place at one time and there were definitely 'material sources' of the radiation that we see.

This would seem to imply it has no temperature.

No so much "no" but "indeterminate"?
I cross posted with @anorlunda there, I see.
Means means Σf_n/n and where n→0 ??

 

[stevendaryl]

Getting back to the original question: In thermodynamics, for every extensive quantity (something that is additive) there is a corresponding intensive quantity that can roughly be thought of as characterizing how willing a system is to give up or acquire more of the extensive quantity. Some examples:
Corresponding to the extensive quantity, "volume" there is a corresponding intensive quantity, "pressure". A gas at high pressure will tend to expand (increase its volume), and a gas at lower pressure will tend to contract. Internal energy and temperature are another example: A system at high temperature tends to lose energy to systems at a lower temperature.

Roughly speaking, systems with higher energy tend to have higher temperature, as well, but that's not always true.

Getting back to empty space. If the space is truly empty, devoid of matter or radiation, then it's temperature will be zero. It has no possibility of giving any energy to any other system. When they talk about the background temperature of space being 4 degrees Kelvin, they're talking about space that isn't completely empty. It has low-intensity radiation, so it has nonzero amount of energy per unit volume.

 

[DrStupid]

How would you measure such a thing?

Bring it in thermal equilibrium (not just steady state!) with a temperature reference.

 

[DLeuPel]

Getting back to the original question: In thermodynamics, for every extensive quantity (something that is additive) there is a corresponding intensive quantity that can roughly be thought of as characterizing how willing a system is to give up or acquire more of the extensive quantity. Some examples:
Corresponding to the extensive quantity, "volume" there is a corresponding intensive quantity, "pressure". A gas at high pressure will tend to expand (increase its volume), and a gas at lower pressure will tend to contract. Internal energy and temperature are another example: A system at high temperature tends to lose energy to systems at a lower temperature.

Roughly speaking, systems with higher energy tend to have higher temperature, as well, but that's not always true.

Getting back to empty space. If the space is truly empty, devoid of matter or radiation, then it's temperature will be zero. It has no possibility of giving any energy to any other system. When they talk about the background temperature of space being 4 degrees Kelvin, they're talking about space that isn't completely empty. It has low-intensity radiation, so it has nonzero amount of energy per unit volume.

And what is this low intensity radiation ?

 

[Vanadium 50]

Bring it in thermal equilibrium (not just steady state!) with a temperature reference.

And then it's not vacuum any more. (Or was this a joke...with your user name, I'm never quite sure)

 

[anorlunda]

And what is this low intensity radiation ?

The CMB. See post #2

 

[Paul Colby]

I don't think vacuum as "a unique state" works. The word state must describe something, not nothing.

In the current theory (The Standard Model), isn't vacuum that state in which every quantum field is in its vacuum state? Moving out of this state implies some quanta of these fields i.e. mater or radiation?

Of course gravitational waves have been observed, so one would need a region of space which is identically flat. That ain't happening just given all the hot matter banging around in stars and such. There should be a cosmic GW background but good luck detecting it.

 

[PeterDonis]

there are other definitions.

Can you give a definition that makes the temperature nonzero when there is no matter, radiation, or anything else present? See further comments below.

The Cosmic Microwave Background (CMB) also has a temperature (roughly 3 degrees K, -270 C, -452 F). It permeates the universe, even when there are no atoms around.

There aren't any atoms around, but that doesn't mean it's a vacuum by the definition the OP is (implicitly) using. The very presence of the CMB everywhere in the universe means the universe is not vacuum by that definition, because there is radiation present.

 

[PeterDonis]

Bring it in thermal equilibrium (not just steady state!) with a temperature reference.

A vacuum can't be brought into thermal equilibrium with anything, because it contains no matter, radiation, or anything else. So this won't work.

In fact, strictly speaking you can't make any measurements at all on a vacuum, because that would require interacting with it, and interacting with it makes it no longer a vacuum.

 

[PeterDonis]

I don't think vacuum as "a unique state" works.

It does if we restrict to inertial observers in flat spacetime. Any QFT in flat spacetime has a unique state that is a ground state--state of lowest energy--with respect to all inertial observers. This state is the vacuum state.

If we allow non-inertial observers, then the Unruh effect shows that these observers--at least, strictly speaking, if they have a constant proper acceleration--will see a different state as the vacuum state. So the notion of "vacuum state" is not unique in this sense.

Also, AFAIK there is not a unique vacuum state in a curved spacetime, even if we restrict to inertial observers.

 

[DrStupid]

And then it's not vacuum any more.

Is water not water anymore if you bring it in thermal equilibrium with a thermometer? If not, why should vacuum be turned into something else by the same procedure?

 

[DrStupid]

A vacuum can't be brought into thermal equilibrium with anything, because it contains no matter, radiation, or anything else.

Vaccum contains no radiation? Please provide a corresponding reference.

 

[Nugatory]

Vaccum contains no radiation? Please provide a corresponding reference.

It depends on how precise we're being when we use the word "vacuum". In casual use, "vacuum" is understand to mean "contains no matter" or "empty space" and interstellar space is an example of a vacuum even though electromagnetic radiation is present. More rigorously, vacuum is the ground state of all fields and there is neither matter not electromagnetic radiation present; this follows from the way that quantum field theories treat everything as a field, with no distinction between those fields that manifest themselves as matter

The casual definition is remarkably unhelpful when trying to understand what temperature is when working with systems that exchange energy by radiation, and this is the source of much of the confusion in this thread. The notion that temperature is the vibration of particles is even less helpful; it may be better to think of the vibration of particles as one of the many ways that a system can store energy, while temperature is a measure of how willing the system is to give up that energy - and @stevendaryl's post #11 above is worth another read.

 

[Klystron]

My working classification of vacuum "vacuum is a condition, not a thing" seems to contradict this dictionary definition

noun: condition; plural noun: conditions
1.the state of something, especially with regard to its appearance, quality, or working order.

 

[PeterDonis]

why should vacuum be turned into something else by the same procedure?

Because vacuum, unlike water or any other substance, cannot exchange heat with anything while still remaining vacuum.

 

[PeterDonis]

this dictionary definition

We're not talking about ordinary language here, we're talking about precise scientific terminology. The precise scientific definition of "vacuum" has already been given in this thread.

 

[PeterDonis]

Vaccum contains no radiation? Please provide a corresponding reference.

Check any QFT textbook for the precise definition of "vacuum".

 

[DrStupid]

In casual use, "vacuum" is understand to mean "contains no matter" or "empty space" and interstellar space is an example of a vacuum even though electromagnetic radiation is present. More rigorously, vacuum is the ground state of all fields and there is neither matter not electromagnetic radiation present; this follows from the way that quantum field theories treat everything as a field, with no distinction between those fields that manifest themselves as matter

Many thanks for this clarification. The "more rigorously" use of the term vacuum seem to be what Wikipedia refers to as QED vacuum whereas the OP refers to the subject of the Wikipedia main entry which refers to vacuum as "space devoid of matter".

The casual definition is remarkably unhelpful when trying to understand what temperature is when working with systems that exchange energy by radiation, and this is the source of much of the confusion in this thread. The notion that temperature is the vibration of particles is even less helpful;

As far as I can see the "notion that temperature is the vibration of particles" is the only source of confusion in this thread. Can you explain what the causual definition of vacuum contributes to the confusion?

 

[Mark Harder]

The precise thermodynamic definition of temperature is T = (∂U/∂S)_v,n,..., where U and S are internal energy and entropy, respectively. Subscripted variables are all independent variables upon which the internal energy depends, volume, no. of particles, etc. The equations of thermodynamics are differential equations, so it best handles changes in its critical variable values. In that spirit, suppose we create a volume of space devoid of matter/energy such as the CMB, perhaps separated from its environment by an adiabatic barrier that keeps out radiation. This volume contains no energy and no entropy (remember, it's only a hypothetical case, so S=0.). Now, remove all barriers between this 'perfect' vacuum and the rest of the universe, thus adding CMB, dark matter, vacuum energy, whatever to our volume. Clearly, internal energy increases. What about entropy? Does it increase? I would argue that it does, because the distribution of the CMB and any other mass/energy that may be present is not uniformly distributed. I base this on observations of the CMB across large volumes of space, which is not perfectly uniform. Also it would be a very peculiar situation if the volume should have photons of CMB moving with identical directions and exact wavelength; and any other candidate for constituent of this vacuum should be particulate as well. Since the energy is not uniformly distributed, the volume can exist in many states consistent with the energy in the volume we have carved out of space. Pick the usual symbol, W, for the number of ways this system's energy can be distributed.Since S=K Ln(W), the volume now contains entropy. We could also base the existence of nonzero entropy on the third law which specifies this fact, but I like the Boltzmann definition better for this purpose. What we have just described is the transition from a true vacuum containing no energy and no entropy to one that contains both. Therefore, T = (∂U/∂S)_v,n,... > 0 and the region of space has nonzero positive temperature.

 

[sophiecentaur]

Is water not water anymore if you bring it in thermal equilibrium with a thermometer? If not, why should vacuum be turned into something else by the same procedure?

I don't think that is any sort of valid argument. You could only draw that parallel if water and vacuum were the same sort of entity. There is a mathematical analogy here. The value Zero follows different rules from all other values. 1, 2 and even π behave the same but Zero is different.

 

[DrStupid]

I don't think that is any sort of valid argument.

It is valid for a classical vacuum. I just missed the point where the discussion drifted into quantum field theroy.

 

[PeterDonis]

It is valid for a classical vacuum.

No, it isn't. 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.

 

[DrStupid]

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.

There can be radiation in a classical vaccuum. That's sufficient.

 

[PeterDonis]

There can be radiation in a classical vaccuum.

Please give a specific reference. My understanding of the term "vacuum" in classical physics is that it means nothing is present--no matter, no radiation, nothing.

 

[Paul Colby]

I certainly agree with what @PeterDonis said in #33. I would add the classical theory of EM in thermal equilibrium fails, badly. This failure was a key starting point of quantum theory (Planck and black body radiation).

 

[DrStupid]

Please give a specific reference.

https://en.wikipedia.org/wiki/Vacuum
https://en.wikipedia.org/wiki/Vacuum#Electromagnetism

All conditions given for the classical vacuum are still valid in presense of radiation.

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

 

[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".