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The standard (the only way I know) to derive the density of electromagnetic mode (per volume and per unit frequency) for a black body consists in modelling it by a cavity with perfectly reflecting walls.

It implies stationnary electromagnetic waves inside. Then, we say that this electromagnetic field is at equilibrium.

After some calculations, we can show that this energy density is:

$$u(\nu)=\frac{8 \pi h \nu^3}{c^3}\frac{1}{e^{\beta h \nu}-1}$$

Using it we can derive the Stefan Boltzmann law that tells us the power radiated by a black body at temperature $T$.

My main question is:

We modelled the black body by saying it is a cavity with perfectly reflecting walls. We computed some properties of this electromagnetic field at equilibrium. This field is by definition inside the cavity. How can those calculations relate of what is going outside the black body (the power radiated for example).

My second question is:

If I take the black body definition from wikipedia page:

A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence (It does not only absorb radiation, but can also emit radiation)

What is the link between this and the cavity model ? I could imagine that the outside of the cavity is made of perfect mirrors. Then it would mean I didn't model a blackbody.

Why is the cavity model ok for black body ?

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    $\begingroup$ try this link hyperphysics.phy-astr.gsu.edu/hbase/bbcon.html $\endgroup$
    – anna v
    Commented Mar 26, 2020 at 17:33
  • $\begingroup$ @annav thank you for the link. There are interesting things inside that made me refine more my question (updated now) $\endgroup$
    – StarBucK
    Commented Mar 26, 2020 at 18:11

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We modelled the black body by saying it is a cavity with perfectly reflecting walls. We computed some properties of this electromagnetic field at equilibrium.

This is an extremely common misconception. The model of the black body isn't the cavity, it's the hole.

If you have a cavity with a small hole in it, then the hole acts like a black body because any incident radiation is extremely unlikely to reflect off of an inner surface and come right back out - the hole therefore acts like a nearly perfect absorber.

Once the radiation enters the cavity, it somehow needs to achieve thermodynamic equilibrium. The radiation being absorbed and immediately re-emitted by the cavity walls is meant to provide a mechanism for this, e.g. by modeling the walls as a system of oscillators which can absorb and emit electromagnetic energy. Ultimately though, it doesn't matter how exactly the radiation in the cavity achieves thermodynamic equilibrium - only that it does. Once that happens, one can calculate how energy is distributed between the various frequency ranges of the radiation inside the box, which results in the Planck distribution.

From there, we simply note that the radiation coming out of the hole is all thermalized (since the probability of light entering the hole and immediately exiting again before being thermalized is vanishingly small). Therefore, the hole itself behaves as a black body would.

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  • $\begingroup$ It is more clear, thanks. Allright so to summarize: a black body is a body that absorbs everything and reflects nothing. We can model it by a tiny hole with a huge cavity behind. The tiny hole combined to the cavity is here to model the fact a light beam entering will necessary go out once it is thermalized (because very little probability that it is directly reflected from pure geometry). However, I have few last questions. So, first, can I say that a good definition for a black body is a material that only emits light which is at thermal equilibrium ? $\endgroup$
    – StarBucK
    Commented Mar 26, 2020 at 21:22
  • $\begingroup$ Indeed, in the end the fact it "absorbs" and doesnt "reflect" is not enough for me. The only point that matters is that it all the light it emits is at thermal equilibrium. Indeed for example with the tiny hole + cavity we could say that the outgoing light is somehow the reflection of an incident one (after many travel in the cavity to thermalize). And what matters in the spectral density is only the fact the light emitted is at thermal equilibrium. Would you agree ? $\endgroup$
    – StarBucK
    Commented Mar 26, 2020 at 21:24
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    $\begingroup$ @StarBucK Yes, I agree with you. If we define a black body as an object which absorbs all incident light, then it is only black bodies in thermal equilibrium which emit the familiar radiation spectrum from Plancks law. The assumption that such objects will always eventually thermalize is a robust one, but it is an assumption nonetheless. One would need to apply non-equilibrium techniques to a specific model of a black body to get a realistic estimate of how long it would take to thermalize. $\endgroup$
    – J. Murray
    Commented Mar 26, 2020 at 21:52
  • $\begingroup$ @StarBucK You maybe interested in this recent paper, which illustrates the point that not all object which emit radiation need to be in thermal equilibrium. $\endgroup$
    – J. Murray
    Commented Mar 26, 2020 at 22:32
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    $\begingroup$ @StarBucK Okay, I see what you mean. We talk about the energy density inside the cavity; the energy flux is what comes out through the hole. You say that observable quantities, such as energy density, are only calculated inside the cavity - I would argue that we only actually measure the outgoing energy flux (per unit frequency), and that everything else in the model, including the energy density inside a box of thermalized electromagnetic radiation, is a vehicle for calculating it. $\endgroup$
    – J. Murray
    Commented Mar 31, 2020 at 6:06
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Reflecting radiation is not the same as absorbing and then emitting. Because in the second case, two processes are independent. The intensity of emitted light doesn't depend on the intensity of absorbed, only on internal properties of the body.

In other words, if you switch off the incident light, the radiation from the mirror will be immediately lower. Whereas the radiation from the black body will stay the same (until the body noticeably cools).

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  • $\begingroup$ But then I guess the difference is a matter of timescale ? I mean if we model the reflected wave on a mirror it will be produced because current or charge density occur on the material. Maybe it is super fast, but it will still take time to occur. So for me there is no fundamental difference with the blackbody instead of a timescale. I guess the physical process are different between the mirror and the black body and maybe it is better way to define it ? $\endgroup$
    – StarBucK
    Commented Mar 26, 2020 at 17:47
  • $\begingroup$ No ok I think I understood what you mean. With a blackbody even I have never emitted light on it it will radiate. For a mirror I will never see any reflection if I never shined on it. I updated my question however to be closer to my exact problem $\endgroup$
    – StarBucK
    Commented Mar 26, 2020 at 18:10
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I concur with J. Murray's answer. The small hole in a large cavity is meant to sample the photons in equilibrium, and the radiation coming out of the hole is the blackbody radiation. The cavity being very large means that the shape of the cavity (or even the exact boundary conditions such as perfectly reflecting or periodic BC) does not matter in obtaining Planck's law.

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My two cents:

Black body radiation, is one of the pillars of the need for quantum mechanics, it showed that classical electrodynamics would lead to distributions radiated from a cavity that were inconsistent with the data. Thus, with quanta introduced, the data could be explained.

This link describes the long process of thinking about heat as radiation

So, we can do this in reverse: have an oven with a tiny hole in the side, and presumably the radiation coming out the hole is as good a representation of a perfect emitter as we’re going to find. Kirchhoff challenged theorists and experimentalists to figure out and measure (respectively) the energy/frequency curve for this “cavity radiation”, as he called it (in German, of course: hohlraumstrahlung, where hohlraum means hollow room or cavity, strahlung is radiation). In fact, it was Kirchhoff’s challenge in 1859 that led directly to quantum theory forty years later!

....

By the 1890’s, experimental techniques had improved sufficiently that it was possible to make fairly precise measurements of the energy distribution in this cavity radiation, or as we shall call it black body radiation. In 1895, at the University of Berlin, Wien and Lummer punched a small hole in the side of an otherwise completely closed oven, and began to measure the radiation coming out.

and they found this:

blackbody

this oven glows deep red.

One minor point: this plot is the energy density inside the oven

goes on to define energy getting out of the hole., see the link for derivations. It is worth reading because it talks of all the other laws coming out of this measurement.

The link also discusses your second question, before delving into the black body formula, emission and absorption..

In fact, we can be much more precise: a body emits radiation at a given temperature and frequency exactly as well as it absorbs the same radiation. This was proved by Kirchhoff: the essential point is that if we suppose a particular body can absorb better than it emits, then in a room full of objects all at the same temperature, it will absorb radiation from the other bodies better than it radiates energy back to them. This means it will get hotter, and the rest of the room will grow colder, contradicting the second law of thermodynamics. (We could use such a body to construct a heat engine extracting work as the room grows colder and colder!)

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