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It makes no sense to me that a cavity with a hole is identical to a black body.

Sure, the cavity will probably be a near perfect absorber, but I don’t see how it will be a perfect emitter as well.

Why exactly does a cavity with a hole behave like a black body? Why can I make such a contraption and simulate and skip considerations of a special chemical makeup that has to occur in a black body?

Another thing that bothers me is that these cavities are not “absorbing” light, they are simply hogging it and letting it bounce inside.

Also, on the typical graphs where intensity is graphed against frequency (or wavelength): this isn’t really intensity, right? I have to integrate this curve in order to get useful information, right?

Some call it the spectral energy density. Is the integral under the curve then the energy density inside the cavity (as in, $J/m^3$)?

TRiG
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DLV
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  • If it is a perfect absorber how else could it be in equilibrium with the surroundings from which it absorbs unless it is also a perfect emitter? – hyportnex Mar 15 '15 at 21:52
  • Well these experiments weren't done in some vacuum, so it could exchange heat energy with the surroundings, right? – DLV Mar 15 '15 at 21:54
  • I added another thing that I think states my question a little bit better : ""Another thing that bothers me is that these cavities are not "absorbing" light, they are just simply hogging it and letting it bounce inside.. "" – DLV Mar 15 '15 at 21:56
  • You can't just plot intensity versus frequency because we usually treat frequency as a continuous variable. It's similar to the way in which if we have a block, we can't talk about the mass at point x,y,z. We have to talk about the density at point x,y,z. A block can't have finite mass at a point because there are infinite points. It's the same for intensity. What's shown is the intensity per unit frequency. It only makes sense to talk about intensity once you've integrated over a frequency interval. – MonkeysUncle Mar 15 '15 at 21:59
  • "Why exactly does a cavity with a hole behave like a black body?" look up the wikipedia link i gave you, it's as good an answer to this question as you will find anywhere, gotta go now, best of luck with it –  Mar 15 '15 at 22:41
  • @MonkeysUncle What about when they say that plancks law is about energy density or "spectral energy density"? Are they still referring to J/m^2/Hz or is it now J/m^3/Hz ? Thanks. – DLV Mar 15 '15 at 22:47
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    I just spent an hour writing a long answer to this, but the question was closed when I tried to post it. If you PM me I can send what I wrote. Shame it is a good set of questions but usually on this site you have to ask only one question. – kotozna Mar 16 '15 at 13:46
  • Yeah its pretty dumb that this should happen. Once I figure out how to PM I will thanks a lot. – DLV Mar 16 '15 at 16:28
  • @kotozna Apparently I cant send you a PM if you havent logged on to chat.stackexchange.com ever before. Tell me if you do so, thanks. – DLV Mar 16 '15 at 16:38
  • @David I have now logged in to chat! – kotozna Mar 16 '15 at 19:56

4 Answers4

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I will address this part of the question:

Why exactly does a cavity with a hole behave like a black body? Why can I make such a contraption and simulate and skip considerations of a special chemical makeup that has to occur in a blackbody?

One starts with a black body, absorbing all radiation and emitting it in thermodynamic equilibrium with the environment.

One should go to the original thought processes about black body radiation and the experimental discrepancy between the calculations with the classical electromagnetic theory that was the only one known at the time:

bbradiation

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.

This prediction is the the divergent line in the plot. How was it arrived at

"Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of radiation which is incident upon it. The radiated energy can be considered to be produced by standing wave or resonant modes of the cavity which is radiating.

The cavity was a reasonable way to model mathematically the electromagnetic field for the radiation observed/measured from all bodies, it simplifies the mathematics, and the small hole allows for the radiation to come out and be measured. It should have the same spectrum as the one radiated from any part of the body. For thermodynamic considerations it makes no difference if the body is hollow or not. Equilibrium is equilibrium and should exist outside and inside. A small hole is the probe of what is happening inside.

In the classical model

A mode for an electromagnetic wave in a cavity must satisfy the condition of zero electric field at the wall. If the mode is of shorter wavelength, there are more ways you can fit it into the cavity to meet that condition. Careful analysis by Rayleigh and Jeans showed that the number of modes was proportional to the frequency squared.

classical bb

From the assumption that the electromagnetic modes in a cavity were quantized in energy with the quantum energy equal to Planck's constant times the frequency, Planck derived a radiation formula.

Planck's radiation formula fitted the experimental curves and showed that the interactions of the waves with the walls of the cavity (the way thermodynamic equilibrium is reached) were quantized and the probability of interaction had to fall for higher frequencies and thus the ultraviolet catastrophy was avoided.

anna v
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  • For thermodynamic considerations it makes no difference if the body is hollow or not......I never thought about that aspect of it before, it was a big part of the op's question about "radiation bouncing around inside"....very nice summary regards –  Mar 16 '15 at 09:00
  • How do we know that the cavity is indeed a reasonable way to model the radiation ? Indeed we need to make use of the fact modes are stationnary inside the cavity to derive everything (the density of modes is related to it). Thus it is a critical part in the proof. How can we know that the end result doesn't actually depend on the fact we needed a cavity ? This is something that disturbs me a lot. – StarBucK Apr 04 '20 at 13:26
  • @StarBucK It fits the data, that is the ultimate "Know" in physics. Since the guess fits the data, it is a reasonable model. Note, the alternative classical model fails to do so. – anna v Apr 04 '20 at 14:23
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To answer a question that the OP put to Irish Physics's answer:

What about the fact that radiation isn't really absorbed in the box, and instead is really just bouncing all over the interior? Why doesn't this matter?

Actually you need the radiation to be absorbed by the box and re-emitted, or at least it must interact with the box material in some way. This is the only way whereby the radiation can "thermalize" as there are no photon-photon interactions (at least not under normal conditions) and the point of a blackbody is that the photon states are distributed by the Boltzmann distribution.

Community
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Selene Routley
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  • Oh ok, Thanks. By "thermalize" do you mean reach a thermal equilibrium? – DLV Mar 15 '15 at 22:45
  • From Wikipedia "The thermal radiation spontaneously emitted by many ordinary objects (me: any box you want to make as long as you keep temperature steady) can be approximated as blackbody radiation. A perfectly insulated enclosure that is in thermal equilibrium internally contains black-body radiation and will emit it through a hole made in its wall, provided the hole is small enough to have negligible effect upon the equilibrium. (me again: so thats why a cavity with a hole behaves like a black body). –  Mar 15 '15 at 22:51
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    @David Yes, sorry for the jargon: that's precisely what it means. – Selene Routley Mar 15 '15 at 22:59
  • @ WetSavannaAnimal aka Rod Vance ...and then they had the perfect light bulb , and something else, some little notion, what was it again, a minor trifle, nothing of consequence , oh yeah quantum theory. thanks for cleaning up my answer regards –  Mar 16 '15 at 08:47
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The whole point of a black body is to make a "box" that shows, when you keep its temperature as constant as possible at some value, say 300 K, how much the intensity of the radiation emitted through a tiny little hole in the box changes with respect to frequency.

So the chemical composition of the box does not really matter as long as it does nothing to stop the temperature staying steady at 300 K.

Same with the shape of the box, as long as the temperature stays steady at 300 K, it can be any shape you want.

here is a link to a good site that should help:

http://en.wikipedia.org/wiki/Black-body_radiation

At any fixed temperature there will no difference between the radiation absorbed by the box walls and the radiation emitted by the walls of the box.

It will be at thermodynamic equilibrium, balanced so that the radiation of energy and the absorption of energy (by the box walls from the heat you are putting in) are exactly the same.

Monkeys Uncle has explained the rest of your question.

  • What about the fact that radiation isn't really absorbed in the box, and instead is really just bouncing all over the interior? Why doesn't this matter? – DLV Mar 15 '15 at 22:10
  • It IS absorbed by the walls of the box, put your hand on the box and you will know its getting absorbed. ie. hotter –  Mar 15 '15 at 22:13
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    the radiation bouncing around inside doesn't matter because the walls are putting out exactly as much radiation as they are taking in. –  Mar 15 '15 at 22:15
  • @David - see my answer: you need the radiation to be absorbed. – Selene Routley Mar 15 '15 at 22:27
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As you know from thermodynamics, heat cannot flow one one body to a hotter body.

Imagine a body with a cavity that is not a perfect black body emitter. Let's put it face to face with a second cavity - but this one is a perfect black body. Initially they are the same temperature. Radiation will flow from the perfect black body to the imperfect one - but a lesser amount will be reflected. The imperfect black body will be getting hotter and we violated thermodynamics. This is an argument that says that emission and absorption coefficients must be the same (at any wavelength: otherwise you could repeat the above thought experiment with appropriate filters in place).

The cavity looks "black" because it reflects not the light we shine at it (for incandescent light this might correspond to a spectral temperature of 3300 K or so) but rather the "thermalized" light - emission corresponding to its temperature (which is much lower).

As for the units plotted in a spectrum - yes the vertical axis has to show "density" of intensity as light from a single (infinitely narrow) wavelength can carry no energy. If you plot wavelength along the X axis the units are intensity/m; is you it frequency, units are intensity * time. Either way $\int{I\cdot dx}$ gives the total intensity in the range over which you integrate (where $x$ is either wavelength or frequency).

Floris
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  • The word density gives me the impression that its m^-3… is calling it density a misnomer in a way? Is this distribution what you would observe if you place a detector somewhere? No matter what the details are, the detector will always see a similar curve? Why does the infinite integral always converge to simgaT^4 though? Doesnt intensity measured depend on distance? – DLV Mar 16 '15 at 05:29
  • Serways book says it is m^-3 – DLV Mar 16 '15 at 05:37
  • When frequency is on the horizontal axis, btw. It even says that considering energy per unit volume is more useful than per unite area. – DLV Mar 16 '15 at 06:09
  • It is. Possible to consider the density of energy inside the cavity - then the inverse square law doesn't come into play but the speed of light does. It is more conventional though to calculate it per unit area of the emitting surface. The total energy a black body emits will be $A\sigma T^4$ - so as you look at a distant point you do need to scale by $1/4\piR^2$, the area of the sphere over which the energy is spread. – Floris Mar 16 '15 at 11:15
  • I am not familiar with the book you quote. – Floris Mar 16 '15 at 11:15