I don't understand why a cavity with a hole is a black body

Question

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, $J/m^3$)?

Answer 1

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:

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.

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.

Answer 2

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.

Answer 3

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.

Answer 4

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).