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I can't understand how a large cavity maintained at a constant inner surface temperature T, with a small opening on its surface behaves like a black body.

How is it a perfect emitter and a perfect absorber?

Please explain in as simple terms as possible.

Urb
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3 Answers3

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The hole is the black body.

The term black body is to do with an object that doesn't reflect anything, and so appears black.

Any light falling on the hole doesn't get reflected, but bounces around inside the box until absorbed. The chance of it re-emerging from the small hole is zero, so it's a perfect absorber.

As there is no reflected light, the only light leaving the hole is due to thermal radiation from the walls of the box, depending only on the temperature.

John Hunter
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We'll take the case of a large cavity and a small hole.

If a photon leaves the small hole, its departure has essentially no effect on the rest of what's going on inside the cavity. It also is unaffected by the walls of the hole. This means that the escaping photon comes flying out the hole towards us just as if it had originated a light-year away from us and was simply zooming through space. In this sense, the hole in the cavity behaves as a photon source with no dynamics of its own: a perfect emitter.

Conversely, if a photon from outside the cavity is aimed just so that it zips right through the hole and into the cavity, its entrance has essentially no effect on what's going on outside the cavity and it quickly gets lost inside- and it doesn't come back out. Furthermore, the details of the hole itself have no influence on the transit of the photon through it, so the photon is not "colored" in any way by its passage, and leaves no trace behind once it goes through the hole. It is as if that photon were zooming away from us and into deep space, never to return. This means that the hole behaves as a perfect absorber of photons.

niels nielsen
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Not a very theoretical explanation, but more of a visualization: what would the tiny hole in an actual big box on your desk look like? Black!

What you can see only in the visible range, is also true for all wavelengths. A photon that enters the hole only has a tiny probability of being reflected or reemitted right back out of the hole, which would make it sensitive to the specific nature of the box inner walls (they could be painted green for example). The vast majority of incidences will end up in the photon being reflected/absorbed/reemitted over and over again inside the box, until it is finally in equilibrium with the temperature of the walls of the box.

If the whole is a perfect absorber, it is invariably also a perfect emitter due to energy conservation.

oliver
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