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Name: Farooq
Status: other
Grade: other
Country: Canada
Date: Fall 2012


Question:
Blackbodies are considered to perfect absorbers and emitters of electromagnetic radiation. If the Sun is a blackbody radiator, it must be a perfect light absorber as well. So any photon "entering" the Sun will never get out. But why doesn't the Sun completely self-absorb its own light? This came up as a remark from the speaker in a science talk in a observatory. His comment was that the "darkest" object in the solar system is the Sun. How come it still appears luminous to us?

Replies:
As you state, black body radiators are (by definition) perfect absorbers AND emitters of radiation. Therefore, they are not dark.

When a black body absorbs a photon the photon's energy is incorporated into the black body, raising its temperature (even if only slightly). If the black body started out at thermal equilibrium with its environment that excess temperature will eventually be lost by radiation to the environment. This radiation will be at a different wavelength and in accord with the laws of black body radiation. With a sensitive enough detector the additional radiation could be detected.

The sun is a black body is far from equilibrium with its surroundings so it is constantly losing energy. Most of the energy is lost in the form of radiation which makes it appear brilliant in the cold dark background of space.

Greg Bradburn


The sun is not the darkest object in the solar system. However, it is the object that most nearly behaves as a "black body", in the narrow physics definition of that term, because the spectrum of light it emits is well described by the black body equation. The terms "black body", "perfect absorber", and "perfect emitter" don't at all mean what a literal interpretation would suggest. Instead, they are a sort of shorthand intended to stand for paragraphs of description of an idealized way light can interact with matter.

"Black body" means that the spectrum of light emitted by the object is completely determined by its temperature, according to a simple equation from quantum mechanics. That is all it means. It does not mean that the object is black. A very cold black body is black, because it has little energy to emit; a warmer black body emits radio waves, or microwaves, or infrared radiation; an even warmer black body emits visible light. "perfect emitter" means the same thing as "black body".

"Perfect absorber", in this context, means that any light landing on a black body is completely distributed among all of the particles that make up the body. The incident light loses all of its characteristics (frequency, incident direction, exact time of incidence) and simply raises the temperature of the black body. That's all it means. It doesn't mean that the incident energy never comes back out; instead it describes how the energy is distributed before it comes back out.

Tim Mooney


Planck’s analysis of blackbodies assumes an enclosed volume. The Sun does not fulfill that requirement. It is in an open system, so electromagnetic radiation escapes from the Sun. If it did not we could not see it.

Vince Calder


What you say is true, to an extent.

In a black body, the wavelengths of energy emitted depend upon temperature. A black body is a perfect absorber of all energies only at absolute zero. Otherwise, there is some amount of energy being emitted by the black body.

In the case of the sun it has an internal energy source. Its temperature is so high the emission overpowers any absorption. Lastly, since the light emitted is traveling away from the sun, it cannot be recaptured.

Hope this helps Bob Avakian Tulsa


Farooq,

“Dark” in this sense means that it interacts with all kinds of light, all electromagnetic waves. If the material is hot, it emits its extra energy easily. If the material is cold, it absorbs the missing energy easily. Light has a speed and direction. When the sun emits light, the light is moving away from the sun. To be absorbed, light must be moving toward the sun. Absorbed light increases the energy, and perhaps the temperature, of the sun. Increased temperature causes the sun to emit more radiation. If a great deal of energy were absorbed by the sun, the sun would get hotter and then brighter. As the extra radiation is emitted as brighter light, the sun cools off to its original temperature and emission rate.

Dr. Ken Mellendorf Physics Instructor Illinois Central College


The confusion you have may be about the difference between a true black body and an object that exhibits behaviors of a black body. For example, an object can exhibit black body radiation without being a true black body in all aspects. As a matter of shorthand you might refer to something as a blackbody for the way it emits radiation even though you did not mean that it exhibits all behaviors of a black body.

Hope this helps, Burr Zimmerman


Hi Farooq,

This does appear as a sort of conundrum. Fortunately it is mostly semantic in nature. Equally fortunate is that our Sun is not a perfect black body.

Black-body electromagnetic radiation is radiation of a body in thermodynamic equilibrium with its local environment. The radiation has a specific spectrum and intensity that depends only on the temperature of the body. A perfectly insulated enclosure in thermal equilibrium contains black body radiation and will emit it through a tiny hole made in its wall if the thermal equilibrium is maintained.

The Sun has layers and flows, the spectrum is not smooth and it is not in thermal equilibrium. The Sun is a close approximation of a black body, but it is not a perfect black body. The Sun has a core, convection and radiative zones, surface flows and most pertinent to this question, a photosphere. The photosphere is where the photons bounce around, but some escape. Those photons are emitted into space, we see those. The energy lost as photons is replaced by core energy, so that the temperature of the photosphere is maintained. The photosphere is reasonably analogous to the enclosure with a small hole in it.

Thank you for your question and keep on shining! Peter E. Hughes, Ph.D. Milford, NH


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