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BLACK BODY RADIATION

In physics, a black body is an object that absorbs all electromagnetic radiation that falls onto it. No radiation passes through it and none is reflected, yet it theoretically radiates every possible wavelength of energy. Despite the name, black bodies are not actually black as they radiate energy as well. The amount and type of electromagnetic radiation they emit is directly related to their temperature. Black bodies below around 700 K (426.85 ºC) produce very little radiation at visible wavelengths and appear black (hence the name). Black bodies above this temperature, however, begin to produce radiation at visible wavelengths starting at red, going through orange, yellow, and white before ending up at blue as the temperature increases.

The term "black body" was introduced by Gustav Kirchhoff in 1862. The light emitted by a black body is called black-body radiation[1].

At 1000 K the radiation appears bright red. At 3000 K the color is orange. By about 6000 K it appears nearly pure white. Above 6000 K the color becomes slowly more blue. At 10000 K the color is a sky-blue.

Contents

Explanation

In the laboratory, the closest thing to black-body radiation is the radiation from a small hole entrance to a larger cavity. Any light entering the hole would have to reflect off the walls of the cavity multiple times before it escaped and is almost certain to be absorbed by the walls in the process, regardless of what they are made of or the wavelength of the radiation (as long as it is small compared to the hole). The hole, then, is a close approximation of a theoretical black body and, if the cavity is heated, the spectrum of the hole's radiation (i.e., the amount of light emitted from the hole at each wavelength) will be continuous, and will not depend on the material in the cavity (compare with emission spectrum). By a theorem proved by Kirchhoff, this curve depends only on the temperature of the cavity walls.

Calculating this curve was a major challenge in theoretical physics during the late nineteenth century. The problem was finally solved in 1900 by Max Planck as Planck's law of black-body radiation. By making changes to Wien's Radiation Law (not to be confused with Wien's displacement law) consistent with Thermodynamics and Electromagnetism, he found a mathematical formula fitting the experimental data in a satisfactory way. To find a physical interpretation for this formula, Planck had then to assume that the energy of the oscillators in the cavity was quantized (i.e., integral multiples of some quantity). Einstein built on this idea and proposed the quantization of electromagnetic radiation itself in 1905 to explain the photoelectric effect. These theoretical advances eventually resulted in the replacement of classical electromagnetism by quantum mechanics. Today, these quanta are called photons. In addition, it led to the development of quantum versions of statistical mechanics, called Fermi-Dirac statistics and Bose-Einstein statistics, each applicable to a different class of particles. See also fermions and bosons.

The temperature of a Pahoehoe lava flow can be approximated by merely observing its colour. The result agrees nicely with the measured temperatures of lava flows at about 1,000 to 1,200 °C.
The temperature of a Pahoehoe lava flow can be approximated by merely observing its colour. The result agrees nicely with the measured temperatures of lava flows at about 1,000 to 1,200 °C.

The wavelength at which the radiation is strongest is given by Wien's displacement law, and the overall power emitted per unit area is given by the Stefan-Boltzmann law. So, as temperature increases, the glow color changes from red to yellow to white to blue. Even as the peak wavelength moves into the ultra-violet enough radiation continues to be emitted in the blue wavelengths that the body will continue to appear blue. It will never become invisible—indeed, the radiation of visible light increases monotonically with temperature.

The radiance or observed intensity is not a function of direction. Therefore a black body is a perfect Lambertian radiator.

Real objects never behave as full-ideal black bodies, and instead the emitted radiation at a given frequency is a fraction of what the ideal emission would be. The emissivity of a material specifies how well a real body radiates energy as compared with a black body. This emissivity depends on factors such as temperature, emission angle, and wavelength. However, it is typical in engineering to assume that a surface's spectral emissivity and absorptivity do not depend on wavelength, so that the emissivity is a constant. This is known as the grey body assumption.

Interestingly, this means that every object around you is emitting electromagnetic waves with wavelengths of all values. Every object in the universe has heat, even the emptiness of space, and when the particles that make up an object vibrate on a microscopic level they radiate electromagnetic waves. These wavelengths are predominantly infra-red (heat), but there is also a minute amount of visible light like red, yellow, green and blue. So, right now, you and everything around you is emitting visible light. The reason this light cannot be seen is that it has a very low intensity so it is overpowered by the light that is reflected by the object.

When dealing with non-black surfaces, the deviations from ideal black body behavior are determined by both the geometrical structure and the chemical composition, and follow Kirchhoff's Law: emissivity equals absorptivity, so that an object that does not absorb all incident light will also emit less radiation than an ideal black body.

In astronomy, objects such as stars are frequently regarded as black bodies, though this is often a poor approximation. An almost perfect black-body spectrum is exhibited by the cosmic microwave background radiation. Hawking radiation is black-body radiation emitted by black holes.

The spectrum of an incandescent bulb in a typical flashlight. Here, the filament temperature appears to be about 4600 kelvins due to a peak emittance of around 630 nanometers. However, the shape shows that the filament is not acting as a true black body.
The spectrum of an incandescent bulb in a typical flashlight. Here, the filament temperature appears to be about 4600 kelvins due to a peak emittance of around 630 nanometers. However, the shape shows that the filament is not acting as a true black body.

Equations governing black bodies

Planck's law of black-body radiation

Main article: Planck's law of black-body radiation
I(\nu) = \frac{2 h\nu^{3}}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1}

where

Wien's displacement law

Main article: Wien's displacement law

The relationship between the temperature T of a black body, and wavelength λmax at which the intensity of the radiation it produces is at a maximum is

T \lambda_\mathrm{max} = 2.898... \times 10^6 \ \mathrm{kelvin-nanometers}. \,

The nanometer is a convenient unit of measure for optical wavelengths. Note that 1 nanometer is equivalent to 10−9 meters.

Stefan-Boltzmann law

Main article: Stefan-Boltzmann law

The total energy radiated per unit area per unit time j^{\star} (in watts per square meter) by a black body is related to its temperature T (in kelvins) and the Stefan-Boltzmann constant σ as follows:

j^{\star} = \sigma T^4.\,

Temperature relation between a planet and its star

Here is an application of black-body laws. It is a rough derivation that gives an order of magnitude answer. See p. 380-382 of Planetary Science, for further discussion.

Assumptions

The surface temperature of a planet depends on a few factors:

  • Incident radiation (from the sun, for example)
  • The albedo effect (the fraction of light a planet reflects)
  • The greenhouse effect (for planets with an atmosphere)
  • Energy generated internally by a planet itself (This is more important for planets like Jupiter)

For the inner planets, incident radiation has the most significant impact on surface temperature. This derivation is concerned mainly with that.

If we assume the following:

  1. The Sun and the Earth both radiate as spherical black bodies in thermal equilibrium with themselves.
  2. The Earth absorbs all the solar energy that it intercepts from the Sun.

then we can derive a formula for the relationship between the Earth's surface temperature and the Sun's surface temperature.

Derivation

To begin, we use the Stefan-Boltzmann law to find the total power (energy/second) the Sun is emitting:

P_{S emt} = \left( \sigma T_{S}^4 \right) \left( 4 \pi R_{S}^2 \right) \qquad \qquad (1)
where
\sigma \, is the Stefan-boltzmann constant,
T_S \, is the surface temperature of the Sun, and
R_S \, is the radius of the Sun.

The Sun emits that power equally in all directions. Because of this, the Earth is hit with only a tiny fraction of it. This is the power from the Sun that the Earth absorbs:

P_{E abs} = P_{S emt} \left( \frac{\pi R_{E}^2}{4 \pi D^2} \right) \qquad \qquad (2)
where
R_{E} \, is the radius of the Earth and
D \, is the distance between the Sun and the Earth.

Even though the earth only absorbs as a circular area πR2, it emits equally in all directions as a sphere:

P_{E emt} = \left( \sigma T_{E}^4 \right) \left( 4 \pi R_{E}^2 \right) \qquad \qquad (3)
where TE is the surface temperature of the earth.

Now, in the first assumption the earth is in thermal equilibrium, so the power absorbed must equal the power emitted:

P_{E abs} = P_{E emt}\,
So plug in equations 1, 2, and 3 into this and we get
\left( \sigma T_{S}^4 \right) \left( 4 \pi R_{S}^2 \right) \left( \frac{\pi R_{E}^2}{4 \pi D^2} \right) = \left( \sigma T_{E}^4 \right) \left( 4 \pi R_{E}^2 \right).\,

Many factors cancel from both sides and this equation can be greatly simplified.

The result

After canceling of factors, the final result is

T_{S}\sqrt{\frac{R_{S}}{2 D}} = T_{E}
where
T_S \, is the surface temperature of the Sun,
R_S \, is the radius of the Sun,
D \, is the distance between the Sun and the Earth, and
T_E \, is the average surface temperature of the Earth.

In other words, the temperature of the Earth only depends on the surface temperature of the Sun, the radius of the Sun, and the distance between the Earth and the Sun.

Temperature of the Sun

If we plug in the measured values for Earth,

T_{E} \approx 14 \ \mathrm{{}^\circ C} = 287 \ \mathrm{K},
R_{S} = 6.96 \times 10^8 \ \mathrm{m},
D = 1.5 \times 10^{11} \ \mathrm{m},

we'll find the surface temperature of the Sun to be

T_{S} \approx 5960 \ \mathrm{K}.

This is within three percent of the standard measure of 5780 kelvins which makes the formula valid for most scientific and engineering applications.

Radiation emitted by a human

In contrast with the above section, black-body laws can also be applied to things not in radiative equilibrium. A great deal of a person's energy is radiated away in the form of electromagnetic radiation. Most of that radiation is in the infrared.

In addition to emitting energy, humans absorb energy from the surrounding environment. The net power (energy/second) of energy radiated away is the difference between what someone absorbs and what they radiate:

P_{net}=P_{emit}-P_{absorb} \,

Plugging in the Stefan-Boltzmann law:

P_{net}=A\sigma \epsilon \left( T^4 - T_{0}^4 \right) \,

The above equation is applicable to any object which behaves similar to a black body. People have an area of about 2 square meters, and emissivity of nearly 1. They also have a skin temperature of about 32 °C (305 K). But clothing reduces the surface temperature a few degrees, so in addition to reducing heat loss through conduction, it reduces loss of heat by radiation. So for surface temperature of people we should use 301 K. The temperature of the surrounding environment varies, but for a rough order of magnitude answer, one can use 20 °C (293 K). Plugging in these values results in a net rate emission of energy for people of about:

P_{net} = 95 \ \mathrm{watts} \,

In this scenario, people are roughly 100 watt light bulbs, except they emit all infrared and longer wavelength light. The amount of energy in a whole day turns out to be almost 9 million joules, or 2,000 (food) calories. Normal rate of metabolism is typically 100-120 watts, and a person losing more than 160 watts (with extra losses by evaporation, convection and conduction) would feel cold and need to increase activity or cover with clothes. In contrast, during physical activity the metabolism is much higher and since the emission is not large enough, the excess heat is carried by sweating.

Also, applying Wein's Law to humans, one finds that the peak wavelength of light emitted by a person is:

\lambda_{peak} = \frac{2.898\times 10^6 \ \mathrm{K} \cdot \mathrm{nm}}{305 \ \mathrm{K}} = 9500 \ \mathrm{nm} \,

This, presumably, would be the wavelength that infrared goggles would be designed to be most sensitive to.

A few historical examples of black body radiation

Blast furnaces before 1700 heated with charcoal could only produce "red hot" pig iron. The introduction of coke for heating in English ironworks in 1709 enabled "yellow hot" iron, required for the more advanced products of the industrial revolution.

See also

Footnotes

  1. ^ When used as a compound adjective, the term is typically hyphenated, as in "black-body radiation", or combined into one word, as in "blackbody radiation". The hyphenated and one-word forms should not generally be used as nouns, however.

References

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