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Since this star has a peak wavelength that is at a shorter wavelength (in the ultraviolet part of the spectrum) than that of our Sun (in the visible part of the spectrum), it should come as no surprise that its surface temperature is much hotter than our Sun’s.

We can also describe our observation that hotter objects radiate more power at all wavelengths in a mathematical form. If we sum up the contributions from all parts of the electromagnetic spectrum, we obtain the total energy emitted by a blackbody. What we usually measure from a large object like a star is the energy flux    , the power emitted per square meter. The word flux means “flow” here: we are interested in the flow of power into an area (like the area of a telescope mirror). It turns out that the energy flux from a blackbody at temperature T is proportional to the fourth power of its absolute temperature. This relationship is known as the Stefan-Boltzmann law    and can be written in the form of an equation as

F = σ T 4

where F stands for the energy flux and σ (Greek letter sigma) is a constant number (5.67 × 10 -8 ).

Notice how impressive this result is. Increasing the temperature of a star would have a tremendous effect on the power it radiates. If the Sun, for example, were twice as hot—that is, if it had a temperature of 11,600 K—it would radiate 2 4 , or 16 times more power than it does now. Tripling the temperature would raise the power output 81 times. Hot stars really shine away a tremendous amount of energy.

Calculating the power of a star

While energy flux tells us how much power a star emits per square meter, we would often like to know how much total power is emitted by the star. We can determine that by multiplying the energy flux by the number of square meters on the surface of the star. Stars are mostly spherical, so we can use the formula 4π R 2 for the surface area, where R is the radius of the star. The total power emitted by the star (which we call the star’s “absolute luminosity”) can be found by multiplying the formula for energy flux and the formula for the surface area:

L = 4 π R 2 σ T 4

Two stars have the same size and are the same distance from us. Star A has a surface temperature of 6000 K, and star B has a surface temperature twice as high, 12,000 K. How much more luminous is star B compared to star A?

Solution

L A = 4 π R A 2 σ T A 4 and L B = 4 π R B 2 σ T B 4

Take the ratio of the luminosity of Star A to Star B:

L B L A = 4 π R B 2 σ T B 4 4 π R A 2 σ T A 4 = R B 2 T B 4 R A 2 T A 4

Because the two stars are the same size, R A = R B , leaving

T B 4 T A 4 = ( 12,000 K ) 4 ( 6000 K ) 4 = 2 4 = 16

Check your learning

Two stars with identical diameters are the same distance away. One has a temperature of 8700 K and the other has a temperature of 2900 K. Which is brighter? How much brighter is it?

Answer:

The 8700 K star has triple the temperature, so it is 3 4 = 81 times brighter.

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Key concepts and summary

The electromagnetic spectrum consists of gamma rays, X-rays, ultraviolet radiation, visible light, infrared, and radio radiation. Many of these wavelengths cannot penetrate the layers of Earth’s atmosphere and must be observed from space, whereas others—such as visible light, FM radio and TV—can penetrate to Earth’s surface. The emission of electromagnetic radiation is intimately connected to the temperature of the source. The higher the temperature of an idealized emitter of electromagnetic radiation, the shorter is the wavelength at which the maximum amount of radiation is emitted. The mathematical equation describing this relationship is known as Wien’s law: λ max = (3 × 10 6 )/ T . The total power emitted per square meter increases with increasing temperature. The relationship between emitted energy flux and temperature is known as the Stefan-Boltzmann law: F = σ T 4 .

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Source:  OpenStax, Astronomy. OpenStax CNX. Apr 12, 2017 Download for free at http://cnx.org/content/col11992/1.13
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