5-4 Wien’s law and the Stefan-Boltzmann law are useful tools for analyzing glowing objects like stars

The mathematical formula that describes the blackbody curves in Figure 5-12 is a rather complicated one. But there are two simpler formulas for blackbody radiation that prove to be very useful in many branches of astronomy. They are used by astronomers who investigate the stars as well as by those who study the planets (which are dense, relatively cool objects that emit infrared radiation). One of these formulas relates the temperature of a blackbody to its wavelength of maximum emission, and the other relates the temperature to the amount of energy that the blackbody emits. These formulas, which we will use throughout this book, restate in precise mathematical terms the qualitative relationships that we described in Section 5-3.

Wien’s Law

Figure 5-12 shows that the higher the temperature (T) of a blackbody, the shorter its wavelength of maximum emission (λ_max). In 1893 the German physicist Wilhelm Wien used ideas about both heat and electromagnetism to make this relationship quantitative. The formula that he derived, which today is called Wien’s law, is

Wien’s law for a blackbody

According to Wien’s law, the wavelength of maximum emission of a blackbody is inversely proportional to its temperature in kelvins. In other words, if the temperature of the blackbody doubles, its wavelength of maximum emission is halved, and vice versa. For example, Figure 5-12 shows blackbody curves for temperatures of 3000 K, 6000 K, and 12,000 K. From Wien’s law, a blackbody with a temperature of 6000 K has a wavelength of maximum emission λ_max = (0.0029 K m)/(6000 K) = 4.8 × 10⁻⁷ m = 480 nm, in the visible part of the electromagnetic spectrum. At 12,000 K, or twice the temperature, the blackbody has a wavelength of maximum emission half as great, or λ_max = 240 nm, which is in the ultraviolet. At 3000 K, just half our original temperature, the value of λ_max is twice the original value—960 nm, which is an infrared wavelength. You can see that these wavelengths agree with the peaks of the curves in Figure 5-12.

Wien’s law is very useful for determining the surface temperatures of stars. It is not necessary to know how far away the star is, how large it is, or how much energy it radiates into space. All we need to know is the dominant wavelength of the star’s electromagnetic radiation.

The Stefan-Boltzmann Law

The other useful formula for the radiation from a blackbody involves the total amount of energy the blackbody radiates at all wavelengths. (By contrast, the curves in Figure 5-12 show how much energy a blackbody radiates at each individual wavelength.)

Energy is usually measured in joules (J), named after the nineteenth-century English physicist James Joule. A joule is the amount of kinetic energy contained in the motion of a 2-kilogram mass moving at a speed of 1 meter per second. The joule is a convenient unit of energy because it is closely related to the familiar watt (W): 1 watt is 1 joule per second, or 1 W = 1 J/s = 1 J s⁻¹. (The superscript −1 means you are dividing by that quantity.) For example, a 100-watt lightbulb uses energy at a rate of 100 joules per second, or 100 J/s. The energy content of food is also often measured in joules; in most of the world, diet soft drinks are labeled as “low joule” rather than “low calorie.”

The amount of energy emitted by a blackbody depends both on its temperature and on its surface area. These characteristics make sense: A large burning log radiates much more heat than a burning match, even though the temperatures are the same. To consider the effects of temperature alone, it is convenient to look at the amount of energy emitted from each square meter of an object’s surface in a second. This quantity is called the energy flux (F). Flux means “rate of flow,” and thus F is a measure of how rapidly energy is flowing out of the object. It is measured in joules per square meter per second, usually written as J/m²/s or J m⁻² s⁻¹. Alternatively, because 1 watt equals 1 joule per second, we can express flux in watts per square meter (W/m², or W m⁻²).

The nineteenth-century Irish physicist David Tyndall performed the first careful measurements of the amount of radiation emitted by a blackbody. (He studied the light from a heated platinum wire, which behaves approximately like a blackbody.) By analyzing Tyndall’s results, the Slovenian physicist Josef Stefan deduced in 1879 that the flux from a blackbody is proportional to the fourth power of the object’s temperature (measured in kelvins). Five years after Stefan announced his law, Austrian physicist Ludwig Boltzmann showed how it could be derived mathematically from basic assumptions about atoms and molecules. For this reason, Stefan’s law is commonly known as the Stefan-Boltzmann law. Written as an equation, the Stefan-Boltzmann law is

Stefan-Boltzmann law for a blackbody

The value of the Stefan-Boltzmann constant σ (the Greek letter sigma) is known from laboratory experiments.

The Stefan-Boltzmann law says that if you double the temperature of an object (for example, from 300 K to 600 K), then the energy emitted from the object’s surface each second increases by a factor of 2⁴ = 16. If you increase the temperature by a factor of 10 (for example, from 300 K to 3000 K), the rate of energy emission increases by a factor of 10⁴ = 10,000. Thus, a chunk of iron at room temperature (around 300 K) emits very little electromagnetic radiation (and essentially no visible light), but an iron bar heated to 3000 K glows quite intensely.

Box 5-2 gives several examples of applying Wien’s law and the Stefan-Boltzmann law to typical astronomical problems.

TOOLS OF THE ASTRONOMER’S TRADE

Using the Laws of Blackbody Radiation

The Sun and stars behave like nearly perfect blackbodies. Wien’s law and the Stefan-Boltzmann law can therefore be used to relate the surface temperature of the Sun or a distant star to the energy flux and wavelength of maximum emission of its radiation. The following examples show how to do this.

EXAMPLE: The maximum intensity of sunlight is at a wavelength of roughly 500 nm = 5.0 × 10⁻⁷ m. Use this information to determine the surface temperature of the Sun

Situation: We are given the Sun’s wavelength of maximum emission λ_max, and our goal is to find the Sun’s surface temperature, denoted by T_⊙. (The symbol ⊙ is the standard astronomical symbol for the Sun.)

Tools: We use Wien’s law to relate the values of λ_max and T_⊙.

Answer: As written, Wien’s law tells how to find λ_max if we know the surface temperature. To find the surface temperature from λ_max, we first rearrange the formula, then substitute the value of λ_max:

Review: This temperature is very high by Earth standards, about the same as an iron-welding arc.

EXAMPLE: Using detectors above Earth’s atmosphere, astronomers have measured the average flux of solar energy arriving at Earth. This value, called the solar constant, is equal to 1370 W m⁻². Use this information to calculate the Sun’s surface temperature. (This calculation provides a check on our result from the preceding example.)

Situation: The solar constant is the flux of sunlight as measured at Earth. We want to use the value of the solar constant to calculate T_⊙.

Tools: It may seem that all we need is the Stefan-Boltzmann law, which relates flux to surface temperature. However, the quantity F in this law refers to the flux measured at the Sun’s surface, not at Earth. Hence, we will first need to calculate F from the given information.

Answer: To determine the value of F, we first imagine a huge sphere of radius 1 AU with the Sun at its center, as shown in the figure. Each square meter of that sphere receives 1370 watts of power from the Sun, so the total energy radiated by the Sun per second is equal to the solar constant multiplied by the sphere’s surface area. The result, called the luminosity of the Sun and denoted by the symbol L_⊙, is L_⊙ = 3.90 × 10²⁶ W. That is, in 1 second the Sun radiates 3.90 × 10²⁶ joules of energy into space. Because we know the size of the Sun, we can compute the energy flux (energy emitted per square meter per second) at its surface. The radius of the Sun is R_⊙ = 6.96 × 10⁸ m, and the Sun’s surface area is 4πR_⊙². Therefore, its energy flux F_⊙ is the Sun’s luminosity (total energy emitted by the Sun per second) divided by the Sun’s surface area (the number of square meters of surface):

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Once we have the Sun’s energy flux F_⊙ we can use the Stefan-Boltzmann law to find the Sun’s surface temperature T_⊙:

T_⊙⁴ = F_⊙/σ = 1.13 × 10¹⁵ K⁴

Taking the fourth root (the square root of the square root) of this value, we find the surface temperature of the Sun to be T_⊙ = 5800 K.

Review: Our result for T_⊙ agrees with the value we computed in the previous example using Wien’s law. Notice that the solar constant of 1370 W m⁻² is very much less than F_⊙ the flux at the Sun’s surface. By the time the Sun’s radiation reaches Earth, it is spread over a greatly increased area.

EXAMPLE: Sirius, the brightest star in the night sky, has a surface temperature of about 10,000 K. Find the wavelength at which Sirius emits most intensely.

Situation: Our goal is to calculate the wavelength of maximum emission of Sirius (λ_max) from its surface temperature T.

Tools: We use Wien’s law to relate the values of λ_max and T.

Answer: Using Wien’s law,

Review: Our result shows that Sirius emits light most intensely in the ultraviolet. In the visible part of the spectrum, it emits more blue light than red light (like the curve for 12,000 K in Figure 5-11), so Sirius has a distinct blue color.

EXAMPLE: How does the energy flux from Sirius compare to the Sun’s energy flux?

Situation: To compare the energy fluxes from the two stars, we want to find the ratio of the flux from Sirius to the flux from the Sun.

Tools: We use the Stefan-Boltzmann law to find the flux from Sirius and from the Sun, which from the preceding examples have surface temperatures 10,000 K and 5800 K, respectively.

Answer: For the Sun, the Stefan-Boltzmann law is F_⊙ = σT_⊙⁴, and for Sirius we can likewise write F_* = σT_*⁴, where the subscripts ⊙ and * refer to the Sun and Sirius, respectively. If we divide one equation by the other to find the ratio of fluxes, the Stefan-Boltzmann constants cancel out and we get

Review: Because Sirius has such a high surface temperature, each square meter of its surface emits 8.8 times more energy per second than a square meter of the Sun’s relatively cool surface. Sirius is actually a larger star than the Sun, so it has more square meters of surface area and, hence, its total energy output is more than 8.8 times that of the Sun.