Analysing spectra

Surface temperature and luminosity

By analysing the spectra of various astronomical objects, astronomers are able to learn a lot about them. For instance, you’ve already seen (in Figure 5-10) that the Sun’s spectrum is approximately equal to a blackbody curve with peak wavelength at 502 nm. With just this bit of knowledge, we can apply Wien’s law and—from the comfort of the Earth, without having to fly 150 billion metres to stick a thermometer in it—determine that the Sun’s temperature is 5777 K.

Learning Activity

Try it! Apply Wien’s law to determine the Sun’s temperature, given λmax = 502 nm.

Answer: From Wien’s Law,
\displaystyle T = \frac{0.0029\, \textrm{m}\cdot\textrm{K}} {\lambda_{\max}} = \frac{0.0029\, \textrm{m}\cdot\textrm{K}} {502 \times 10^{-9}\,\textrm{m}} = 5777\, \textrm{K}

And with this result, we can use the Stefan-Boltzmann law to determine how many Watts of energy the Sun emits per unit area.

Learning Activity

Try this one, too! Apply the Stefan–Boltzmann law to determine the Sun’s luminosity per unit area.

Answer: The Stefan–Boltzmann law states:
\displaystyle \frac{L}{A} = \sigma T^{4} = 5.7\times 10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}} \cdot (5777\,\mathrm{K})^{4} = 6.35\times 10^{7}\,\mathrm{W\,m^{-2}}

The answer you just calculated is the energy released by a single square-metre on the Sun’s surface. The power released through a single square-metre of Sun is equal to that of a million light bulbs! (i.e. your answer, multiplied by 1 m2, divided by a million should be 60 W).

The next thing we can use astronomical measurements for is to determine just how much power is emitted over the whole Sun. First of all, we’ll need to know how far away the Sun is, plus its angular diameter, so we can work out its radius. This is done by measuring the Sun’s parallax. The first accurate Solar parallax measurement was made by Christiaan Huygens in 1659, but that has since been refined through modern techniques. The value is stated above: 150 billion metres. Given that the Sun’s angular diameter is 0.5°, we can then use the arclength formula,

D = rθ,

to determine the diameter of the Sun.

Learning Activity

You know the drill! Apply the arclength formula to determine the physical diameter of the Sun in metres.

Answer:
\displaystyle D = r\theta = 150 \times 10^{9}\,\mathrm{m} \cdot \left(\frac{0.5^{\circ} \cdot \pi}{180^{\circ}}\right) \approx 1.3 \times 10^{9}\,\mathrm{m}

Then we divide the diameter by 2 to work out the Sun’s radius and use

A = 4πR2

for the area of a sphere of radius R. Finally, multiplying the Sun’s entire surface area by its luminosity per unit area gives the total luminosity—i.e. the total power output—of the Sun.

Learning Activity

Bring ’er home! Determine the surface area of the Sun and use your previous result to calculate the Sun’s total energy output.

Answer:
\displaystyle A = 4\pi R^{2} = 4\pi\left(\tfrac{1.3\times 10^{9}\,\mathrm{m}}{2}\right)^{2} \approx 5.3\times 10^{18}\,\mathrm{m^{2}}
\displaystyle L = A \cdot 6.35\times 10^{7}\,\mathrm{W\,m^{-2}} \approx 3.4\times 10^{26}\,\mathrm{W}

This is a really huge number. It’s also really important, since it basically led physicists to sort out how stars work, which led to even more amazing quantum physics discoveries (which you’ll learn all about in a stellar astronomy course), including the one that earned the Physics Nobel Prize in 2015.

Another interesting thing that this number can be used for is to work out how hot that cast iron frying pan I talked about earlier in the module will get when it’s left out in the Sun. In order to work this out, we need to know how much solar power per unit area is received at Earth. This means we need to divide the total luminosity from the Sun by the area of a sphere whose radius is 150 billion metres. Following through from our previous results, you should get an answer of 1200 W/m2.

Learning Activity

Try it! Divide the Sun’s total energy output by the area of a sphere of radius 1 AU (150 billion m) to determine the total energy per unit area that the Earth receives from the Sun.

Answer:
\displaystyle \frac{L}{A} \;=\; \frac{3.4\times 10^{26}\,\mathrm{W}} {4\pi (150\times 10^{9}\,\mathrm{m})^{2}} \;\approx\; 1.2\times 10^{3}\,\mathrm{W\,m^{-2}}

The luminosity per unit area from the Sun is another important number, since it determines Earth’s total energy budget. It even has a special name: the solar constant. More accurate satellite measurements that do not assume a perfect blackbody spectrum but actually add up all the radiation received by the Sun over all wavelengths give a value of 1362 W/m2. The fact that our calculation gave a result so close to this indicates that the assumption that the Sun is a perfect blackbody is not a bad one to make.

Now, back to that cast iron pan we left out in the Sun. We might as well use the more accurate value of the solar constant to calculate its temperature. If the pan is a perfect blackbody—and if we correctly understand what that means—then all the energy it receives from the Sun per second per unit area should be simultaneously absorbed in whatever wavelength it comes in at, and then re-emitted as perfect blackbody spectrum with its temperature determined by the Stefan-Boltzmann law.

Learning Activity

Calculate how hot the pan is!

Answer: The Stefan–Boltzmann law states:
\displaystyle \frac{L}{A} = \sigma T^{4}
Dividing both sides by σ and taking the fourth root gives:
\displaystyle T = \left(\frac{L/A}{\sigma}\right)^{1/4}
Inserting the solar constant for L/A:
\displaystyle T = \left(\frac{1362\,\mathrm{W\,m^{-2}}}{5.7\times 10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}}\right)^{1/4} \approx 393\,\mathrm{K} \;\approx\; 120^{\circ}\mathrm{C}

Chemical composition

Energy output and surface temperature are just two of the quantities that can be determined by analysing an object’s spectrum. Another that you’ve encountered is its chemical composition. By illuminating various sample elements and molecules in the lab, scientists are able to determine their characteristic spectral lines. Then, by comparing the spectra of astronomical objects to tables of known spectral lines, astronomers are able to determine the chemical composition of their atmospheres, where diffuse gas absorbs escaping light at certain wavelengths.

By determining the chemical compositions of stars, nebulae, and other astronomical objects, astronomers have found that hydrogen is by far the most abundant element in the Universe, accounting for 74% of the mass of all elements, while helium, at 24%, is the second most abundant. All the other elements combined make up just 2% of visible mass of the Universe. In fact, since hydrogen and helium are also the two lightest elements, these relative abundances are even more stark when put in terms of numbers of atoms: our observations show that 92% of the atoms in the Universe are hydrogen and 8% are helium, with only trace amounts of the rest.

Learning Highlight

Did you know that one of the pioneers in the field of atomic and molecular spectroscopy was Gerhard Herzberg? He was awarded the Nobel Prize in Chemistry in 1971 for his contributions to the knowledge of electronic structure and geometry of molecules. He was also a physics professor at the University of Saskatchewan from 1935-1945, where he did much of the work that earned him the Nobel Prize.

The Doppler effect

One final important application of spectroscopy that is used in all areas of astronomy is the Doppler effect. The Doppler effect is a shift in the wavelengths of light (or any other wave) emitted by a source that is moving relative to an observer. Or, in other words,

There is a good chance that if you’ve never heard of the Doppler effect before, Sheldon’s explanation hasn’t really done you much good. However, you should not be discouraged because it is actually not a difficult concept to understand. In fact, the Doppler effect is something you’re probably already familiar with. It’s what causes the sound coming from a car to change from a high pitch to a low pitch as it drives past you. The cause of this change in pitch comes from the fact that sound, like light, is a wave. When the source of a wave moves in one particular direction, it also closes the distance between subsequent crests so that they end up closer together.

Similarly, since a moving source actually makes up extra distance between wave crests it emits when moving in the opposite direction, those waves end up with longer wavelength and lower frequency. This is actually best understood when seen in a video—and not the one above. However, in case you did find that funny, the following actually starts with that clip from The Big Bang Theory as well:

The Doppler effect is measured in astronomical objects through shifts in the positions of their prominent absorption lines. Generally, the change in wavelength, called the Doppler shift, is measured through the prominent hydrogen line pattern (see Figure 5-11) that is present in most astronomical spectra because of the great abundance of hydrogen in the Universe. Since Doppler shifts emerge as a direct result of an object’s relative velocity, there is a direct relation between the two.

Measuring relative motion has a number of applications in astronomy, from the detection of extra-solar planets through slight wobbles they exert on the stars that they orbit, to the rotation rates of any object that is resolved well enough to measure the rate at which one side is moving toward us while the other moves away.