The Interaction of Light and Matter

In order to understand how starlight is produced we also need a number of basic ideas about the motion and energy of atoms. Perhaps the most basic of these ideas is that of temperature.

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Temperature

Temperature is a measure of the average (kinetic) energy of a gas. A gas in which the atoms or molecules are moving rapidly has a higher temperature than the same kind of gas in which the atoms move sluggishly. The video clips in Table 6.2 illustrate this. Also note that the temperature is measured in units of Kelvin degrees (named after Lord Kelvin). The Kelvin temperature scale is the one most commonly used in astronomy and differs only a little from the Celsius (or centigrade) scale that you are familiar with. The Kelvin scale begins at absolute zero which is about -273 C. Room temperature is about 300 K, water freezes at 273 K and boils at 373 K.

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Temperature vs Heat

The water is "only" at 100 C while the steel may be 1500 C or higher yet the answer is simple. Neither would be pleasant but the kettle would be far more dangerous. The kettle has more heat which is a measure of the total energy of all the atoms in the system. In future discussions we will encounter situations in which we will talk about very high temperatures which may or may not involve a correspondingly high heat content.

• Heat measures the energy contained in a body
• Temperature measures the intensity of motion (speed) of the particles in a body

Example 6.1 If temperature is related to average energy of motion what does this suggest about the motion of a gas at a temperature of 0 K?

Solution:In classical physics the concept of temperature was directly related to the idea of atomic motion. At absolute zero it was believed all motion would cease. In quantum physics there will still be subtle "vibratory" motion but motion would be at its minimum possible.

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Incandescence - The Black Body Spectrum

 Why do hot objects glow? This is a simple question that has a surprisingly subtle and complex answer that literally transformed physics. The answer embraces two important ideas:

1. atoms are in continuous motion - colliding with each other "billions" of times each second. The electrons in these materials live an even more frenetic life and are speeding-up and slowing down countless times each second. The hotter the object the more vigorous the motion and collisions.
2. when an electron accelerates it emits light (more correctly electromagnetic waves). The hotter the object the more frequent and more extreme the accelerations (stopping and starting) of the electrons will be.

A hot, glowing object gives off light in a continuous spectrum called a black body spectrum. The term black body refers to a body that will completely absorb (hence black) and then re- radiate energy that falls on it. If you measure the intensity of light for different wavelengths you get a graph like the one shown in Figure 6.8. Spend some time experimenting with this applet - adjust the temperature control slider (panel on right side) by sliding to the left (lower temperature) or to the right. Pay attention to the shape of the black body curve and note where the maximum intensity (highest point) on the curve occurs.  (link to nice image by Aakash!)

Figure 6.8 The Black Body Explorer. You can use this applet to investigate the black body curve as well the idea of band-pass filters.

Why is the Black Body Curve Important?

Our main purpose in studying the black body curve is because - to a first approximation - stars behave like black body radiators. That means that the physics of black bodies can be used to tell us a great deal about the stars. There are two very important properties of black bodies that are directly applicable here:

Wien's Law

There is a precise relationship between the temperature of the black body and the colour of the object. This is usually expressed by noting the wavelength an object emits most of its light. The mathematical relationship is called Wien's Law after Wilhelm Wien who fist published this law in 1893. This can be expressed in the following simple formula:

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. In this formula if T is expressed in degrees K then the wavelength will be in nanometers. The applets in both Figure 6.8 and 6.9 can be used to illustrate this idea.

Example 6.2 Use the applet in Figure 6.8 to determine the wavelength of maximum emission for a star that has a surface temperature of 10 000 K. Confirm this by using the formula and then use the applet in 6.9 to determine the colour that this star would be expected to have.

Solution The applet shows that the peak emission occurs at 290 nm. You can verify this quite easily by using the formula . Note that the values differ by a few percent. The applets use the more accurate version of Wien's Law. Using the applet in Figure 6.9 you can see that this star would have a distinctly bluish color.
	Figure 6.9 StarColor - Applet showing how the colour of a star changes with temperature.

Example 6.3 Suppose we determine that a star emits the maximum amount of light at 700 nm. What is a good estimate for the surface temperature of the star?

Solution: Just re-arrange Wien's Law to read The star is cooler than our sun.

Stefan's Law

Another useful idea relates the temperature of an object to the amount of energy it will emit. This is called Stefan's Law. In words... the intensity of light emitted by an object is proportional to the fourth power of the Temperature of that object, in symbols...

The symbol "E" denotes the energy emitted each second per square meter. The symbol 's' is called the Stefan-Boltzmann Constant and has a numeric value of 5.67 X 10^-8 J/m^sK⁴.

Another way to express this is to use the concept of Flux which is a more specific term that combines
energy per second" as power which is measured in Watts (W) emitted per square meter. The black body applet uses Flux.

You should notice that the energy emitted by an object depends very strongly on its temperature. A hot star emits much more energy per unit area than a cool one. We can illustrate this in the next example:

Example 6.4 Stars A and B are shown in Figure 6.10. How much more energy does star B emit than star A? (Assume that the stars are identical in size).

Solution: The easiest way to approach this is to realize that star B is "3 times hotter" than star A. Since the energy emitted increases with the 4th power of temperature this means that star B will emit (3 x 3 x 3 x 3) = 81 times as much energy! You can also do this with the equations if you set up ratios: Notice how big a difference there is in energy emitted. As we will see in a future chapter, big hot stars emit much, much more energy each second than smaller stars and as such, have much shorter life spans.

Figure 6.10

The applet Black Body Curve (provided in Figure 6.8) illustrates Stefan's Law by enabling you to shade in the area under the black body curve. This area represents the total energy emitted per square meter and calculated by Stefan's Law. To see how you could use this compare the energy emitted by the star Vega (T_surface = 9600 K) and our Sun (T_surface = 5800 K).

The red shaded region represents the light energy emitted by Vega. The green curve (much lower on the scale) shows how much less energy our Sun emits. Also - notice that the hotter star outshines the cooler star at all wavelengths. To compare the amount of energy emitted per square meter look carefully at the right hand panel of the applet. Vega emits 4.81 x 108 W/m2 while the Sun emits 6.42 x 107 W/m2 . If you express these as a ratio then you will see that Vega emits about 7.5 times as much energy per square meter as does our Sun.

Figure 6.11 Area under the Black Body curve indicates the total energy emitted. This is what the Stefan-Boltzmann formula calculates.

Taking a Star's Temperature - Color Indices

In the late fall the bright star Aldebaran climbs above the eastern horizon. It is both a harbinger of winter and a wonderful example of how the colour of a star tells us the temperature. Aldebaran has an easily discernible orange hue. As an experiment see if you can adjust the temperature on the applet StarColor (Figure 6.9) to match the color you see in Figure 6.12. If you do this then you will find that Aldebaran has a surface temperature of about 3500 K. If you used a spectroscope and measured carefully where the peak emission occurs for Aldebaran you would, after applying Wien's Law, get much the same answer. While you could, in principle, do this to measure a star's surface temperature it would actually be quite time consuming. What if you wanted to measure the temperature of thousands of stars? A far simpler way to "take a star's temperature" is to measure how bright it is (in magnitude units) in specific parts of the visible spectrum.

Figure 6.12

To do this astronomers have developed special sets of filters called broadband filters which transmit only specific parts of the visible spectrum. The filters are given the designations U,B,V and R. Figure 6.13 shows how the applet BLack Body Explorer can be used to investigate how these filters work and Table 6.2 shows the wavelength region for each filter.

Figure 6.13 Black Body Explorer in "filter mode" - click on image to launch applet in separate window and then select the filters tab in the upper right corner.

Example 6.5 Compare - qualitatively - the U,B and V magnitudes that you would expect for a star of surface temperature 12 000 K. Use Black Body Explorer to assist you.

Solution: At 12 000 K peak emission will be around 250 nm - in the ultra-violet so you should expect that this star will be quite different than Aldebaran. The applet Black Body Explorer produces the following graph which shows clearly that the amount of light in the U region is greater than the B region which is greater than the V region. This means that the U magnitude will be smallest and the V magnitude greatest for this star.

Color Index - An Approximate Stellar Temperature Indicator

Example 6.xx and the example of Aldebaran provide examples of how one can quickly estimate stellar temperature. Obtain the magnitude of the star in different spectral bands and compare the magnitudes by subtracting them. The difference in magnitudes is called the colour index for the star and this turns out to be a good temperature indicator. Table 6.3 summarizes this.

Practice

1. A star is observed to emit most of its light at a wavelength of 550 nm. Assume that the star is emitting as a black body. Estimate the surface temperature of this star.
2. Look up the following stars in Stellarium and find their B-V colour index and put them in order of increasing temperature: Castor, Pollux, Aldebaran, Vega, Arcturus, Mintaka, Antares
3. Use the applet BlackBody Explorer to estimate the temperatures of the stars in question #2 based on their B-V indices.
4. Four stars are numbered and shown in the figure below. Put the stars in order from hottest to coolest. Which of these stars would have the largest (B-V) index?
   
   
   
5. Star A has a U-B value of -0.3 while star B has a U-B value of 0.6. Sketch a black body curve and use this to explain which of the two stars is hottest.
6. Betelgeuse is a bright star in Orion and is a distinctly reddish colour while Sirius (in Canis Major - just east and a bit lower) looks bluish. Which of the stars has the largest B-V index?
7. Star A is twice as big as Star B but the same temperature as Star B. Star C is half the size of star A but twice the temperature of both A and B. Which of the the three stars is brightest and by how many times compared to the other two?
8. You have discovered a star that is 100 times brighter than the sun. Its colour suggests that it has the same surface temperature as our Sun. From this information estimate the star's diameter. Express your answer in solar units (ie - multiples of the Sun's diameter)

The formula given here is an approximation often used to make quick, "back-of-the-envelope" calculations. A more accurate statement of Wien's Law is:

The following unit conversions:

1 nm = 1 x 10^-9 m

so, 300 nm is also the same as 300 x 10^-9 m = 3 x 0^-7 m

What's a Watt?

A Watt (abbreviated W) is a unit of power. By definition power is just energy used or produced per second. So ... in units 1W = 1 J/s