Photons, the Spectra of Stars, and the Doppler Effect

How do we know what stars are made from? By the light they emit.


Photons

Here by "light" we actually mean electromagnetic radiation. We see with our eyes only part of the electromagnetic radiation. Electromagnetic radiation (often abbreviated in em) consists of oscillating electric and magnetic fields. Electromagnetic radiation is a wave that propagates, in vacuum, with the speed of light, c=300,000 Km/sec (1 Km=1,000 meters). Contrary to sound waves, which are longitudinal, em waves are transverse, which means the electric and magnetic fields are perpendicular to the direction of propagation of the wave. In sound waves, the compression/rarefaction of air is parallel to the direciton of propagation.
A simple sinusoidal wave is characterized by its wavelength l (lambda), which is the length between two consecutive crests, and its frequency n (nu), which is the number of oscillations per seconds. l and n are related to c in the following way:

c = l n ,

if we want to get n from l and c, then

n = c / l ,

if we want to get l from n and c, then

l = c / n .

A photon can have an arbitrary large value of its frequency, as well as an arbitrary small one. There is no minimum or maximum frequency, and the same holds for the wavelength, of course. Visible light is composed of photons within a certain values of frequencies (and wavelengths). Specifically, light visible to human eyes has wavelengths between 4,000 Angstrom (1 Angstrom is 10-8 cm)(this is red light) and 7,000 Angstrom (this is violet light). It should be clear that there are no sharp boundaries; indeed electromagnetism (the study of electromagnetic phenomena) is very democratic and all electromagnetic waves have the same properties, except for different values of wavelengths.(Bear in mind that, since c = l n wavelength and frequency are related, where c, the speed of light, is a physical constant; thus if the wavelength changes, so does the frequency in order to keep the product the same).

Gamma rays and Radio waves

What distinguishes a gamma ray from a radio wave is the different value of the wavelength. They have different energies too, since E=h n , where h = Planck's constant (6.62 10-34 Joule x sec) and n is the frequency and E is the energy. (The unit if energy is the Joule, if distances are given in meters, mass in Kg and time in seconds.)
What distinguishes gamma rays from radio waves besides the wavelength (and, thus, its frequency and energy), is their interactaction with matter.
Since c=l n, we can write E= h c /l. The lower the wavelength, the higher the energy. Gamma rays can deposit a considerable amount of energy in human tissues and disrupt cell functioning. Radio waves bounce off most obstacles and the energy deposited is small.

The Inverse Square Law

Suppose you are near a lightbulb and that you have measured the energy that crosses the glass envelope in one second. The glass envelope, which we assume to be spherical - to make things easy, is at one inch (2.54 cm) from the center of the filament. Let's assume we build a larger glass envelope 10 inches away. What's the energy that goes through that in one second? It is still the same, since all the energy that crosses the first envelope crosses also the second.
Let's ask now what is the energy that crosses a square of 1 inch side in the small envelope. The answer is: if the surface area of the glass envelope is 12.5 square inches (i.e., 4 p R2, where R=1 inch), then 1 square inch is the 1/12.5 fraction of the total area of the glass envelope. Thus the energy that goes through 1 square inch in one second is 1/12.5 of the total energy.
What is the amount of energy that goes through 1 square inch of the large glass? Now the energy goes through an area of 1250 square inches (i.e., 4 p R2, where R=10 inch) in one second. The energy going through 1 square inch of this envelope will be 1/1250 of the total energy.

Conclusion: The energy going through a surface of area S diminishes as the distance R from the source R decreases: specifically, it goes as 1/R2.
- or the inverse square law. It is then straightforward to extend this discussion to stars: replace the lightbulb with a star and the inverse square law still holds.

Luminosity, Radius and Temperature

Luminosity (energy irradiated at all wavelengths per second), radius and temperature of a star are related by the following relation:
L = area of star x s x T4,
where area= 4 x 3.14 x R2, R = radius of the star, and s = the Stefan-Boltzman constant. We don't have to worry about the value of this constant, since we can refer everyting with respect to the luminosity of the Sun, LSun:

Apparent and Asbolute Visual Magnitude

Absolute brightness tells us how luminous, or bright, a star is; however, we usually see stars at different distances; two stars with equal absolute brightness but at different distances will appear with different apparent brightness; the closer one will appear brighter than the one further away. Because of the inverse square law mentioned above, the apparent brightness (the one measured by a telescope) will decrease as 1/R2, where R is the distance between the star and the telescope.

For historical reasons, astronomers don't give the brightness of a star in energy ejected per second, but rather they use a somewhat arcane way of classifying the brightness of stars.
Introducing absolute visual magnitude and apparent visual magnitude. The word "magnitude" could be exchanged for "brightness" in the paragraph above, except "magnitude" is measured in a peculiar way.
If we know the apparent visual magnitudes, i.e. the brightness of different stars as we see them, what is the relation to the absolute visual magnitudes?

m = MV -5 x (1 - log10(d))

where d is the distance (in parsec), and MV is the absolute visual magnitude. At 10 parsecs, apparent and absolute magnitude concide (use the formula above to check it is true; log10 is the logarith in base 10).

Finally, absolute visual magnitude and luminosity are related. There is a correction factor that takes into account that visual magnitude gives the luminosity in the visible part of the spectrum, while the star also emits at both lower and higher wavelengths.