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CEPHEID VARIABLES AND THE SIZE OF THE UNIVERSE
Henrietta Swan Leavitt's discovery that Cepheid variables could be used as standard candles set the stage for Edwin Hubble's discovery of the size scale of the universe. A Cepheid variable can be recognized by the distinctive shape of its periodic pattern; approximately, * * * * * * * * * * * * * * * (etc.) * * * * * * * * * * * * * * * * * * --------------------------------------------------------------- ---> time (a typical period is a few days to a few weeks) H. S. Leavitt found that the length of the period - the time between one peak and the next - is proportional to the luminosity of that particular Cepheid. For example, a recently measured Cepheid "period-luminosity relationship" published in 1999 shows: Period Luminosity ------ ---------- 3 days 2000 times that of the Sun 10 days 7500 times that of the Sun 30 days 25000 times 100 days 95000 times The luminosity of the Sun, by the way, is about 4.0 * 10^26 Watts. (Remember that 1 Watt = 1 joule/sec, and 1 joule is 1 kg m / sec^2.) Note that this is a logarithmic relationship; the logarithm of a Cepheid's period is directly proportional to the logarithm of its luminosity. On a log-log graph, then, the period-vs-luminosity graph would follow a straight line. As an example of how to use Cepheids as a standard candle, remember that Edwin Hubble found a Cepheid with a period of 31 days in the Andromeda Galaxy back in 1923. If he used this P-L relation above, he would have concluded that the Cepheid he measured had a luminosity of 27000 times that of the Sun, or L = (2.7 * 10^4) * (4.0 * 10^26) = 1.08 * 10^31 W So if Hubble had measured the flux of that Cepheid to be, for example, 2.0 * 10^(-15) W/m^2, which is about the flux a typical faint star in our own galaxy would produce on Earth, we use the equation F = L / (4 * pi * r^2) to get 1.08 * 10^31 r^2 = L / (4 * pi * F) = ------------------------- = 4.3 * 10^44 m^2 4 * 3.14 * 2.0 * 10^(-15) Now take the square root of r^2 to get r = sqrt(4.3 * 10^44 m^2) = 2.1 * 10^22 m 2.1 * 10^22 m = ------------------ = 680000 pc or 680 kpc. 3.1 * 10^16 m/pc Since he knew that the disk of the Milky Was was only 30 kpc across, Hubble had shown that the Andromeda Galaxy was far outside the Milky Way - and that the Universe was much larger than our galaxy alone. |