        CEPHEID VARIABLES AND THE SIZE OF THE UNIVERSE

 motion: TOC for Knowledge Concepts, Exercises, and Solutions

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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.

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