THE DISTANCE OF DELTA CEPHEI

The first Cepheid variable star to be discovered was the star Delta (the fourth brightest star in the constellation Cepheus). It was discovered in 1784 by the deaf and dumb John Goddricke, who lived to the ripe old age of twenty-two years. He was spot-on when he ascribed the variability of the brightness of the star to its pulsating. He had also previously correctly ascribed the variability of Algol to its being an eclipsing binary!

I was very interested to see on page 8 of the November 1993 issue of Sky and Telescope, that the Multichannel Astrometric Photometer on the Thaw refractor at Allegheny Observatory has enabled astronomers to rectify the distance of Delta Cephei (the prototype of the Cepheid Variables) to a value of 1100 light years, instead of the formerly accepted value of 630 light years.

In 1990 while writing my book "Ontsluier die Heelal" ("Unveil the Universe"), I used the very valuable list of Cepheids compiled by A. Sandage and G.A. Tammann, as published in the Astrophysical Journal, 1969: 157.683. I used the 12 stars ranging in period from 1.95 days to 13.62 days. (I did not include the star of period 41.38 days because its period is much longer and there are no stars of periods intermediate between 13.62 and 41.38 days in the list). Looking at the graph of absolute magnitude M against log of the period logP, one sees that the brighter stars (of absolute magnitude higher negative numbers) have greater values for logP and vice versa. There is an apparent proportionality between absolute magnitude and log of the period. By eye one can draw a straight line which most nearly fits the data. But what is the equation to the straight line of best fit?

see diagram Ceph1.doc and Ceph2.doc

The equation to the straight line of best fit can be found by the method of least squares. The equation will have the form: y = a + bx, where a is the intercept on the y-axis (the M-axis) and b is the slope of the line. The x-scale is obtained by subtracting the mean of the logs of P, i.e. l from each of the l-values. When this is done, the total of the x-column must be equal to zero, i.e. x will equal 0. This has the same effect as moving the y-axis to pass through the mean of the l-values.

Now, a will be equal to y n, i.e. the sum of the y- or M-column divided by the number of cases, which is 12. Also, b will be equal to xy x2 , i.e. the sum of the products, x times y, divided by the sum of the squares of the x-values. This is worked out in the table below:

see diagram Ceph3.doc

Since l (the mean of the logs) has been taken as 0.766 instead of 0.76616, the amount subtracted on each line from l was 0.00016 too small. Therefore, corrections have to be made as follows : x has to be decreased by 12 times 0.00016, i.e. by 0.002, thus making x = 0, as it should be. xy has to be corrected by –42.60 times –0.00016, i.e., by +0.007. The correct value of xy is thus -1.651 + 0.007 = -1.644. The correction in x2 does not affect the third decimal place. Now, a = y n = -42.60 12 = - 3.55; and b = xy x2 = -1.644 0.608 = -2.70. Therefore, the equation to the straight line of best fit y = a + bx, becomes :
 

y = M = -3.55 - 2.70x, or y = M = -3.55 –2.70 log(P – 0.766)
y = M = -3.55 –2.70(-0.766) – 2.70 logP,

i.e.     y = M = -1.48 –2.70logP.

To find the 95% confidence limits within which the values of M will fall, the Variance, Sd2, has to be calculated, i.e. the mean of the squares of the deviations of the plotted points from the straight line. The square root of the Variance is the Standard Deviation. Twice the standard deviation gives the limits within which 95% of observations will fall in a normal distribution. This has been worked out in the last three columns.

Firstly, substitute each of the x-values in the fourth column in the equation M = -3.55 -2.70x. These substituted values give the computed values = c, i.e. the ordinates of the points on the straight line vertically above or below the plotted points.

Secondly, substract each of the computed values c, from the corresponding values of y in the y-column to give the vertical deviations d, of the plotted points from the calculated straight line. The sum of these deviations must be zero. We get –0.01 because of the approximation in the value of l which we used. But it is close enough to zero as to make no difference to the final 95% limits.

Thirdly, square each of the deviations and find their sum d2 = 0.3081. The Variance Sd2 is defined as d2 divided by n, i.e. 0.3081 12 = 0.025675. The square root of this is 0.160 and this is, therefore, the Standard Deviation. Twice this value (= 0.32) gives the vertical widths of the confidence limits above and below the straight limits. Notice how the plotted points on the graph fall just within these limits.

We can, therefore, state the Period Luminosity Law for Cepheid Variables of periods between 2 days and 14 days, by means of the equation : M = -1.48 –2.70logP ± 0.32

We can now apply this equation to the case of Delta Cephei. The period of Delta Cephei is very well known as 5.366 days. Substituting this value in the equation, we get:
 

M = -1.48 –2.70log(5.366)

i.e.     M = -1.48 –2.70(0.72965)
          M = -3.45

The absolute magnitude of Delta Cephei is, therefore, -3.45 ± 0.32.

Now substitute this value of M in the equation for distance :
 

M = m + 5 – 5logD

D is the distance in parsecs and m is the apparent visual magnitude. The apparent visual magnitude of Delta Cephei varies between 3.78 and 4.63. The mean of this is 3.78 x 4.63 = 4.18.
 

Therefore M = 4.18 + 5 –5logD = -3.45

      logD = 4.18 + 5 + 3.45 = 12.63 = 2.53 5 5 D = antilog 2.53 = 338.8

The distance of Delta Cephei is, therefore, 338.8 parsecs. In light years this equals 338.8 times 3.26 = 1104, which is 1100 light years, correct to the third significant figure, in complete agreement with the result found at Allegheny Observatory.

When the 95% confidence limits are applied, it works out that the distance of Delta Cephei lies between 930 and 1250 light years.

To draw the straight line on the graph: (i) substitute the value 0.3 for logP in the equation M = -1.48 –2.70logP. We get –2.29. This is the ordinate of the point A on the graph. (ii) substitute 1.1 for logP. We get –4.45. This is the ordinate of point B. (iii) join these two points by means of a straight line. (iv) draw two parallel dotted lines, one 0.32 absolute magnitude units above and the other 0.32 units below the straight line of best fit.

Any amateur astronomer can use the equation M = -1.48 –2.7logP to determine the absolute magnitude of a Cepheid of period 2 to 14 days and he can then calculate its distance from the formula:
M = m + 5 –5logD, where M is the absolute magnitude and m is the apparent visual magnitude. The distance D is in parsecs. This can be converted to light years by multiplying by 3.26.

Eben van Zyl