126 



Dr. Frank Watson Dyson 



[April 24, 



the shaded portion).* How must the stars be distributed in distance 

 to make these laws harmonize ? This is a mathematical problem 

 which is solved fairly easily, and the answer is that the stars must 

 be distributed in distance as shown graphically in Fig. 4. 



14. In the diagram distances are measured horizontally, the unit 

 of distance being that at which a star's parallax equals 1". This 

 distance is 206,265 times the distance of the sun, and is a convenient 

 unit for purposes of calculation. It is convenient to have a name 

 for the unit, and I shall adopt jparsec, a composite word suggested 

 to me by Prof. Turner. With this unit, a distance of 100 in the 

 diagram denotes 20 million times the distance of the sun from the 

 earth. It may be verified by the diagram (Fig. 4) that : — 



6 per cent of the stars are between 

 15 „ 

 40 „ 

 33 „ „ ,, 



6 „ ,, greater than 



and 100 parsecs 



100 „ 200 „ 



200 „ 400 „ 



400 „ 700 „ 



700 „ 



It follows that 88 per cent of the stars in Carrington's catalogue, 

 that is 88 per cent of all stars brighter than about lO'^'S, lie between 

 20 and 150 million times the distance of the sun from the earth. 

 This law of the distribution of the stars is at first sight rather sur- 



kO 



lOQO TAKSECi 



prising. It must be remembered that the only stars at a great 

 distance which are included are those which are intrinsically . very 

 bright, and these form only a small proportion of all the stars. Prof. 

 Eddington has found that a similar law holds for stars brighter than 

 6™-0. 



* The number of stars out of a total N with velocities between v and 



■y + rf y is N e—^^ dv ; the number of proper motions between t and 



a/it 



T + <Zt is N ^ (l + ^o)~ ^' The solution of an integral equation, gives for 



the number between r and r + d r, "^ 2 a^ h- r e - a^ ^^ »'^ d r. 



