302 Mr Glazebrooh, On the isockromatic curves of [Jan. 29, 



Consider any line through the origin making an angle a sup- 

 pose with the initial line. Then along this line 



. 77T 2 / f cos 2 7 — cos 2 fi 



sin —«- = f 



c 2 ~~ V (sin 2a sin 2 (a — 7) 



vrr 

 Let e be the least value of — ^ Avhich satisfies this equation. 



itr 1 

 Then s- = nir ± e. 



c 



ttt 2 

 The circles of uniform intensity p 2 cos 2 7 are given by —5- = 7i7r. 



Thus between each two such consecutive circles a curve of given 

 intensity p 2 cos 2 (3 cuts any radius vector twice in the points given 



7JT 2 



by -jr = mr + €, 



77T 



and — j- = (n + 1) ir — e, 



If however e = - so that a is such that 



2 



sin 2a sin 2 (a — 7) = cos 2 7 - cos 2 /3, 



the two points coincide. 



. Each curve of constant intensity p 2 cos 2 /3 consists therefore of a 

 series of closed loops lying between the consecutive circles of the 

 series 



irr z 



—5- = 7177% 



■ c 2 

 This had been noticed by Mr W. D. Niven in a paper treating 

 of the same subject in the Quarterly Journal of Mathematics, 

 Vol. xiil. 



Moreover the straight lines given by 



sin 28 sin 2 (8 — 7) = cos 2 7 — cos 2 (3 



meet each of the loops which make up the curve of intensity 

 p 2 cos 2 /3 in two coincident points. 



Again, taking logarithmic differentials of the equation we find 



( ) d8 2irr irr- 



Jcot 20 + cot 2 (0 - 7) [ ^- + ^ cot ~ = 0. 



I "J rtr c' c' 



2 



77"/ 77* 



Hence, when — g ~= (2?2 + 1) - the radius vector and the tangent 



C Ji 



to the curve coincide. 



