Stoney — Cause of Double Lines in Spectra. 589 



in the plane of xy, and with its axis major inclined at an angle ^ to Ox: u, v, 

 e, and t, being determined in the same way as a, b, a and /8 in Problem A. 

 Thus, taking Ox', Oy', and Oz as axes, we find the motion represented by 



x' = u cos {dt + e), \ 



y' = V sin {et + e), I (^3) 



z = Gcoset+ C sin 6L J 



Let us, by equating coefficients of cos 6t and of sin 6t, determine Jf and iV, such that 



C cos et + C sin et = M cos {Ot + e) + N sin {et + e), 

 whereby equations {b^) become 



x' = u cos [et + e), 



y' = v sin {Qt + e), \ (b^) 



z =MGOs{et+ £) + JSfsm{et+ e). , 



Now, it is possible to identify this with the pendulous elliptic motion 



x" = a cos{et + e + a), 



(^0 



y" = b sin (^^ + e + a), 



having the same frequency, and lying in a position which can be determined. 

 For— 



Let OX be the intersection of the plane x'y' (which is identical with xy), with 

 the plane x'y ; and let 13 be the angle x"OX and y the angle z'OX. Then equa- 

 tions (^4) are equivalent to 



Xi = u cos {dt + e) cosy — V sin {et + e) sin y, ' 

 Yi = u cos {et + e) sin y + v sin {et + e) cos y, 

 Z, = Mcos{et+e) + I^sm{et+e), 



the two former being in the plane xy, and Xi being along OX, the line of inter- 

 section. 



Again, equations (ci) are equivalent to 



X" = a cos {et+e + a) cos ft - b sin {et + e + a) sin ft, 



{h) 



Y" = a cos {et + e + a) sinft + b sin {et + e + a) cos ft, 



in the plane x"y", X" being along OX, the line of intersection. This, if w be the 



angle at which the planes xy and x"y" are inclined to one another, is equivalent 



to— 



Xi = X" along the line of intersection, 



F2 = Y" cos oi, in the plane xy, and perpendicular to the line of intersection. 



Zi = Y" sin CD, along Oz. 



4 2 



