402 PROFESSOR STOKES, ON THE COMPOSITION AND RESOLUTION OF 



similar formulae whereby x it y\, &c. are expressed, and then let the sines and cosines of 

 + e, (j> + e lt and (p + e 2 be developed. In order that equations (3) may be satisfied inde- 

 pendently of (p, the coefficients of sin <p and cos <p must separately be equal to zero, so that 

 each of these equations will split into two. We shall thus have four equations to determine 

 the four unknown quantities c l , c 2 , e p and e 2 . For the sake of shortness, let 



c cos e = gt c sin e = h, 



whether the letters be or be not affected with suffixes ; and further put 



a i -a = 7i' a2-a = 72 5 



then our four equations become 



cos ft cos 7j . gx + sin ft sin 7, . A, + cos ft cos y 2 . g t + sin ft sin y 2 .h 2 = cos /3 . g,' 



cos ft cos 7, . h x — sin ft sin 7, . g, + cos /3 2 cos y 2 . h 2 — sin ft sin y 2 .g 2 = cos /3 . A, 



cos ft sin y x . g t - sin ft cos 7, . A, + cos ft sin y 2 . g 2 — sin ft cos 73 . A 2 = - sinftA,f 



cos ft sin 7, . h x + sin ft cos y x . g t + cos ft sin y 2 .h 2 + sin /3 2 cos 7 2 . £2 = sin /3 . g.\ 



Multiplying the first and second of these equations by 1, y/ - 1 , r and adding, then multi- 

 plying the third and fourth by -\/-l, 1, and adding, and putting generally 



g + x /~ih = G, (5) 



we have 



(4) 



;} 



(6) 



(cos ft cos 7, - \/ - 1 sin ft sin 7^ G x + (cos /3 2 cos y 2 - \/ - 1 sin ft sin 72) G 2 = cos j3 . G, 

 (sin ft cos7! -\/ - 1 cos/3, sin 7,) G x + (sinftcos7 2 -\/- 1 cos ft sin 72) G 2 = sin ft G 

 which two equations are equivalent to the four (4). 



Putting for shortness p x , p 2 , <?,, q 2 for the coefficients in the left-hand members of equations 

 (6), we have 



g. „ G * G (7) 



<j 2 cos /3 - p 2 sin /3 p t sin /3 - gi cos )8 Pi<7 2 -p 2 g , i" 



On substituting for p,, q 2 , &c. their values, we find 



Prfi -Prfi = cos ( 7l - 73) sin (ft - ft) + v 7 - 1 sin (7, - 7 2 ) cos (ft + ft). 

 Now the equations (6) cannot be incompatible or identical unless the above quantity vanish. 

 But this can only take place when 



sin (ft - ft) = and sin (7! - 72) = 0, 

 or else 



cos (ft + ft) = and cos (7, - 72) = 0, 



or lastly 



sin (ft - ft) = and cos (ft + ft) = 0. 



The first case gives ft = ft, 71 - 72 - a, - a 2 = 0, or ±180°, so that the two streams 

 into which it was proposed to resolve the first are of the same nature. The second case gives 

 ft = 90 - ft, 7, - 72 = cti - a 2 = ± 90°, which shews that the two streams are identical in 

 their nature, only the first and second principal planes of the first of these streams are 

 accounted respectively the second and first of the second stream. The third case gives 

 ft = ft = ± 45°, so that the two streams are circularly polarized and of the same kind, which 

 is a particular instance of the first case. 



