210 , M. Lorenz on the Theory of Light. 



We will now investigate specially two cases, namely those in 

 which the exponents p have the respective values and 2. It 

 will then be seen that the result in both these cases agrees with 

 experiment, since the non-periodic portions of the excursion are 

 in both cases in the same plane through the perpendicular to 

 the plane of the wave. Let this plane be determined by the 

 equation 



Since it passes through the perpendicular to the plane of the 

 wave and the non-periodic part of the excursion, we have 



Au + Bv + Cw=0; 

 and for jo =0, 



Af o +B%+Cg,=:0. 



For jo=2, we get 



f=i?=i[io+s|l(± P! ,)]c+... 



If, therefore, the non-periodic part of the component of the excur- 

 sion is here denoted by £ C, equation (4) gives 



fo=^ [fo-^fo + ^o + ^fo)]- 



By multiplying this equation by A, and forming two analogous 

 equations by putting in one case r), B, v, and in the other f, C, w 

 for f , A, u, we get by addition of all three equations, 

 A£ + B*, +Cr o =0. 



Hence this excursion also lies in the plane D, which was to be 

 proved. 



It will consequently make no essential difference whether we 

 take p = or p = 2, for in both cases the plane of polarization is 

 the same. In the first case the components of the excursion are 

 determined by (9), and accordingly do not lie in the plane of the 

 wave \ on the contrary, it is easily proved that they are perpen- 

 dicular to the ray, which in doubly refracting media differs, as 

 is well known, from the perpendicular to the plane of the wave. 

 In the other case, however, p = 2, we have 



fo=^fo=^[?o" w ( w ?o + ^o + Ko)]^ 



and this equation, in conjunction with the analogous equations 

 for rj and ? , gives 



6 2 - 



— — ? =— — 77 = — — f = Afo + ^Wo + c^o. (11) 

 But since we also find wf + vfjQ+w^—0, the vibrations lie in 



