1076 
: 2x 
has along the Zaxis a component also with the period —, but 
bi 
alone the rotating axes in the plane of the path components propor- 
tional to cos (stp) cos (wl Hg), which may be split up into terms 
; 2 23 É ; 
with the periods ——— and ——. The electric moment which the mole- 
s+o s—Ww 
cule has under the influence of the electric force will depend linearly 
: As Meee 
on 7, B and p. If we write D for the sign of differentiation re the 
a 
equations (4) and (5) for these quantities may be brought into the form 
(D? +. n° pi) (D? — n*p2) B = 0 
(D* + n° pi) (D? — n’*p2) p= 0 
Eder 0. 
(The integration constants in the solutions 6 and ~ must then be 
determined with regard to the equations (4)). 
For the forced vibrations we obtain instead of these equations: 
(D? + n°pi) (D? — n* po) B= Fe + Fe (7) 
(D? + n° pi) (D? — n° pa) p= F, a Fy ee 2) U 
(Bz oo n” 2) y= F, 
where Ha, Fe, PF, Fy, #, are homogeneous, linear functions of the 
components of the electric force in the beam of light, taken with respect 
to the rotating system of coordinates, so that Pe and /’, have the 
2a 2m 2 
period — —, /’,' and // the period - and F, the period —. 
stow s—w 8 
In virtue of the linear character of the equations (7) we may add 
the solutions that would be obtained, first if on the right hand sides 
Fz and F, stood alone, secondly if only Fy and 4 occurred there. 
In the first case we assume that 8 and ~ contain the time only in 
the factor est+)', in the seeond that they are proportional to els” 
and so we obtain 
; Fs Fy 
3— eae ur iad = = = = a oll = 
\n? pr— (s + w)?}{n? po 4- (sw) fn? pi — (s—w)*} fn? po + (s—w)*} 
Fy, Fy 
(00) Ze = — Ees — == = an —— = Lb == — 
/ in? py — (s+o) Un po + (st) } n° pr (s—w)*} pa +- (s—o)} 
and in the same way for y: 
REE 
n° B—s* 
In the expression for the moment of a molecule we obtain 
terms corresponding to each of these, and the first term in 2 can 
