451 
At last we have to substitute the first term of (2) in the seeond 
part of 2G;;. This yields 
2x? > all germ Cig fel = hi EE ) x 
lm miel b |e mal l Ie 
2 a) ee | OB 03, l OB in leap Bus OB OBim 
Se Pap nL he — ; ry oO : 
hi Or, Oa, On, Oa, vn. On; 
OFim , OFjm — OBij ) tee pence + ur (eij + dij) = OAT 
zijn . Sy Ag = TX Nij + OFF) = 
Ow; Ow; Orie dir, \ Ou; Ox; (i) \ Ow, 
Omitting terms of the first order, we so find for the left hand 
© 
le 
member of the equations of the field 
- ; | Dy; ie O2; 074 Vil 
wthe CAs An is 4: Ee J os j 
fe) Ow,” On, Ou Ta vj 0. v; 
a 7? hal „ZU Oyu ; d 078 4 ( 07 4;4 . 0? Ojs 
we tt + Oj, % -— x13 de a) tae 
l Ow; Ow; j 3 Oar. 5 ÒayOr, Ow 402; 
ERE) 25") ene 03 0? 03 
— 2% EO |I Vijn pecs 
Ow; oe ih a (D a an On; dar 
f i , ck 7 OF cy) 070; 073 
inthe case 1 =— 4. = Ae gly» — = : Se. - 
Cy Oe Oa; OEE ven Ou Oa, 
E ; i ; 5 O78 a) J a 0? Os; 
in the ene t= 9 = 4: z A ope tee eee ee i Dae == 
a (ij) Oa’ Oe, 
) 03 
agen) oa ape )+ ons de: 
(1) Oa] Om, il) Ovy 
From these expressions we see that the third must serve for the 
calculation of y,,, the second for that of o4,, the first for that of 
the six quantities y;;(@@==4, )='=4) after substitution of the values, 
found for o4,; and y4; in some terms of it. AS we want only the 
terms in £ up to the second order Hdd the case =|=4, j=l=d 
may be omitted. 
The last expression can be reduced a little further. First we have 
0 03 
= —— 1 22 
(L) Oa (255) mn (B) 
dB \? | 
2E (=) — A (8?) — 2848 = A (8) — 202, 
and further 
so that we obtain 
2 i oe Dart 0*B Ges Ou aes 0714 
— x* A (y,, — B) — 2x og — Bm on rar aut 
The quantities 7’ in the right hand member of the equations of 
the field are to be calculated only up to terms of the first order 
inclusive. Consequently 
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