D O C U M E N T 2 5 5 A U G U S T 1 9 2 8 2 4 9 This now proceeds as follows:[4] Apart from sign changes for the index 4, we are dealing here with the—only in the first approximation coupled—equations: (1) , and this time, we take generally: (2) without summation over . (1) We then w[ithout] l[oss] o[f] g[enerality][5] integrate them, obtaining (3) , always summing over  For , we find in particular (3*) , so that (3) can now be written as: (3 * ) naturally, (3*) + (3 * ) is equivalent to (3). We now take in general: (4) (3*) and (3 * ) are now converted to: (4*) (4 * ) We will denote the symmetric part of as , the nonsymmetric as . Our equations then become (5) (5*) and thus also, by combination: (5 * ) (5) and (5*) complete the problem. Now I set: (6) (therefore harmonic) from which h  v  v h  vv  h v v  – h  v  v – + 0 = h  v H () v  2    x x ---------------- =  v H     H   v  – H     H v    – +   vv 0 =   = H     H v    –   vv 0 = H   v  H   v  – H   v  H     – +   vv 0 = H () () H () () w/o sum. H  a  + = H    w/o sum. 0 = H v vv    0 = H  v  H  v  – H    +   vv 0 = H  () H    +   vv 0 = H    H   U   H v  U v  +   vv H  v U  v –   vv 0 = = H   w/o summ. U   = = H  v H  v – H   U  v U  v – U   + + +   vv = H v  H v  – H   U v  – U v  U   – + +   vv 0 = = H     vv 0 = H     vv 0 = 2U  v 2U  v – U   +   vv 0 = U  vv v H  vv v H  = =