2 5 0 D O C U M E N T 2 5 5 A U G U S T 1 9 2 8 (6*) and furthermore, for the as the single condi- tion, we obtain: (6*) , i.e., . Again I set further: (7) , where is again antisymmetric, and I determine the by means of (7*) , after which one still finds from (6*): (7 * ) , i.e., is harmonic. (6) together with (7*) yields: (8) I write (8*) (8*) . Now, (6), (7), (7*), (8*), (8 * ) completely finish the problem for harmonic and its solution is found as follows: furthermore: 1) Choose four harmonic functions with (6*), i.e., . 2) Determine the six symmetric harmonics , according to (6), i.e., , which can be carried out without difficulties. 3) Determine four from (7*), that is [because of (7*), (6 * )], , with a harmonic . 4) I determine the six antisymmetric harmonics , , according to (8 * ), which again can be carried out without difficulties. Now, everything is settled. II. For each solution of Maxwell’s equations (B), there again exists a fundamen- tal solution of the original equations (A).[6] III. Owing to the linear character of the latter, the general solution in the electric case is obtained from the general solution of the pure gravitation case by addition of a particular electric solution to the Maxwell equations being considered. U  vv  H  H  –  –2 = H  H    0 = H   2  2 x -  ----------- 0 = U   2 F  F  –  – u   + = u   F  F  vv H  = u  vv  0 = u   2F v vv  – 2H  u  vv v + + H  = F v vv  F = u  vv v 2F H  – = H   H   (w/o summ.) 0 = H  H 2    2 x v ----------- 0 = H      H   2 v v 2 x v -------------- H  = F  F   2 v 2 x v ----------- H  = F v vv  F = u     