4 0 2 D O C U M E N T 4 5 1 M A R C H 1 9 2 9 We now replace the first parenthesis in (1), before going to the limit , with , and write the equation in the form If we now perform the operation on this equation, then owing to (3), we find that . Now we go to the limit then the second term vanishes, because goes to zero as . We thus obtain the electromagnetic equations . I believe that we must first investigate this preferred case , although it may be that the case is in fact implemented in nature. ————— The symmetry of obtains as is symmetric, in any case. Only the symmetry of the second term needs to be proved.  0 = 2G * 2Q * – 2G *  G ** G ** –   2------------- Q *  – + 0 = D  D  f  2------------- Q *  – 0 =  0 = Q *  2 D  f    0 =  0 =  0  G  G  H  H     / – = H  H h 1 2 --       1 2 --            – + = H H  h 2   1 2 --    2      – +    + = and thus 1 h --H   1 2 --    1 2 --    – 1 2 --        –     + + = H     / h         –      +   / = 1 h -- H   H   –  =