D O C . 4 4 3 G E N E R A L R E L AT I V I T Y A N D M O T I O N 4 2 9 , (5) which satisfies (3). This solution is not strictly centrally symmetric, as Weyl has demonstrated [18] however, the smaller m is, the closer the solution approaches the central symmetric solution. Applying (4) yields (6) Hence, γ vanishes on the positive as well as on the negative z-axis, as demanded. At infinity the metric is Euclidean.[19] We now consider the case where in addition to the field produced by the singu- larity just considered,[20] there is another “external” field. We express this by set- ting . (5a) Let likewise be a function of r and z alone, let it satisfy equation (3), and let it be regular in the vicinity of . Then, owing to the axial symmetry, we can set , (7) where G contains the terms of second and higher orders in r and z. Equation (4) de- termines γ. Equation (4) implies that γ is constant along the z-axis, as long as we stay on one side of the singularity located at .[21] Therefore, we can set on the negative z- axis, as demanded above. However, in order for the solu- tion to be regular except at the point , γ must also vanish along the positive z-axis. This will be the case if and only if the integral over the infinitely small half circle K ( ), indicated in the sketch to the left, vanishes. The calculation yields the condition:1) , (8) whereas no constraint results for G. 1) The value for γ on the upper side yields . In all cases m and are very small compared to 1. If they are called quantities of first “order,” then the quantity that determines the violation of the regularity of the metric in the general case is determined by a quantity of second order. ψ m r2 z2 + ------------------- -–= γ m2 2 ------ r2 r2 z2)2 + ( ----------------------. - –= [p. 6] ψ m r2 z2 + ------------------- -– ψ + = ψ r z 0 = = ψ α0 α1z G + + = z 0= γ 0= r z 0 = = ³dγ r2 z2 + const. = 4α1m α1 α1 0=