2 4 8 D O C U M E N T 2 5 5 A U G U S T 1 9 2 8 form of the is conserved and the transform as tensors.[4] The field equation for the symmetric part of the actually agrees with in first ap- proximation for . This is proved in my second note.[5] I cannot understand what fault you can find with my simple calculation. The fact that is also included among the invariants of first order & second rank was known to me, too.[6] But I was convinced that these invariants could not play a role in the field equations. In your work I found your construction of the Riemann from new n-Bein frames particularly interesting. The invariant appears [to be] more complicated relative to A,B and as a [quantity of] 2nd order. I therefore believe that—if the new formal point of view is found to be at all productive, then the first three invariants will suffice. But up to now, I have not succeeded in convincing myself of the plausibility of the validity of the theory for it would appear that with my attempted interpretation of the as electric potentials the equations do not permit the existence of electric masses.[7] One must, however, be cautious in making a final judgment, since the limits of va- lidity of the Maxwell equations are still an unresolved problem. In any case, the combination of a parallel displacement that can be integrated with a metric seems to me to be very natural, since even the assumption of the met- ric at one point in the continuum overdetermines the metric when the displacement law is known. However, the metric need not be defined in terms of a quadratic func- tion but the principle of the constancy of the velocity of light speaks in favor of this. Kind regards, your A. Einstein 255. From Chaim Herman Müntz [Berlin-Nikolassee,] 16 August 1928 Dear, esteemed Professor, After receiving your kind reply,[1] for which I thank you very much, I tried to pursue your various suggestions and proposals, and now I would like to report the results. First of all, calculating the special case , with the “second” invariant[2] gave a physically unsatisfactory result (see below),[3] so that I returned to the “first” invariant and attempted to carry out the integration (to first approximation). a k a – k a k a R ikml 0 = a 0 = 4 j r =