2 5 6 D O C U M E N T 2 4 5 A P R I L 1 9 2 6 op[inion], it is more advantageous to trace the properties of particles back to the makeup of the field surrounding them then it becomes possible to derive the rela- tions between two particles from the properties of their shared field.[3] Far be it from me to claim that all the difficulties are thus already eliminated. Rather, a mathematical problem (or a series of problems) still remains that is diffi- cult to solve, but the necessity to modify the field equations—the necessity inher- ent in the problem you indicated—does not seem to exist anymore.[4] 2. First I want to elaborate on how I imagine charges to be given by the field. You have for the field inside matter the equation . . . . . . . . . . . . . . . . . . . . (1) and define the density scalar by the equation[5] . . . . . . . . . . . . . . . . . . . (2) I prefer to work in regions free of matter there we have . . . . . . . . . . . . . . . . . . . . (1′) (and, of course, ). Instead of considering the charge density, I consider the charge ρ contained within a closed surface S. The square of this charge is given by the expression . . . . . . (2′)[6] Here is the dual or reciprocal tensor to , and I integrate over S. 3. We now speak of particles and their charges a particle’s charge is the charge—given by (2′)—inside a surface that contains only this particle. (The inde- pendence of this quantity from the surface—inasmuch as it encompasses only the particle under consideration—is ensured by Maxwell’s equations.) 4. Let us now consider two particles T1 and T2. Their charges, taken individually, are given by the values A1 and A2, which the expression A in (2′) assumes if we choose the surface S such that it encompasses T1 on its own, or, T2 on its own. We thereupon consider a surface S3 that simultaneously encompasses T1 and T2. What is the charge inside this surface? Experience tells us that the total charge is either equal to the sum of or the difference between the two individual charges. If we don’t want to use the charges themselves, which are only given by their squares, we can then say that iν ∂f ∂xν -------- J i = σ2 gμνJμJν = iν ∂f ∂xν -------- 0 = ∂xk ∂fij ∂xi ∂fjk ∂xj ∂fki + + 0 = ρ2 A dxμ)dyν]2 ³fμν( [ xμdyν]2 ³rμνd [ + = = rij fij