D O C U M E N T 2 5 3 A P R I L 1 9 2 6 2 6 7 One would certainly not need angular momentum of the quantum for this. This thing only gets difficult when the quantum is emitted in the direction of the mag- netic field then an eccentric emission is useless. II. The whole difficulty is deeply rooted, because ¢the law of² angular momen- tum isn’t a quantity independent of the choice of the coordinate system’s origin. That’s why this quantity cannot easily be represented as the sum of tensors. In classical mechanics it is like this: let be the momentum of a particle with coordinates then the angular momentum is This expression only has the character of a vector (or tensor) if , i.e., if the total momentum vanishes. Otherwise it changes with a parallel shift of the co- ordinate system. In the theory of relativity one correspondingly has These expressions are independent of for a closed (complete) system. Hence, a conservation law for angular momentum exists here, too. (The theorem only loses its meaning for the ’s in the general theory of rel.) But it doesn’t help here to con- sider the case in which the sum of the momenta , etc., vanishes, because cannot vanish in any case. Angular momentum isn’t tensorial in any direct way. It does appear possible, however, to describe the rotating electron by its energy momentum + angular momentum tensor (antisymmetric) if one in fact assigns an- gular momentum to the latter structure, but not to either energy or momentum. Then this is an antisymmetric tensor, which at rest only has the components different from zero. p1, p2, p3 x1, x2, x3 I23 x2p3 x3p2) – ¦( = — — — — — — — — — — — — — — — — ¦pν 0= I23 x2T34 x3T24) – x1 x2 x3 ddd ³( = — — — — — — — — — — — — — — — — — — — — — — — — I41 x4T14 x1T44) – x1 x2 x3 ddd ³( = x4 xν Vd ³T41 Vd ³T44 I23, I31, I12 I23 I23)0 ( = I14 0= I31 I31)0 ( 1 v2 – ----------------- -= I24 v I ( ) 1 v2 – -----------------012 -=