D O C U M E N T 3 2 3 J U L Y 1 9 2 6 3 3 3 1) Every excited atom emits undamped sinusoidal oscillations up to the moment of the jump into the unexcited state. (Bohr-Kramers-Slater)[3] 2) Every excited atom emits damped sine waves. (Classical theory) You plead for indistinguishability. I have now done the calculations, in the fol- lowing manner: For case 1) the visibility curve for the wave of one atom consists of separate tri- angular pieces, as you also said. The maxima all lie at the same height, namely, at the values for of the path- (or better yet) time-difference δ, where n is a positive whole number.[4] Very approximately, we now describe the probability of the ces- sation at with the function , where τ means the relaxation time of the in- tensity (for hydrogen, measured by Wien[5] ). Then we have to redraw the depicted curve in such a way that we shift the maxima to the curve . Strictly speaking, these “triangles” aren’t rectilinearly limited, but the rise and fall occurs along slightly bent curves, which presumably escapes observation. For case 2), the oscillation of the particle is described by the function in which τ has exactly the same meaning as above. By means of the Fourier inte- gral, the associated spectral distribution F(s) of the energy is obtained, and from that the integral , which, according to the known calculation by Michelson, if I am not mistaken, is equal to the “visibility,” with the exception of a constant factor.[6] [ ] Thereby another curve results, again of separate trian- gular parts. But this time the maxima lie on the curve . Hence, one can decide between the possibilities mentioned if time τ is known from other measurements. This, as mentioned, is the case with hydrogen. Thus, in principle, this decision can be reached I obviously cannot know whether the accuracy of Rupp’s measure- ments suffices, because I’m not well enough informed. But perhaps you can send him this letter so that he can consider the matter once and, as the case may be, can supplement his observations, since his apparatus is probably still set up. I did not deem it right for me to write to him myself, in order not to intervene in the exchange of correspondence between him and you. nϑ nϑ e nϑ τ ------ -– e δ τ /– f t) ( 0 for t 0 e t 2τeivt ----- -– for t 0 , = E(δ) F(ε) cosεδdε +∞ ³–∞ = ε μ ν –= J(δ) F(ε) sinεδ dε –∞ +∞ ³ 0 = = e δ 2τ ----- -–