D O C U M E N T 1 3 6 D E C E M B E R 1 9 2 5 1 5 7 to have an excellent wife and healthy children, but better not for another 2 or 3 years. Perhaps the miserable economic situation will be a blessing to him.[6] An- schütz inquired about him in a fatherly manner.[7] He is touchingly attached to the “un¢washed²licked bear,” as he calls him. Warmest regards from your Albert 136. From Rudolf Ladenburg[1] Dahlem, Berlin, Faradayweg 4–8, 23 December 1925 Dear Professor, May I return, in a few words, to the question of the damping measurements of spectral lines? Unfortunately, I cannot see why the damping should not be measur- able—at least when applying the classical theory of dispersion—from absorption measurements at a greater distance away from the eigenfrequency—given a suffi- ciently large layer thickness. For, according to that theory, the extinction coeffi- cient at a large distance from ν0 is , where N is the number of atoms per ccm, f is the “strength of the oscillators,” and δ is the “damping constant,” which is defined as the coefficient of the “friction term” of the equation of the oscillators’ vibration. The amplitude of the oscillators and the refraction at a greater distance from ν0 are obviously independent of δ, just as is but from the absorption itself, for a narrow frequency region at a greater distance from ν0, one should be able to measure δ, as Minkowski has done,[2] given sufficiently long layers. Given a small sodium vapor pressure and the absence of foreign gases, there thus results for δ quite precisely the value for the classical radiation damping above 10–2 mm Na pressure, a pressure broaden- ing becomes noticeable, but the frequency dependence of nκ now no longer follows the above formula, but rather nκ diminishes more slowly with increasing n, propor- tional to . I would be very grateful, if you would tell me sometime what you think of this or, otherwise, what my misunderstanding is based on. With kind regards, sincerely, R. Ladenburg nκ π e2 m ---- -Nf--------------------------- δ ν0( ν0 ν)2 – = n2 1– dν ³nκ 1 ν0 ν)2 – ( --------------------- -