3 9 8 D O C U M E N T 4 2 0 N O V E M B E R 1 9 2 6 Abstract: ¢Of course not! Provided no calculation errors are made, at least.² No. We want to show in what follows: Even when one decides to apply Bose-Einstein statistics to the translational energy of force-free molecules (which one ought per- haps better refrain from!), one does not arrive at that degenerative condensation (decay in two phases) which Mr. Einstein derives from a sufficiently high concen- tration and a sufficiently low temperature.[2] The fact that Mr. Einstein arrives at condensation is based on the circumstance that he uses an approximation in his cal- culation (substitution of an integral for a sum) which is legitimate for not-too-high concentrations and not-too-low temperatures in that domain, but which leads to re- sults that are quite interesting albeit quite incorrect.— If one dispenses with this ap- proximation, one can see immediately that no decay in two phases occurs at all. § 1. Whereas, according to regular statistics, one would obtain the distribution law 1) for the somehow quantized translational energies ܭs of the force-free gas molecule, as Einstein shows, the distribution law 2) follows from the Bose-Einstein statistics. This is directly contrasted against the distribution law that Fermi and Dirac de- rived out of the ெPauli exclusion.”[3] 3) ¢Hence all three can be subsumed in:² The quantity of α) in (2) is determined from the total number n of the molecules in volume V by 4) One sees:[4] For a given V and T, one can conform α to an arbitrarily large value for n by al- lowing it to approach from large values arbitrarily close to the value . That means approaching the value 1, if one joins Einstein in assuming that the lowest quantum state possesses the energy = zero, or approaching the value , if one joins Schrödinger in forbidding the value zero.[5] ns Ae–βεs = β kT¹ -----·1 -= © § ns 1 αeβεs 1– --------------------- -= ns 1 αeβεs 1+ ---------------------- = n ∞ ¦----------------------s1–1βεeαs0 = e–βε0 ε0 e–βε1