1 2 4 D O C U M E N T 1 0 3 N O V E M B E R 1 9 2 5 The first three terms agree, except for the notorious , about which Planck would say that we had wrongly set the weight of any “state” equal to N! we would have had to set it = 1 (omit N! in this letter’s first formula). Now about .[6] I shall set . Then measures the energy deviation based on degeneracy meas- ures the pressure deviation. Therefore, . I want to have the relative deviations, so I divide by , or . I find . At first one sees that for these corrections are very vanishingly small. I find there: (XX) This is at most of the order of magnitude . If one then generally considers the , i.e., those { } in (X), one sees that the integral jumps explosively from very small values to near 1 when β exceeds the value l. The ϑ term, however, con- stantly stays small it never exceeds the order of magnitude , even if at it has a very sharp peak. thus drops very suddenly at from 1 to very small values.[7] In the beginning I thought that this sudden variation of indicates a very sudden onset of strong degeneracy. But perhaps this is not the case. If the said vari- ation occurs within a tiny temperature range, the critical β-range still has the order of magnitude , and since the entire change of amounts to just one unit, in this range would not exceed the order of magnitude , either.— On the basis of (X), no more can be said about what happens afterward, because the mathematical correction factor ϑ will surely depend strongly on β as soon as Rlg§ N ---· - © klgf(β) klgf(β) ΔΨ = T2----------- ∂ΔΨ ∂T T----------- ∂ΔΨ ∂V ΔE k----------- f β) ( f(β) - T2----- β∂ ∂T - = Δp k----------- f β) ( f(β) - T------ β∂ ∂V = E 3RT 2 ---------- = p RT V ------ -= ΔE E ------- Δp p ------ - β N - f β) ( f(β) - –-------------- = = β l « =------· 3N 2 - © ¹ § f β) ( f(β) ----------- - f′ β) ( ϑ------------------ βl 1– e–β l 1– ( )! - 1 ϑ)------------· ( βle–β l! + © ¹ –§ = = 1 l ----- f(β) 0 β ³ 1 l ----- β l = f(β) β l = f(β) l f(β) f′ β) ( f(β) 1 l
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