The Big Bang — Question 8

Section 44-13 · Physics pages 7546–7547

Question 8 Entropy transfer heats photons relative to decoupled neutrinos Assume instantaneous neutrino decoupling before electron-positron annihilation. The coupled electromagnetic entropy degrees of freedom drop from 11/2 to 2. Its comoving entropy obeys gsT 3 γ a3 = constant, while Tνa = constant. photons + pairs heated photons decoupled neutrinos Tνa = const. Schematic; labels, not drawing lengths, specify the data. (a) Derive the final neutrino-to-photon temperature ratio. (b) Find the neutrino temperature for Tgamma=2.725 K and the photon-temperature enhancement relative to pure a inverse cooling.
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Question 8 - Solution Entropy transfer heats photons relative to decoupled neutrinos photons + pairs heated photons decoupled neutrinos Tνa = const. (a) Method and calculation gs,i(Tiai)3 = gs,f(Tγ,faf)3. Tν/Tγ = (2/(11/2))1/3 = (4/11)1/3 = .713766 . (b) Interpretation and further calculation Tν = .713766(2.725) = 1.94501 K . (Tγa)f/(Tγa)i = (11/4)1/3 = 1.40102 . Physical check Finite-duration decoupling gives small corrections. The exercise isolates entropy transfer and explains why all relic species need not share one temperature.

Question 8

Entropy transfer heats photons relative to decoupled neutrinos

Assume instantaneous neutrino decoupling before electron-positron annihilation. The coupled electromagnetic entropy degrees of freedom drop from 11/2 to 2. Its comoving entropy obeys gsTγ3a3=g_sT_\gamma^3a^3= constant, while Tνa=T_\nu a= constant.

Diagram shown in the original SVG worksheet.

  1. Derive the final neutrino-to-photon temperature ratio.

  2. Find the neutrino temperature for Tgamma=2.725 K and the photon-temperature enhancement relative to pure a inverse cooling.

Question 8 - Solution

Entropy transfer heats photons relative to decoupled neutrinos

Diagram shown in the original SVG worksheet.

(a) Method and calculation

gs,i(Tiai)3=gs,f(Tγ,faf)3.g_{s,i}(T_i a_i)^3=g_{s,f}(T_{\gamma,f}a_f)^3.Tν/Tγ=(2/(11/2))1/3=(4/11)1/3=.713766.\boxed{T_\nu/T_\gamma=(2/(11/2))^{1/3}=(4/11)^{1/3}=.713766}.

(b) Interpretation and further calculation

Tν=.713766(2.725)=1.94501K.\boxed{T_\nu=.713766(2.725)=1.94501\ {\mathrm{K}}}.(Tγa)f/(Tγa)i=(11/4)1/3=1.40102.\boxed{(T_\gamma a)_f/(T_\gamma a)_i=(11/4)^{1/3}=1.40102}.

Physical check

Finite-duration decoupling gives small corrections. The exercise isolates entropy transfer and explains why all relic species need not share one temperature.