A Summing Up — Question 4

Section 44-14 · Physics pages 7558–7559

Question 4 A small matter excess can survive almost complete pair annihilation A toy comoving volume initially contains Nb = 109 + 1 baryons and N¯b = 109 antibaryons. Each annihilation removes one of each and produces two photons. Ignore all initial photons and entropy redistribution. b + ¯b 2γ one excess baryon survives Schematic; labels, not drawing lengths, specify the data. (a) Find surviving baryon count and produced photon count after complete annihilation of the minority species. (b) Calculate final baryon-to-photon ratio and the initial fractional baryon excess relative to total massive particle count.
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Question 4 - Solution A small matter excess can survive almost complete pair annihilation b + ¯b 2γ one excess baryon survives (a) Method and calculation Nb −N¯b = 1 is preserved, Nb,f = 1 . Nγ,f = 2(109) = 2 × 109 . (b) Interpretation and further calculation ηf = Nb,f/Nγ,f = 5.0 × 10−10 . Nb −N¯b Nb + N¯b = 1 2 × 109 + 1 ≃5.0 × 10−10 . Physical check The model demonstrates survival of a pre-existing asymmetry, not its origin. Real annihilation products and photon entropy require a thermal calculation.

Question 4

A small matter excess can survive almost complete pair annihilation

A toy comoving volume initially contains Nb=109+1N_b=10^9+1 baryons and Nb‾=109N_{\bar b}=10^9 antibaryons. Each annihilation removes one of each and produces two photons. Ignore all initial photons and entropy redistribution.

Diagram shown in the original SVG worksheet.

  1. Find surviving baryon count and produced photon count after complete annihilation of the minority species.

  2. Calculate final baryon-to-photon ratio and the initial fractional baryon excess relative to total massive particle count.

Question 4 - Solution

A small matter excess can survive almost complete pair annihilation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Nb−Nb‾=1is preserved,Nb,f=1.\boxed{N_b-N_{\bar b}=1\ \text{is preserved},\quad N_{b,f}=1}.Nγ,f=2(109)=2×109.\boxed{N_{\gamma,f}=2(10^9)=2\times10^9}.

(b) Interpretation and further calculation

ηf=Nb,f/Nγ,f=5.0×10−10.\boxed{\eta_f=N_{b,f}/N_{\gamma,f}=5.0\times10^{-10}}.Nb−Nb‾Nb+Nb‾=12×109+1≃5.0×10−10.\boxed{\frac{N_b-N_{\bar b}}{N_b+N_{\bar b}}=\frac1{2\times10^9+1}\simeq5.0\times10^{-10}}.

Physical check

The model demonstrates survival of a pre-existing asymmetry, not its origin. Real annihilation products and photon entropy require a thermal calculation.