The Big Bang — Question 6

Section 44-13 · Physics pages 7542–7543

Question 6 The deuterium bottleneck depends on the abundance of energetic photons Use the rough criterion η−1e−BD/(kBT) = 1, where baryon-to-photon ratio η = 6.0 × 10−10 and deuterium binding energy BD = 2.225 MeV. Ignore the polynomial factor in the thermal tail. E nγ(E) BD Schematic; labels, not drawing lengths, specify the data. (a) Solve for the thermal energy at which this criterion is met. (b) Compare with the binding energy and explain why temperature much below BD is needed.
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Question 6 - Solution The deuterium bottleneck depends on the abundance of energetic photons E nγ(E) BD (a) Method and calculation e−BD/(kBT) = η ⇒BD/(kBT) = ln(1/η). kBT = 2.225/ ln[1/(6.0 × 10−10)] = .104784 MeV . (b) Interpretation and further calculation BD/(kBT) = 21.2341 . nγ,>BD/nb ≃η−1e−BD/(kBT). Physical check The enormous photon-to-baryon ratio keeps a destructive high-energy tail even when the mean thermal scale is far below the binding energy. This is a scale estimate, not a reaction network.

Question 6

The deuterium bottleneck depends on the abundance of energetic photons

Use the rough criterion η−1e−BD/(kBT)=1\eta^{-1}e^{-B_D/(k_BT)}=1, where baryon-to-photon ratio η=6.0×10−10\eta=6.0\times10^{-10} and deuterium binding energy BD=2.225B_D=2.225 MeV. Ignore the polynomial factor in the thermal tail.

Diagram shown in the original SVG worksheet.

  1. Solve for the thermal energy at which this criterion is met.

  2. Compare with the binding energy and explain why temperature much below BD is needed.

Question 6 - Solution

The deuterium bottleneck depends on the abundance of energetic photons

Diagram shown in the original SVG worksheet.

(a) Method and calculation

e−BD/(kBT)=η⇒BD/(kBT)=ln⁡(1/η).e^{-B_D/(k_BT)}=\eta\Rightarrow B_D/(k_BT)=\ln(1/\eta).kBT=2.225/ln⁡[1/(6.0×10−10)]=.104784MeV.\boxed{k_BT=2.225/\ln[1/(6.0\times10^{-10})]=.104784\ {\mathrm{MeV}}}.

(b) Interpretation and further calculation

BD/(kBT)=21.2341.\boxed{B_D/(k_BT)=21.2341}.nγ,>BD/nb≃η−1e−BD/(kBT).n_{\gamma,>B_D}/n_b\simeq\eta^{-1}e^{-B_D/(k_BT)}.

Physical check

The enormous photon-to-baryon ratio keeps a destructive high-energy tail even when the mean thermal scale is far below the binding energy. This is a scale estimate, not a reaction network.