A Summing Up — Question 5

Section 44-14 · Physics pages 7560–7561

Question 5 Extra relativistic species shift a weak freeze-out temperature Assume weak rate Γ ∝T 5 and expansion rate H ∝√g∗T 2. All coefficients are fixed. Effective relativistic degrees of freedom rise from 10.75 to 14.25. ln T ln(rate) Γ larger H Schematic; labels, not drawing lengths, specify the data. (a) Derive how freeze-out temperature depends on gstar and find its fractional increase. (b) Using n/p = e−∆/Tf , derive the sign of the neutron-ratio change and explain the direction of the approximate helium response.
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Question 5 - Solution Extra relativistic species shift a weak freeze-out temperature ln T ln(rate) Γ larger H (a) Method and calculation T 5 f ∝√g∗T 2 f ⇒Tf ∝g1/6 ∗ . T ′ f/Tf = (14.25/10.75)1/6 ≃1.0481 . (b) Interpretation and further calculation d ln(n/p) dTf = ∆/T 2 f > 0. Y ≃ 2r 1 + r, dY dr = 2 (1 + r)2 > 0 ⇒Y increases . Physical check The sign estimate holds other inputs fixed and omits changes in elapsed neutron-decay time. Precision abundances require coupled expansion and nuclear kinetics.

Question 5

Extra relativistic species shift a weak freeze-out temperature

Assume weak rate Γ∝T5\Gamma\propto T^5 and expansion rate H∝g*T2H\propto\sqrt{g_*}T^2. All coefficients are fixed. Effective relativistic degrees of freedom rise from 10.75 to 14.25.

Diagram shown in the original SVG worksheet.

  1. Derive how freeze-out temperature depends on gstar and find its fractional increase.

  2. Using n/p=e−Δ/Tfn/p=e^{-\Delta/T_f}, derive the sign of the neutron-ratio change and explain the direction of the approximate helium response.

Question 5 - Solution

Extra relativistic species shift a weak freeze-out temperature

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Tf5∝g*Tf2⇒Tf∝g*1/6.T_f^5\propto\sqrt{g_*}T_f^2\Rightarrow\boxed{T_f\propto g_*^{1/6}}.Tf′/Tf=(14.25/10.75)1/6≃1.0481.\boxed{T_f^\prime/T_f=(14.25/10.75)^{1/6}\simeq1.0481}.

(b) Interpretation and further calculation

dln⁡(n/p)dTf=Δ/Tf2>0.\frac{d\ln(n/p)}{dT_f}=\Delta/T_f^2>0.Y≃2r1+r,dYdr=2(1+r)2>0⇒Y increases.Y\simeq\frac{2r}{1+r},\quad\frac{dY}{dr}=\frac2{(1+r)^2}>0\Rightarrow\boxed{Y\text{ increases}}.

Physical check

The sign estimate holds other inputs fixed and omits changes in elapsed neutron-decay time. Precision abundances require coupled expansion and nuclear kinetics.