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Sandbox Physics
The first minutes and the light elements

How baryon density and extra radiation move the light elements

Sandbox Physics ·

Near the standard point each light-element abundance responds to its inputs like a power law, and the exponents decide what each observation can measure. Deuterium scales as η−1.6\eta^{-1.6}, so the observed D/H fixes the baryon density to better than one percent. Helium barely responds to η\eta but grows by about 0.013 per extra neutrino-like species, so it bounds extra radiation. Nuclear rate errors add in quadrature: in the tutorial network they give 0.7 percent on D/H, set mainly by d(p,γ)3Hed(p,\gamma){}^3\mathrm{He}, and 1.8 percent on lithium-7, far too little to explain the factor of three between predicted and observed lithium.

This chapter builds on When neutrons and protons stop turning into each other and Why deuterium waits for the universe to cool. You need logarithms and the idea of a derivative as the ratio of two small changes. Standard deviations and adding independent errors in quadrature are explained where they are used. By the end you should be able to compute a logarithmic sensitivity from two runs, invert an observed abundance into a baryon density with an error bar, turn a helium measurement into a bound on extra radiation, and build an error budget from rate uncertainties.

Logarithmic sensitivity

Let XX be an abundance and pp an input: the baryon-to-photon ratio η\eta, the number of neutrino-like species NνN_\nu, the neutron lifetime or a reaction rate. The logarithmic sensitivity

SX,p=∂ln⁡X∂ln⁡pS_{X,p}=\frac{\partial\ln X}{\partial\ln p}

is the percentage change in XX per percent change in pp, all other inputs fixed. If SS is constant over a range, X∝pSX\propto p^{S} there. Sensitivities are convenient because they do not depend on units, and because they turn small fractional changes into a linear sum:

δXX≃∑pSX,p δpp.\frac{\delta X}{X}\simeq\sum_pS_{X,p}\,\frac{\delta p}{p}.

The method guide computes each SS with a central difference. It runs the network with the input multiplied by 1.011.01 and by 0.990.99, and divides the change in ln⁡X\ln X by ln⁡(1.01/0.99)\ln(1.01/0.99).

Worked example: the neutron lifetime from two runs

The saved runs at τn=860 s\tau_n=860\,\mathrm s and 900 s900\,\mathrm s give Yp=0.23861Y_p=0.23861 and 0.246750.24675. Then

SYp,τn≈ln⁡(0.24675/0.23861)ln⁡(900/860)=0.03350.0455=0.74.S_{Y_p,\tau_n}\approx\frac{\ln(0.24675/0.23861)}{\ln(900/860)}=\frac{0.0335}{0.0455}=0.74.

The 1 percent central difference gives 0.7380.738. Fields, Olive, Yeh and Young list 0.7290.729 in Table II of their 2019 preprint, computed with their full code at η10=6.129\eta_{10}=6.129.

Table 4.1 Logarithmic sensitivities at η10=6.129\eta_{10}=6.129 and τn=879.4 s\tau_n=879.4\,\mathrm s. The first number in each cell is the tutorial network, the second is Fields et al., Table II.
inputYpY_pD/H\mathrm{D/H}3He/H{}^3\mathrm{He/H}7Li/H{}^7\mathrm{Li/H}
η\eta0.0390.039 / 0.0390.039−1.618-1.618 / −1.597-1.597−0.597-0.597 / −0.583-0.5832.0582.058 / 2.0942.094
NνN_\nu0.1650.165 / 0.1630.1630.4000.400 / 0.3960.3960.1420.142 / 0.1390.139−0.269-0.269 / −0.280-0.280
τn\tau_n0.7380.738 / 0.7290.7290.4100.410 / 0.4090.4090.1460.146 / 0.1440.1440.4520.452 / 0.4340.434
d(p,γ)3Hed(p,\gamma){}^3\mathrm{He}0.0000.000 / 0.0000.000−0.333-0.333 / −0.315-0.3150.3760.376 / 0.3760.3760.5880.588 / 0.5890.589
3He(α,γ)7Be{}^3\mathrm{He}(\alpha,\gamma){}^7\mathrm{Be}0.0000.000 / 0.0000.0000.0000.000 / 0.0000.0000.0000.000 / 0.0000.0000.9630.963 / 0.9640.964

The agreement to within 0.040.04 is a check of the tutorial network, not new evidence: both calculations share the same physics, and the tutorial's rates are older. The table already says what each element is good for. Deuterium has the largest η\eta exponent. Helium has a tiny η\eta exponent but a sizeable one for NνN_\nu. Lithium-7 rises with η\eta, because at high density it is made as beryllium-7, which later captures an electron.

Weighing the baryons with deuterium

The 2025 Review of Particle Physics recommends, from twelve quasar absorption systems,

D/H∣p=(25.08±0.29)×10−6(Eq. 24.2).\mathrm{D/H}\big|_p=(25.08\pm0.29)\times10^{-6}\quad\text{(Eq. 24.2)}.

Worked example: invert the deuterium abundance

The network gives D/H=2.548×10−5\mathrm{D/H}=2.548\times10^{-5} at η10=6.12\eta_{10}=6.12. Treat the sensitivity as constant over the small step:

η10=6.12×(2.5482.508)1/1.618=6.12×1.0098=6.180.\eta_{10}=6.12\times\left(\frac{2.548}{2.508}\right)^{1/1.618}=6.12\times1.0098=6.180.

A direct bisection on the network gives 6.1806.180 as well. The error has two independent parts. The observational error is 0.29/25.08=1.160.29/25.08=1.16 percent of D/H, or 1.16/1.618=0.711.16/1.618=0.71 percent of η\eta. The rate error of the prediction, derived below, is 0.690.69 percent of D/H, or 0.430.43 percent of η\eta. Independent errors add in quadrature, as the square root of the sum of squares, because their variances add:

δηη=0.712+0.432  percent=0.83  percent,η10=6.18±0.05.\frac{\delta\eta}{\eta}=\sqrt{0.71^2+0.43^2}\;\text{percent}=0.83\;\text{percent},\qquad \eta_{10}=6.18\pm0.05.

The review gives (η10)SBBN=6.040±0.118(\eta_{10})_{\mathrm{SBBN}}=6.040\pm0.118 (Eq. 24.5), a range it describes as essentially determined by D/H, and 6.104±0.0586.104\pm0.058 from the microwave background without BBN. The tutorial value sits 1.2 standard deviations above the review's BBN value. The difference comes from the predicted D/H at a given η\eta, so from the nuclear rates: the tutorial keeps the 2004 Serpico fits, while the review's value comes from a later analysis that this chapter has not reproduced. Table 4.1 shows why the rates matter so much here: a 1 percent change in d(p,γ)3Hed(p,\gamma){}^3\mathrm{He} moves D/H by a third of a percent, and so the inferred η\eta by a fifth of a percent.

Bounding extra radiation with helium

Any extra relativistic energy, counted as ΔNν\Delta N_\nu additional neutrino species, speeds up the expansion. The network's SYp,Nν=0.165S_{Y_p,N_\nu}=0.165 at Nν=3N_\nu=3 converts into an absolute slope

dYpdNν=SYp,Nν YpNν=0.165×0.24243=0.0133,\frac{\mathrm dY_p}{\mathrm dN_\nu}=S_{Y_p,N_\nu}\,\frac{Y_p}{N_\nu}=0.165\times\frac{0.2424}{3}=0.0133,

in line with ΔYp≃0.013 ΔNν\Delta Y_p\simeq0.013\,\Delta N_\nu in the review. The direct run with ΔNν=1\Delta N_\nu=1 gives 0.254850.25485, a rise of 0.01250.0125; the slope falls slowly because the sensitivity is logarithmic in NνN_\nu.

Worked example: a one-measurement bound

The review's helium abundance is Yp=0.245±0.003Y_p=0.245\pm0.003 (Eq. 24.3). The full-physics prediction with three species and the microwave-background baryon density is Yp=0.24691±0.00018Y_p=0.24691\pm0.00018 (Fields et al., Eq. 22). Then

ΔNν=0.245−0.24690.0133=−0.14,σΔNν=0.0030.0133=0.23.\Delta N_\nu=\frac{0.245-0.2469}{0.0133}=-0.14,\qquad \sigma_{\Delta N_\nu}=\frac{0.003}{0.0133}=0.23.

The two-sided 95 percent range is roughly −0.59<ΔNν<0.30-0.59<\Delta N_\nu<0.30. One extra fully populated neutrino species lies about five standard deviations away. The review's joint analysis of helium and deuterium gives Nν=2.898±0.141N_\nu=2.898\pm0.141 (Eq. 24.9), which is ΔNν=−0.10±0.14\Delta N_\nu=-0.10\pm0.14.

Notice what would happen with the tutorial's own prediction, Yp=0.2424Y_p=0.2424. The same arithmetic gives ΔNν=+0.20\Delta N_\nu=+0.20. The shift of 0.340.34 species is entirely the weak-rate corrections the tutorial omits, discussed at the end of the first chapter. A bound on new physics is only as good as the absolute prediction it is compared with.

Propagating rate errors

Each nuclear rate fit comes with a one-standard-deviation fractional error σi=δfi/fi\sigma_i=\delta f_i/f_i that depends on temperature. To first order, and if the rate errors are independent,

(δXX)2=∑i(SX,i σi)2.\left(\frac{\delta X}{X}\right)^2=\sum_i\left(S_{X,i}\,\sigma_i\right)^2.

The method guide does slightly better than multiplying one number by one sensitivity: it shifts each rate by its full temperature-dependent error and reruns the network. The neutron lifetime is included with an assumed error of 0.5 s0.5\,\mathrm s.

Table 4.2 One-standard-deviation error budget at η10=6.12\eta_{10}=6.12, in percent. Only the largest contributions are listed.
sourceD/H\mathrm{D/H}7Li/H{}^7\mathrm{Li/H}
d(p,γ)3Hed(p,\gamma){}^3\mathrm{He}0.670.671.211.21
d(d,n)3Hed(d,n){}^3\mathrm{He}0.170.170.220.22
3He(α,γ)7Be{}^3\mathrm{He}(\alpha,\gamma){}^7\mathrm{Be}0.000.001.051.05
7Be(n,p)7Li{}^7\mathrm{Be}(n,p){}^7\mathrm{Li}0.000.000.400.40
all, in quadrature0.690.691.761.76

As a check that does not rely on the linear formula, the guide also draws 200 random sets of rates and repeats the calculation. The spread is 0.72 percent for D/H and 1.78 percent for lithium-7, close to the quadrature values.

The lithium column answers a question raised in the first article. The review quotes Li/H=(1.45±0.25)×10−10\mathrm{Li/H}=(1.45\pm0.25)\times10^{-10} for metal-poor halo stars (Eq. 24.4) and a BBN-plus-CMB prediction of (4.72±0.7)×10−10(4.72\pm0.7)\times10^{-10} (Eq. 24.10), a factor of 3.3 apart. With S=0.963S=0.963, closing that gap through 3He(α,γ)7Be{}^3\mathrm{He}(\alpha,\gamma){}^7\mathrm{Be} alone would need its rate to fall by a factor of about 3.31/0.963≈3.43.3^{1/0.963}\approx3.4. The fit's quoted error is 1 to 2 percent at the temperatures where beryllium-7 forms. A drop of about 70 percent is 35 to 70 times that error.

Lab task: weigh the baryons, then test the radiation

  1. Predict. Before moving anything, estimate from Table 4.1 how far you must move the baryon slider to change D/H by 10 percent, and how far the radiation slider to change YpY_p by 0.0030.003.
  2. Change. Drag the baryon slider until the deuterium curve crosses the centre of the observed band. Then return it to the default and drag the radiation slider until YpY_p reaches 0.2450.245.
  3. Compare. The first value should be close to η10=6.18\eta_{10}=6.18; the Lab's coarser time step can move it by about 0.010.01. For the second, compare your ΔNeff\Delta N_{\mathrm{eff}} with the +0.20+0.20 from the worked example.
  4. Explain. Explain why the radiation slider moves D/H as well as helium, and why the second reading cannot be quoted as a measurement of extra radiation.

Exercises

  1. Use Table 4.1 to predict the change in 7Li/H{}^7\mathrm{Li/H} when η10\eta_{10} goes from 6.126.12 to 1010. The saved run gives 1.09×10−91.09\times10^{-9} from 4.51×10−104.51\times10^{-10}. Why does the prediction miss?
  2. A future survey halves the error on D/H to 0.145×10−60.145\times10^{-6} while the rate errors stay the same. What is the new error on η10\eta_{10}, and which source of error would then dominate?
  3. Extra radiation also raises D/H. Use Table 4.1 to find the Δη/η\Delta\eta/\eta that cancels the D/H change from ΔNν=0.5\Delta N_\nu=0.5. Then find the resulting helium change. Explain why deuterium alone cannot measure both η\eta and NνN_\nu.
  4. Transfer. A colleague's code gives Yp=0.2440Y_p=0.2440 for the standard case. They compare it with Yp=0.245±0.003Y_p=0.245\pm0.003 and report evidence for ΔNν≈0.08\Delta N_\nu\approx0.08. List what you would check before believing the number.
Answers
  1. ln⁡(10/6.12)=0.491\ln(10/6.12)=0.491, and 2.058×0.491=1.0102.058\times0.491=1.010, so the predicted factor is e1.010=2.75\mathrm e^{1.010}=2.75 and Li/H≈1.24×10−9\mathrm{Li/H}\approx1.24\times10^{-9}. The run gives 1.09×10−91.09\times10^{-9}, a factor of 2.412.41. The sensitivity is a local slope. Over this range the average exponent is only ln⁡2.41/0.491=1.79\ln2.41/0.491=1.79, because the slope itself changes with η\eta: lithium-7 made directly falls while lithium made as beryllium-7 rises, and their mix keeps shifting.
  2. The observational part becomes 0.58/1.618=0.360.58/1.618=0.36 percent of η\eta, the rate part stays 0.430.43 percent, so δη/η=0.362+0.432=0.56\delta\eta/\eta=\sqrt{0.36^2+0.43^2}=0.56 percent, or ±0.034\pm0.034 in η10\eta_{10}. The nuclear rates, mainly d(p,γ)3Hed(p,\gamma){}^3\mathrm{He}, would then dominate.
  3. ΔNν=0.5\Delta N_\nu=0.5 is Δln⁡Nν=ln⁡(3.5/3)=0.154\Delta\ln N_\nu=\ln(3.5/3)=0.154; D/H changes by 0.400×0.154=0.0620.400\times0.154=0.062 in ln⁡\ln. Cancelling needs Δln⁡η=0.062/1.618=0.038\Delta\ln\eta=0.062/1.618=0.038, a 3.8 percent higher η\eta. Helium then changes by 0.165×0.154+0.039×0.038=0.0270.165\times0.154+0.039\times0.038=0.027 in ln⁡\ln, about +0.0066+0.0066. A higher η\eta with extra radiation can reproduce the same D/H, so D/H constrains a combination of the two. Helium, which mostly tracks NνN_\nu, or the microwave background, which measures η\eta independently, breaks the degeneracy.
  4. Check the inputs: the baryon density, the neutron lifetime and its error, and the rate set. Check the weak rates: are radiative and finite-mass corrections included? Without them a code sits about 0.0050.005 low, which by itself fakes ΔNν≈+0.3\Delta N_\nu\approx+0.3. Check whether YpY_p is the true mass fraction or four times the helium number fraction. Check numerical convergence by halving the time step. Finally, 0.08/0.230.08/0.23 is a third of a standard deviation: even with a perfect code that sits well inside the noise.

What these numbers do and do not show

All model numbers in this chapter come from the tutorial network and are saved in teaching-values.json and expected.json. They show how sensitivities and error budgets are built; they are not a new determination of η\eta or NνN_\nu. Published determinations use a fuller weak-rate treatment, newer rates and joint likelihoods with the microwave background. The guide Reading the BBN evidence on extra radiation and the lithium problem walks through those analyses table by table.

Try it in the Lab

Drag the baryon and radiation sliders one at a time and compare each curve with its observed band. Open the light-element network Lab

Next: Recompute a small network from a frozen rate grid and propagate rate errors

Sources

  1. Fields, Olive, Yeh and Young
  2. 2025 Review of Particle Physics

Keep reading

What this page links to directly

The first minutes and the light elements · Reading path

From the first second to the light elements

  1. ArticleHow temperature and expansion speed set helium and deuterium
  2. ArticleWhen neutrons and protons stop turning into each other
  3. ArticleWhy deuterium waits for the universe to cool
  4. You are hereHow baryon density and extra radiation move the light elements
  5. ArticleRecompute a small network from a frozen rate grid and propagate rate errors
  6. ArticleReading the BBN evidence on extra radiation and the lithium problem
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