5
$\begingroup$

This question is related to the Big Bang Nucleosynthesis of light elements, more concretely I do not understand some features of the graph

BBabundances

  • Why do the $^3$He and D abundances go down with increasing baryon to photon fraction while the Li abundance goes up?

  • And what is that strange dip in the Li abundance?

$\endgroup$

2 Answers 2

3
$\begingroup$

Note first of all that the deuterium and $^3$He are three order of magnitude smaller than $^4$He. For $^7$Li the difference is about nine order of magnitude! The features in these elements are therefore enhanced compared to $^4$He.

An important picture to keep in mind in order to understand what is going on is the following describing the reactions that are going on during BBN:

Reactions

The main features of the graph are these:

  1. In contrast to the lighter nuclei the abundance of $^4$He is relatively insensitive to the baryon density. The abundance of $^4$He increases slightly with the baryon to photon ratio $\eta$. This is because at the onset virtually all available neutron are being locked into $^4$He, its abundance depends critically on the competition between the weak interaction rates and the expansion rate of the universe. The higher the nucleon density, the earlier the Deuterium bottleneck$^\dagger$ can be breached. Since at earlier times there are more neutrons, more $^4$He will be made. This result in a very slow increase with $\eta$.

  2. On the other hand D and $^3$He abundances decrease with baryon-to-photon ratio. This is related to the previous point: at a higher the nucleon density more D and $^3$He are burned into $^4$He, leaving behind fewer of these elements.

  3. The behavior of $^7$Li is trickier. At lower $\eta$ values this element is synthesized by the $^3$H$(\alpha,\gamma)^7$Li reactions, but they are easily destroyed in collisions with protons. Hence, with increasing baryon-to-photon ratio (keeping it however in the low regime $\eta<3$ ) the destruction of $^7$Li goes faster and the abundance decreases. At higher abundance however, about $\eta>3$, the lithium is largely indirectly synthesized as berilium. This mass-7 element is made via the reaction $^3$He$(\alpha,\gamma)^7$Be. Since $^7$Be is more tightly bound it is harder to destroy than $^7$Li and the abundance of $^7$Be increases. Later, when the universe is cooler and neutral atoms can begin to form, $^7$Be will capture an electron and decay to $^7$Li.

The dip in between is caused by the destruction reaction involving proton capture and decay into two $^4$He nuclei.

$\dagger$ Nucleosynthesis can only start once the binding energy of D (which has the lowest binding energy of about 2.2 MeV) can be breached. All other nuclei requires 3-body collision to be formed if D is not yet available. Once D is formed they are immediately processed into $^4$He.

Note that the Lithium abundance observed, unlike that of the lighter nuclei, by direct observation does not agrees with the theoretical prediction inferred from the baryon density measured from the CMB.

graph

Source: Big Bang Nucleosynthesis: Probing the First 20 Minutes by G. Steigman

$\endgroup$
2
$\begingroup$

I found this link for the plot.

The predicted abundance of deuterium, helium and lithium depends on the density of ordinary matter in the early universe, as shown in the figure at left. These results indicate that the yield of helium is relatively insensitive to the abundance of ordinary matter, above a certain threshold. We generically expect about 24% of the ordinary matter in the universe to be helium produced in the Big Bang. This is in very good agreement with observations and is another major triumph for the Big Bang theory.

However, the Big Bang model can be tested further. Given a precise measurement of the abundance of ordinary matter, the predicted abundances of the other light elements becomes highly constrained. The WMAP satellite is able to directly measure the ordinary matter density and finds a value of 4.6% (±0.2%), indicated by the vertical red line in the graph.

abundancenucleosynthesis

This leads to predicted abundances shown by the circles in the graph, which are in good agreement with observed abundances.

So it is evident that the curves are results of theoretical calculations, validated by satellite measurements at the circles.

Now the answer to "why" becomes "because that is what the theoretical calculation predicts" and it is important that it be tested.

This is an important and detailed test of nucleosynthesis and is further evidence in support of the Big Bang theory. Had the results been in conflict, it would point to 1) errors in the data, 2) an incomplete understanding of the process of Big Bang nucleosynthesis, 3) a misunderstanding of the mechanisms that produce fluctuations in the microwave background radiation, or 4) a more fundamental problem with the Big Bang theory.

One can hand wave, looking at the way the reactions go :

One second after the Big Bang, the temperature of the universe was roughly 10 billion degrees and was filled with a sea of neutrons, protons, electrons, anti-electrons (positrons), photons and neutrinos. As the universe cooled, the neutrons either decayed into protons and electrons or combined with protons to make deuterium (an isotope of hydrogen). During the first three minutes of the universe, most of the deuterium combined to make helium. Trace amounts of lithium were also produced at this time.

From this plot

creation of nuclei

Why do the 3He and D abundances go down with increasing baryon to photon fraction while the Li abundance goes up?

I would guess because protons/neutrons are tied up increasingly into He so less deuterium and Tritium can be fused out of them.

And what is that strange dip in the Li abundance?

We see that Lithium is the end product of either decay of the unstable 7Be or the fusion of tritium with 4He. With competing chains of build up it is not out of the ordinary to see a dip in the 7Li, from the diminishing of one path (tritium, see a) , as helium is constant) while the other takes over( the scale is logarithmic, the slow increase of He4 could do it) .

In any case, one needs to have the exact model to check the individual behaviors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.