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I have read this question:

Over the next 380,000 years, the universe gradually cooled down enough for the sub-atomic particles to condense and form the first Hydrogen atoms

After the big bang

As far as I understand, heavier elements are formed in the core of super-heavy stars, where there is extreme pressure.

enter image description here

https://en.wikipedia.org/wiki/Stellar_nucleosynthesis

But the answer says that the early universe was too hot (and I guess this means too dense and too much pressure) for atoms to form.

So basically what I am asking is, why was the pressure in the early universe conceptually different (hindering atoms from forming) from the extreme pressure required at the core of the stars to form atoms (which on the other hand was the reason for heavier atoms to form)?

You need pressure to form atoms, and more pressure to form heavier atoms, but too much pressure means no atom can form. Why? What happens above certain energy (pressure) levels that hinders atoms from forming, when the very driving force of atom formation is pressure itself?

Question:

  1. If for heavier atoms to form, you need a lot of pressure, then why do we say that the early universe was too hot for atoms to form?
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    $\begingroup$ Because pressure and temperature are different things? $\endgroup$
    – Jon Custer
    Aug 26, 2021 at 23:46

2 Answers 2

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TL;DR: Your question should really be asking why Big Bang Nucleosynthesis (BBN) does not produce heavier nuclei (not about the formation of atoms). There's no fundamental reason BBN couldn't produce nuclei up to Iron, if the Universe stayed at a sufficiently high temperature for long enough; but since BBN only lasted for 3 minutes, it could only get so far up the periodic table. Stars get up to Iron because they have so much more time; also, metallicity increases over the history of the Universe so later generations of stars get a "head start" over earlier ones). To get beyond Iron, you need an extremely violent non-equilibrium process, which you find in supernovae and neutron-star mergers.


There are quite a few things to unpack here.

First, let's take pressure out of the game, since I think it is just confusing things, and focus on temperature.

Second, there is a difference between the formation of atoms and the formation of nuclei. The quote you give is about the formation of atoms, but the image you show is about the formation of nuclei.

Before going further, let's define some terms. The formation of atoms (in cosmology) is called recombination. I think the reasoning behind this name is that electrons and protons are "combining" to form atoms. This is a bad name since atoms had not previously formed in the Universe, so nothing is recombining, but we are stuck with the terminology. Meanwhile, the formation of nuclei is called nucleosynthesis. Nucleosynthesis can be further broken down into the region of spacetime where it occurs: primordial nucleosynthesis or Big Bang Nucleosynthesis (BBN), which happened everywhere in the very early Universe (about 10 seconds to three minutes after the Big Bang); stellar nucleosynthesis occurs in the cores of stars when the Universe is sufficiently old that stars are able to form; and then what I'm going to call nuclear capture nucleosynthesis as a grab bag of other processes; the $s$-process ("slow" process) which occurs in supernovae, and the $r$-process ("rapid" process) which is now believed to occur in the collisions of neutron stars -- these processes happen as some stars reach the end of their lifetime.

The reason that recombination cannot happen until the Universe has cooled to a sufficiently low temperature, is that Hydrogen has a binding energy of $13.6\ {\rm eV}$. If a typical photon in the primordial plasma has enough energy to ionize Hydrogen, then Hydrogen won't form (I'm focusing on Hydrogen since the majority of the nuclei were Hydrogen at this stage). You might think that recombination would happen at a temperature of $kT=13.6\ {\rm eV}$, but in fact the recombination was around $3000\ {\rm K}$, for which $kT=0.26\ {\rm eV}$! The reason is that the number of photons in the Universe vastly exceeds the number of baryons, so for recombination to occur the temperature needs to cool down enough that the ionization of a Hydrogen atom is extremely rare, accounting for the enormous number of photons per each Hydrogen nucleus.

I think your question is also asking about nucleosynthesis. This story is quite involved; here I will just try to summarize a few of the main points.

First, BBN took place over the course of about 3 minutes (hence Weinberg's famous book, "The First Three Minutes"); it begins when free neutrons and protons can form, and ends when the rate of nuclear reactions becomes negligibly small. This timescale is set by the Universe's expansion rate; as the Universe expands, it cools, and over the course of three minutes it cools enough that nuclear fusion processes are no longer thermodynamically favorable. The time scale is crucial; BBN reaches several "bottleneck" processes which are very rare, and so have a very small rate, that essentially stop BBN from producing nuclei heavier than Lithium. On the other hand a main sequence star's lifetime is about 10 billion years. So there is much more time for nucleosynthesis to occur, and generate heavier elements, than in BBN.

Second, the conditions of BBN and stellar nucleosynthesis are quite different. At the start of BBN, the Universe consisted of a gas of protons and neutrons that had to fuse into Hydrogen, and work up to heavier elements one step at a time. Meanwhile, there have been a few (2-3) generations of stars; each generation undergoes nuclear fusion and populates the interstellar medium with heavy elements. These heavy elements in turn fuse future generations of stars, which are not "starting from scratch".

Finally, even in stellar nucleosynthesis, there is an upper limit on what fusion can do. Nuclei heaver than iron have a lower binding energy than iron, and so in thermal equilibrium will tend to disassociate and become iron. Therefore, non-equilibrium processes need to come into play to generate nuclei heavier than iron. This is the role of the $s$-process and $r$-process I mentioned earlier. In a violent event such as a supernova or a neutron star merger, nuclei that are ejected with very high energies will fuse, form stable heavy nuclei, and leave the cataclysm from whence they came to be distributed safely in the cosmos. These processes are responsible for the elements colored purple, green, and yellow on your chart.

To summarize:

  • BBN takes place approximately 10 seconds to 3 minutes after the big bang, and takes the free protons and neutrons from the early Universe and converts them into Hydrogen and Helium nuclei, with some trace amounts of nuclei as heavy as Lithium.
  • Recombination takes place much later, when the Universe is about 380,000 years old, and refers to the process where Hydrogen atoms are able to form, because the Universe has cooled to the point where photons in the cosmic soup do not have enough energy to ionize Hydrogen.
  • Stellar nucleosynthesis, over several generations of stars, takes the primordial Hydrogen and Helium and produces heavier nuclei, up to Iron.
  • Non-equilibrium neutron capture processes, such as the $s$-process and $r$-process, take place in cataclysmic events like supernovae and neutron star mergers, and are responsible for producing nuclei heavier than iron.
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  • $\begingroup$ Thank you so much! Can you please elaborate on this one "to the point where photons in the cosmic soup do not have enough energy to ionize Hydrogen."? $\endgroup$ Aug 27, 2021 at 4:06
  • $\begingroup$ @ÁrpádSzendrei Here's an order of magnitude estimate. The key parameter is the baryon-to-photon ratio, $\eta \sim 5 \times 10^{-10}$ (this is very small! -- showing that there are many more photons than baryons). We would naively guess the temperature of the Universe at recombination should be $13.6\ {\rm eV}$. In fact, accounting for the photon to baryon ratio, we instead guess $-13.6\ {\rm eV}/\ln \eta \approx 0.63\ {\rm eV}$, which is within a factor of a few of the number I quoted in the answer. $\endgroup$
    – Andrew
    Aug 27, 2021 at 4:44
  • $\begingroup$ A slightly more careful calculation involves the use of the Saha equation; and at an even higher level of sophistication there are full-blown simulations that evolve all the degrees of freedom in the primordial plasma. $\endgroup$
    – Andrew
    Aug 27, 2021 at 12:23
  • $\begingroup$ The factor $\ln \eta$ from the order of magnitude estimate can be understood like this. For each Hydrogen atom, there are $1/\eta$ photons. We want each Hydrogen to interact with, on average, fewer than one photon with enough energy to ionize it. The probability that a photon has enough energy to ionize a Hydrogen atom is approximately proportional to $e^{-(13.6\ {\rm eV})/kT}$. Setting $\eta^{-1} e^{-(13.6\ {\rm eV})/kT}=1$ and solving for $kT$ yields the estimate. $\endgroup$
    – Andrew
    Aug 27, 2021 at 12:40
  • $\begingroup$ Thank you, can you please elaborate on this :"it begins when free neutrons and protons can form", why were they not able to form before? $\endgroup$ Aug 27, 2021 at 17:29
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In thermodynamic terms, which is why temperature is used as an axis for displaying the Big Bang, one can model the universe as a black body, at least by the time it reaches proton formation.

histunivbb

The T on the y axis is the temperature describing this black body radiation

Black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, emitted by a black body (an idealized opaque, non-reflective body). It has a specific spectrum of wavelengths, inversely related to intensity that depend only on the body's temperature, which is assumed for the sake of calculations and theory to be uniform and constant

The radiation , as seen here, in spectra for stars depends on the temperature, as seenin this plot for temperatures of stars, where nucleo-synthesis has high probability :

enter image description here

But in the Big Bang, there are exponential temperatures, as seen above. This means that the energy and momentum of the constituent particles is such, the wavelengths so short, that continuous scatterings (interactions at high energy in the soup of particles and radiation of the black body), destroy any bound states that might form with some probability in the tails of the black body distribution. To form nuclear bound states one needs energies and momenta of the order of mev, nuclear bound state. Only when the BB temperature reaches the star temperatures one can start comparing, and then the continuous expansion does not have enough density for creating more complicated atoms. It is then the role of gravitational attraction of the light nucleons that will create gravitational wells with enough strength for synthesizing heavier nuclei.

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  • $\begingroup$ Thank you so much! Can you please elaborate on this: "the energy of the constituent particles is such that continuous scatterings destroy any bound states that might form"? $\endgroup$ Aug 27, 2021 at 4:41

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