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I'm studying the nuclear reactions involved in nucleosynthesis. As I understand it, the primary variables are the density of the particles, the cross section of the particle, and the average velocities. At around 1 MeV, we assume that the velocities are relativistic (e.g. $v\approx c$) and the cross section, as I understand it, is derived from quantum mechanics. However, I'm curious about pressure.

We know that pressure is a critical variable in chemical reactions. We also know that pressure was a great deal larger in the past universe than it is now. How do nuclear reactions change with pressure, if at all?

Quark Soup
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    At around 1MeV, electron velocities are relativistic, protons not really, and bigger nuclei even less so. As for any pressures obtainable on Earth, they make no difference. In a neutron star, well, yes they did. – Jon Custer Jan 06 '22 at 14:23
  • I mean...the sun's nuclear reactions can occur solely because of the pressure due to gravity. No pressure, no sun fusion. And larger stars = more gravity = more pressure = burn out faster, presumably from much faster fusion since larger stars should have even more fuel present than smaller stars. – DKNguyen Jan 07 '22 at 03:58
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    @DKNguyen, the pressure of the sun helps keep the density high and makes interactions more frequent, but it doesn't drive the reaction. To say the sun reactions are "solely due to the pressure" is incorrect. https://physics.stackexchange.com/questions/281082/nuclear-fusion-with-extremely-high-pressure-and-low-temperature – BowlOfRed Jan 07 '22 at 06:24

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Essentially all of chemistry is determined by interactions between the outermost electrons surrounding a nucleus. The size of that electron cloud is larger than the size of the nucleus it surrounds by about the same ratio as the size of a football field is to a pea on the 50-yard line.

It takes truly gigantic pressure to get two nuclei close enough to trigger any sort of nuclear reaction between them, which means that the pressure ranges which strongly affect chemical reactions are many power of ten too small to have any effect on nuclear reactions.

Furthermore, the energies required to achieve pressures which would essentially push all the electrons aside and force nuclei into physical contact are so great that the usual rules of chemistry simply do not apply any more in those conditions.

niels nielsen
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  • He is studying nucleosynthesis. Under the relevant conditions yes, pressure is important. – Dan Jan 09 '22 at 20:54
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Remember your ideal gas law.

$$ P V = n R T $$

You have the temperature and desnity as variables. Temperature is in the equation. Volume is just the size of the object you are considering. Density is just $ \rho =\frac{ n}{ V}$. So pressure is not independent. You can pick two of them. Which two you pick will depend on the context of the particular problem you are interested in.

Now in a relativistic system the equation of state will not be so simple as the ideal gas law. It will depend on many different things. For example, if you increase the temperature and density you may allow certain reactions to move forward, and your system may move over into a new phase. The value of $n$ may change radically, or the types of particles around may suddenly change. Or you may need to specify the density in terms of partial densities of each species of particle.

And if you are dealing with stellar masses (as you need to be if you are studying nucleosynthesis) you start to get things like gravity being very important. And in things like a supernova you get whacky things like shockwaves and drastic fluxes of neutrinos and so on. These all make the equation of state very complicated.

But in the usual case you will have some function of temperature and density that gives you the pressure. In some circumstances it might be convenient to choose pressure and temperature, for example. Or pressure and density. But you usually get to choose only two of the three.

Dan
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