Early universe: equilibrium before nucleosynthesis

[Davide82]

Hi.

First of all I apologize because I already posted this topic in the "Homework & Coursework Questions > Advanced Physics" but since it exquisitely concerns astrophysics and it is not getting many answers, I believe it's better to post it here. If a moderator wants to merge the two conversation, I would be glad.


I am studying the evolution of the universe.
In particular, I am reading the history of the universe happening just under a temperature of 100 MeV.
At this time, it is said that neutrons and protons are present along with some other particles: electrons, positrons, photons, neutrinos and anti-neutrinos.
They say the reactions which maintain kinetic equilibrium are:
e^- + e^+ \longleftrightarrow \gamma +\gamma
e^\pm + \gamma \longleftrightarrow e^\pm + \gamma
while the reactions which are responsible for both kinetic and chemical equilibrium are:
e^- + e^+ \longleftrightarrow \nu + \bar\nu
\nu + e^- \longleftrightarrow e^- + \nu
n + \nu_e \longleftrightarrow p + e^-
n + e^+ \longleftrightarrow p + \bar\nu_e
n \longleftrightarrow p + e^- + \bar\nu_e
I am wondering why reactions such as:
p + e^+ \longleftrightarrow p + e^+
p + e^- \longleftrightarrow n + \nu_e + e^+ + e^-
are not taken into account.

 

[Drakkith]

How could a proton and an electron have a reaction that produced a proton and an electron??

Also, if i had to guess i would say that the reverse beta decay reaction doesn't happen as much because it is moderated by the weak force instead of the electromagnetic or strong forces. Just a guess though.

 

[Davide82]

How could a proton and an electron have a reaction that produced a proton and an electron??

I believe it is perfectly legal and it is simply a momentum/energy exchange which maintains kinetic equilibrium. I think it is like compton scattering, where a photon and an electron hit and particles type are the same incoming and outcoming.

 

[Drakkith]

I believe it is perfectly legal and it is simply a momentum/energy exchange which maintains kinetic equilibrium. I think it is like compton scattering, where a photon and an electron hit and particles type are the same incoming and outcoming.

I could see that. I would say that the transfer of kinetic energy from a moving proton to an electron or vice versa probably isn't the kind of "reaction" they meant. But that's just a guess.

 

[Davide82]

But they in the very beginning are writing a similar reaction:

e^\pm + \gamma \longleftrightarrow e^\pm + \gamma

So I believe these kind of reactions are considered, too.

 

[Drakkith]

But they in the very beginning are writing a similar reaction:


So I believe these kind of reactions are considered, too.

Ah, ok. Now I understand the confusion. And have joined in too!

 

[twofish-quant]

The reason that p+e^+ -> p + e^+ isn't considered is because of the mass difference between positrons and protons.

The mass between the proton and the positron is so large that it's an elastic collision and there isn't any kinetic energy change. You can think of it as bouncing a ball off a mountain. The momentum changes, but the energy doesn't. By contrast, electrons are light. If you hit the electron with a photon or neutrino there is energy transfer.

(I edited the following a few times.)

The second equation can be seen as the combination of the equations that did get listed. Take the first equation in the first section and add it to the third equation in the second.

 

[Davide82]

Thank you twofish-quant.
I understand your point with elastic collision. It didn't come to my mind.

But, regarding the second equation, I don't understand well what you mean...
because of course the processes are related:
for example the two processes:
e^- + e^+ \longleftrightarrow \nu + \bar\nu
\nu + e^- \longleftrightarrow e^- + \nu
have the same interaction vertex, as far as I know, because you can take an antiparticle to the other side of the reaction (this means temporal inversion).
But the two precesses, if I am not mistaken, are different due to the phase space of the final states... This is why we are considering both.
I never heard about this "additivity" of reactions that you are telling about.

 

[twofish-quant]

What we have is something different. There are two reactions

p + e- <-> n + ve

gamma + gamma <-> e+ + e-

The reaction rate for

p + e- + gamma + gamma <-> n + ve + e+ + e-

Can be decomposed into the reaction rates for those two processes.

 

[Davide82]

Ok. I am understanding.