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I Nuclear statistical equilibrium

  1. Oct 4, 2017 #1
    Sorry, I have never found what does it mean Nuclear statistical equilibrium. It is used in any text but exact explanation nowhere.
    Please explain a physical meaning of it.

    Thank you.
  2. jcsd
  3. Oct 4, 2017 #2


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    Hm, although working in nuclear physics, I've never heard the expression "nuclear statistical equilibrium". Do you have a concrete reference, where the term is used? There is a statistical hadronization model in relativistic heavy-ion collisions. Maybe you are refering to that?
  4. Oct 4, 2017 #3
    I deal with this paper... http://adsabs.harvard.edu/abs/1991NuPhA.535..331L
    I don't know if is possible to access online, but... This is stated in the introduction....for example
    " upload_2017-10-4_16-51-39.png

    Attached Files:

  5. Oct 4, 2017 #4


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    I see. This paper is about dense (and not too hot) nuclear matter in thermal equilibrium as it occurs in neutron stars. To understand neutron stars you need to know the equation of state, i.e., the relation between pressure and energy density. That's what this paper is about.

    Note that there has been a lot of more research about this still very interesting issue!
  6. Oct 4, 2017 #5
    Oki I know it is about EoS in neutron star. But I'm not sure how to explain "nuclear statistical equilibrium". Does it mean that the reactions are in equilibrium? Why statistical. I understand thermodynamical equilibrium. What is the difference?
  7. Oct 23, 2017 #6


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    Nuclear equilibrium means the same as in thermodynamics (thermodynamics for chemists, because particles can decay or scape/dissapear as neutrinos).

    So, in a neutron star (beta-stable matter) you have
    mu_n = mu_p + m_e

    Inside a nucleus (beta-stable) you have
    mu_n = mu_p (i
    f it is not beta-stable, the chemical potentials will be different and it will try to decay to beta-stability)

    In white dwarfs, electrons keep the pressure and nuclei the mass. If you want to compute the partition function of electrons you can do it normally, taking into account the electrons are relativistic.

    The EoS is just the next step (application in astrophysics) of the research in theoretical models of nuclear force. As far as I know, the flow is something like:
    -Model of nuclear force interaction (effective forces)
    -Refine the model with experimental data (3000 isotopes sounds good, but not enough)
    -Derive EoS
    -Astrophysical part

    You can find more information about the nuclear statistical equilibrium in books about "nuclear astrophysics", in the section about compact objects. Sorry, at this moment I don't remember any author/title.

  8. Oct 24, 2017 #7


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  9. Oct 24, 2017 #8


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    Very nice indeed!
    Thank you :)

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