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Boltzmann equation for the early Universe

  1. Dec 30, 2014 #1
    I'm currently reading about the Boltzmann equation, used for the early Universe.
    The equation I end up with, after some simplifications is the following:

    \begin{equation}
    a^{-3}\frac{d}{dt}\left(n_1a^3\right) = n_1^{(0)}n_2^{(0)}\langle\sigma v\rangle\left[\frac{n_3 n_4}{n_3^{(0)}n_4^{(0)}} - \frac{n_1 n_2}{n_1^{(0)}n_2^{(0)}}\right]
    \end{equation}

    My problem is, I'm not actually sure what it tells me? Does it just tell me how species 1 evolves with time, when I'm taking interactions into account, or...?
    I mean, it's used for non-equilibrium phenomenon, but does it tell me whether or not some particle is in, or out of equilibrium? Basically, I'm not quite sure what it tells me. So I was hoping someone could give a brief and clear explanation perhaps?


    Thanks in advance.
     
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  3. Dec 30, 2014 #2

    George Jones

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  4. Dec 30, 2014 #3
    Ahhh, a quite good note actually. A bit better explained than in Dodelson :)
    I do have a couple of questions if that is okay?

    1. The left-hand side of my equation tells me how the number density of the chosen particle evolves with time. So if the right-hand side is equal to zero (As is the case with the Friedmann equation/Fluid equation), well, then we have equilibrium. So by putting the collision terms on the right-hand side, i.e. "creating" the Boltzmann equation, they are no longer in equilibrium, right? So it just tells me how it evolves in time with respect to the creation and annihilation rates in my reaction? Or am I missing something? So, when you enforce the Saha equation, making the bracket (again in my equation above) zero, well, you get zero on the right-hand side, and then I'm at equilibrium again, or...?

    2. The above leads to my next question: So in order for me to know whether a particle is in equilibrium or not, I just have to see if the left-hand side of the Boltzmann equation is equal to zero (i.e. Saha equation), and if not, then it is out of equilibrium?

    Sorry for the question in advance, but that I didn't get from the notes, or Dodelson. So I'm hoping it's okay :)
     
  5. Jan 1, 2015 #4
    Given that I haven't seen this before, but also that it is a holiday and long response times [and that I am procrastinating from a course ;)], let me try. Let the reader beware of the result...

    1. "So by putting the collision terms on the right-hand side, i.e. "creating" the Boltzmann equation, they are no longer in equilibrium, right?"

    No, the notes uses an annihilation process as demonstration, and it can be put in equilibrium with as many i = 1, 2, ... 4 particles annihilated as created. That is explicitly done in eq. (3.3.84).

    But I think you know that, and put it poorly/I am reading poorly.

    2. "So in order for me to know whether a particle is in equilibrium or not, I just have to see if the left-hand side of the Boltzmann equation is equal to zero (i.e. Saha equation), and if not, then it is out of equilibrium?"

    Yes, for the particle species _under that process_.

    Note how they go on to discuss non-equilibrium freeze-out, the point of the exercise. It happens as the reaction rate slows under dilution so that the process becomes ineffective: "the r.h.s. of (3.3.86) gets suppressed and the comoving density of particles approaches a constant relic density". But the system (the universe) is in non-equilibrium after decoupling. (I kind of like that, seeing how it gets us an interesting physics with stars and planets and New Year's headaches...)

    So we can't look at a known differential of a species and assume it tells us everything about the system. We need to know the dominant processes the species partake in.

    But again, I think you know that.
     
  6. Jan 1, 2015 #5

    ChrisVer

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    Looking at the equation, it deals with [itex]1+2 \rightarrow 3 + 4 [/itex] right?
    I'll write [itex]N_i = a^3 n_i [/itex] (volume * density)
    In general you see how [itex]1[/itex] (or [itex]2[/itex]) changes with time indeed. If you are at equilibrium, this means that as many 3,4 you create, so many 3,4 are annihilated and 1,2 are created too (both side rates are equal). At equilibrium [itex]N_i = N_i^{(0)}[/itex] (the particle numbers don't change - are constant/stable) and the quantity in the bracket is zero , as it should be (that's equilibrium).
    This equation tells you how,after equilibrium, your [itex]1[/itex] changes.

    Yes one way to say that you are at equilibrium is to write:
    [itex]\frac{dN}{dt}=0[/itex].
    If [itex]\frac{dN}{dt} \ne 0 [/itex] then it means that the number of the particle in consideration doesn't remain the same, so there are non-equilibrium processes that lead it to change.
     
    Last edited: Jan 1, 2015
  7. Jan 3, 2015 #6
    Again, I'm not sure if there is some language difficulty on either side, or if I'm too strict/dumb.

    Yes, if we measure the total differential for a particle species, we can (by definition) see whether or not there is an increase/decrease and a lack of equilibrium. But the equation doesn't describe that, it pertains to a specific process that we attempt to model.

    So whether it "deals with" (describes) equilibrium is a matter of scope. Yes, if the model includes all the dominant processes. No, if it is misses one or more processes* that can take it out of equilibrium.

    * Or constraints - but as it happens, one can identify freeze-out, the point of the exercise.
     
  8. Jan 3, 2015 #7

    ChrisVer

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    But Boltzmann equation is not complete until you write all the set of the dynamical equations that govern/dominate the process...
    In your model as I said you take the [itex]1+2 \leftrightarrow 3+4[/itex]...my left-right arrow doesn't necessarily mean equilibrium...just that the process can go both ways at some rate...
    If you had some other additional process let's say [itex]X \leftrightarrow 1+Y[/itex] , then you have to put additional terms in the RHS of the equation you gave, to take into account the raise of the 1's coming from this process... except for if you can "drop them away" = neglect them...
    As long as there is an overall equilibrium, 1+2 and 3+4 won't change the number of 1s, neither will X's decay do that...

    the names/terms used in the Boltzman equations for the derivative to be zero or not is collisionless or not respectively...In the collisionless case, the particles evolve "freely" and their number densities scale with the scaling of the universe.
     
    Last edited: Jan 3, 2015
  9. Jan 3, 2015 #8
    Ah, yes, thanks! You are correct, the equation should have all terms so you can predict thermodynamic quantities and conservation laws, because it is based on probabilities of finding particles. [ http://en.wikipedia.org/wiki/Boltzmann_equation ]

    I used to know this... Worse, I didn't consider the physical context, my bad.

    Whether an example is complete is another question, which is what I was trying to address. Sorry for the confusion! :s
     
  10. Jan 3, 2015 #9

    ChrisVer

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    To me it sounds really fun that Boltzmann equations (something that is totally classical) is used in such a way when describing particle interactions (quantum mechanics). o0)
     
  11. Jan 4, 2015 #10
    hehe, thank you for the answers :)
    It really made it a bit easier to understand.
     
  12. Jan 4, 2015 #11
    Statistical physics _is_ fun!:w
     
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