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Boltzman equation on cosmology

  1. May 22, 2013 #1
    I need integrate "over heavy momenta" the following equation
    \sigma v>=
    [tex]<\sigma v>= \frac{1}{n^{(0)}_1n^{(0}}_2 \int \frac{d^3p_1}{(2Pi)^3 2 E_1} \int \frac{d^3p_2}{(2Pi)^3 2 E_2} \int \frac{d^3p_3 }{(2Pi)^3 2 E_3} \int \frac{d^3p_4}{(2Pi)^3 2 E_4} e^{-(E_1+E_2)/T} \\

    \cdot (2 Pi)^4 \delta^3 (p_1+p_2-p_3-p_4) \delta (E_1+E_2-E_3-E_4) M^2[/tex]

    where [tex]n+ \nu \rightarrow p+ e^-[/tex]

    I need to find that_

    [tex]n_{\nu}^{(0)}< \sigma v>= \frac{Pi}{4m^2}\int \frac{d^3p_\nu}{(2Pi)^3 2 p_\nu} e^{-p_\nu/T}\int \frac{d^3p_e}{(2Pi)^3 2 p_e} \delta (Q-p_\nu-p_e) M^2 [/tex]

    where [tex]Q=m_n-m_p[/tex]

    I tried very ways, for example using [tex]\delta^3 (p_1+p_2+p_3+p_4)= \delta(p_1)\delta(p_2)\delta(p_3)\delta(p_4) [/tex] but i have very problems with the integration and with the propierties of Dirac Delta...

    Help please
     
    Last edited: May 22, 2013
  2. jcsd
  3. May 23, 2013 #2
    You are confusing your notation here. pn are not different components of a 4-vector, they are 3-vectors corresponding to different particle species.


    Now, let's start from the beginning. What are Ep, En, Ev and Ee? Replace them in your equation, and you should only have particle momenta left. Then consider integrating.
     
    Last edited: May 23, 2013
  4. May 23, 2013 #3
    i think that if the particle is heavym (proton and neutron)

    [tex]E=\sqrt{m^2+p^2} = m\sqrt{1+\frac{p^2}{m^2}} \approx m[/tex]

    [tex]\delta^3 (p_1+p_2-p_3-p_4) \delta (m_n+E_\nu-m_p-E_e) M^2[/tex]

    [tex]\int \frac{d^3p_e}{(2Pi)^3 2 E_e} e^{-(E_n+E_\nu)/T} \delta^3 (p_n+p_\nu-p_p-p_e) \delta (Q+E_\nu-E_e) M^2[/tex]

    but ¿how i use de Dirac Delta? ¿what is the propierty?
     
  5. May 23, 2013 #4

    DEvens

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  6. May 23, 2013 #5
  7. May 23, 2013 #6
    That's good! And also for the relativistic particles, Ee = pe and same for neutrino.

    Now you notice that your integrand does not depend on the heavy momenta at all. So let's look at the standard delta function identity
    [tex] \int d^3 x \delta^3(x-a) = 1 [/tex] if a is in the integral domain. Use this to get rid of one of the integrals (it's really that easy!).

    Then you're left with another integral over a heavy momentum, but the integrand still doesn't depend on the integral variable. This integral looks a lot like a phase space number density, so maybe you can find a nice identity for that one somewhere from your material.
     
  8. May 23, 2013 #7
    very thanks, but if Ee = pe then my equation is


    [tex]\int \frac{d^3p_e}{(2Pi)^3 2 p_e} e^{-(E_n+E_\nu)/T} \delta^3 (p_n+p_\nu-p_p-p_e) \delta (Q+p_\nu-p_e) M^2[/tex]

    but, the integrand [tex] \frac{1}{p_e}[/tex] itself depends on the heavy momentum pe..

    then [tex]\int \frac{d^3p_e}{(2Pi)^3 2 p_e} \delta^3 (p_n+p_\nu-p_p-p_e) \neq 1 [/tex]

    ¿ how i use [tex]\int d^3 x \delta^3(x-a) = 1 [/tex] ?
     
  9. May 24, 2013 #8
    Assume that proton and neutron are heavy, p << m, and that electron and neutrino are light, p >> m.
     
  10. May 24, 2013 #9
    thank, but my problem is the integral of last post,
     
  11. May 25, 2013 #10
    First of all, i have problems here regarding the reaction.It should be probably
    n→p+e- +v-(however it does not seem from the delta function you have written).Anyway,you are probably calculating cross section,right?So why are there 4 phase space integrals.There should be only two for your reaction and three for mine but there are four?the way to handle the space δ is to integrate over any one of the three variable,which will left only two momentum integral(according to me) then you have to move forward to time delta function with relation E=√p2+m2.
     
  12. May 28, 2013 #11
    You are not asked to calculate that integral in this problem.
     
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