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Boltzmann equation and Kinetic equilibrium

  1. Mar 26, 2015 #1
    In section 3.1 of Dodelson's "Modern Cosmology", after introducing the Boltzmann equation, in the second paragraph of page 60 the author states that:

    "The first, most important realization is that scattering processes typically enforce kinetic equilibrium.
    That is, scattering takes place so rapidly that the distributions of the various species take on the generic Bose-Einstein/Fermi-Dirac forms."


    QESTIONS:
    1. What is the definition of kinetic equilibrium?
    2. When achieved, why does kinetic equilibrium it imply that the distributions take on the equilibrium forms?
    3. Why does scattering processes typically enforce kinetic equilibrium?
    In Cercignani's "The Relativistic Boltzmann Equation: Theory and Applications" section 2.7 he derives so called equilibrium distribution functions by demanding that the entropy production rate must vanish. Is this the same as kinetic equilibrium? Is the above somehow related to the socalled H-theorem (which i dont know much about)?

    To the moderator: I first posted this post in the cosmology forum. However, since I got no replies there, I reposted it here. If you must delete one, delete the one in the cosmology forum,
     
  2. jcsd
  3. Mar 26, 2015 #2

    vanhees71

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    Cercignani's book is brillant. As you can read there, you'll see that the stationarity of entropy implies local thermal equilibrium. For a simplified introduction, see my lecture notes about the subject (not yet completed):

    http://fias.uni-frankfurt.de/~hees/publ/roorkee.pdf
     
  4. Mar 26, 2015 #3
    Ah, I actually already had a look at them!
    Since you are here, might I take the opportunity to ask you to elaborate on how you obtained equation 1.4.10?
     
  5. Mar 26, 2015 #4

    vanhees71

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    First of all, I've updated the manuscript to the newer version due to another lecture week in Kolkata:

    http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

    To get (1.4.10), write (1.4.9) in the form
    $$\psi+\psi_2-\psi_1'-\psi_2'=0$$
    and note that ##\psi=\psi(x,p)##, ##\psi_2=\psi(x,p_2)##, etc.

    Further since this equation holds under the constraint of energy-momentum conservation, i.e.,
    $$p+p_2-p_1'-p_2'=0$$
    we take the independent variation with respect to the four-momenta of the expression
    $$\psi+\psi_2 - \psi_1'-\psi_2' - \lambda \cdot (p+p_2-p_1'-p_2').$$
    Then setting ##\delta p_2=\delta p_1'=\delta p_2'=0##, you get
    $$\delta p^{\mu} \left (\frac{\partial \psi}{\partial p^{\mu}}-\lambda^{\mu} \right)=0.$$
    Now ##p^0=E=\sqrt{\vec{p}^2+m^2}##, so that ##\delta p^0=\delta \vec{p} \cdot \vec{p}/E##, and thus
    $$\delta \vec{p} \left (\frac{\vec{p}}{E} \frac{\partial \psi}{\partial p^0}+\frac{\partial \psi}{\partial \vec{p}} \right )= \delta \vec{p} \left (\lambda^0 \frac{\vec{p}}{E}-\vec{\lambda} \right),$$
    which is (1.4.10).
     
  6. Mar 27, 2015 #5
    Thanks! I'll certainly check them out.

    So in the light of local entropy always increasing, is the statement from Dodelson above a reference to equilibrium in the sense that the entropy has reached its maximum? Is this the definition of kinetic equilibrium? How does this tie in with "scattering takes place so rapidly that the distributions of the various species take on the generic Bose-Einstein/Fermi-Dirac forms"?
     
  7. Mar 27, 2015 #6

    vanhees71

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    I don't know, which statement from Dodelson you mean. Local thermal equilibrium means that your phase-space distribution function takes the Boltzmann/Bose/Fermi-Jüttner form with a given flow-velocity four-vector field ##u(x)##, a temperature field ##T(x)##, and a chemical potential ##\mu(x)##:
    $$f(x,p)=\frac{1}{\exp[(p \cdot u-\mu)/T] + \eta},$$
    with ##\eta=0## for classical (Boltzmann), ##\eta=1## for Fermi-Dirac, and ##\eta=-1## for Bose-Einstein statistics.

    This is a good description of an off-equilibrium situation, if the mean free path of the particles is very small compared to the typical length scales over which the macroscopic fields, ##u##, ##T##, and ##\mu## change. Then the Boltzmann equation is well approximated by the relativistic Euler equations of an ideal fluid.

    The next approximation steps are then given by viscous hydrodynamics. For a good review on these issues, look at the following PhD thesis of one of my colleagues:

    http://arxiv.org/abs/1205.0782
     
  8. Apr 16, 2015 #7


    Why does it make sense to use Lagrange multipliers here? Lagrange multipliers implies that we are extremizing under the energy-momentum constraint, but how do we know that the extremization of this function corresponds to $$\psi+\psi_2-\psi_1'-\psi_2'=0$$? Does it make sense that this function can not attain negative values?
     
  9. Apr 17, 2015 #8

    vanhees71

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    If the constraint is fulfilled
    $$\psi+\psi_2 - \psi_1'-\psi_2' - \lambda \cdot (p+p_2-p_1'-p_2')=0$$
    is of course equivalent to
    $$\psi+\psi_2 - \psi_1'-\psi_2' =0.$$
     
  10. Apr 22, 2015 #9
    Thanks a lot for your help vanhees71! Might I ask you one(three) last questions?

    My motivation for reading up on the Boltzmann equation is related to the study of cosmology.
    When computing first order versions of the Boltzmann equations for the CMB temperature fluctuations, one assumes a form for the photon distribution function given by
    $$f(\vec x, \vec p) = \frac{1}{e^{\frac{p}{T(t)( 1+\Theta(\hat p, \vec x,t))}} - 1},$$
    where $\Theta$ is said to represent the fluctuations in temperature ##\delta T/T## of the CMB and is stated to be independent of the magnitude of the photon momentum ##p##, only its direction ##\hat p##.

    Now, the equilibrium distributions of the Boltzmann equation for bosons that you derive in your note are on the form
    $$ f(\vec x, \vec p) = \frac{1}{e^{\frac{E(x,p) - \mu(x)}{T(x)}} - 1},$$
    and since ##\mu \leq m##; ##\mu = 0## for photons.

    Q1: Is the photon distribution defined above used to study the CMB a local equilibrium distribution?
    Q2: If so, is there any justification by appeal to the Boltzmann equation that ##\Theta## is not a function of the momentum magnitude ##p##, but only of its direction?
    Q3: If not, how can we assume that it has the same functional form as an equilibrium distribution?
     
  11. Apr 22, 2015 #10

    vanhees71

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    This is a quite delicate question, and I'm not an expert in cosmology.

    Usually one assumes that as long as the matter in the universe is in a plasma state, i.e., consists of charged particles it's confining for electromagnetic radiation (photons), which implies that the photons are kept in thermal equilibrium with this matter. At the Mott transition, where the protons and light nuclei combine with electrons to neutral atoms, the matter consists of neutral atoms, and the universe becomes transparent for photons, which thus decouple. However, the Hubble expansion doesn't change the equilibrium Planck distribution of the photons but only changes their temperature due to the Hubble redshift of the photons.

    The anisotropies come from fluctuations in the gravitational field (deviations of the metric from the Friedmann-Lemaitre-Robertson-Walker metric) and is described as part of the left-hand side of the Boltzmann equation and from scatterings of the photons in the medium which also deviates from equilibrium due to the fluctuations.

    This should answer Questens 1 and 3. I've not a good answer for Question 2. I can only refer to the empirical fact that the CMBR is the most accurate black-body spectrum ever measured, and I guess that motivates the ansatz for the solution of the (linearized) Boltzmann equation.
     
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