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I Thermalisation of particles

  1. Aug 12, 2016 #1
    I've recently been reading a bit into the thermalisation of a system of particles and I'm unsure on a couple of concepts.

    Firstly, if a system of particles are out of mutual thermal equilibrium, does this essentially correspond to the particles in the system having randomly distributed momenta, such that each has a different magnitude and a different direction?!

    Secondly, once the system has reached thermal equilibrium, does this then correspond to each of the particles having the same magnitude of momentum (although their momentum will in general be different as even though each as the same magnitude, they will all point in different directions)?!

    Finally, is the process of thermalisation then literally a redistribution of momentum between the particles through mutual interactions, such that the magnitudes of their momenta are all equal?!
     
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  3. Aug 12, 2016 #2

    A. Neumaier

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    No, it means that there are deviations form the equilibrium distribution. These can be random or nonrandom. If you stir a glass of water using a rotating spoon it gets out of equilibrium and there will be a net vorticity and hence systematic position-dependent mean momenta. After a bit of waiting, the water regains equilibrium and the directions are uniformly distributed.
    No. In the rest frame of the thermal system the momenta will have a 3D Gaussian distribution.
     
  4. Aug 12, 2016 #3
    So what is the reason why, when particles are produced due to some decay process, they will generally be out of thermal equilibrium initially? Is it simply they will be produced with random momenta and then thermal equilibrium is attained through mutual interaction until the distribution of momentum between the particles is Gaussian?

    Will the particles not have equal average kinetic energy (as a result of the equipartition theorem)?
     
    Last edited: Aug 12, 2016
  5. Aug 12, 2016 #4

    A. Neumaier

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    If one particle decays into two, they will have correlated momenta in random but opposite directions in the rest frame of the decaying particle, hence in the lab frame they have a momentum distribution different from the decaying material. This is enough to destroy the equlibrium. After sufficiently many randomizing collisions, equilibrium is restored.
     
    Last edited: Aug 13, 2016
  6. Aug 12, 2016 #5
    Apologies, I'm probably just being a bit thick here, but how will the momentum distribution differ? Also, is the reason why thermal equilibrium is destroyed because the decaying particle was originally in thermal equilibrium with itself, and so if the momentum distribution of the created particles differs from that of the decaying particle, they will necessarily be out of mutual thermal equilibrium?
     
  7. Aug 13, 2016 #6

    A. Neumaier

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    After a single decay, the distribution in the bulk changes only marginally. But many decays happen more or less simultaneously, and in the coarse-graining necessary anyway to talk about equilibrium, averaging over these decays leads to a rate of change of the distribution. The details are nontrivial statistical mechanics - one gets reaction rates that go into kinetic equations, and the H-theorem then tells how equilibrium is reestablshed with time.

    Yes for your second question.
     
  8. Aug 13, 2016 #7
    I'm slightly unsure why the distribution in the bulk changes due to the decay. Is it because the original particle has some momentum and then when it decays the momentum is redistributed between the two particles that the original particle has decayed into?
     
  9. Aug 13, 2016 #8

    A. Neumaier

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    The momentum is distributed according to the mass proportions of the decay products, but there is a mass difference that is also converted into kinetic energy, which adds to the momentum of the decay products in random but opposite directions.
     
  10. Aug 13, 2016 #9
    Ah ok. So, say that that a particle decays into two particles of equal mass, will the small mass difference between the original particle and the sum of the masses of the created particles add to the momentum of each of the particles and push them out of thermal equilibrium?

    What about the case in which the decaying particle is a lepton (and as such has no internal structure), there won't be any mass deficit (since the lepton will not have a binding energy), so how do the created particles start out in thermal non-equilibrium?
     
  11. Aug 13, 2016 #10

    A. Neumaier

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    If there is no binding energy there is no decay hence no particles are created.
     
  12. Aug 13, 2016 #11
    So for a system of particles initially in thermal equilibrium, as each of them decays, the binding energy released causes the produced particles to gain some additional momentum. Thus, with each decay, the momentum distribution (of the whole system) is very slightly shifted, and hence, overall the momentum distribution is shifted enough, that averaging over all of these decays leads to the system falling out of thermal equilibrium. Would this be correct?
     
  13. Aug 14, 2016 #12

    A. Neumaier

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    At this level of description, yes.
     
  14. Aug 14, 2016 #13
    Ok cool, thanks for your help.
     
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