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How energy divided up in early Universe.

  1. May 23, 2010 #1
    Say we are in the early Universe where energies are large compared with any rest mass in the standard model.

    At this time how will the energy in the electron field compare with the energy in the photon field?

    Thanks for any help!
    Last edited: May 23, 2010
  2. jcsd
  3. May 24, 2010 #2


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    Well, what happens in that sort of situation is that the entire system is thermalized: when you have temperatures far above the rest mass energies of various particles, then collisions will frequently produce particle/anti-particle pairs of them. So you end up with energy equally distributed throughout the various degrees of freedom of the system. Photons have two available spin states, and thus two degrees of freedom. Electrons have two available spin states plus anti-particles, so between the positrons and electrons you have four degrees of freedom. Thus the electrons and positrons will between them have twice the energy as the photons, as long as the temperature remains far above the rest mass of the electron.

    As the universe cooled, however, the rest mass of the electron started to become important, as did the slight overabundance of electrons with respect to positrons.
  4. May 24, 2010 #3


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    There's also a factor of 7/8 that comes into play: Ignoring internal states, the energy density of a relativistic species of fermions will be 7/8 the energy density of a relativistic species of boson.
  5. May 24, 2010 #4


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    Ah, yeah, I thought I was missing a factor somewhere, but couldn't find it quickly to make sure. Thanks.
  6. May 24, 2010 #5
    Stuff every school child should know.

    Thanks to the both of you!
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