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Usefulness of SUSY models when it cant exist at non-zero temperatures

  1. May 30, 2013 #1
    Unlike other symmetries (like electroweak symmetry), SUSY is spontaneously broken at any non-zero temperature due to some variation of the fact that the boundary conditions on bosons and fermions in thermal QFT are different. If this is the case, what is the rationale for considering SUSY phenomenological models? ie. how valid is the assumption that temperature remains zero up until the point that the energy in the initial universe falls below the SUSY breaking energy scale?

    For more on thermal SUSY: A. Das, "Supersymmetry and finite temperature"
     
  2. jcsd
  3. May 30, 2013 #2
    I know nothing about thermal SUSY, but I don't except anyone makes this assumption. The usually assumption as I understand it is the reverse, that SUSY is preserved at high energy and spontaneously broken at low energy. Since the article you cite is from 1989 and people still study SUSY, I can only guess that their conclusion that "supersymmetry is spontaneously broken at finite temperature independent of whether supersymmetry is broken at zero temperature or not" is not as damning as it sounds... perhaps it doesn't really matter if SUSY ever actually existed in the universe? Perhaps it could still help with the hierarchy problem etc regardless of whether it was an exact symmetry of the early universe.

    But yes I too would be interested to hear more on this, if someone here knows it.
     
  4. May 30, 2013 #3
    I too think, for the same reason (that people are still studying SUSY), thermal effects probably dont upset the advantages of SUSY much. The section above that paper's conclusion heuristically discusses how the goldstino from thermal breaking of SUSY mixes with the goldstino from the usual SUSY breaking (at zero-temperature).

    I saw a recent paper on this - http://arxiv.org/abs/1206.2958 - which talks (in the two paragraphs preceding acknowledgements) about possible cosmological implications of this mixing.

    But it means that unbroken SUSY could never really have existed anytime in our universe's past.
     
  5. May 31, 2013 #4

    Demystifier

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    Bose-Einstein and Fermi-Dirac distributions are different at any (including zero) given temperature, which is the PHYSICAL reason why SUSY may seem "broken" in thermal field theory. The different boundary conditions is only a convenient mathematical tool to describe this physical fact. But the SUSY is still there, even at a non-zero temperature, in the sense that the numbers of bosonic and fermionic degrees of freedom are equal. So there is no reason to expect that temperature remains zero.
     
  6. May 31, 2013 #5
    PS: I have also posted this question at the physics stackexchange

    You should read that review paper I posted above. Under non-zero temperatures, SUSY is always (spontaneously) broken, along with a goldstino associated with this breaking. But yes, since this is spontaneous breaking, the symmetry itself is not explicitly broken.
     
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