# Usefulness of SUSY models when it cannot exist at any non-zero temperature

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"

edits: Thanks Sujeet, for the link to the scanned version of that paper. I have also posted the same question at the physicsforums

• Is there a free arxiv version of the paper or something? It is behind a paywall :-/ – Dilaton May 30 '13 at 21:53
• That paper is pre-arXiv. You can find a scanned document from KEK: ccdb5fs.kek.jp/cgi-bin/img/allpdf?198811283 ymmv – sujeet May 31 '13 at 2:56
• Real-world susy theories don't rely on this effect to break supersymmetry. Real-world susy is supposed to be broken by other effects, e.g. see arxiv.org/abs/hep-th/0509029 – Mitchell Porter Aug 20 '13 at 6:46

Your other question about assuming that $T=0$ for every renormalization scale $Q$ above $Q_c$, the SUSY breaking scale, might be coming from mistaking finite temperature / a bath of thermal energy with the renormalization scale. These are not the same thing.
As an additional note, you should understand that temperature is not Lorentz-invariant. We can write QFT with Poincare symmetry in a thermal bath even if the finite $T$ is not Lorentz-invariant. It boils down to developing all the symmetry structures, e.g. covariant derivative, central charges, etc. within the thermal formalism. This has actually been done -- we have a formalism for thermal superspace with super-Poincare algebra and it is useful even if the symmetry is broken.
• I am not sure if I am understanding your remark about $T$ and $Q$. I thought I understood their difference when I kept $T$ at zero, while bringing $Q$ down from Planck scale, $Q_{pl}$. Also, if SUSY is compulsorily broken at any $T \ne 0$ and if unbroken SUSY is not even relevant, why are there papers that discuss SUSY breaking scale being much smaller than Planck scale ($Q_c << Q_{pl}$) ? – GuSuku May 31 '13 at 20:41
• (contd..) and papers discussing string completion of unbroken SUSY for $Q > Q_c$ ? – GuSuku May 31 '13 at 20:49