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# Gravitino

[grav-i-tee-noh]
The gravitino is the supersymmetric partner of the graviton, as predicted by theories combining general relativity and supersymmetry; i.e. supergravity theories. If it exists it is a fermion of spin 3/2 and therefore obeys the Rarita-Schwinger equation.

The plural of gravitino is gravitini according to the rules of Italian declension

The gravitino field is conventionally written as $psi_\left\{mualpha\right\}$ with $mu=0,1,2,3$ a four-vector index and $alpha=1,2$ a spinor index. For $mu=0$ one would get negative norm modes, as with every massless particle of spin 1 or higher. These modes are unphysical, and for consistency there must be a gauge symmetry which cancels these modes: $deltapsi_\left\{mualpha\right\} = partial_muepsilon_alpha$ where $epsilon_alpha\left(x\right),$ is a spinor function of spacetime. This gauge symmetry is a local supersymmetry transformation, and the resulting theory is supergravity.

Thus the gravitino is the fermion mediating supergravity interactions, just as the photon is mediating electromagnetism, and the graviton is presumably mediating gravitation. Whenever supersymmetry is broken in supergravity theories, it acquires a mass which is directly the supersymmetry breaking scale.

As a proposed solution to the fine tuning problem of the Standard Model, and in order to allow grand unification, the supersymmetry breaking scale needs to be pushed down to the TeV range. Therefore the gravitino mass needs to be of this order (unless we have intermediate scale SUSY breaking), much lower than Planck scale, which is the natural scale for gravity interactions. This difference in energy scales is known as the hierarchy problem.

## Gravitino cosmological problem

If the gravitino indeed has a mass of the order of TeV, then it creates a problem in the standard model of cosmology, at least naïvely .

One option is that the gravitino is stable. This would be the case if the gravitino is the lightest supersymmetric particle and R-parity is conserved (or nearly so). In this case the gravitino is a candidate for dark matter; as such gravitinos will have been created in the very early universe. However, one may calculate the density of gravitinos and it turns out to be much higher than the observed dark matter density.

The other option is that the gravitino is unstable. Thus the gravitinos mentioned above would decay and will not contribute to the observed dark matter density. However, since they decay only through gravitational interactions, their lifetime would be very long, of the order of $M_\left\{pl\right\}^2/m^3$ in natural units, where $m$ is their mass and $M_\left\{pl\right\}$ is the Planck mass. For a mass of the order of TeV this would be $10^5$ seconds, much later than the era of nucleosynthesis. At least one possible channel of decay must include either a photon, a charged lepton or a meson, each of which would be energetic enough to destroy a nucleus if it strikes one. One can show that enough such energetic particles will be created in the decay as to destroy almost all the nuclei created in the era of nucleosynthesis, in contrast with observations. In fact, in such a case the universe would have been made of hydrogen alone, and star formation would probably be impossible.

One possible solution to the cosmological gravitino problem is the split supersymmetry model, where the gravitino mass is much higher than the TeV scale, but other fermionic supersymmetric partners of standard model particles already appear at this scale.