# Does the gravitino contribute to the gravitational interaction?

I have a very basic question with respect to supersymmetry. Actually, I no clear idea at all what the effect of the superpartners (called gauginos??) of the exchange particles of interactions (photons, gravitons, gluon, W+-s and Z0s) on the corresponding interaction is. Example: Gravitation: According to supersymmetry apart from the graviton there is the gravitino. Does the gravitino contribute to the gravitational interaction ? As it is supposed to have high mass (compared to a proton for instance) its possible contribution should be only short-range. I would appreciate to learn more about this. Thank you

• The gravitino is a fermion and hence no force carrier particle. – pfnuesel Sep 8 '14 at 14:30
• I am not sure how this would be with the gravitino, but one of the very much stressed feature of SUSY models is the cancellation of divergent loops by supersymmetric partners. But since we assume broken supersymmetry, the result would be a "natural cut-off" at the scale of energy/length of the gravitino mass providing a non-Einstein regularized gravitational interaction. But I'm just extrapolating my already shaky knowledge. – Void Sep 8 '14 at 17:09
• So far, there are no signs of supersymmetry anywhere. Even if there were signs of supersymmetry, it wouldn't imply that there would be anything resembling a graviton or a gravitino. – CuriousOne Sep 8 '14 at 17:09
• @pfnuesel: is that true? (I'm asking, not doubting you.) Wikipedia describes the gravitino as a gauge fermion and says the gravitino is the fermion mediating supergravity interactions, just as the photon is mediating electromagnetism, and the graviton is presumably mediating gravitation. – John Rennie Sep 8 '14 at 17:18
• @JohnRennie Interesting quote. It is new to me, that a fermion could mediate a force, be it super or not. So I don't know. Hope someone can clarify this. – pfnuesel Sep 8 '14 at 18:03

Let us take a few steps back and try to understand why that statement from Wikipedia is correct. To see this we have to first understand the coupling of the graviton to matter particles. The graviton couples to matter via its energy-momentum tensor $\sim g_{\mu\nu}T^{\mu\nu}.$ The energy momentum tensor is a response to the variations of the metric (spacetime tells matter how to move). It is the (covariantly) conserved quantity corresponding to the general coordinate invariance (diffeomorphism invariance) of general relativity and can be computed directly from the action via: $$T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu\nu}}.$$ Likewise in supergravity, in addition to diffeomorphism symmetry there is another symmetry. Guess what? Supersymmetry! It is a local symmetry in supergravity and has an associated conserved current that can be obtained analogously by varying the gravitino $\psi_{\alpha\mu}$ (where $\alpha$ is a spinor index and $\mu$ as usual is a vector index) in the supergravity action: $$\mathcal{S}_{\alpha\mu}\sim \frac{\delta S}{\delta \psi_{\alpha\mu}}.$$ $\mathcal{S}_{\alpha\mu}$ is called supercurrent. The gravitino couples to this conserved supercurrent. Therefore it is a gauge particle associated to local supersymmetry transformations. At scales below the supersymmetry breaking scale the gravitino acquires a mass however the supercurrent ceases to exist. So it doesn't couple to anything. Thus in broken supersymmetry it does not contribute to (super) gravitational interaction.