Sneutrino mass measurements ate+e−linear colliders

Abstract

It is generally accepted that experiments at an ${e}^{+}{e}^{\ensuremath{-}}$ linear collider will be able to extract the masses of the selectron as well as the associated sneutrino with a precision of $\ensuremath{\sim}1%$ by determining the kinematic end points of the energy spectrum of daughter electrons produced in their two body decays to a lighter neutralino or chargino. Recently, it has been suggested that by studying the energy dependence of the cross section near the production threshold, this precision can be improved by an order of magnitude, assuming an integrated luminosity of $100 {\mathrm{fb}}^{\ensuremath{-}1}.$ It is further suggested that these threshold scans also allow the masses of even the heavier second and third generation sleptons and sneutrinos to be determined to better than 0.5%. We reexamine the prospects for determining sneutrino masses. We find that the cross sections for the second and third generation sneutrinos are too small for a threshold scan to be useful. An additional complication arises because the cross section for sneutrino pairs to decay into any visible final state(s) necessarily depends on an unknown branching fraction, so that the overall normalization is unknown. This reduces the precision with which the sneutrino mass can be extracted. We propose a different strategy to optimize the extraction of $m({\stackrel{\ifmmode \tilde{}\else \~{}\fi{}}{\ensuremath{\nu}}}_{\ensuremath{\mu}})$ and $m({\stackrel{\ifmmode \tilde{}\else \~{}\fi{}}{\ensuremath{\nu}}}_{\ensuremath{\tau}})$ via the energy dependence of the cross section. We find that even with an integrated luminosity of $500 {\mathrm{fb}}^{\ensuremath{-}1},$ these can be determined with a precision no better than several percent at the 90% C.L. We also examine the measurement of $m({\stackrel{\ifmmode \tilde{}\else \~{}\fi{}}{\ensuremath{\nu}}}_{e})$ and show that it can be extracted with a precision of about 0.5% (0.2%) with an integrated luminosity of $120 {\mathrm{fb}}^{\ensuremath{-}1} (500 {\mathrm{fb}}^{\ensuremath{-}1}).$