A numerical study of non-Gaussianity in the curvaton scenario
By Karim A. Malik, D. Lyth. 2006
148 citations recorded by the source corpus.
We study the curvaton scenario using gauge-invariant second order perturbation theory and solving the governing equations numerically. Focusing on large scales we calculate the non-linearity parameter fNL in the two-fluid curvaton model and compare our results with previous analytical studies employing the sudden decay approximation. We find good agreement of the two approaches for large curvaton energy densities at curvaton decay, Ωσdec, but significant differences of up to 10% for small Ωσdec.
Subfields: Astronomy and Astrophysics, Nuclear and High Energy Physics
View Paper: https://www.semanticscholar.org/paper/4c78b74f3c03b1d89ef47b2968b8457b8959f7cf
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Citations in the corpus
- Planck 2018 results. IX. Constraints on primordial non-Gaussianity (2019, background): “In the minimal adiabatic curvaton scenario f localNL = (5/4rD) − 5rD/6 − 5/3 (Bartolo et al. 2004d,c), in the case when the curvaton field potential is purely quadratic (Lyth & Wands 2002; Lyth et al. 2003; Lyth & Rodriguez 2005; Malik & Lyth 2006; Sasaki et al. 2006).”
- Non-linear isocurvature perturbations and non-Gaussianities (2008, methodology): “The precise density perturbation produced after the curvaton decays can be calculated numerically [27, 28, 23], but it can also be estimated analytically using the sudden-decay approximation [6], which assumes that the curvaton decays suddenly on a spatial hypersurface of uniform total energy densit”
- Review of local non-Gaussianity from multi-field inflation (2010, methodology): “[31] with a sudden decay approximation and also by Malik and lyth [32] using a fully numeric approach, which accounted for both the sudden decay and nonsudden decay approximations.”
- Planck 2015 results. XVII. Constraints on primordial non-Gaussianity (2015, unknown): “(11), we will search for the following equilateral feature ansatz: Bfeat−eq(k1, k2, k3) ≡ S eq(k1, k2, k3) × Bfeat(k1, k2, k3) (16) = 6A2 f feat−eqNL (k1k2k3)2 k̃1k̃2k̃3 k1k2k3 sin [ ω(k1 + k2 + k3)+φ ] .”