Reducing the Tension Between the BICEP2 and the Planck Measurements: A Complete Exploration of the Parameter Space

TL;DR

This paper investigates how allowing certain cosmological parameters to vary can reduce the tension between BICEP2 and Planck measurements of the tensor-to-scalar ratio, using comprehensive parameter space exploration.

Contribution

It performs a complete MCMC analysis of the cosmological parameter space, identifying the role of $N_{eff}$ in alleviating measurement tensions between BICEP2 and Planck.

Findings

$N_{eff}$ significantly reduces the tension.

Constraints on $r_{0.002}$ are consistent across datasets.

Results are similar with or without the consistency relation.

Abstract

A large inflationary tensor-to-scalar ratio $r_\mathrm{0.002} = 0.20^{+0.07}_{-0.05}$ is reported by the BICEP2 team based on their B-mode polarization detection, which is outside of the $95\%$ confidence level of the Planck best fit model. We explore several possible ways to reduce the tension between the two by considering a model in which $\alpha_\mathrm{s}$, $n_\mathrm{t}$, $n_\mathrm{s}$ and the neutrino parameters $N_\mathrm{eff}$ and $\Sigma m_\mathrm{\nu}$ are set as free parameters. Using the Markov Chain Monte Carlo (MCMC) technique to survey the complete parameter space with and without the BICEP2 data, we find that the resulting constraints on $r_\mathrm{0.002}$ are consistent with each other and the apparent tension seems to be relaxed. Further detailed investigations on those fittings suggest that $N_\mathrm{eff}$ probably plays the most important role in reducing the tension. We also find that the results obtained from fitting without adopting the consistency relation do not deviate much from the consistency relation. With available Planck, WMAP, BICEP2 and BAO datasets all together, we obtain $r_{0.002} = 0.14_{-0.11}^{+0.05}$, $n_\mathrm{t} = 0.35_{-0.47}^{+0.28}$, $n_\mathrm{s}=0.98_{-0.02}^{+0.02}$, and $\alpha_\mathrm{s}=-0.0086_{-0.0189}^{+0.0148}$; if the consistency relation is adopted, we get $r_{0.002} = 0.22_{-0.06}^{+0.05}$.