# Strong-Interaction Matter under Extreme Conditions from Chiral Quark Models with Nonlocal Separable Interactions

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## Abstract

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## 1. Introduction

## 2. Two-Flavor Nonlocal NJL Models

#### 2.1. Two-Flavor nlNJL Model at Vanishing Temperature and Chemical Potential

#### 2.1.1. Effective Action

#### 2.1.2. Mean Field Approximation and Chiral Condensates

#### 2.1.3. Meson Properties

#### 2.2. Extension to Finite Temperature and Chemical Potential. Inclusion of the Polyakov Loop

#### 2.3. Form Factors, Parameterizations and Numerical Results for $T=\mu =0$

- In parameterization PA, we consider the simple case in which one takes$$g\left(p\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}exp(-{p}^{2}/{\Lambda}_{0}^{2})\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}f\left(p\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}0\phantom{\rule{0.166667em}{0ex}}$$
- In the second parameterization, PB, it is assumed that both $g\left(p\right)$ and $f\left(p\right)$ are given by Gaussian functions, namely$$g\left(p\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}exp(-{p}^{2}/{\Lambda}_{0}^{2})\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}f\left(p\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}exp(-{p}^{2}/{\Lambda}_{1}^{2})\phantom{\rule{4pt}{0ex}}.$$Note that the range (in momentum space) of the nonlocality in each channel is determined by the parameters ${\Lambda}_{0}$ and ${\Lambda}_{1}$, respectively. From Equation (12), one has$$\begin{array}{ccc}\hfill M\left(p\right)& =& Z\left(p\right)\phantom{\rule{0.166667em}{0ex}}\left[{m}_{c}+{\overline{\sigma}}_{1}\phantom{\rule{4pt}{0ex}}exp(-{p}^{2}/{\Lambda}_{0}^{2})\right]\phantom{\rule{0.166667em}{0ex}}\hfill \\ \hfill Z\left(p\right)& =& {\left[1-{\overline{\sigma}}_{2}\phantom{\rule{4pt}{0ex}}exp(-{p}^{2}/{\Lambda}_{1}^{2})\right]}^{-1}\phantom{\rule{4pt}{0ex}}.\hfill \end{array}$$
- Finally, the third parameterization, PC, is taken from Refs. [31,32]. In this case, the functions $M\left(p\right)$ and $Z\left(p\right)$ are written as$$\begin{array}{ccc}\hfill M\left(p\right)& =& {m}_{c}+{\alpha}_{m}\phantom{\rule{0.166667em}{0ex}}{f}_{m}\left(p\right)\phantom{\rule{0.166667em}{0ex}}\hfill \\ \hfill Z\left(p\right)& =& 1+{\alpha}_{z}\phantom{\rule{0.166667em}{0ex}}{f}_{z}\left(p\right)\phantom{\rule{0.166667em}{0ex}}\hfill \end{array}$$$${f}_{m}\left(p\right)={\left[1+{\left({p}^{2}/{\Lambda}_{0}^{2}\right)}^{3/2}\right]}^{-1}\phantom{\rule{4pt}{0ex}},\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}{f}_{z}\left(p\right)={\left[1+\left({p}^{2}/{\Lambda}_{1}^{2}\right)\right]}^{-5/2}\phantom{\rule{4pt}{0ex}}.$$The analytical expression for ${f}_{m}\left(p\right)$ was originally proposed by [82,83], while that of ${f}_{z}\left(p\right)$ has been chosen so as to reproduce lattice results, ensuring at the same time the convergence of quark loop integrals. Some alternative parameterization of this type, suggested from vector meson dominance in the pion form factor, can be found in Ref. [84]. In terms of the functions ${f}_{m}\left(p\right)$ and ${f}_{z}\left(p\right)$ and the constants ${m}_{c}$, ${\alpha}_{m}$ and ${\alpha}_{z}$, the form factors $g\left(q\right)$ and $f\left(q\right)$ are given by [see Equation (12)]$$\begin{array}{c}\hfill g\left(p\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\frac{1+{\alpha}_{z}}{1+{\alpha}_{z}{f}_{z}\left(p\right)}\phantom{\rule{4pt}{0ex}}\frac{{\alpha}_{m}{f}_{m}\left(p\right)-{m}_{c}\phantom{\rule{0.166667em}{0ex}}{\alpha}_{z}{f}_{z}\left(p\right)}{{\alpha}_{m}-{m}_{c}\phantom{\rule{0.166667em}{0ex}}{\alpha}_{z}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}f\left(p\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\frac{1+{\alpha}_{z}}{1+{\alpha}_{z}{f}_{z}\left(p\right)}\phantom{\rule{4pt}{0ex}}{f}_{z}\left(p\right)\phantom{\rule{0.166667em}{0ex}}\end{array}$$$${\overline{\sigma}}_{1}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\frac{{\alpha}_{m}-{m}_{c}{\alpha}_{z}}{1+{\alpha}_{z}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}{\overline{\sigma}}_{2}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\frac{{\alpha}_{z}}{1+{\alpha}_{z}}\phantom{\rule{4pt}{0ex}}.$$