Effect of magnetic field on jet transport coefficient $\hat{q}$

Abstract

We report the estimation of jet transport coefficient $\hat{q}$ for quark- and gluon-initiated jets using a simple quasiparticle model in the absence and presence of magnetic field. This model introduces a temperature and magnetic field-dependent degeneracy factor of partons, which is tuned by fitting the entropy density of lattice quantum chromodynamics data. At a finite magnetic field, $\hat{q}$ for quark jets splits into parallel and perpendicular components whose magnetic field dependence comes from two sources: the field-dependent degeneracy factor and the phase-space part guided by the shear viscosity-to-entropy density ratio. Due to the electrically neutral nature of gluons, the estimation of $\hat{q}$ for gluon jets is affected only by the field-dependent degeneracy factor. In the presence of a finite magnetic field, we find a significant enhancement in $\hat{q}$ for both quark- and gluon-initiated jets at low temperature, which gradually decreases towards high temperature. We compare the obtained results with the earlier calculations based on the anti-de Sitter/conformal field theory correspondence, and a qualitatively similar trend is observed. The change in $\hat{q}$ in the presence of magnetic field is, however, quantitatively different for quark- and gluon-initiated jets. This is an interesting observation which can be explored experimentally to verify the effect of magnetic field on $\hat{q}$.

Appendix A

In terms of the Fermi–Dirac (FD) distribution function of quarks and the Bose–Einstein (BE) distribution function of gluons, the energy density ($\epsilon$) of the QGP system can be expressed as

$$\begin{aligned} \epsilon _{\textrm{QGP}} ={}&\frac{g_{g}}{(2\pi )^{3}} \int _{0}^{\infty } \frac{\omega _{g}}{\textrm{e}^{\beta \omega _{g}} - 1} {\textrm{d}^{3}\vec {k}}\nonumber \\&+ \frac{g_{Q} }{(2\pi )^{3}}\int _{0}^{\infty } \frac{\omega _{Q}}{\textrm{e}^{\beta \omega _{Q}} + 1} {\textrm{d}^{3}\vec {k}}. \end{aligned}$$

(A.1)

Here $\omega _{g}$ and $\omega _{Q}$ are energies and can be expressed as $\omega _{g,Q} = \sqrt{\vec k^2+m_{g,Q}^2}$ and $\beta = 1/T$. Here $m_{g}$ and $m_{Q}$ are masses of quarks and gluons. However, for massless QGP, $m_{g,Q} = 0$. Therefore, for massless QGP, $\omega _{g,Q} = \vec k_{g,Q}$. If one converts the volume integral to line integral, $\int _{0}^{\infty } {\textrm{d}^{3}\vec {k}} \rightarrow 4\pi \int _{0}^{\infty }\vec {k}^{2} {\textrm{d}\vec {k}}$.

Equation (A.1) can be expressed as

$$\begin{aligned} \epsilon _{\textrm{QGP}} ={}&\frac{g_{g}}{(2\pi )^{3}} \int _{0}^{\infty } \frac{{\vec {k}_{g}}}{\textrm{e}^{\beta \vec {k_{g}}} - 1} {4\pi {\vec {k}_{g}}^{2} \textrm{d}{\vec {k}_{g}}}\nonumber \\&+ \frac{g_{Q}}{(2\pi )^{3}} \int _{0}^{\infty } \frac{{{\vec {k}_{Q}}}}{\textrm{e}^{\beta {\vec {k}}_{{Q}}} + 1} {4\pi {{\vec {k}}_{Q}}^{2} \textrm{d}{\vec {k}}_{{Q}}}\nonumber \\ ={}&\frac{g_{g}}{2\pi ^{2}} \int _{0}^{\infty } \frac{{\vec {k}_{g}^{3}}}{\textrm{e}^{\vec {k}_{{g}}/T} - 1} {\textrm{d}{\vec {k}}_{{g}}}\nonumber \\&+ \frac{g_{Q}}{2\pi ^{2}} \int _{0}^{\infty } \frac{{\vec {k}_{Q}^{3}}}{\textrm{e}^{\vec {k}_{{Q}}/T} + 1} {\textrm{d}\vec {k_{Q}}} \hspace{0.5cm} [\beta = 1/T]\nonumber \\ ={}&\frac{g_{g}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } \frac{x^{3}}{\textrm{e}^{x} - 1} {\textrm{d}x}\nonumber \\&+ \frac{g_{Q}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } \frac{y^{3}}{\textrm{e}^{y} + 1} {\textrm{d}y} \nonumber \\&\hspace{0.5cm} {[}\vec {k_{g}}/T = x, \vec {k_{Q}}/T = y]\nonumber \\ ={}&\frac{g_{g}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } \frac{x^{3}}{\textrm{e}^{x} (1 - \textrm{e}^{-x})} {\textrm{d}x}\nonumber \\&+ \frac{g_{Q}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } \frac{y^{3}}{\textrm{e}^{y} (1 + \textrm{e}^{-y})} {\textrm{d}y}\nonumber \\ ={}&\frac{g_{g}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } x^{3}\textrm{e}^{-x} (1 - \textrm{e}^{-x})^{-1} {\textrm{d}x}\nonumber \\&+ \frac{g_{Q}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } y^{3}\textrm{e}^{-y} (1 + \textrm{e}^{-y})^{-1} {\textrm{d}y}\nonumber \\ ={}&\frac{g_{g}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } x^{3}\textrm{e}^{-x} \left[ \sum _{n = 0}^{\infty } \textrm{e}^{-nx} \right] {\textrm{d}x}\nonumber \\&+ \frac{g_{Q}T^{4}}{2\pi ^{2}} \int _{0}^{\infty } y^{3}\textrm{e}^{-y} \left[ \sum _{n = 0}^{\infty } (-1)^{n}\textrm{e}^{-ny} \right] {\textrm{d}y}\nonumber \\ ={}&\frac{g_{g}T^{4}}{2\pi ^{2}} \sum _{n = 0}^{\infty } \int _{0}^{\infty } x^{3}\textrm{e}^{-(1+n)x} {\textrm{d}x}\nonumber \\&+ \frac{g_{Q}T^{4}}{2\pi ^{2}} \sum _{n = 0}^{\infty } (-1)^{n} \int _{0}^{\infty } y^{3}\textrm{e}^{-(1+n)y} {\textrm{d}y}. \end{aligned}$$

(A.2)

If one considers $(1+n)x = a$ and $(1+n)y = b$, then eq. (A.2) can be represented as

$$\begin{aligned} \epsilon _{\textrm{QGP}} ={}&\frac{g_{g}T^{4}}{2\pi ^{2}} \sum _{n = 0}^{\infty } \frac{1}{(n+1)^{4}} \int _{0}^{\infty } a^{3}\textrm{e}^{-a} {\textrm{d}a}\nonumber \\&+ \frac{g_{Q}T^{4}}{2\pi ^{2}} \sum _{n = 0}^{\infty } (-1)^{n} \frac{1}{(n+1)^{4}} \int _{0}^{\infty } b^{3}\textrm{e}^{-b} {\textrm{d}b.} \end{aligned}$$

(A.3)

Simplification of $\int _{0}^{\infty } t^{3}\textrm{e}^{-t} {\textrm{d}t} = \Gamma (4) = 6$ and expanding binomially one can get

$$\begin{aligned} \sum _{n = 0}^{\infty } \frac{1}{(n+1)^{4}} = \xi (4) = \frac{\pi ^{4}}{90} \end{aligned}$$

(A.4)

and

$$\begin{aligned} \sum _{n = 0}^{\infty } (-1)^{n} \frac{1}{(n+1)^{4}} = \frac{7}{8} \xi (4) =\frac{7}{8} \frac{\pi ^{4}}{90}. \end{aligned}$$

(A.5)

$$\begin{aligned} \epsilon _{\textrm{QGP}}&= \frac{g_{g}T^{4}}{2\pi ^{2}} \frac{6\pi ^{4}}{90}+ \frac{g_{Q}T^{4}}{2\pi ^{2}}\frac{7}{8} \frac{6\pi ^{4}}{90}\nonumber \\&= \Bigg [g_g+g_{Q}\Bigg (\frac{7}{8}\Bigg )\Bigg ]\frac{3\pi ^2}{90}T^4\approx 15.6~T^4. \end{aligned}$$

(A.6)

Pressure (P) of the QGP system follows similar prescription as the energy density ($\epsilon$).