strange quark flavor is here supposed to be equal to the light quark contribution up to a substitution of $\mu\rightarrow0$ and $N_{q}\rightarrow N_{s}$ in the following formulae.

Taking the trace in Minkowski space, the gluonic part decomposes into three contributions for one longitudinal and two (equivalent) transverse polarizations, while the quark contribution becomes the sum of the normal and the abnormal quark branch (positive and negative chirality over helicity ratio, respectively) when taking the Dirac trace. The remaining traces only give overall factors: the color trace $(N_{c}^{2}-1)$ for the gluons and $N_{c}$ for the quarks, and the spin traces for quarks an additional $2$. Defining the prefactors $d_{g}=N_{c}^{2}-1$, $d_{q}=2N_{c}N_{q}$ and $d_{s}=2N_{c}N_{s}$ and introducing the abbreviation $\int_{{\mathrm{d}^{n}k}}=\int{\mathrm{d}^{n}k}/(2\pi)^{4}$ the HTL grand canonical potential then reads$$\begin{aligned}
\frac{\Omega}{V} & = & d_{g}\int_{{\mathrm{d}^{4}k}}\!\!{n_\text{B}}\,\Big\{2{\text{Im}}\!\left(\ln{D_\text{T}}^{-1}-{D_\text{T}}{\Pi_\text{T}}\right)+{\text{Im}}\!\left(\ln\left(-{D_\text{L}}^{-1}\right)+{D_\text{L}}{\Pi_\text{L}}\right)\!\Big\}\\
 &  & +2\sum_{i=q,s}d_{i}\int_{{\mathrm{d}^{4}k}}\!\!{n_\text{F}}\,\Big\{{\text{Im}}\!\left(\ln S_{+}^{-1}-S_{+}\Sigma_{+}\right)+{\text{Im}}\!\left(\ln\left(-S_{-}^{-1}\right)+S_{-}\Sigma_{-}\right)\!\Big\}-\frac{T}{V}\Gamma_{2}.\nonumber \end{aligned}$$

Differentiating the thermodynamic potential with respect to the temperature at constant chemical potential gives the entropy. In contrast to the pressure, which is influenced by vacuum fluctuations, the entropy is sensitive to thermal excitations and therefore manifestly ultraviolet (UV) finite. As such, it is ideally suited to investigate the properties of the QGP [@BIR01].

Due to the stationarity of the thermodynamic potential with respect to the full 