that later on we will use a phenomenological linear collision operator that will suffice for our present purposes.

Conserved currents {#currents}
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We now introduce the matter entropy and the conserved currents of the microscopic theory. The nonabelian current reads \_=gDpp\_ \[10\] where $Dp=d^4p\delta\left(p^2\right)/\left(2\pi\right)^3$.

We have not written down an explicit equation for the matter stress-energy tensor $T^{\mu\nu}_{m}$, but we know that since the total stress-energy must be conserved, we must have T\^\_[m;]{}=-T\^\_[YM;]{}= \_\^ \[ecT\] We get this by writing T\^\_[m]{}=\^\_[m]{} \[11\]

\^\_[m]{}=Dpp\^p\^ \[12\] In Eq. (\[ecT\]), the semicolon stands for an ordinary derivative. We shall drop the subindex $m$ for $\mathbf{T}^{\mu\nu}_{m}$ and write $\mathbf{T}^{\mu\nu}$ in what follows.

Eqs. (\[5\]) and (\[8\]) are identically satisfied provided

&Dp(p\^0)(\_a\_[col]{})=\
&Dp(p\^0)p\^\_[col]{}=0 \[13\]

The entropy current is

S\^=Dpp\^(p\^0){(+)(+)-} \[14\] leading to the entropy production

S\^\_[;]{}=Dp{\_[col]{}\^[-1]{}(+)} \[15b\]

We note that to go from Eq. (\[14\]) to Eq. (\[15b\]) one must assume that $[\mathbf{f},\mathbf{D}_\mu \mathbf{f}]=0$ (see [@localeq]). If this condition is not imposed on the distribution function, the entropy production contains terms of the form $\rm{tr} \{\mathbf{F}_{\mu\nu},\mathbf{f}\}$ which contribute to entropy production even in mean field. Assuming $[\mathbf{f},\mathbf{D}_\mu \mathbf{f}]=0$ then corresponds to assuming that entropy is produced solely by collisions among the particles. We shall stick 