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We go back to the entropy four-flow (7.1), which we decompose as
follows:
$$ \bea S^\mu=&-&\sum
\limits_{k} \int d\om_k\, v^\mu_k \left[f_k \log f^{(0)}_k - \et(1
+ \et f_k)\log (1 + \et f^{(0)}_k)\right] \nn \\ &-&
K^\mu\left[f,f^{(0)}_k\right]{\rm{ }}{\rm{.}} \la{7.1} \eea $$
The last term is simply the difference of (7.1) and the first
terms at the right-hand side of (8.1). The quantity $K^\mu$
vanishes when $\dl \overline{f}_k$ vanishes. In fact, an expansion
of the integrand shows that $K^\mu$ is quadratic in the deviation
from local equilibrium:
$$ \be K^\mu = O\left[ \left(
\dl \overline{f}_k \right)^2 \right] {\rm{ }}{\rm{.}} \la{7.2} \ee
$$
Like before such terms will be neglected. So we are left with the
first term at the right-hand side of (8.1).
Let us substitute (7.5) with (7.6). We then recognize the
expression for the current density (3.5) and energy-momentum
tensor (3.13), so that we can write
$$ \be S^\mu (x) = \sum
\limits_{\rm A} \al_{\rm A}(x) J^\mu_{\rm A}(x) +\bt_\nu
(x)T^{\mu\nu}(x) +\bt^\mu (x)P^{(0)}(x) {\rm{ }}{\rm{,}} \la{7.3}
\ee $$
with $\bt^\mu =(\bt,\vecbt)$. In equilibrium this expression
reduces to the familiar Euler relation of thermodynamics.
The last term is the local-equilibrium pressure defined by
$$ \be \bt (x)
P^\nul(x)=\et\sum \limits_{k} \int d\om_k\, \log \left[1 + \et
f^\nul_k(x,p)\right]{\rm{ }}{\rm{.}} \la{7.4} \ee $$
Notice that the integral is invariant for a shift $\vecp
\rightarrow \vecp + m_k \vecv$ of the integration variable. Using
this fact, it follows by explicit calculation that we may write
$$ \be \dl \left(\bt^\mu
P^\nul \right)= - \sum \limits_{\rm A} J^{\nul\mu}_\rA \dl
\al_\rA- T^{\nul\mu\nu} \dl \bt_\nu {\rm{ }}{\rm{,}} \la{7.5} \ee
$$
which may be regarded as a generalisation of the thermodynamic Gibbs-Duhem
relation. The superscript $(0)$ indicates that the currents
and energy-momentum tensor are calculated with $f^{(0)}_k$.
Because of the "conditions of fit" (7.10)-(7.12), we have
$$ \be N_\rA -N^\nul_\rA = 0
{\rm{ }}{\rm{,}} \la{7.6} \ee $$
$$ \be T_{\mu 0} -T^\nul_{\mu
0} = 0 {\rm{ }}{\rm{.}} \la{7.7} \ee $$
No such restrictions are imposed on the other components, and
these determine, in fact, the entropy production as we shall see
now.
Exercise 9
- Make sure that the thermodynamic Gibbs and the Gibbs-Duhem
relations are valid outside equilibrium if quadratic
deviations from local equilibrium are neglected.
We calculate the entropy production by taking the four-divergence
of (8.3). On account of the conservation laws (3.6) and (3.12),
and the above equations (8.5) through (8.7), we obtain
$$ \bea \pt_\mu S^\mu =
&\sum \limits_{\rA}& \left( \vecJ_\rA
-\vecJ^\nul_\rA\right) \cdot \nabla\al_\rA + \left(\vecJ_E -
\vecJ^\nul_E \right) \cdot \nabla \bt \\ &-& \bt\left({\sf
P} -P^\nul {\sf I}\right) : \nabla\vecv {\rm{ }}{\rm{.}} \la{7.8}
\eea $$
Hence, the entropy production is a bilinear expression in terms
of gradients and so-called irreversible flows
$$ \be \vecI_\rA = \vecJ_\rA
-\vecJ^\nul_\rA\ {\rm{ }}{\rm{,}} \la{7.9} \ee $$
$$ \be \vecI_q = \vecJ_E -
\vecJ^\nul_E {\rm{ }}{\rm{,}} \\ \la{7.10} \ee $$
$$ \be {\sf \Pi} = {\sf P }
-P^\nul {\sf I} {\rm{ }}{\rm{.}} \la{7.11} \ee $$
In equilibrium and local equilibrium these flows are zero. Note
especially the definition of the viscous pressure tensor (8.11)
which is the difference between the pressure tensor (3.11), and
the local equilibrium pressure as defined in (8.4).
Exercise 10
- The requirement that the viscous pressure tensor (8.11)
vanishes in local equilibrium may be used to define the
`hydrostatic' pressure $P^\nul$. Is this definition compatible
with the `thermodynamic' definition (8.4)?