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The irreversible flows may be expressed in terms of $\dl f_k$,
the deviation from global equilibrium. Take e.g. the heat flow
(7.10). The first term can be written, see (2.8),
$$ \be \vecJ_E = \sum
\limits_{k} \int d\om_k\,\ep_k \vecv_k \dl f_k {\rm{ }}{\rm{,}}
\la{8.1} \ee $$
since the energy flow vanishes in equilibrium. Furthermore, the
zeroth-order flow may be calculated by a shift of integration
variables as
$$ \be \vecJ^\nul_E= (E+P)
\vecv {\rm{ }}{\rm{,}} \la{8.2} \ee $$
where the factor in front is the equilibrium enthalpy density.
For the hydrodynamic velocity we can write
$$ \be \rh \dl \vecv =\dl \vecG
= \sum \limits_{k} \int d\om_k\,\vecp \, \dl f_k {\rm{ }}{\rm{.}}
\la{8.3} \ee $$
So we finally arrive at the heat flow
$$ \be \vecI_q = \sum
\limits_{k} \int d\om_k\, \vecj_{qk} \,\dl f_k {\rm{ }}{\rm{,}}
\la{8.4} \ee $$
expressed in terms of $\dl f_k$ and the so-called dissipative
current
$$ \be \vecj_{qk} = \left[
\ep_k - m_k \rh^{-1} (E+P)\right]\, \vecv_k {\rm{ }}{\rm{,}}
\la{8.5} \ee $$
in which the enthalpy flow is subtracted from the energy flow.
With regard to the viscous pressure we follow a similar
reasoning and we find the sum
$$ \be {\sf \Pi} =
\overset{\circ}{\sf \Pi} + \Pi \,\,\sf I \la{8.6} \ee $$
of a shear tensor part
$$ \be \overset{\circ}{\sf
\Pi} = \sum \limits_{k} \int d\om_k\, \overset{\circ}{\sf \pi}_k
\,\dl f_k {\rm{ }}{\rm{,}} \la{8.7} \ee $$
$$ \be \overset{\circ}{\mathsf
\pi}_k = \vecp \vecv_k -\frac{1}{3} (\vecp \cdot \vecv_k) \sf I
{\rm{ }}{\rm{,}} \la{8.8} \ee $$
and the scalar volume, or bulk, viscous pressure
$$ \be \Pi = \frac{1}{3} \sum
\limits_{k} \int d\om_k\, \vecp \cdot \vecv_k f_k - P^\nul {\rm{
}}{\rm{,}} \la{8.9} \ee $$
i.e. the difference between the `mechanical' and the
`hydrostatic' pressure. In general, $\Pi$ is non-vanishing for
fluids, but for the ideal non-relativistic gases we consider here,
one finds
$$ \be \Pi = \frac{2}{3}
\left( E - E^\nul \right) = 0 {\rm{ }}{\rm{,}} \la{8.10} \ee $$
in virtue of the condition of fit (7.7), and the equation of
state $P^\nul = \twothird E^\nul$.
Exercise 11
- Show that ultra-relativistic ideal gases have no volume
viscosity.
Lastly, there are the flows (7.9) which describe diffusion.
Without derivation we write
$$ \be \vecI_\rA = \sum
\limits_{k} \int d\om_k\, \vecj_{\rA k} \, \dl f_k {\rm{
}}{\rm{,}}\\ \la{8.11} \ee $$
$$ \be \vecj_{\rA k} = \left(
q_{\rA k} - m_k \rh^{-1} N_\rA \right) \vecv_k {\rm{ }}{\rm{.}}
\la{8.12} \ee $$
The thermodynamic quantities between the parenthesis refer to
equilibrium.
Exercise 13
- Show that $\vecI_\rA = 0$ for a simple non-relativistic gas.
The three dissipative currents (8.5), (8.8), and (8.12) have one
property which is very important for the following, namely, they
are orthogonal to the collision invariants in the sense
$$ \be \sum \limits_{k} \int
d\om_k\,\ps_k \left( \vecj_{\rA k},\,\, \vecj_{q k},\,\,
\overset{\circ}{\sf \pi}_k \right)n_k(1 + \et n_k) = 0 {\rm{
}}{\rm{.}} \la{8.13} \ee $$
For $\ps_k =q_{\rA k} ,\ep_k$ the proof is trivial. To see the
truth of the statement for $\ps_k = \vecp$ one needs the
identities below, which are proved with the aid of a partial
integration.
$$ \be N_\rA = \frac{1}{3} \bt
\sum \limits_{k} q_{\rA k} \int d\om_k\, \vecp \cdot \vecv_k\,
n_k(1 + \et n_k) {\rm{ }}{\rm{,}} \la{8.14} \ee $$
$$ \be E+P= \frac{1}{3} \bt
\sum \limits_{k} \int d\om_k\, \ep_k \,\vecp \cdot \vecv_k \,n_k
(1 + \et n_k) {\rm{ }}{\rm{,}} \la{8.15} \ee $$