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After these preliminaries, we shall now construct a solution of
the linearised kinetic equation (6.8) for the purpose of
calculating the irreversible flows, which, as we recall, are
nothing but the non-equilibrium parts of the charge flows and the
energy-momentum tensor. We follow the Chapman-Enskog method and
write
$$ \be \left( \ptt + \vecv_k
\cdot \nabla\right) f^{\nul}_k = -L_k [\phi] {\rm{ }}{\rm{,}}
\la{9.1} \ee $$
with the linearised collision operator given by (6.9), here and
henceforth denoted as $L_k [\phi]$. At the left-hand side we have
replaced $f_k$ by its zeroth-order approximation. However, one
should note that for consistency with (6.10) we must have
$$ \be \sum \limits_{k} \int
d\om_k\, \ps_k \left( \ptt + \vecv_k \cdot \nabla\right)
f^{\nul}_k=0 {\rm{ }}{\rm{.}} \la{9.2} \ee $$
With the help of this equation the time derivative can be
eliminated in favour of the gradients of the thermodynamic
parameters. The details of this procedure, which may be pictured
as a projection orthogonal to the collision invariants, are rather
involved. Nevertheless, the final answer is easy to guess. Let us
take a look at the gradient part in (10.1). Neglecting as usual
quadratic terms, we get with (7.5), (7.6)
$$ \be \vecv_k \cdot \nabla
f^{\nul}_k =- n_k (1 + \et n_k) \left[ \sum \limits_{\rA} q_{\rA
k} \vecv_k \cdot \nabla \al_{\rA} + \ep_k \vecv_k \cdot \nabla \bt
- \bt \vecp \vecv_k : \nabla \vecv \right] {\rm{ }}{\rm{.}}
\la{9.3} \ee $$
If we now compare the various factors multiplying the gradients
with the dissipative currents (9.5), (9.8), and (9.12), we observe
that we can satisfy the orthogonality requirement by merely
replacing these factors with the appropriate dissipative currents.
This suggests that we should write
$$ \be n_k (1 + \et
n_k)\left[ \sum \limits_{\rA} \vecj_{\rA k} \cdot \nabla \al_{\rA}
+ \vecj_{\rA k} \cdot \nabla \bt - \bt \overset{\circ}{\sf \pi}_k
: \nabla \vecv \right] = L_k[\phi]{\rm{ }}{\rm{.}} \la{9.4} \ee $$
- Exactly the same answer is obtained by the standard
elimination procedure.
We can now cast the kinetic equation (10.4) into an elegant and
convenient form by defining the inner product
$$ \be (\chi|\phi)= \sum
\limits_{k} \int d\om_k\,n_k (1 + \et n_k)\chi_k\phi_k \la{9.5}
\ee $$
for two sets of momentum dependent functions $\chi_k$ and
$\phi_k$. One may verify that this definition fulfils the basic
axioms of an inner product. Furthermore, we define a so-called collision
bracket
$$ \be (\chi|L|\phi) = \sum
\limits_{k} \int d\om_k\,\chi_k \, L_k [\phi] {\rm{ }}{\rm{,}}
\la{9.6} \ee $$
for arbitrary $\phi$'s and $\chi$'s.
It is tempting to read (10.6) as the matrix element of an
operator $L$ in an abstract Hilbert space of states $|\phi)$, and
nothing prevents us from doing so. The kinetic equation can then
be represented as
$$ \be |X) = L\,|\ph){\rm{
}}{\rm{,}}\\ \la{9.7} \ee $$
$$ \be |X) = \sum
\limits_{\rA} |\vecj_\rA)\cdot \nabla \al_\rA +|\vecj_q) \cdot
\nabla \bt - |\overset{\circ}{\sf \pi} ) :
\overset{\circ}{\overline{\nabla \vecv}}{\rm{ }}{\rm{,}} \la{9.8}
\ee $$
where the last symbol denotes the traceless strain rate
$$ \be
\overset{\circ}{\overline{\nabla \vecv}}=\frac{1}{2} \left(\nabla
\vecv + \vecv \overleftarrow{\nabla} \right) - \frac{1}{3}
\left(\nabla \cdot \vecv\right) {\sf I} {\rm{ }}{\rm{.}} \la{9.9}
\ee $$
In the new notation the irreversible flows appear as
$$ \be \vecI_q =
(\vecj_q|\ph){\rm{ }}{\rm{,}} \\ \la{9.10} \ee $$
$$ \be \vecI_\rA =
(\vecj_\rA|\ph) {\rm{ }}{\rm{,}}\\ \la{9.11} \ee $$
$$ \be \overset{\circ}{\sf
\Pi} = (\overset{\circ}{\sf \pi} | \ph) {\rm{ }}{\rm{.}} \la{9.12}
\ee $$
Exercise 14
- Check the equivalence of (10.4) and (10.7).
Now suppose that the kinetic equation has a solution, which means
that the collision operator is invertible:
$$ \be |\ph) = L^{-1} |X)
{\rm{ }}{\rm{.}} \la{9.13} \ee $$
This implies then for the flows
$$ \be \vecI_q = (\vecj_q
|L^{-1} |X){\rm{ }}{\rm{,}} \la{9.14} \ee $$
etc.
Let us first assume that there is only a temperature gradient in
the system. Then we get for the heat flow the linear law
$$ \be \vecI_q = - \lm
\frac{\nabla T}{T} {\rm{ }}{\rm{.}} \la{9.15} \ee $$
In deriving this, one uses that a gas in equilibrium is isotropic
so that inner products between vectors and tensors can only be
expressed in terms of the unit tensor, e.g.
$$ \be \bt (\vecj_q | L^{-1}
|\vecj_q ) =\lm \, {\sf I} {\rm{ }}{\rm{.}} \la{9.16} \ee $$
This relation defines the heat conductivity coefficient
which also can be written as
$$ \be \lm =\frac{1}{3}
\,\bt (\vecj_q | L^{-1}|\vecj_q) {\rm{ }}{\rm{,}} \la{9.17} \ee $$
where now a contraction of vector indices is understood.
Similarly, one finds for the viscous pressure
$$ \be \overset{\circ}{\sf \Pi}
= -2\et\, \overset{\circ}{\overline{\nabla \vecv}} {\rm{
}}{\rm{,}} \la{9.18} \ee $$
with the viscosity coefficient given by
$$ \be \et = \frac{1}{10}
\bt (\overset{\circ}{\sf \pi}| L^{-1} |\overset{\circ}{\sf
\pi}){\rm{ }}{\rm{,}} \la{9.19} \ee $$
where again a contraction of tensor indices is meant.
Because of the isotropy of the system only the phenomena with the
same tensor character couple to each other. This is called Curie's
principle. In the present case this implies that there is no
coupling between viscosity, on the one hand, and heat conduction
and diffusion, on the other. However, the latter are both vector
phenomena so that diffusion couples to heat flow and vice versa.
These phenomena are called the Soret effect (thermal
diffusion) and Dufour effect, respectively.
Exercise 15
- Show that the complete set of linear laws for heat
conduction and diffusion has the form $$ \bean \vecI_q
&=& \lm \frac{\nabla \bt}{\bt} + \sum \limits_{\rB}
D_{q\rB} \nabla \al_\rB \\ \vecI_\rA &=& \sum
\limits_{\rB} D_{\rA\rB} \nabla \al_\rB + D_{\rA q}
\frac{\nabla \bt}{\bt} \eean $$ and write down expressions for
the transport coefficients figuring here.
Without going into more detail, the general structure of an
arbitrary transport coefficient should be clear by now. Apart from
a numerical factor, they have the form
$$ \be \kp = \bt (j| L^{-1}
|j) \la{9.20} \ee $$
of the inverse collision operator sandwiched between the relevant
dissipative currents. We can also write
$$ \be \kp = \bt \int_0^\infty
dt \, (j| e^{-Lt} |j){\rm{ }}{\rm{.}} \la{9.21} \ee $$
The exponential operator is the evolution operator
associated with the linearised kinetic equation. This suggests
that we write the integrand as
$$ \be (j|e^{-Lt} | j) =
(j(t)|j) {\rm{ }}{\rm{.}} \la{9.22} \ee $$
Such an inner product of dynamical variables at different times
is called a time-correlation function. It is a measure for
the memory of the system.
- Hence, transport coefficients are time integrals of
time-correlation functions. General expressions of the kind
(10.21) are referred to as Green-Kubo formulae [5].