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A gas left to itself, like any macroscopic system, will tend to a
state of thermal and chemical equilibrium. In this state the
entropy of the system reaches its maximum value. A necessary
condition for complete equilibrium is, therefore, that the entropy
production vanishes everywhere in space-time.
If we define an entropy density and flux $S^\mu =
(S,\vecJ_S)$ according to
$$ \be S^\mu=-\sum \limits_{k}
\int d\om_k\, v^\mu_k \left[f_k \log f_k - \et(1 + \et f_k)\log (1
+ \et f_k)\right]{\rm{ }}{\rm{,}} \la{6.1} \ee $$
we can write the entropy law (2.1) as
$$ \be \pt_\mu S^\mu (x)=
\sg (x) {\rm{ }}{\rm{,}} \la{6.2} \ee $$
and with the help of the kinetic equation (4.2) we find for the
entropy production
$$ \be \sg = - \sum
\limits_k \int d\om_k\, C_k \log \frac{f_k}{1+\et f_k} {\rm{
}}{\rm{.}} \la{6.3} \ee $$
Since the establishment of statistical equilibrium in a gas is
brought about by collisions, the increase of entropy arises from
the collision integral which must be such that $\sg \geq 0$. The
proof that this is indeed the case is known as the H-theorem.
- The H-theorem was the great success of Ludwig Boltzmann who in
1872 provided the proof for the collision integral (5.2). By an
analogous reasoning the H-theorem for the Uehling-Uhlenbeck
collision integral (5.4) can be established.
Exercise 7
- Write down the entropy production (7.3) in the linear
approximation, and verify the H-theorem for this case.
It is remarkable that the condition $\be \sg(x) = 0 \ee $ is in
itself not sufficient for complete equilibrium. The reason is,
what else, the existence of conservation laws. Indeed, if we
substitute in (7.3) a distribution function $f^{(0)}_k (x,p)$ such
that
$$ \be \log
\frac{f^{(0)}_k}{1+\et f^{(0)}_k}= -\psi_k \la{6.5} \ee $$
with $\psi_k$ a linear combination of collision invariants
$$ \be \psi_k (x,p) = \sum
\limits_{\rm A} q_{\rm A k} \al_{\rm A}(x) +\bt(x)\ep_k -
\vecbt(x) \cdot \vecp + \gm (x){\rm{ }}{\rm{,}} \la{6.6} \ee $$
then equation $\be \sg(x) = 0 \ee $ is satisfied identically in
virtue of the integral constraints (4.3) through (4.5). The
parameters $\al_{\rm A}(x)$ etc. can be arbitrary functions of
space and time. One calls $f^{(0)}_k$ a local-equilibrium
distribution function.
By using the fact that distribution functions transform as
scalars under Galilei (as well as Lorentz) transformations, one
may show that the vector field
$$ \be
\vecv(x)=\vecbt(x)/\bt(x) \la{6.7} \ee $$
transforms as a velocity (see exercise 4), and that the parameter
$\gm$ is equal to $\bt\half mv^2$. For convenience we shall always
neglect such non-linear terms.
It may seem puzzling that there exists a distribution function
$f^{(0)}_k$ different from $n_k$, for which the entropy production
vanishes. However, one should keep in mind that $f^{(0)}_k$ is not
a solution of the kinetic equation and, as a consequence, does not
represent a real state of the system. The actual distribution
function always deviates from $f^{(0)}_k$, although close to
equilibrium the deviation may be small. Therefore, a clear
distinction must be made between the deviation from total (global)
equilibrium:
$$ \be \dl f_k = f_k - n_k=
n_k(1+\et n_k)\ph_k{\rm{ }}{\rm{,}} \la{6.8} \ee $$
and the deviation from local equilibrium
$$ \be \dl \overline{f}_k =
f_k - f^{(0)}_k= n_k(1+\et n_k)\overline{\ph}_k{\rm{ }}{\rm{.}}
\la{6.9} \ee $$
If we assume that deviations like $\dl \al_{\rm A}(x) = \al_{\rm
A}(x) - \al_{\rm A}$, $\dl\vecv (x)=\vecv(x)$, etc. are small, we
find that the two deviation functions are connected by
$$ \be \overline{\ph}_k
(x,p)= \ph_k (x,p) + \sum \limits_{\rm A} q_{\rm A k} \dl \al_{\rm
A}(x) + \dl \bt(x)\ep_k - \bt \dl \vecv(x) \cdot \vecp {\rm{
}}{\rm{.}} \la{6.10} \ee $$
- The difference is a linear combination of collision
invariants.
The local-equilibrium distribution function may be regarded as a
zeroth-order approximation in an expansion with respect to
$\overline{\ph}_k $, called the Chapman-Enskog expansion.
This raises the question what significance should be attributed to
the arbitrary functions $\al_{\rm A}(x)$ etc. How are these functions related to the
actual state of the system? In analogy with equilibrium
thermodynamics, it seems natural to consider these functions as
Lagrange multipliers, that is, to choose them such that the
conserved densities are locally determined by $f^{(0)}_k$:
$$ \be N_{\rm A}(x)= \sum
\limits_k q_{{\rm A}k} \int d\om_k\, f^{(0)}_k (x,p) {\rm{
}}{\rm{,}}\\ \la{6.11} \ee $$
$$\be E(x)= \sum \limits_k
\int d\om_k\, \ep_k f^{(0)}_k (x,p) {\rm{ }}{\rm{,}}\\ \la{6.12}
\ee $$
$$ \be \vecG(x)= \sum
\limits_k \int d\om_k\, \vecp f^{(0)}_k (x,p) {\rm{ }}{\rm{.}}
\la{6.13} \ee $$
Of course, other quantities, especially the flows, are not
correctly given by $f^{(0)}_k$. One may look upon the above
equations as definitions of the local chemical potentials
$\mu_{\rm A}(x)$, temperature $T(x)$, and hydrodynamic velocity
$\vecv(x)$ in terms of the hydrodynamic variables. By this
construction the non-equilibrium thermodynamic parameters are
uniquely defined.
- Since $f^{(0)}_k$ has the equilibrium form, local
thermodynamic relationships have the same form as in full
equilibrium, provided that the non-equilibrium entropy density
can also be defined in terms of $f^{(0)}_k$. This will be
investigated next.
Exercise 8
- Show that $\vecv (x)$ is indeed the hydrodynamic velocity:
$\vecG (x) = \rh (x)\vecv (x)$.