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The collision integral (5.4) is a non-linear functional of the
distribution functions, and usually too difficult to handle in
practical calculations. The standard procedure is to linearise
around equilibrium by substitution of (2.3) and by expansion up to
first order in $\dl f$. In this way one gets:
$$ \be C_k[f]=\int \left[n_i
n_j (1+\et n_k)(1+\et n_l) - n_k n_l (1+\et n_i)(1+\et
n_j)\right]W_{ij|kl} -L_k[\dl f]{\rm{ }}{\rm{,}} \la{5.1} \ee $$
if also use is made of the detailed balance property
$$ \be
W_{ij|kl}=W_{kl|ij}{\rm{ }}{\rm{,}} \la{5.2} \ee $$
which need not always be valid, but usually is due to
time-reversal invariance.
We first consider the term written out explicitly. It should
vanish identically because in equilibrium the collision integral
must be zero. This implies the relationship
$$ \be \frac{n_i}{1+\et
n_i}\frac{n_j}{1+\et n_j}=\frac{n_k}{1+\et n_k}\frac{n_l}{1+\et
n_l}{\rm{ }}{\rm{.}} \la{5.3} \ee $$
We know that these special combinations of equilibrium
distribution functions have the form of exponentials
$$ \be \frac{n_i}{1+\et
n_i}=e^{-\psi_i}{\rm{ }}{\rm{.}} \la{5.4} \ee $$
Hence, condition (6.3) implies
$$ \be \psi_i +\psi_ j=\psi_k
+\psi_l{\rm{ }}{\rm{,}} \la{5.5} \ee $$
which is the generic form of a conservation law. As a
consequence, $\psi_k$ can only be a linear combination of the
collision invariants
$$ \be \psi_k = \sum
\limits_{\rm A} q_{\rm A k} \al_{\rm A} +\bt\ep_k - \vecbt \cdot
\vecp +{\gm}{\rm{ }}{\rm{.}} \la{5.6} \ee $$
For a system in rest one may show that $\vecbt$ and $\gm$ are
zero. Then the equilibrium distributions take the standard
Bose-Einstein or Fermi-Dirac form
$$ \be
n_k=\frac{1}{e^{\psi_k}-\et }{\rm{ }}{\rm{.}} \la{5.7} \ee $$
- The equilibrium distribution functions are characterized by
the inverse temperature $\bt=1/T$ , and the independent
chemicals potentials $\mu_{\rm A}=-\al_{\rm A}/\bt$
corresponding to the conserved charges.
- The above derivation shows that there exists an intimate
connection between the equilibrium distribution function and the
form of the collision integral.
Exercise 4
- Under a Galilean transformation $\vecp \rightarrow \vecp +
m\vecv$ the distribution function transforms as a scalar. Use
this fact to show that $\bt$ behaves as a scalar, $\vecbt$ as
a velocity, and that $\gm=\half m_k \vecbt^2 /\bt$.
With the result obtained above the linearised kinetic
equation for the deviation function $\dl f_k$ becomes:
$$ \be \left( \ptt + \vecv_k
\cdot \nabla\right) \dl f_k =-L_k[\dl f] {\rm{ }}{\rm{.}}
\la{5.10} \ee $$
We write $\dl f_k = n_k(1+\et n_k)\ph_k$ in the collision term.
We then get, collecting terms linear in $\ph_k$, using also
detailed balance (6.2) and the product rule (6.3), the linearised
collision operator
$$ \be L_k[\dl f_k]=\half \sum
\limits_{l,i,j}\int d\om_l d\om_i d\om_j\, n_k n_l (1+\et
n_i)(1+\et n_j)\left(\ph_k +\ph_l-\ph_i -
\ph_j\right)W_{ij|kl}{\rm{ }}{\rm{,}} \la{5.8} \ee $$
where the summations and integrations have been made explicit.
This symmetrical form makes it immediately clear that the
collision integral vanishes for any deviation function $\ph_k$
that satisfies (6.5), i.e. for any collision invariant $\ps_k$. By
interchanging dummy variables, it is also easy to show that one
has
$$ \be \sum \limits_{k}\int
d\om_k \ps_k L_k[\dl f] =0 \la{5.9} \ee $$
- As it should be, the linearised collision operator preserves
the conservation laws.
Exercise 5
- Verify equation (6.9) and show that the linearised collision
operator indeed preserves the conservation laws.
The linearised equation (6.8) is still rather hard to solve, in
general. For the purpose of a qualitative treatment it is
convenient to estimate the collision operator by means of the mean
free time $\ta_k$ which is defined as follows:
$$ \be (1+\et n_k)
{\ta_k}^{-1}=\half \sum \limits_{l,i,j}\int d\om_l d\om_i d\om_j\,
n_l (1+\et n_i)(1+\et n_j)W_{ij|kl}{\rm{ }}{\rm{.}} \la{5.11} \ee
$$
For a first estimate we can then put
$$ \be L_k[\dl f] \sim
{\ta_k}^{-1}\, \dl f_k{\rm{ }}{\rm{.}} \la{5.12} \ee $$
This is called the relaxation-time approximation because
$\ta_k$ acts as a relaxation time for the establishment of
equilibrium in each volume element of the gas.
Exercise 6
- Consider a one-component system and ignore the statistics
and spin factors. Calculate ${\ta_k}^{-1}$ given that $$ \be
W_{ij|kl}=\frac{(2\pi)^6}{m} \left| \vecp_i -\vecp_j
\right|\,\sg\, \dl^{(4)} \left( p_i +p_j -p_k -p_l \right) \ee
$$ with the so-called cross-section $\sg$ a constant.
The relaxation time still depends on the energy $\ep_k$. However,
various average relaxation times can now be defined by
$$ \be <\ta>^r = \left(
N^r \right)^{-1} \sum \limits_{k} \int d\om_k \, n_k (1+\et n_k)
(\ep_k)^r \,\ta_k {\rm{ }}{\rm{,}} \la{5.13} \ee $$
where the normalization factor $N^r$ is the same integral without
$\ta_k$.
- These averages, when employed with care, are useful for
estimating the value of transport coefficients; the value of $r$
varies according to which transport phenomenon is under
consideration.
We close with a few general remarks about the limitations of the
kinetic equation (6.8) with (6.9).
- In the first place it is obvious that only two-particle events
have been taken into account which seems to restrict the
treatment to dilute gasses. However, there is a way around this
problem which in practice works quite well. It consists in using
for the transition rate $W$ an effective transition rate which
describes scattering of particles in a medium rather than in
vacuum. In this way many-body correlation effects can be taken
into account.
- A second remark is that we have regarded collisions as
essentially instantaneous and occurring at a particular point in
space. It is therefore clear that the equation allows us in
principle to follow the variation of the distribution function
only over times long compared with the duration and extension of
a collision. The typical distance is of the order of the range
of the interaction. Such a value gives a lower limit of
distances that can be dealt with by means of the transport
equation. In practice, however, there is usually no need for a
more detailed account of the behaviour of the system.
- Finally we mention that there exist collision integrals other
than the Boltzmann one, to suit special purposes especially in
plasma physics. For the following this is hardly relevant
because the only essential property we will use is (6.10), which
is the immediate consequence of the existence of conservation
laws.