We left off with:

<dv(t) )>_{u } =

(e
E o/ m _{e }) ^{2 }½ k^{2 } __v___{ o} ^{2 }** ****| **2* *| cos (k__ v___{ o }t) –1] +

k__ v___{ o }t sin
(k__ v___{ o }t) ]

And restate the aim here is
to obtain the change in energy

< E o > x o = m _{e }v_{ o }d v x o

and integrate over all velocities v_{ o } in the lab frame to find the total change in
energy of the particles. Before
proceeding we evaluate the top most equation (we start with) at a time such
that k__ v___{ }t <<1.

We make use of the following approximation (to series) formulae:

sin x » x - x^{3}/ 6 + ……

cos x » 1 – x ^{2}/ 2
+ x^{4}/ 24
+ ….

And we find:

2 (cos x – 1) + x sin x
» – x ** ^{4}**/1 2

Using this, our original eqn. becomes:

<d v(t) )>_{u } = (e
E o/ m _{e }) ^{2 } k^{2 } __v___{ o} t ^{4}/ 24 ^{ }** ** (k__ v___{ }t <<1}

Which shows that beams with __v__ >
0 (i.e. faster than the wave) are
slowed down at early times, in the sense of an average over x_{ }_{o }. Beams moving slower than the wave are
conversely sped up. Now we use the
original equation to find the total energy change W(t):

W(t) = ò ^{¥}_{-}_{¥} dv_{ o} f_{ o} (v_{ o} ) (DE)xo =

ò ^{¥}_{-}_{¥} dv_{ o} f_{ o} (v_{ o} ) m _{e }v_{ o }d v (x o)

Here we make the change of
variable: ~v_{ o} = v_{ o } - v _{f}

Then:

W(t) = m _{e } ò ^{¥}_{-}_{¥} d(~v_{ o}) ~f_{ }(~v_{ o}) ( ~ v_{ o }+ v _{f}
) (Dv )x o

Where: ~f(~v_{ o}) __=__
f_{ o}__ __

We expect the major
contribution to W(t) to come from particles with velocities close to ~ v_{ o }» 0._{ }

This is given the original
equation for dv x o

varies as (~v_{ o})^{ -3}

Therefore we expand:

~f (~v_{ o}) » ~f (0) + (~v_{ o}) ~f (0)

After taking the product

~f (~v_{ o}) ( ~ v_{ o }+ v _{f}
)

And working through the algebra, one arrives at:

W(t) =

m _{e }v _{f} (~f (0))/ 2 k^{2 }(e E o/ m _{e }) ^{2 } ò ^{¥}_{-}_{¥} d(~v_{ o})/ (~v_{ o})^{2 }

[**| **2* *| cos (k ~v_{ o }t)
– 1] + k ~v_{ o }t
sin (k ~v_{ o }t)]

With
change of variable, i.e. x = k ~v_{ o }t

The
integral in the preceding equation can be written:

kt ò ^{¥}_{-}_{¥} dx /x_{ }^{2 }[2 (cos x
– 1) + x sin x

From a table of integrals:

ò ^{¥}_{-}_{¥} dx sin x/ x = p

While the other term yields
(- 2 p)

The
preceding integral can be evaluated using contour integration, i.e. by first
moving the contour off the nonexistent pole at z = 0. The sine is then expanded
in terms of exponentials. We obtain the
result:

W(t) = - p / 2 (m _{e }v _{f} /k) f’(0)
(e Eo/ m _{e}) ^{2 }t

Or:

W(t) = - p / 2 (m _{e }w_{ r }/k) f o’(v _{f}) (e E o/ m _{e }) ^{2 }t

Next,
identify f o with no g and
take: w_{ r }» _{ e}

So
that:

W(t) =
- (w_{ e})_{ }^{2}/ k_{ }^{2} g’(v _{f}) E o ^{2 }t

The
last eqn. shows the total particle energy is changing as the 1^{st}
power of time t and is positive when g’(v _{f}) < 0 and negative when g’(v _{f}) >
0. The energy gained or lost must come from the wave. The rate of change of the wave energy W _{wave} must be equal and
opposite to the rate of change of particle energy. Thus:

d/dt (W _{wave} ) = - d W(t) / dt = (w_{ e})_{ }^{2}/ k_{ }^{2} g’(v _{f})

The total wave energy
averaged over a wavelength is:

W _{wave} = p (w_{ e})_{ }^{3}/ k_{ }^{2} [ g’(v _{f}) W _{wave} ]

The E-field varies with time as:

E o(t) ~ exp
(g t)

Comparing
the two previous results we find:

g _{L}_{ } = (p/ 2) w _{e}_{ }^{3}_{ }_{ }/ k ^{2} [g' (v _{f} )]

Which
is the *Landau damping rate*.

*Note*: My thanks to Prof. J.R. Kan of Univ. Of Alaska -Fairbanks for the Parts (2), (3) derivations from his class notes in Advanced Space Physics, recorded in December, 1985.

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