**Analytic graph for Landau damping- applied to Langmuir waves.**

Having seen how Landau damping scuttles the notion of ATTIP military -sighted UFOs as holograms (designed to confuse pilots), e.g.

It is instructive to see how the Landau damping rate is derived. We recall here that Landau damping is a result
of a solution to the dispersion relation involving w and k, i.e. the phase
velocity resulting is v _{f } = w /
k. We proceed then by considering the
effect of a small oscillating (*Langmuir*) wave associated with a perturbed plasma distribution f1(x,v,t) = 0.

Consider then a linear wave of form:

E1 (x’,t) =
E_{ o} sin (kx’ -
w _{r} t)

Where x’ denotes the lab frame of
reference. E_{ o} is a small constant for amplitude. In the frame of reference x moving with phase
speed v _{f } = w / k > 0 with respect to the laboratory frame the
wave field is independent of time t and is given by: E1 (x’,t) = E_{ o} sin (kx’)
This is represented in the graphic below. (E_{ o} amplitude position marked by bold horizontal line)

Note that all the particles in the background distribution plasma (f _{o} (v)) are affected by this non-time dependent
E-field. Some are speeded up, others slowed down.

For simplicity, we focus on only those that
have speed v_{ }_{o } in the lab frame at t = 0. Hence, they have speed v_{ }_{o } - v _{f} in the frame moving with the wave phase speed v _{f } = w / k. (See top graphic.)

Change in energy (d E ) of a given particle:

d E = ½ (m v_{ }_{o }+ ** _{ }**d v)

^{2}- ½ m v

_{ }_{o}

^{2}

_{ }= m v_{ }_{o }d v** _{ }**+ O (d v

^{2})

^{2}

Averaging the
energy change over one wavelength:

< d E >xo** ** _{ }= m v_{ }_{o
}< d v > _{xo }

Since d v can be calculated in any frame, we work in the wave frame. Then the change in velocity is:

d v =
v(t) - v_{ }_{o }= (-e) E_{ o }/ m ò^{t}_{ o} dt’ sin kx(t) dt’

x(t) must be along particle orbit so:

x(t’) =
x_{ }_{o }+ ò^{t}_{’ o} v(t”) dt”

Examine again the equation of motion:

mx” = -
e E_{ o} sin kx

For which we may write:

mx’ x”
= - e E_{ o} sin kx (dx/ dt)

d/dt (½ m x’^{2})** _{ }**= -e E

_{ o}/ k [ d/dt (cos kx)]

x’
= ** Ö 2** e** **E_{ o}**/ **km [cos kx
- cos kx_{ }_{o } ]

Now integrate both sides:

dx/[cos kx
- cos kx_{ }_{o } ] ^{½} = ** Ö (2** e** **E_{ o}**/ **km) dt

We subst. x = x_{ }_{o} + v_{ }_{o} t"' into integral:

ò^{t’’}_{ o} sin (kx_{ }_{o} + k v_{ }_{o} t"') dt’" =

- 1/k v_{ }_{o} [cos (kx_{ }_{o} + k v_{ }_{o} t"') - cos kx_{ }_{o} ]

And:

v(t") = v_{ }_{o }+ e E_{ o }/n_{e} kv** _{ }** [cos (kx

_{ }_{o}+ k v

_{ }_{o}t"') - cos kx

_{ }_{o}]

The preceding parameter to be subst. into:

x(t') = x_{ }_{o} + ò^{t’}_{ o} v (t") dt"

(*To be continued*)

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