Deriving The Landau Damping Rate Of An E-Field (1)

 

             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.

• Are Laser Plasma Holographs Created By "Foreign A...

 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 _o²     

=   m v _o  d v  +  O (d v ²) 

  where we ignore the term in d v 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)