Derivation of Landau Damping Rate - Conclusion

 We left off with:

  <dv(t) )>_u   =  

(e E o/ m _e ) ²  ½ k²  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 -  x³/ 6  +  ……

cos x  »  1 – x ²/ 2  +  x⁴/ 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   k²  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 k²  (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  »  w _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.