## Monday, September 20, 2021

### Derivation of Landau Damping Rate - Conclusion

We left off with:

<dv(t) )>u   =

(e E o/ m e ) ½ k2  v o  | 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    -  f

Then:

W(t) = m e  ò ¥-¥  d(~v o) ~f  (~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 f )

And working through the algebra, one arrives at:

W(t)  =

m f (~f (0))/ 2 k(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 f /k) f’(0)  (e Eo/ m e) t

Or:

W(t)  = - p / 2  (m e  w r  /k) f o(f(e E o/ m e ) 2 t

Next, identify f o  with no g  and take: w »  w e

So that:

W(t)  =  -  (w e)  2/ k 2 g’(f)  E o 2  t

The last eqn. shows the total particle energy is changing as the 1st power of time t and is positive when  g’(f)   < 0 and negative when g’(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’(f)

The total wave energy averaged over a wavelength is:

W wave    =  p (w e) 3/ k  2 [ g’(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.