Friday, November 23, 2018

Space Plasma Physics Revisited (3)

A space (or laboratory) plasma may be regarded as “charge neutral” if it exists or occurs on spatial scales that exceed the Debye length.  This is a characteristic length for a plasma defined by:

l D =[kT  eo / 4p  n e  e2 ] 1/2


Where k is Boltzmann’s constant and e is the electronic charge. Substituting the values for those two constants one can write a simpler numerical form:


l D =10 0.84   (ÖT/ Ön)

 
Technically, this applies only to electrons. The ions will have a Debye length peculiar to the species of ion, e.g. for helium ions one will use Z = 2 and Lithium ions Z = 3.  For this ion-species based Debye length we may write:

 
l D,s =[kT s eo / 4p  Z s2 n s  e2 ] 1/2
 
 
Where the subscript ‘s’ refers to the ion species.

 
Example Problem: A helium plasma is at a temperature of  10 6 K, and exhibits a number density of  10 10 /m3. Find the Debye length.

 
Solution:

 For helium plasma we have Z s =2.   T s= T He = 10 6 K


 n s  =  n He  =  10 10 /m3

 
l D,s   =[(1.38 x 10-23  (10 6 K) (8.85 x 10 -12 F/m)   / (4p)  22 (10 10/m3)  e2 ] ½

 
l D,s   =  0.30    m

 
Note that in the case of a ‘mixed’ plasma, electron and ion components, the charge imbalance between the electron and ion clouds shields the potential of the test charge outside the Debye sphere (i.e. a sphere with the radius r =  l D,s  )   Thus for r << l D,s     the electric field of the test charge is unaffected by the presence of the plasma.

However, for  r >> l D,s    the E-field is attenuated by an exponential factor (exp –r/ l D) The test charge is then said to be shielded by the plasma with the shielding distance measured by  l D.
 
For a system to be regarded as a plasma we require the medium be almost neutral, nearly all the time. Given electron and ion distributions:
 
n e = n o exp [e j/ Te ]
 
and:    n i = n o exp [e j/ Ti ]


We have:     Ñ 2 F      =  e (n e  -   n i)/  e o


Which is Poisson’s equation, so the potential  F  must be satisfied in S.I. units..

 Adjusting the potential for the case  for  r >> l D,s   we have:


j D  = q / 4p eo r  (exp –r/ l D)


Which also goes by the name, “Yukawa potential”. It is of interest to note that because of the exponential factor it decreases much more rapidly than the Coulomb potential. This means the E-field from the test charge q is effectively shielded at distances much larger than  l D .

In addition to the Debye length scale we also need to use the plasma parameter, L, defined:

  L  =  n o  l3 D

 
This definition implies:


 L  =  n o  l3 D  >> 1


Which in turn implies that the number N of particles in a Debye sphere:

N  =   4p n o  l3 D  / 3 

 
Is much larger than unity. This is consistent with the principle of shielding we have laid out. Hence, a considerable shielding of individual charges only occurs on the Debye length scale if there are sufficient charges in the Debye sphere of each individual particle.

 
Problems :

1)For the helium plasma in the example problem, compute the plasma parameter, and the number of particles in the Debye. sphere.

2)Compare your values with the same ones for the solar atmosphere, for which:
  n o = 10 20 /m3  and T = 10,000K.

3)Compare the values for a laser fusion plasma  with a particle density n o = 10 29 /m3  and  temperature T = 10 7   K)  to the plasma of the tail of Earth’s magnetosphere with:
 n o = 10 6 /m3  and T = 10 7   K

 
4)Can a plasma with n o = 10 6 /m3    be maintained at an electron temperature of 100K?  (Hint: Calculate the density limit using the plasma parameter).


5) In the limit 1/N << 1 and 1/N ® 0  we say a plasma is “collisionless”. Do any of the plasmas cited in the previous problems qualify as collisionless?

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