Solutions to Collisionless Shocks Problems

1) The Mach number M = 80 correlates to a lab –shocked hydrogen plasma at T =   10 ⁵    K and ion gyroradius of 0.22 cm  with  shock value Ω _es  >  Ω _e    (which can be shown on working out Ω _es   based on the result below).

Now, the magnitude of the associated  B-field before the shock is obtained via: 

 Ω _e  =     qB/ m _e

So:  B =    Ω _e   m _e  /  q 

=   (1.9 x 10 ⁷  /s  )  (9.1 x 10 ^-31 kg)/   (1.6 x 10 ^-19 C)

=   10 ^-4 T

This is the equilibrium value of B before the shock is applied. But from the shock equation:


B _m =  (2M – 1) B_o   =    (2(80)  – 1) 10 ^-4 T   =     0.015 T

   Which is the magnetic field associated with the collisionless shock.

2)The pure electron plasma frequency is:  

n   =   9  Ö N   =  9  Ö (10 ¹⁶ /m³)  =   9 x 10 ⁸  /s 

Then:  w_e  =   2p n  =  2p(9 x 10 ⁸  /s ) =   5.6 x 10 ⁹  /s  

Debye length l _D,  =

[(1.38 x 10^-23  )(10 ⁶K) (8.85 x 10 ^-12 F/m)  / (4p)  (10 ¹⁶/m³)  e² ] ^½

= 1.9 x 10 ^{- 4}   m

Plasma parameter:    L  =  n _o  l³ _D

=   ( 10 ¹⁶ /m³) (1.9 x 10 ^{- 4}   m) ³

L  =  7.4 x 10  ⁴      

The radio emission must exceed this 900 MHz frequency for propagation in the medium, so we need:

f >  9 x 10 ⁸  Hz = 900 MHz

3)  Let u   »  w_e/ k _w   =  10 ⁵ m/ sec

w_e =   2p (9 x 10 ⁸  /s )  =    5.6 x 10 ⁹ /s  

k _w   =   w_e/ u   »  (5.6 x 10 ⁹  /s )/ 10 ⁵ m/ s

k _w =  5.6 x 10 ⁴  m^-1

4)  We need to re-arrange the equation to solve for velocity v _o:

Using algebra we find:

v _o    =  w/ k _w   -   1/ k _w  [1/(w_p ² -  (m/M)/ w²] ^1/2

Substitute values:     v _o    =

 (10 ¹⁰ s^-1 )/ 5.6 x 10 ⁴  m^-1     - 

 1/(5.6 x 10 ⁴ m^-1) [1/(5.6 x 10 ⁹ s^-1)  ² -  (5.4 x 10 ⁴ )/ (10 ¹⁰ s^-1) ²] ^1/2

v _o    =   7.8 x 10 ⁴  ms^-1

5)  The shock equation is:            B _m =  (2M – 1) B_o

We can estimate the value of M from the ratio of the velocities: u/ v

Where u/ v   =    (10 ⁵  ms^-1)   / ( 7.8 x 10 ⁴  ms^-1 )  »   1.3

 Then:    B _m =  (2M – 1) B_o   »    (2(1.3) -1 ) B_o    »  (2.6 - 1) B_o   »   1.6 B_o

6)   The form of the result yields 9 as a constant outside the root sign so we must have:

9 »  e/2p   Ö( 1 / ε_o m _e )

Compute:     e/2p   Ö( 1 / ε_o m _e )  =

 1.6 x 10 ^{- 19}   C/ 2p  Ö( 1 / (8.5 x 10 ^-12 F/m)( 9.1 x 10 ^{- 31}   kg) 

=  8.973 Ö N   or  n   »    9  Ö N