Expert: Philip A. Stahl - 1/4/2012

Question
Your assistance in the previous self-study problem I had was so much appreciated, I wondered if you could help me with this one?

An isotropic particle distribution of 10^13 particles is suddenly introduced into an auroral magnetic bottle which has a mirror ratio Rm = 20. Find: a) the bounce period of a particle trapped in the mirror system, and b) find the fraction of and also the number of particles that will be lost.

c) If a cosmic ray proton of energy 1 keV becomes trapped in the bottle at its midpoint and if v_perp  = v _par  there, find its gyro-radius if B’ = 0.00001T if B’ denotes the magnetic field intensity at the bottle’s midpoint. What would then be the maximum field intensity for the auroral magnetic bottle, the loss cone angle and the condition for particles to escape? The condition for particles to be reflected?

Thanks!

Answer
Hello,

Okay, like the previous solution I provided for your self-study this is fairly straightforward. You are given the mirror ratio, which we know is:

R_m = [B (max)/ B(min)] = 20   e.g. B(max) = 20 B(min)

Since we also know:

R_m = 1/ (B(min) / B (max)]

Then the bounce period is:

t(b) = 4 INT (0 to Θ) [ds/ v‖ ]

where ds is an element of arc length along the field line (B) and an integration is performed between 0 and Θ, for which we find:

v ‖ =   v⊥ [1 – B(min)/B_max]^½  

The preceding integral can’t be solved analytically but can be done numerically. You do not specify any values other than the mirror ratio, and the particles aren’t identified. But the numerical integrations yield for electrons and protons, respectively (cf. ‘Introduction to Plasma Physics’, by Richard Fitzpatrick, p. 42):

t(b) e = (0.0056)  x L/ [E(Mev)]^½  {1 – 0.43 sin φ} s

where L is the length of the mirror system, E the energy of the elections in MeV and φ, the "equatorial" pitch angle. (I.e. φ' = arc sin [B(eq)/B]^½  φ)

Recall the generic pitch angle is defined φ = arc tan { v⊥/v‖}

Meanwhile, for the protons, the numerical integration yields (ibid.):

t(b)p = (2.41)  x L/ [E(Mev)]^½  {1 – 0.43 sin φ} s

Again, unless specific values are known, i.e. for L, E etc. this is all that can be done, or just leaving t(b) in the original integral format.

The fraction of and number of particles lost is easier.  If an isotropic particle distribution  is introduced at a position, say L = 0,  the fraction of particles that will be lost to the mirror system is:

f(L) =  1/ 2 π  INT (0 to Θ (L)) 2 π sin(Θ) d Θ


= f(L) =  INT (0 to Θ (L)) sin(Θ) d Θ

=  1 - cos (Θ)_L

And we already know (from previous problem worked):

Θ_L =  ± arc sin  [ (B(min) / B (max)]^½  

Or (using data on mirror ratio):

Θ_L =  ± arc sin  [ 1/20]^½  = ± arc sin  [ 0.05]^½  

Θ_L =  ± 2.8 degrees

Then the fraction of particles that will be lost is:

f(L) =  1 - cos (Θ)_L = 1  - cos (2.8 ) = 1 – 0.998 = 0.002

And the number of particles lost, predicated on the number initially introduced:

N(L) = 0.002 (10^13) = 2 x 10^10


Now, in the second part of your problem reference is made to  “a cosmic ray proton of energy 1 keV” – but this has to be an *initial* energy.  This energy will increase because while trapped in the auroral bottle it is going to ‘bounce’ and that means reflection between mirror points (e.g. at +/- L) Therefore in each reflection the velocity changes by 2 v_m, where v_m is the initial velocity at the midpoint.  From the data provided this will be: v_m = 4.3 x 10^5 m/s (e.g. since the initial energy is 1 keV = 1.6 x 10^-16 J, given that 1 eV = 1.6 x 10^-19 J)

Then we have, from your problem info: v⊥ = v ‖ = 4.3 x 10^5 m/s

The gyro-radius for the particle is then:

r = m/ q [v⊥ / B] = v⊥ / (qB/m)

where the quantity (qB/m) is the gyro-frequency. In this case:

qB/m = [(1.6 x 10^-19 C) (0.00001T)/ (1.7 x 10^-27 kg) =  941 /s

Therefore: r = (4.3 x 10^5 m/s)/ 941 /s = 456 m

Now, since we know B(max) = 20 B(min) and B(min) = 0.00001T then the maximum field intensity for the magnetic bottle is:

B(max) = 20 (0.00001T) =  0.00020 T

The loss cone angle was already computed earlier:

Θ_L =  ± arc sin  [ 1/20]^½  = ± arc sin  [ 0.05]^½  

Θ_L =  ± 2.8 degrees

The condition for escape is based on the condition already given in my original response which actually provided the condition for trapping in the magnetic bottle, or:

(v ‖  / v⊥ ) < (B (max) / B(min) - 1)^½


Then the condition for escape would be:

(v ‖ / v⊥ ) >  (B (max) / B(min) - 1)^½

Or in the case of this magnetic bottle:

(v ‖ / v⊥ ) >  (20 - 1)^½

Or : (v ‖  / v⊥) >   4.35

Lastly, as I showed in the solution to the previous problem, the condition to undergo reflection is just that the sine of the pitch angle exceed the square root of the mirror ratio or:

sin (φ) > [ (B(min) / B (max )]^½  so: sin (φ) > [ (1/20) ]^½  

I.e. sin (φ) > [ 0.05]^½    

sin (φ) > 0.223  

In other words, we require φ  > 12.8 degrees

Hopefully, this further assists your self-study program!