## Wednesday, June 2, 2021

### Looking Again At Adiabatic Invariants

In space plasma physics, an "adiabatic invariant"   refers not to an absolute invariant but to a quantity which is approximately invariant if the rates of change  of the parameters of the system are sufficiently  slow and the variation  is  aperiodic.  We begin by considering the first adiabatic invariant, the magnetic moment  m   for which slow changes in 2 external parameters (space, time) leave the value effectively constant. We can write in the first instance :

m=  ½  [m v ^2  /B]

And the equation of motion (applicable to the guiding center):

m dv/dt =  q(E + v X B)

Whence:

mv  · dv/dt  =   q v ·  E + 0

So that the work done on a particle moving in E, B fields comes from those fields.  We may then write:

v     =      w +   v ^

With w  the guiding center velocity.  Therefore:

v 2   =    v ^2     +   w 2   +   2 w · v ^

Whence:

< v 2   >   =    < v ^2  >   +  w 2   +   0

Then we proceed as follows:

d/ dt<½ m v ^2  >  =  q< v ^ ·  E >  +  q< w  ·  E  >

-          d/ dt <  1/2 m w 2  >  =

q < v ^ ·  E >  =    q/ T  ò  v ^ ·  E dt  =  q/T  ò E · ds

Where T is the gyration period. (T = 2 p  /  )  and ds is a line element along a circular path (which is not the actual path of the particle)   Thus:

q/ T   ò E · ds  =   q   / òA Ñ x  E  · da

where A is the surface area enclosed by the path of integration.  Using Maxwell’s curl E equation:

Ñ X E  =  - B / t

We can therefore write:

q/T  ò E · ds  =    -  q   / 2 p   [B / t] p r2    =

-         (  q  r2 / 2)  B / t  =   - m  B / t

Where:   m     =    q  r2 / 2

Thence:

d/dt <½ m v ^2  >  =   - m  B / t  +  w · (q<E> - mw’)

It can further be shown that the perpendicular energy of the particle changes with time such that:

d/ dt(½ m v ^2 )  =  v ^ F ^   =    -q/ c  v ^  v ||   B r

Where  B r   denotes the radial magnetic field intensity and:

F ^   =    - q/ c  (v ||   B r  )

With r = r     the Larmor radius, i.e. at the location of the particle, we get for the total perpendicular energy:

d/dt  (W ^ ) =  q/ c  v ^   v ||       r/2  B x / x   =

½ m v ^2  v ||     1/B   B  / x      Or:

d (W ^ ) /dt =   W ^   v ||     1/B  ( B  / x )

The preceding equation can then be combined with F  = - m Ñ  B    to show the particle energy  W in a time-independent B-field is conserved

Thus:  dm/ dt     =    d/dt (W ^   / B)

Or:  dm/ dt     =      1/B  d (W ^ ) /dt   -   1/B 2  ( B  / t )

= W ^   v ||    1/B 2  ( B  / x ) -  1/B 2  ( B  / x) W ^   v ||

=  0

So that the rate of change of B along the particle orbit, i.e. dB/ dt  =   v ||   ( B  / x ) leading to:

m     =    0,    a constant of the motion

A second  adiabatic invariant can also be obtained, associated with the bounce period between magnetic mirrors..  Here, we can apply Hamilton’s action principle.   This usually starts by examining plasma configurations to see where they conform to the typical magnetic mirror profile used in standard plasma physics. In space physics, one uses the sine of the loss cone angle to obtain the mirror ratio:

sin (q L ) = ± Ö (Bmin / B max )

If one finds that there are particles within the “mirrors” for which the “pitch angle” (a) has:

sin (a )  >  Ö (Bmin / B max )

then these will be reflected within the tube, On the other hand, those particles for which the “less than” condition applies will be lost, e.g. on transition out of the mirror configuration.

Since the original  adiabatic invariant for particle motion is a constant of the motion

m=  ½  [mv 2 /B

we have:

[v ^2 /Bmax]  =     [v || 2 /Bz] =    const.  or

[v ^2 / v || 2  ]  =  Bz /Bmax

where we take  Bz  = Bmin

A simple model for a magnetic mirror system is shown below with M1 and M2 the magnetic mirrors separated by a distance L.

If we are careful to change L slowly, then one finds:

v || L = const.

Now, assume M1 is stationary and M2 moves toward M1 at a velocity,  v m  , then the incident velocity relative to the wall is:

[(v ||   +  v m)   -   v m ]

And:

D v ||   =  - [- (v ||   +  v m)   -   v m ]  + v |    =   2 v m

Thus, with each reflection, the velocity changes by 2 v m and the number of reflections per second will be:

v ||  / 2L

Further:

dv ||  /dt  =   2 v m  (v ||  / 2L)  =  v ||  /L (-dL/dt)

=    - v ||  /L (dL/dt)

So that: L (dv ||  /dt )  +   v ||  (dL/dt)  =  0

Whence:  d/dt (v ||  L)  = 0

For kinetic energy of particles we must have:

E   =  ½ m (v ^2  + v || 2 ) = const.

But based on the Hamiltonian action S   we can write, say for the motion between M1 (= x1) and M2 (= x2):

S    =  ò p  dq  =      mv x  dx

And the integral on the right is taken by following the particle through one oscillation.  Then we can also assert the action S  is a constant of the motion and effectively a second adiabatic invariant.

Note also that:  v/ Ω  =   v/(qB/m) = m v/qB

Let the guiding center lie on the z-axis (for simplicity) then the average force experienced will be written:

F z = + ½ q  v r L   (B z/ z) = + ½ q  v / Ω B z/ z

Since:   v   = r c     so:  r L   =  v/c

Where r L   is the guiding center (Larmor) radius.

With appropriate substitutions we get:

F z = + ½ q  v r L   (m v/qB) (B z/ z) =

+ ½ m v2 /B  (B z/ z)

But by previous definition for the adiabatic invariant:

m m  = m(v)2/ 2B  = const .

So:   F z =   m  B z/ z

Now consider a symmetric configuration for a magnetic bottle with an  isotropic particle distribution    suddenly injected  at the  coordinate point  x = x o.  Then  fraction of particles lost will be given by:

F L  =  1/ 2 π   ò o Θ(L)  2 π sin(Θ) d Θ

=   1 - cos(Θ) L   =   1 -  Ö [1    (Bmin / B max )]

Suggested Problems

1) Consider a plasma mirror machine of length 2L with a mirror ratio of 10 so that B(L) = B(-L) = 10 B(0). A group of N (N > 1) electrons with an isotropic velocity distribution is released at the center of the machine. Ignoring collisions and the effect of space charge, how many electrons escape?

2) Consider a similar mirror configuration for a solar coronal loop for which  Bmin =   0.2 Bmax . Find the loss cone angle for this loop and also determine whether particles will remain within it. Find the velocity  ratio :

v ^ / v ||   if    Bmax = 0.95 B z