Phase Spaces In Plasma Physics - The Klimontovich Equation & The Liouville Equation (Part 1 Basics)

In addition to the Klimontovich equation, e.g.

 Examining Aspects Of The Klimontovich Equation Used In Plasma Physics

Plasma physicists also work with the Liouville equation, which provides an exact description of a plasma.  However, like the Klimontovich equation it is of no direct use, instead giving us a starting point for the development of approximate statistical theories.  These latter begin with distinguishing the behavior of particles from the behavior of systems. Specifically the Klimontovich equation describes the former, while the Liouville equation describes the latter.

To fix ideas, consider a basic basic 'system' of one charged particle. How do we go about its description? Let us say we measure the particle's position in a coordinate system using x₁   then the orbit of the particle - designated  X₁(t) is the set of positions x_i occupied by the particle at consecutive times t. Similarly, in velocity space one can denote the orbit of the particle as V_i(t)  which is the set of velocities of the particle at consecutive times t. But these are measured in a coordinate system v_i  for velocity space.  Then we end up with a phase space denoted: (x₁, v₁)  = ( _x1 , y₁ , z₁, _vxi, _vyi,  _vzi)

Meanwhile, the density of systems in this phase space is:

N(x₁ ,v₁ ,t) = d (x₁ -  X₁ (t)) d (v₁ -  V₁ (t))

Next, we consider a system of two particles. Particle 2 now has coordinate axes (x₂, v₂) which overlay the (x₁, v₁) coordinate axes.  In addition the orbit X₁ (t),  V₁ (t) (i.e. of particle 1), is measured with respect to the (x₁, v₁) coordinate axes while the orbit X₂ (t),  V₂ (t) (i.e. of particle 2) is measured with respect to the (x₂, v₂) coordinate axes

Plasma statistics, basically, reference a 6-D phase space.  But in Liouville's Theorem the approach is very different. Specifically, it is applicable to a 6No  - D phase space, where No is the number of particles in the system.  For a one particle system then No  = 1  so the system phase space is just 6D.   I.e. the system is represented by only one phase space point.

For a 2-particle system (No = 2) so the phase space is:

6 (2) D = 12D

And again there is only one phase space point for the system. In general for an No -particle system, though with 6No  - D phase space, there is still only one phase point. 

Recall the number density for a 1-particle system is written:

N = d (X₁   -    X₁ (t)) d (V₁   -   V₁ (t))

And for a 2-particle system:

N( X₁, X₂ ,V₁ , V₂ ) =

d (X₁   -   X₁ (t)) d (V₁   -  V₁ (t))d (X₂   -    X₂ (t)) d (V₂   -   V₂ (t))

And in general:

N( X_i.... X_n ,V_{i ....}V_n ) = å ^No _i=1      d (X_i   -   X_i (t)) d (V_i  -   V_i (t))

If no source or sink exists in the phase space then the number density must be conserved.  The conservation of N in the 6No - D phase space leads to:

¶ N / ¶ t +  å ^No _i=1  Ñ x  (X_i  · N) +å ^No _i=1  Ñv  (V_i  · N) = 0 

Where:

 Ñ x i  ·(X_i  · N)  =   X'_i  · N Ñ x i  N +   N Ñ x i  ·X'_i 

And:

Ñv i  (V_i  · N)= V'_i · Ñv i  N + N Ñ v i  · V'_i 

If we then identify:   X'_i  = V_i 

We arrive at:

 V'_i  =   q/ m [ E (X_i ) + V_i   x  B( X_i )

which is just Newton's 2nd law of motion (F = ma) in an EM -field.  

Suggested Problems:  

1. (a)Write the phase space dimension for a 3-particle system.

(b) Write out the number density for this system.

2. We saw  (x₁, v₁)  = ( _x1 , y₁ , z₁, _vxi, _vyi,  _vzi)  is a 6-dimensional phase space.  What would be the number of dimensions for the phase space:

 (x₁, v₁, x₂, v₂ )?

Write them all out in a bracket.

3.  Show how V'_i  =   q/ m [ E (X_i ) + V_i   x  B( X_i )

translates to Newton's 2nd law of motion.

(Hint: You need to incorporate the Lorentz force.)