Friday, August 28, 2026

Phase Spaces In Plasma Physics (Part 2): The Liouville Equation

 Recall in Part 1, we saw the density of systems for the (x1, v1) phase space is:

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

Where:  x1 = X1 (t)),  v1  V1 (t)) 

Analogously, in the 12-dimensional phase space (two particles considered) we saw from the Part 1 solutions (#2):  

(x1, v1, x2, v2 ) =  ( x1 , y1 , z1, vx1, vy1,  vz1, x2 , y2 , z2, vx2, vy2,  vz2,

And there is one system occupying the point:

x1 = X1 (t)),  v1  = V1 (t), x= X2 (t)),  v2  = V2 (t) i.e. at time t.  The density of systems in this phase space is then:

N( x1, x2 ,v1 , v2 ) =

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

Basically, then there is one system in 6No  -dimensional space, so by analogy with the density of systems expression e.g. for N( x1, x2 ,v1 , v2 ), we can write:

N(x1, x2 ,v1 , v2 .... xNo ,vNo t) = åNo i=1      d (x  -   Xi (t)) d (vi  -   Vi (t))

As with the Klimontovich equation, the Liouville equation is found by taking the time derivative of the appropriate density.  Given the already defined  density of systems is the product of 6No  terms then the time derivative must involve the product of  6No  terms.  We can use the expression:

/ t   d [ (x  -   X1 (t)]  =  -      Xi/ ·  Ñ x i  d [ (x  -   X1 (t)] 

The time density will be:

N / t +  å No i=1  Vi (t)  ·  Ñ x i   åNo j=1      d (x  -   Xj (t)) d (vj  -   Vj (t))  +

åNo i=1    V'i åNo j=1      d (x  -   Xj (t)) d (vj  -   Vj (t))  = 0

The next standard step employed by plasma physicists is to use the identity:  ad (a  -  b)  =  bd (a  -  b) to replace Vi   by vi

Leading to:

V'i (t) = s/ m s [ E (xi,t) + V/c  B( xi ,t)]

Given that the products are just the density of systems, N, the equation for the time derivative of N(x1, x2 ,v1 , v2 .... xNo ,vNo , t) 

becomes:

N / t +  å No i=1  V· Ñ x N +å No i=1  V'i (t)  ·  Ñvi N  = 0

which is the Liouville equation, and when combined with Maxwell's equations, viz. 


i)  Ñ X H  J    + D / t

ii)             Ñ X E  - B / t

iii)           Ñ ·0   

iv)       Ñ ·r    


is an exact description of a plasma.

Algebra II Review: Solutions To Algebraic Fraction Problems

 Simplify each of the following:

1) x2  - y2/ y2  - x2

Solution:

Factor numerator and denominator:

(x + y) (x - y)/ (y + x) (y - x)

Change sign of (y - x) in denominator to - (x - y):

Þ  

(x + y) (x - y)/ - (x - y) (y + x)

We can then cancel the (x - y)  factors in numerator and denominator:

=  -  (x + y) / (x + y)  =   -1


2)  27 a3 x25  / 36 a 2 x4y4

Solution:  

Divide, first reducing 27/36 to 3/4 and paying attention to properties of exponents (the realm of Algebra I)

a y  / 4 x2


3) 2a 1/2 (x + y)/  5 x a 3/2

Solution:  

Apply quotient rule for exponents (from Algebra I):

a m /a n     =  a m - n

In this case, m = 1/2 and n = 3/2 so m - n =  1/2 - 3/2 =  -1

So:  a- 1  =  1/ a

Then the fraction simplifies to: 

2 (x + y)/ 5 a x


4) (x -1) (2 - x) (3 - x)/ (1 - x) (x - 2) (x - 4)

Solution:  

Notice that the numerator contains (x - 1) and (2 - x), while the denominator contains (1 - x) and (x - 2). This indicates application of sign rule.  So, following the process in initial post, we rewrite the factors in the numerator to match those in denominator by factoring out (-1) in each, viz.

(x -1)  =  (-1) (1 - x)

(2 - x) =   (-1) (x - 2)

Substituting back into the numerator we obtain:

(1 - x) (x - 2) (3 - x)   (Rem: (-1) (-1) = +1)

So the algebraic fraction becomes:

(1 - x) (x - 2) (3 - x) /  (1 - x) (x - 2) (x - 4)

After canceling common factors:

(3 - x) / (x - 4)

Thursday, August 27, 2026

Solutions To Phase Space Plasma Physics Problems

The  Problems:

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

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

Solutions:

a) We need the phase space formula: 6No  - D where No is the number of particles in the system.  In this case  No   = 3, so:

 6No  - D =   6 (3) -D  or 18 dimensions

b)  The number density for the system is:

N( Xi.... Xn ,Vi ....Vn ) = å No i=1      d (X  -   Xi (t)) d (Vi  -   Vi (t))

Where  No    = 3,  so:

N( X1, X2 X3 ,V1 , V2.V3 =

d (X -  X1 (t)) d (V  -  V1 (t))d (X -  X2 (t)) d (V  -   V2 (t)) d (X -   X3 (t)) d (V  -  V3 (t)

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 12-D  phase space:

 (x1, v1, x2, v2 )?

Write them all out in a bracket.

Solution:

(x1, v1, x2, v2 ) =  ( x1 , y1 , z1, vx1, vy1,  vz1, x2 , y2 , z2, vx2, vy2,  vz2,


3.  Show how V' =   q/ m [ E (Xi ) + V  x  B( Xi )]

translates to Newton's 2nd law of motion.

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

Answer:

Given the nature of the plasma, charged particles of the system moving in magnetic field (B) the Lorentz force is:  

F = q ( E  + v x   B)

Or, in terms of Newton's 2nd law:

m (dv/dt) = ma = q ( E  + v x   B)

Or, to conform with the plasma expression:

V' =   q/ m ( E  + v x   B)

Where: E =  E (Xi )       and: v x   B  = V  x  B( Xi )

Where the  Xi   are position dependent in the coordinate system.