2.3 Moments of the distribution function 33For example, if the energyEof particles is constant along their orbits thenf=f(E)is
a solution to the Vlasov equation. On the other hand, if both the energy and the momen-
tumpare constant along particle orbits, then any distribution function with thefunctional
dependencef=f(E,p)is a solution to the Vlasov equation. Depending on the situation
at hand, the energy and/or canonical momentum may or may not be constant along an or-
bit and so whether or notf=f(E,p)is a solution to the Vlasov equation depends on the
specific problem under consideration. However, there always exists at least one constant of
the motion for any trajectory because, just like every human being has an invariant birth-
day, the initial conditions of a particle trajectory are invariant alongits orbit. As a simple
example, consider a situation where there is no electromagnetic field sothata=0in which
case the particle trajectories are simplyx(t) =x 0 +v 0 t,v(t) =v 0 wherex 0 ,v 0 are the
initial position and velocity. Let us check to see whetherf(x 0 )is indeed a solution to the
Vlasov equation. Writex 0 =x(t)−v 0 tsof(x 0 ) =f(x(t)−v 0 t)and observe that
∂f
∂t+v·∂f
∂x+a·∂f
∂v=−v 0 ·∂f
∂x+v·∂f
∂x= 0. (2.9)
x
v
Figure 2.3: Moments give weighted averages of the particles in the shaded vertical strip
2.3 Moments of the distribution function
Let us count the particles in the shaded vertical strip in Fig.2.3. The number of particles in
this strip is the number of particles lying betweenxandx+ dxwherexis the location of
the left hand side of the strip andx+dxis the location of the right hand side. The number
of particles in the strip is equivalently defined asn(x,t)dxwheren(x)is thedensityof