Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation 19. Deriving the Vlasov Equation From the Klimontovich Equation

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1 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation 19 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation Topics or concepts to learn in Chapter 2: 1. The microscopic plasma distribution: the Klimontovich equation 2. The statistic plasma distribution: the Boltzmann equation and the Vlasov equation Suggested Reading: (1) Chapter 3 in Nicholson (1983) 2.1. Klimontovich Equation Let us define a microscopic distribution function of the αth species in the six-dimensional phase space N α = δ[x x k (2.1) where x k and v k satisfy the following equations of motion dx k dv k = v k (2.2) = e α {E m [x k,t] + v k B m [x k,t]} (2.3) in which E m (x,t) and B m (x,t) are the microscopic electric field and magnetic field, respectively. The Klimontovich equation can be obtained by evaluating the time derivative of N α. Taking time derivative of Eq. (2.1) and making use of Eqs. (2.2)~(2.3) and aδ(a b) = bδ(a b), it yields

2 20 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation N α = δ[x x k = { δ[x x k }δ[v v k + δ[x x k { δ[v v k } = { δ[x x k [ dx N k 0 ]} δ[v v k + δ[x x k { δ[v v k = δ[v v k [ v k δ[x x k + δ[x x k [ e α {E m [x k,t] + v k B m [x k,t]}] δ[v v k = δ[v v k [ v] δ[x x k + δ[x x k ( e α )[E m (x,t) + v B m (x,t)] δ[v v k = [ v] {δ[x x k } + ( e α )[E m (x,t) + v B m (x,t)] = v N α or N α {δ[x x k } e α [E m (x,t) + v B m (x,t)] N α + v N α [ dv k ]} [E m (x,t) + v B m (x,t)] N α = 0 (2.4) Eq. (2.4) is the Klimontovich equation of the microscopic distribution function N α. Exercise 2.1 Show that δ[x x k [ dx k ]

3 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation 21 Answer to Exercise 2.1 δ[x x k = {δ[x x k δ[y y k δ[z z k } = {δ[x x k }δ[y y k δ[z z k {δ[y y k }δ[z z k δ[y y k {δ[z z k } = { dδ[x x k d[x x k { dδ[y y k d[y y k [x x k }δ[y y k δ[z z k δ[y y k { dδ[z z k d[z z k = { δ[x x k x { δ[y y k y [y y k }δ[z z k [z z k } ( dx k )}δ[y y k δ[z z k δ[y y k { δ[z z k z where δ[x x k and dx k = ˆx d x k ( dx k ( d y k )}δ[z z k ( dz k )} ) = { x δ[x x k } ( dx k ) = {( ˆx x + ŷ y + ẑ z )(δ[x x k δ[y y k δ[z z k )} + ŷ d y k + ẑ dz k Exercise 2.2 Show that δ[x x k x [ dx k ]

4 22 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation Answer to Exercise 2.2 Let f be a functional of a function W (x,t), i.e, f = f = d f W f [W (x,t)]. Then If W / x =1, then f x = d f W x = d f Thus, for W / x = 1, we have f = d f W = f W x This is the reason why δ[x x k [x x k x x [ dx k ] Exercise 2.3 Show that δ[x x k ( v k B m [x k,t]) δ[v v k = δ[x x k [v B m (x,t)] δ[v v k 2.2. Vlasov Equation Let f α, E(x,t), and B(x,t) be the ensemble average of N α, E m (x,t), and B m (x,t), respectively. Let N α = f α + δ N α E m (x,t) = E(x,t) + δ E m (x,t) B m (x,t) = B(x,t) + δ B m (x,t) If we use A to denote the ensemble average of A, then we have N α = f α E m (x,t) = E(x,t)

5 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation 23 B m (x,t) = B(x,t) and δ N α = 0 δ E m (x,t) = 0 δ B m (x,t) = 0 Taking the ensemble average of Eq. (2.4), it yields N α + v N α [E m (x,t) + v B m (x,t)] N α = 0 or f α + v f α [δ E m (x,t) + v δ B m (x,t)] δ N α [E(x,t) + v B(x,t)] f α = 0 (2.5) Let Df α / Dt denote the time derivative of the distribution function f α along its characteristic curve in the (x,v) phase space, then Eq. (2.5) can be rewritten as Df α Dt = e α = f α + v f α [δ E m (x,t) + v δ B m (x,t)] δ N α [E(x,t) + v B(x,t)] f α = δ f α δ t collision (2.6) For e α [δ E m (x,t) + v δ B m (x,t)] δ N α = δ f α δ t collision = 0, the Boltzmann equation, Eq. (2.6), is reduced to the Vlasov equation (Vlasov, 1945): f α + v f α [E(x,t) + v B(x,t)] f α = 0 (2.7) References Nicholson, D. R. (1983), Introduction to Plasma Theory, John Wiley & Sons, New York. Vlasov, A. A. (1945), J. Phys. (U.S.S.R.), 9, 25.

6 24 Chapter 2. Deriving the Vlasov Equation From the Klimontovich Equation