11.D.2. Collision Operators

Transcription

1 11.D.. Collision Operators (11.94) can be written as + p t m h+ r +p h+ p = C + h + (11.96) where C + is the Boltzmann collision operator defined by [see (11.86a) for convention of notations] C + g(p) = d p' d Ω v σ(b, v) f 0 ' g f - g + g f ' - g' g (11.97) Next, we define an inner product between functions on the momentum space by 3/ φ, χ d p e - p m φ(p) χ(p) (11.98) = V N d p f 0 (p) φ(p) χ(p) [ (11.87) used. ] (11.98a) As usual, the adjoint O of the operator O on our inner product space is defined by O φ, χ = φ, O χ (11.99a) Putting (11.97) into (11.98) gives φ, C + χ = 4 V and Now, C + φ, χ = 4 V N d p d p' d Ω v σ(b, v) f 0 f 0 ' φ χ f - χ + χ f ' - χ' (11.99b) N d p d p' d Ω v σ(b, v) f 0 f 0 ' φ f - φ + φ f ' - φ' χ (11.99c) I = d p d p' d Ω v σ(b, v) f 0 f 0 ' is invariant under the following interchange of variables: 1. ( p, p' ) ( p', p ) & ( p f, p f ' ) ( p f ', p f ) [ incident target ]. ( p, p' ) ( -p f, -p f ' ) & ( p f, p f ' ) ( -p, -p' ) [ reverse scattering ] 3. ( p, p' ) ( -p f ', -p f ) & ( p f, p f ' ) ( -p', -p ) [ 1+ ] Furthermore, since I is invariant under p -p & p' -p' the reverse scatterings can be written as a. ( p, p' ) ( p f, p f ' ) & ( p f, p f ' ) ( p, p' ) 3a. ( p, p' ) ( p f ', p f ) & ( p f, p f ' ) ( p', p ) Therefore, (11.99b) can be written as φ, C + χ = V N d p d p' d Ω v σ(b, v) f 0 f 0 ' φ χ f - χ + χ f ' - χ' + φ' χ f ' - χ' + χ f - χ + φ f χ -χ f +χ' -χ f ' + φ f ' χ' -χ f ' +χ -χ f

2 11.D.._CollisionOperators.nb = - V N d p d p' d Ω v σ(b, v) f 0 f 0 ' (11.99d) φ f - φ +φ f ' - φ' χ f - χ + χ f ' - χ' = C + φ, χ (11.99) (11.99d) gives φ, C + φ = - V N d p d p' d Ω v σ(b, v) f 0 f 0 ' φ f - φ +φ f ' - φ' = N V 3 d p d p' d Ω v σ(b, v) e - p +p ' m φ f - φ +φ f ' - φ' (11.101) where (11.87) was used. Since every term in the integrand of (11.101) is non-negative, we have φ, C + φ 0 φ (11.101a) with φ 0, C + φ 0 = 0 φ 0 f -φ 0 +φ 0 f ' -φ 0 ' = 0 p, p' (11.101b) There are 5 independent scalar solutions to (11.101b): 1. φ = A = const φ 0 f -φ 0 +φ 0 f ' -φ 0 ' = A - A + A - A = 0. φ = B p with B = const φ 0 f -φ 0 +φ 0 f ' -φ 0 ' = B (p f - p + p f ' - p' ) = 0 [ momentum conservation ] 3. φ = C p with C = const φ 0 f -φ 0 +φ 0 f ' -φ 0 ' = C p f - p + p f ' - p' = 0[ energy conservation ] The solution of (11.96) can be obtained using the eigenfunctions of C + defined by the eigen-equation C + Ψ = λ Ψ (11.100) Taking the inner product gives Ψ, C + Ψ = λ Ψ, Ψ From (11.98), we have Ψ, Ψ 0 Using (11.101a) on (11.100) then gives λ 0 (11.100a) which means C + is a negative semidefinite (or nonpositive) operator. Furthermore, its λ = 0 eigenvalue has a 5-fold degeneracy [see solutions of (11.101b)]. Since d p' d Ω v σ(b, v) f 0 ' is invariant under any rotation R in the momentum space, (11.97) indicates that C + g(p) transforms like g(p) under R. In other word, C + & R commute and share the same eigenfunctions. Hence, the eigenfunctions Ψ of C + take the form Ψ λ l m (p) = R λ l (p) Y l m (p ) (11.10) where Y l m (p ) = Y l m (θ, ϕ) are the spherical harmonics (eigenfunctions of R ).

3 11.D.._CollisionOperators.nb 3 As an example, consider now a spatially homogeneous system with no external forces. (11.96) simplifies to = C + h + t = d p' d Ω v σ(b, v) f 0 ' h f + - h + + h f + ' - h + ' (11.103) The solutions to (11.103) can be written as h + (t) = A λ l m e λ t Ψ α l m (p) (11.104) where A λ l m are constants. The proof is simple. Taking the time derivative of (11.104) gives = λ A λ l m e λ t Ψ α l m (p) t Operating C + on (11.104) gives QED. C + h + = A λ l m e λ t C + Ψ α l m (p) = λ A λ l m e λ t Ψ α l m (p) [ (11.100) used. ] The expansion coefficients A λ l m can be obtained from the initial condition Since λ 0, h + (0) = A λ l m Ψ α l m (p) h + ( ) = A 0 l m Ψ 0 l m (p) Similarly, the Lorentz-Boltzmann equation (11.95) can be written as h - + p t m h- r + p h- p = C - h + (11.105) where C - is the Lorentz-Boltzmann collision operator defined by C - g(p) = d p' d Ω v σ(b, v) f 0 ' g f - g g (11.106) The analog to (11.101a-b) is with ψ, C - ψ 0 φ (11.106a) ψ 0, C - ψ 0 = 0 ψ f 0 - ψ 0 = 0 p, p' (11.106b) The only solution to (11.106b) is ψ = A = const ψ f 0 - ψ 0 = A - A = 0 The λ = 0 eigenvalue of C - is therefore non-degenerate. Exercise Write χ, C + χ in a form that makes explicit the conservation of kinetic & momentum during the scattering process.

4 4 11.D.._CollisionOperators.nb Answer Let the initial & final relative velocities be v 0 = v 0 - v 10 & v f = v f - v 1 f, respectively. For convenience, we have reproduced Fig.11.6(b) of 11.C.1.1 below. ϕ f v f incoming outgoing b v 0 ϕ f r min 0 Θ Let e be the unit vector in the direction of r min. Since v f = v 0, we have v f = v 0 - e e v 0 v f v 0 = v 0 cos Θ = v 0 cos(π - ϕ f ) = -v 0 cos ϕ f = v cos ϕ f (1) Our task is to rewrite χ, C + χ = - V N d p d p' d Ω v σ(b, v) f 0 f 0 ' χ f - χ +χ f ' - χ' (1b) (1a) in a form that display explicitly the conservation of kinetic & momentum. To begin, since p & p' in (1b) denote the initial momenta, we shall drop the subscript 0 in (1) & (1a) to make our notations consistent. Next, we rewrite d Ω v σ(b, v) in terms of first P f & p f, then p f & p f '. Thus, v d Ω = 1 μ d Ω d p f p f δ( p f - p ) [ v = p μ used. ] Using δ p f - p 1 = [ δ( p f - p ) + δ( p f + p ) ] p f and that p f, p f 0, we have v d Ω = μ d Ω d p f p f δ p f - p = 1 μ d Ω d p f p f δ ( E f -E) [ E = p μ used. ] = 1 μ d Ω d p f p f δ ( E f -E) d P f δ(p f - P) = 1 μ d Ω d p f p f δ ( E f -E) d P f δ(p f - P)

5 11.D.._CollisionOperators.nb 5 Since E CM = P is conserved, we can write M E f -E = E f + E CM -E -E CM Also, = E tot, f - E tot = 1 m p f + p f ' - p - p' P f - P = p f + p f ' - p - p' d p f = d Ω d p f p f v d Ω = m μ d p f δ( E f - E ) d P f δ(p f - P) = 8 m d p f d p f ' δ p f + p f ' - p - p' δ(p f + p f ' - p - p') (4) (1b) thus becomes χ, C + χ = - 8 V m N d p d p' d p f d p f ' δ p f + p f ' - p - p' δ( p f + p f ' - p - p') σ(b, v) f 0 f 0 ' χ f - χ +χ f ' - χ' = - m N V 3 d p d p' d p f d p f ' δ p f + p f ' - p - p' δ( p f + p f ' - p - p') σ(b, v) e - p +p ' m χ f - χ +χ f ' - χ' (6)