Kinetic Theory Total force on particle " " (Newton's second law): dv dx F m , and v dt dt Sum of forces: electromagnetic + gravity + Coriolis + ... In space plasma, the electromagnetic (Lorentz) force is most important: F q E q v B A full description of a real plasma system with N particles 10 30 requires 6N numbers at each time t. But N is large, 10 to1 10 . Particle Distribution Function "Phase space" encompasses ordinary space (x) and velocity space (v), i.e., independent coordinates (x,y,z,v x , v y ,v z ): 6-D space, plus time: 7-D We consider only s-type particles (for example: electrons) # particles in V f s x, v, t lim for volume element V 0. V f s x, v, t is the # of all type-s particles between x and x x, y and y y, and z and z z that have velocities beween vx and vx vx , v y and v y v y , and vz and vz vz , Phase space volume: V xyzvx v y vz 2 Density Function ns If we add up all s-particles from all "velocity volume" elements d 3 v dvx dv y dvz at location x, we get the # of s-particles per unit spatial volume: ns (x, t )
f s x, v, t dvx dv y dvz f s x, v, t d 3 v Number density, mass density, charge density 3 The Boltzmann Equation: time variation of a distribution function Distribution function is a function of time, space, and velocity: f (t , r (t ), v(t )) Time variation of it is: chain rule for a composite function f f r f v f f f df dt dt dt dt vdt adt t r t v t t r v Df s f s fs v f s a v f s Dt t t collision Approximate form of the collision term on the right side of the Boltzmann equation is f s f sM fs coll t coll
Here f sM is the "Maxwellian", and coll the average collision time. 4 Total Derivative in Phase Space f s Df v f s a v f s s t Dt So we can write as Df s f s Dt t collision a F / ms Considering only the em force per unit mass (acceleration), we get; f s q fs v f s s E v B v f s . t ms t collision Neglecting the collision term, we get the Vlasov equation: f s q v f s s E v B v f s 0 t ms Liouville's Theorem: f is constant as it is convected with the particles: f s F Df v f s s v f s s 0 t
ms Dt 5 Homework The magnetic field of a magnetic mirror varies as Bz B0 (1 z 2 ) along the axis. (a) at z=0 an electron has a speed of v 2 3v||2 1.5v2 where does reflection occur? (b) Determine the motion of the guiding center. (c) Show that the motion is sinusoidal. Determine the frequency. (d) Determine the longitudedinal invariant belonging to this motion. Problem 3.6, 3.7, 3.8 6 Examples of Distribution Functions Any number of distribution functions satisfies the Boltzmann equation. 0. -function: ( x 0) 0, (0) , but ( x)dx 1 - 1. Consider a gas of electrons all moving with v u x : x0 f e x, v, t no vx u x 0 v y vz 2. Consider a unifom gas n x, t n0 of ions, all having the same speed v v0 : f i x, v, t A v v0 2 n0 f i x, v, t dv f i x, v, t v 2 sin dvd d 0 0 0 2 A v v0 v 2 sin dvd d Av02 4 0 0 0
A n0 n0 f x , v , t v v0 i 2 2 v0 4 v0 4 7 Maxwellian Distribution Functions The "Maxwellian": 1 2 m v s 2 ms f sM x, v, t ns x, t exp k T B s
2 k BTs Note that in general: Ts Ts x, t ; v 2 vx2 v y2 vz2 or v 2 sin dv d d 32 Drifting Maxwellian for a gas moving as a whole in direction v 0 : 2 1 m v v 0 2 s ms f sM x, v, t ns x, t exp 2 k T k T B s B s Example: Find the one - dimensional Maxwellian distribution function 32 g sM x,v x , t for the s-paraticles with x-velocities between v x and v x +dv x : g sM x,v x , t f sM x, v, t dv y dvz - -
ms ns x, t 2 k BTs 32 1 1 1 2 2 2 m v m v m v s x s y s z 2 2 exp 2 exp dv exp y dvz
k T k T k T B s B s B s 8 One-dimensional Maxwellian, contd 1 2 m v 2 s z ms 2 vz2 exp dv e dv z
z k T 2k BTs B s 1 1 1 2 2 2 32 m v m v m v 2 s x 2 s y 2 s z ms g sM x, vx , t ns x, t exp exp dv y exp dvz 2 k BTs k BTs k BTs k BTs
1 2 32 32 m v 2 2 s x ms ms vx ms ns x, t exp n x , t exp s 2 2 k T 2 k T 2
k T k T B s B s B s B s ms 2 k BTs 12 g sM ms ms vx2 x, vx , t ns x, t exp 2 k T B s 2k BTs 9 Thermal speed 12 ms ms vx2 g sM x, vx , t ns x, t exp
2 k T 2 k T B s B s 2k BTs 1 We define the thermal speed: vth or ms vth2 k BTs . ms 2 With this definition, the one-dimensional Maxwallian can be written: 1 g sM x, vx , t ns x, t exp vth The 3-dimensional Maxwellian: f sM 1 x, vx , t ns x, t 3 2 3 exp vth 2 vx vth v vth
2 10 Equilibrium Distribution A Maxwellian distribution in thermodynamic equilibrium does not change with time. It can be expressed as a function of the total energy E of a particle: f sM x, v ns 0 exp E x k BTs Assume the particle moves in a static electric field: E V x where V is the electric potential. Then charge x voltage = energy 1 E Ekin E pot ms v 2 qV x 2 qV x ms v 2 f sM x, v ns 0 exp exp k T 2 k T B s B s qV x ms v 2 ns x exp where ns x ns 0 exp 2 k T k
T B s B s 11 Velocity Moments "Velocity moments" can be used to define the macroscopic variables of a fluid. The nth moment of the distribution function is defined as: M n f s v n f s dv The 0th moment is simply the density: f dv n x, t s from the definition of f s . The 1st moment is the particle flux: xvvs x, t v f s dv m 3 The dimension of the flux is m m 2 s 1. The x-component is s sx vx f s dvx dv y dvz 12 Bulk flow velocity The "average velocity u" is define by: u s x, t v s u s x, t v s f s dv
ns x, t or xvv s x, t ns x, t 13 Pressure tensor The pressure tensor is proportional to the 2nd moment of f s . If the fluid has a bulk velocity u, we call the velocity relative to the bulk flow, c v u, the random (or peculiar) velocity. (dyad ab is a tensor) Ps x, t ms ns cc ms ns cc f s dv Pxx Pxy Pxz P Pyx Pyy Pyz where P P P zz zy zz Psxx ms ns cx cx f s dv, Psxy ms ns cx c y f s dv, etc. When carrying out the integration set dc dv (since u is a constant). The trace of a matrix is defined as the sum of the diagonal elements: Tr P Pxx Pyy Pzz . 14 Scalar pressure We define the scalar pressure ps as 1 1 ps (x, t ) Tr P Pxx Pyy Pzz . 3 3 This is compatible with our standard definition of pressure, which assumes isotropic velocity distribution. For isotropic distribution: 2 cx cx 0 0 cx 0 0 2 cc 0 c y c y 0 = 0 c y 0 . Then
0 0 0 c 2 0 c c z z z 2 1 1 ps (x, t ) Tr P ms cx2 c y2 cz2 f s c 2 sin dcd d 3 3 0 00 4 ps (x, t ) ms c 4 f s x, c, t dc 3 0 It is easy to show that for a Maxwellian distribution ps (x, t ) ns k BTs 15 Heat Flux The 3rd moment is related to the heat flux vector: 1 1 2 Q s x, t ms cc ms cc 2 f s x, v, t dv 2 2 By now we have defined the most important macroscopic variables: density ns , pressure tensor P s , and the heat flow vector Q. To find these variables, it seems we need to solve the Boltzmann equation to get f s . There is a trick to avoid this: use moments of the Botzmann equation to arrive at simpler equations. 16 Continuity equation: 0th moment of B-equation
The nth moment of the Boltzmann equation: f s n fs v t v f s a v f s dv v t coll dv The 0th moment yields the continuity equation: n f s fs t v f s a v f s dv t coll dv I1 I 2 I 3 I 4 f s ns I1 dv f s dv t t t I 2 v f s dv v s f s dv vs f s dv nsus s f f f I 3 a v f s dv ax ay az dvx dv y dvz x y z 17 f First term = [ ax
dvx ]dv y dvz vx If we assume that a x does not depend on v x : f ax dvx ax f 0 since f 0. vx Therefore I 3 0. The term on the right hand site is the net source per unit volume, Ss . Then ns ns u s S s t 18 The integral form of the continuity equation ns ns u s dx S s dx t vol vol ns t vol N s = rate of change of total number of particles in vol dx t
n u dx n u s s s vol s dS (Gauss's theorem) S net particle flux out of surface enclosing vol ~ S dx S s s net total production inside vol vol N s ~ S s xvv s dS t 19 Momentum Equation: 1st moment of B-equation The 1st moment of the Boltzmann equation leads to the momentum equation. f s fs v v
f a f d v v s v s dv t t coll f s v t dv t vf s dv t nsu s 3 v v f s dv v1 v j f s dv v1v j f s dv; c v u 1 j 1 x j x j j c1 u1 c j u j f s dv j x j j c c
f d v u u f d v c1u j u1c j f s dv 1 j s 1 j s x j j x j j x j i ii iii 20 v v f s dv 1 c c f d v
u u f d v c1u j u1c j f s dv 1 j s 1 j s j x j j x j j x j i i c1c j f s dv ns ii c1c j iii 1 Ps ,1 j by definition ms ii u1u j f s dv u1u j f s dv nsu1u j iii c1u j u1c j f s dv u j c1 u1 c j 0, since c1 0, c j 0 cj = v j - uj = v j - u j = v j - u j 1 v v f s dv P1 j ns u1u j
1 j x j ms 21 1 v v f s dv P1 j ns u1u j 1 j x j ms P1 j 1 1 u1 u u n P u u n u n 1 j s s ,1 1
j s 1 s ms j x j ms j x j j x j j x j = 1 Ps ,1 u1 uns ns u1u1 ms where Ps,1 Ps ,11 , Ps ,12 , Ps ,13 Same for components 2 and 3. Then in vector notation 1 v v f s dv ms Ps u u ns 1 Ps u uns ns u u ms In case the different species move with different average velocities, replace u by u s . 22 The 1-component of the 3rd term is 3 v a v f s dv v1 a j f s dv 1 j 1 v j f s f s f s a1v1 dv a2v1 dv a3v1 dv v1
v2 v1 f s a1v1 f s dv2 dv3 a1 f s dv v1dv1dv3 a2 dv2 .... v2 (in , I used the fact that a1 does not depend on v1 ) v a v f s dv ns a1 0 0 (see problem 2.5) 1 or in vector notation: v a f dv n v s s a 23 The momentum equation The momentum equation now becomes: 1 f nu P unu n u u n a v dv t m t coll or
M u P u u u u a t t coll or M u P uu a t t coll q a E x, t u B x.t a ext m 24 Pressure Gradient Force The momentum equation can be written Du M P a muS Dt t coll The rate of change of momentum in a fluid parcel V with density due to a pressure gradient D momentum in V PG P dx P dS Dt vol S For isotropic pressure (same in all direction) the pressure gradient force per unit mass, i.e., the accelaration due to a pressure gradient is: a PG
1 1 P p. 25 Pressure as Thermal Energy Density Density and flow speed u can be obtained from the continuity and momentum equations. The momentum equation contains the pressure: Du M p a muS . Dt t coll Dimension analysis: force p a 3 l force l energy J p 3 3 energy density. 3 l l m The 2nd moment of the Boltzmann equation leads to the energy equation. f s fs vv t v f s a v f s dv vv t coll dv etc.... 26 Charge and Current Densities Charge density c qs ns x, t Cb / m3
s i, e s Current density J x,t qs ns x, t u s x, t A m 2 s i, e Multiply the conitnuity equations for electrons and ions each by corresponding charge, and add the two equations up. Current Continuity Equation c J 0 t 27 Moments Measurements in Space Physics Particle detector measures electric current produced by the charged particles The electric field current, count-rate, is converted to number of density flux according to calibration The flux can be converted to distribution function The detector can select the energy range with energy resolution, as well as type of the particle species Each detector has a finite field angle which determines the spatial resolution The detector(s) scan over directions Fluxes from all energy channels and all looking directions can be added together. Moment factor can be added in the integration process to obtain moments, which are used as macroscopic quantities, such as density, velocity, temperature, (temperature) anisotropy Satellite charge is a major problem in deriving correct distribution function In the existence of multiple populations, moments may mean different things. Distribution functions have to be examined in this case. 28
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