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1 projectile,Ib Impact Parameter,Figure 3 1 Geometry of the collision orbit. Angular momentum is conserved,m r28 const m bvl 0 clockwise from symmetry. Substitute u,then 8 3 u2 bv,Then radial acceleration is. This orbit equation has the elementary solution, The sin0 term is absent by symmetry The other constant of integration C must be deter. mined by initial condition At initial far distant angle 01 u l oo. sin O1 1 Cb b,cosO1 V 42 C, Notice that tan81 1 when b b g This is when O1 45 and x 90 So particle.
emerges at 90 to initial direction when,b bgO 90 impact parameter 3 16. 3 1 1 Frames of Reference, Key quantity we want is the scattering angle but we need to be careful about reference. Most natural frame of ref is Center of Mass frame in which C of M is stationary C of. M has position,and velocitv in lab frame, So motion of either particle in C of M frame is a factor times difference vector r. Velocity in lab frame is obtained by adding V to the C of M velocity e g V. Angles of position vectors and velocity dzfferences are same in all frames. Angles i e directions of velocities are n o t same. 3 1 2 Scattering Angle,In C of M frame is just the final angle of r. 81 is negative,Figure 3 2 Relation between O1 and x.
But scattering angle defined as exit velocity angle relative to initial velocity in lab frame. is different,Final velocity in CM frame,COS x sin xc vl cos X C sin x 3 27. xC x and vl is initial relative velocity Final velocity in Lab frame. So angle is given by,cot XL v G G,cos Xc V ml m2,cosecx cot X 3 29. For the specific case when m2 is initially a stationary target in lab frame then. cot XL cosecx cot X,This is exact,Small angle approximation cot x i i. cosecx i 1gives, So small angles are proportional with ratio set by the mass ratio of particles. Center of Mass Frame,Particle 1,Laboratory Frame,Stationary Target.
Particle 1 X,Particle 2, Figure 3 3 Collisions viewed in Center of Mass and Laboratory frame. 3 2 Differential Cross Section for Scattering by Angle. Rutherford Cross Section,By definition the cross section 0 for. any specified collision process when a particle is passing. through a density n 2 of targets is such that the number of such collisions per unit path length. Sometimes a continuum of types of collision is considered e g we consider collisions at. different angles x to be distinct In that case we usually discuss differential cross sections. e g defined such that number of collisions in an angle element d x per unit path length. is n2 dx Note that 4 is just notation for a number Some authors just write x but I. find that less clear, Normally for scattering angle discrimination we discuss the differential cross section per unit. solid angle, This is related to scattering angle integrated over all azimuthal directions of scattering by. Figure 3 4 Scattering angle and solid angle relationship. dCl 2 rrsinxdx,So that since, Now since x is a function only of the impact parameter b we just have to determine the.
number of collisions per unit length a t impact parameter b. Figure 3 5 Annular volume corresponding to db, Think of the projectile as dragging along an annulus of radius b and thickness db for an. elementary distance along its path d It thereby drags through a volume. Therefore in this distance it has encountered a total number of targets. d 2 rrbdb n2, at impact parameter b db By definition this is equal to d dbn2 Hence the differential. cross section for scattering encounter at impact parameter b is. Again by definition since x is a function of b, dbldx is negative but different ial cross sect ions are posit ive. Substitute and we get,This is a general result for classical collisions. For Coulomb collisions in C of M frame,b 90 c o t,dff b90 cot b90 2x.
dfl sinx 2 2,bgo cos sin 1 5,2 2sin cos sin2 5,This is the Rutherford Cross Section. for scattering by Coulomb forces through an angle x measured in C of M frame. Notice that i i as x i 0, This is because of the long range nature of the Coulomb force Distant collisions tend to. dominate X i 0 H b i a,3 3 Relaxation Processes, There are 2 main different types of collisional relaxation process we need to discuss for a. test particle moving through a background of scatterers. 1 Energy Loss or equilibrium,2 Momentum Loss or angular scattering. The distinction may be illustrated by a large angle 90 scatter from a heavy stationary. If the target is fixed no energy is transferred to it So the energy loss is zero or small if. scatterer is just heavy However the m o m e n t u m in the x direction is completely lost in. this 90 scatter, This shows that the timescales for Energy loss and momentum loss may be very different.
3 3 1 Energy Loss, For an initially stationary target the final velocity in lab frame of the projectile is. So the final kinetic energy is,m1v 1 2 sin,Hence the kinetic energy lost is AK K Kt. 1 4mlm2 2 XC,m1V2 sin 3 53,1 4mlm2 1 b,2 m l m2 2 1 using cot. exact For small angles x 1 i e blbgo 1 this energy lost in a single collision is. approximately, If what we are asking is how fast does the projectile lose energy Then we need add up the. effects of all collisions in an elemental length d a t all relevant impact parameters. The contribution from impact parameter range db at b will equal the number of targets. encountered times AK,n2d 2rbdb 1 l v 21 4m1m2,encounters m l m2.
Loss per encounter A K, This must be integrated over all b t o get total energy loss. Kn2 8nb In b, We see there is a problem both limits of the integral b i 0 b i oo diverge logarithmically. That is because the formulas we are integrating are approximate. 1 We are using small angle approx for AK, 2 We are assuming the Coulomb force applies but this is a plasma so there is screening. 3 3 2 Cut offs Estimates, 1 Small angle approx breaks down around b bgO Just truncate the integral there. ignore contributions from b bgO, 2 Debye Shielding says really the potential varies as.
P 2 instead of K, so approximate this by cutting off integral at b AD equivalent to. b b90 b AD,So Coulomb Logarithm is 1nA, Because these cut offs are in in term result is not sensitive t o their exact values. One commonly uses Collision Frequency Energy Loss Collision Frequency is. Substitute for bgO and m in bgO,Collision time TK TK 1 uK. Effective Energy Loss Cross section ffKn2,3 3 3 Momentum Loss. Loss of x momentum in 1 collision is, small angle approx Hence rate of momentum loss can be obtained using an integral.
identical t o the energy loss but with the above parameters. Note for the future reference,Therefore Momentum Loss. Collision Frequency,Collision Time T l vp,Cross Section effective 0 vp n2vl. Notice ratio,Energy Loss v m l m2 2ml, Third case e g electrons i shows that mostly the angle of velocity scatters Therefore. Momentum Scattering time is often called 90 scattering time to diffuse through 90 in. 3 3 4 Random Walk in angle, When m l m 2 energy loss momentum loss Hence v i vl All that matters is the. scattering angle X L X 2bso b, Mean angle of deviation in length L is zero because all directions are equally likely.
Mean square angle is, Spread is all round when Act2 1 This is roughly when a particle has scattered 90 on. average It requires,Ln2 8 rrb in A 1 33 4, So can think of a kind of cross section for ugO 90 scattering as such that. n2L ogo 1 when Ln2 8 b 1nA 1 3 85,i e ffgo 8 b 1nA 2oP 3 86. This is 8 in larger than cross section for 90 scattering i n single collision. Be Careful gg0 is not a usual type of cross section because the whole process is really. diffusive in angle, Actually all collisio lprocesses due t o coulomb force are best treated in a Mathematical. way as a diffusion in velocity space,i Fokker Plmck equation.
3 3 5 Summary of different types of collision, The Energy Loss collision frequency is to do with slowing down to rest and exchanging. energy It is required for calculating,Equilibration Times of Temperatures. Energy Transfer between species, The Momentum Loss frequency is to do with loss of directed velocity It is required for. calculating,Mobility Conductivity Resistivity,Particle Diffusion. Energy Thermal Diffusion, Usually we distinguish between electrons and ions because of their very different mass.
Energy Loss Stationary Targets Momentum Loss, Sometimes one distinguishes between transverse diffusion of velocity and momentum loss. The ratio of these two is,1 like particles,ZZ Kvii vii Like Ions. But note ions are slowed down by electrons long before being angle scattered. 3 4 Thermal Distribution Collisions, So far we have calculated collision frequencies with stationary targets and single velocity. projectiles but generally we shall care about thermal Maxwellian distributions or nearly. thermal of both species This is harder to calculate and we shall resort to some heuristic. calculations, Very rare for thermal ion velocity to be electron So ignore ion motion. Average over electron distribution, Momentum loss to ions from assumed drifting Maxwellian electron distribution.
Each electron in this distribution is losing momentum to the ions at a rate given by the. collision frequency,424 4 me mi,4 7 mimzv3, so total rate of loss of momentum is given by per unit volume. To evaluate this integral approximately we adopt the following simplifications. 1 Ignore variations of 1nA with v and just replace a typical thermal value in A. 2 Suppose that drift velocity vd is small relative to the typical thermal velocity written. v t m and express f in terms of u Zle to first order in ud Zle. taking x axis along ud and denoting by f the unshifted Maxwellian. Then momentum loss rate per unit volume, To evaluate this integral use the spherical symmetry o f f to see that. Thus the Maxwell averaged momentum loss frequency is. where p m vdn is the momentum per unit volume attributable to drift. substituting for thermal electron velocity v and dropping order term where Ze qi. This is the standard form of electron collision frequency. Ion momentum loss to electrons can be treated by a simple Galilean transformation of the. e i i case because it is still the electron thermal motions that matter. Ions Electrons Ions Electrons, Figure 3 6 Ion electron collisions are equivalent to electron ion collisions in a moving refer. ence frame,Rate of momentum transfer 2 is same in both cases. since drift velocities are the same, Ion momentum loss to electrons is much lower collision frequency than e i i because ions.
possess so much more momentum for the same velocity. Ion ion collisions can be treated somewhat like e i i collisions except that we have to. account for moving targets i e their thermal motion. Consider two different ion species moving relative to each other with drift velocity vd the. targets thermal motion affects the momentum transfer cross section. Using our previous expression for momentum transfer we can write the average rate of. transfer per unit volume as see 3 74 note for future reference. where v is the relative velocity vl v2 and bgOis expressed.

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