Collisions in Fully-Ionized Plasmas We now look at a fully-ionized plasma. In this case all collisions are Coulomb collisions between charged particles. It turns out that there is a major difference between like-particle collisions (ion-ion or electron-electron) and unlike-particle collisions (electron-ion or ion-electron) Look at two identical particles making a collision In a head-on collision the particles emerge with their velocities reversed so the particles simply interchange their orbits and the two guiding centers remain in the same traces. The very worse case is a 90 collision in which the velocities are changed 90 in direction

Now, after the collision – the guiding centers have indeed shifted – but the “center of mass” of the guiding centers has remained steady. Now in collisions with neutrals all we cared about was the motion of the charged particle. Here, we must consider both particles – and there’s detailed balance. For each ion that moves outward, there’s one that moves inward. Now the situation changes completely for unlike particle collisions. It turns out that the worst case is the 180 collisions. Here the particles emerge with their velocities reversed – for simplicity we’re looking at equal mass particles. Since both particles must continue to gyrate in proper sense, the displacement of the guiding centers is in the same direction. So unlike particle collisions give rise to diffusion.

Left: Schematic showing particle displacements in direct Coulomb collisions between like species in a magnetized plasma. Right: Collisions between unlike particles effectively displace guiding centers Because of the mass disparity, electrons bounce off almost stationary ions and execute a random walk of step length rL. In general, ions only move slightly, but very often. However, conservation of momentum implies that the diffusion rate for ions and electrons is the same (no charge separation, no electric fields).

The trajectory of a neutral particle in a partially ionized gas exhibits “straight-line" motion between abrupt atomic collisions.

The trajectory of a “test" charged particle (electron) in a partially ionized gas exhibits continuous small-angle deflections or scatterings of its direction of motion. The largest deflections occur when a charged particle passes close to another charged particle.

Coulomb collisions An ion nearing another will feel a repulsive force. What is the cross-section for a collision? The answer is ∞, since no matter how far apart the ions are, they will be deflected. However, we can define a cross-section for 90o scattering - cross-section for momentum transfer. Sydney University

We can derive an approximate expression for the 90o scattering cross-section. The correct result is similar but the derivation is much more complicated, it takes into account the accumulated effect of many small angle collisions.

Assume ion remains at rest and the electron is incident with impact parameter, b, such that it is deflected 90o (b1=b2, v1=v2). The attractive force is given by: F varies with time because r varies with time, so change in momentum is given by:

so get when the integral is done properly the answer is: The important point to note is that b∝v-2. If an electron is moving slowly, the force acts for a long time so the electron will experience a larger deflection than a fast electron at the same b.

The cross-section for deflections larger than 90o can be written as: And so the collision frequency for Coulomb collisions of an electron with ions is:

Coulomb collisions Coulomb collisions between free particles in a plasma is an elastic process. Let us consider the Coulomb force between two test charges q and Q: This is a long range force and the cross-section for interaction of isolated charges is infinity! It is quite different from elastic “hard-sphere” encounters such as that which can occur between electrons and neutrals for example. In a plasma, however, the Debye shielding limits the range of the force so that an effective cross-section can be found. Nevertheless, because of the nature of the force, the most frequent Coulomb deflections result in only a small deviation of the particle path before it encounters another free charge. To produce an effective 90 degree scattering of the particle (and hence momentum transfer) requires an accumulation of many such glancing collisions. The collision cross-section is then calculated by the statistical analysis of many such small-angle encounters. University of Alaska, Antonius Otto

Consider the Coulomb force on an electron as it follows the unperturbed path shown in the picture, only the perpendicular component matters because the parallel component of the force reverses direction after q passes Q. Thus, = b/r or where b is the impact parameter for the interaction. Since x (the Parallel coordinate) changes with time, we integrate along the path to obtain the net perpendicular impulse delivered to q

Now so that We consider a statistical average over a random distribution of such small angle collisions. For a random walk with step length s, the total displacement after N steps is (i.e. ) where N is the number of steps. The total change in velocity is thus

Now integrate over the range of impact parameters b to estimate the number of glancing collisions in time t. Small angle collisions are much less likely because of the geometrical effect We combine the three equations above and integrate over impact parameter: bmin is the closest approach that satisfies the small deflection hypothesis. We obtain this be setting

Outside D the charge Q is not felt. We thus take bmax =D and ln is called the Coulomb logarithm and is a slowly varying function of electron density and temperature. For fusion plasma, ln  6 - 16. One usually sets ln = 10 in quantitative estimations.

For electron-ion encounters, q = -e and Q = Ze so We are finally in a position to find the elapsed time necessary for a net 90 degree deflection, i.e. and For electron-ion encounters, q = -e and Q = Ze so and ei = nieiv implies that

Derivation Review Perpendicular impulse  1/vb Angular deflection Random walk to scatter one radian Integrate over The dependence has very important ramifications. A high temperature plasma is essentially collisionless. This means that plasma resistance decreases as temperature increases. In some circumstances populations of particles can be continually accelerated, losing energy only through synchrotron radiation (for example runaway electrons in a tokamak).

Sketch of the collisional drag Fc acting on electrons as a function of their velocity vd for E > ED. The black arrows indicate the overall acceleration or deceleration.

Melting damage to the upper inner wall of JET, thought to be caused by run-away electrons.

Energy transfer in electron-ion collisions A pervading theme in plasma physics is me  mi. This has consequences for collisions between the species in that we expect very little energy Transfer between the species. To illustrate this, consider such a collision in the centre-of-mass frame (a direct hit).

Collision between ion and electron in center of mass frame eliminate V, V0:

Now translate to frame in which ion M is at rest (initially) Now translate to frame in which ion M is at rest (initially). In this frame the initial electron energy is and the final ion energy is Their ratio is given by and thus

For glancing collisions the energy transfer between ions and electrons is even less. Coulomb collisions result in very poor energy transfer between electrons and ions. The rate of energy transfer is roughly (me/mi) slower than the e-i collision frequency. On the other hand, energy transfer rate and collision frequency are the same for collisions between ions ii  ei(me/mi) : That is,

Coulomb Cross section for Momentum Exchange Scattering of momentum reduces the momentum of a stream of particles and is therefore generates resistance for a stream particles. To measure the resistance we need the total cross section for the momentum scattering. Assume scattering of a particle with mass m , charge q , and velocity v by a particle with mass and charge q0 . The impact parameter is b . Antonius Otto

Mechanics: The momentum along x after the scattering is such that the momentum change is Let N be the number of particles per time unit and unit area: Then particles are scattered into the range

Definition: Differential cross section for momentum exchange

total cross section: with and we obtain

and for Finally, we can express  through the scattering parameters by noting that for small values of  we have such that Here, the expression is called the Coulomb logarithm.

Note that p,tot  electrical resistance For Conclusion: The simple model of Coulomb scattering yields an inconsistency. The total cross section diverges through the contribution of large numbers of particle with large impact parameters. However, at distance larger than the Debye length the coulomb potential is shielded. Let us also consider a hydrogen plasma with

=> Coulomb logarithm is the logarithm of the plasma parameter! No coincidence: The plasma parameter (and powers thereof) is the only possibility to create a dimensionless parameter that is a function of m , e , n , and T .

Collision frequency The collision frequency is equal to the flux of particles vn0 multiplied with the total cross section p,tot : Consider a Maxwell distribution function:

Integrating over velocity space yields the average such that It is interesting to compare collision and plasma frequency: Binary collisions are less important than collective plasma effects!!

Collisions in Plasmas 3.1 Binary collisions between charged particles Hutchinson

Figure 3.1: Geometry of the collision orbit

3.1.1 Frames of Reference

3.1.2 Scattering Angle Figure 3.2: Relation between 1 and  .

Figure 3.3: Collisions viewed in Center of Mass and Laboratory frame.

3.2 Differential Cross-Section for Scattering by Angle Rutherford Cross-Section

Figure 3.4: Scattering angle and solid angle relationship.

Figure 3.5: Annular volume corresponding to db.

2 Coulomb collisions H. J. de Blank PLASMA PHYSICS Lecture Notes

A Appendix: Coulomb collisions H. J. de Blank PLASMA PHYSICS Lecture Notes

Figure 38: Motion around a fixed center of force

Coulomb Collisions* *Ref. Jackson Classical Electrodynamics, Chapter 13 We will now look at small-angle coulomb scatterings since these are what is responsible for the resistivity of a plasma. We consider a particle with charge q1 and initial momentum mvo scattered by a fixed particle of charge q2. We assume that the impact parameter is large enough that the angle of deflection of particle 1 is small . The geometry is as shown.

Now for x << 1 we have:   Now so and is the force acting on particle 1. From the diagram you can see that Now we can also write using the fact that the angular momentum of particle 1 about 2 is given by so or

Now taking electrons and singly charged ions we can write: Then:   Now taking electrons and singly charged ions we can write: We recall that the differential scattering crossection is defined by: so Note the exact Rutherford scattering crossection is

But using we have   Now, from what we know of plasma physics so far we suspect that plasma effects – i.e. Debye shielding may turn out to alter this result somewhat. For impact parameters the order of or greater than the Debye length, the potential falls off rapidly so the scattering angle will tend to vanish much more rapidly with increasing impact parameter than we have indicated. So the differential scattering cross section should flatten out rather than increasing with x-4. Therefore, we put in a cutoff:

To calculate xmin we let b = D in the expression for x.   and letting There is also a maximum value for x. Quantum mechanically

Summary 1. Want to calculate ei [i.e. shielded Coulomb collisions].   2. where x is the scattering x = x(b). 3. 4. where

Scattering Cross Section All these land here N0: The number of particles incident on the target foil per unit area per unit time. Any incident particles entering with impact parameter b and b+db will scatter to the angle q and q-dq. In other words, they scatter into the solid angle dW (=2psinqdq). So the number of particles scattered into the solid angle dW per unit time is 2pN0bdb. PHYS 3446, Fall 2006 Jae Yu

Scattering Cross Section For a central potential Such as Coulomb potential Which has spherical symmetry The scattering center presents an effective transverse cross-sectional area of For the particles to scatter into q and q+dq

Scattering Cross Section In more generalized cases, Ds depends on both q & f. With a spherical symmetry, f can be integrated out: Why negative? Since the deflection and change of b are in opposite direction!! What is the dimension of the differential cross section? Differential Cross Section reorganize Area!!

Scattering Cross Section For a central potential, measuring the yield as a function of q, or differential cross section, is equivalent to measuring the entire effect of the scattering So what is the physical meaning of the differential cross section? Measurement of yield as a function of specific experimental variable This is equivalent to measuring the probability of occurrence of a physical process in a specific kinematic phase space Cross sections are measured in the unit of barns: Where does this come from? Cross sectional area of a typical nucleus!

Total Cross Section Total cross section is the integration of the differential cross section over the entire solid angle, W: Total cross section represents the effective size of the scattering center at all possible impact parameter

Cross Section of Rutherford Scattering The impact parameter in Rutherford scattering is Thus, Differential cross section of Rutherford scattering is

Rutherford Scattering Cross Section Let’s plug in the numbers ZAu=79 ZHe=2 For E=10 keV Differential cross section of Rutherford scattering

Total X-Section of Rutherford Scattering To obtain the total cross section of Rutherford scattering, one integrates the differential cross section over all q: What is the result of this integration? Infinity!! Does this make sense? Yes Why? Since the Coulomb force’s range is infinite. Is this physically meaningful? No What would be the sensible thing to do? Integrate to a cut-off angle since after certain distance the force is too weak to impact the scattering. (q=q0>0)

Measuring Cross Sections Rutherford scattering experiment Used a collimated beam of a particles emitted from Radon A thin Au foil target A scintillating glass screen with ZnS phosphor deposit Telescope to view limited area of solid angle Telescope only need to move along q not f. Why? Due to the spherical symmetry, scattering only depends on q not f.

Measuring Cross Sections With the flux of N0 per unit area per second Any a particles in b and b+db will be scattered into q and q-dq The telescope aperture limits the measurable area to How could they have increased the rate of measurement? By constructing an annular telescope By how much would it increase? 2p/df

Measuring Cross Sections Fraction of incident particles approaching the target in the small area Ds=bdfdb at impact parameter b is –dn/N0. dn particles scatter into R2dW, the aperture of the telescope This ratio is the same as The sum of Ds over all N nuclear centers throughout the foil divided by the total area (S) of the foil. Probability for incident particles to enter within the N areas of the annular rings and subsequently scatter into the telescope aperture So this ratio can be expressed as

Measuring Cross Sections For a foil with thickness t, mass density r, atomic weight A: Since from what we have learned previously The number of a scattered into the detector angle (q,f) is A0: Avogadro’s number of atoms per mol

Measuring Cross Sections Detector acceptance Number of detected particles Projectile particle flux Density of the target particles Scattering cross section This is a general expression for any scattering process, independent of the existence of theory This gives an observed counts per second

Coulomb collisions An ion nearing another will feel a repulsive force. What is the cross-section for a collision? The answer is ∞, since no matter how far apart the ions are, they will be deflected. However we can define a cross-section for 90o scattering - cross-section for momentum transfer. We can derive an approximate expression for the 90o scattering cross-section. The correct result is similar but the derivation is much more complicated, it takes into account the accumulated effect of many small angle collisions. University of Sydney

Assume ion remains at rest and the electron is incident with impact parameter, b, such that it is deflected 90o (b1=b2, v1=v2). The attractive force is given by: F varies with time because r varies with time, so change in momentum is given by:

So we get: when the integral is done properly the answer is: The important point to note is that b∝v-2. If an electron is moving slowly, the force acts for a long time so the electron will experience a larger deflection than a fast electron at the same b.

The cross-section for deflections larger than 90o can be written as: And so the collision frequency for Coulomb collisions of an electron with ions is:

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