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The situation is totally different ina plasma, which has charged particles. As these charges move around, they cangenerate local concentrations of positive or negative charge, which give rise toelectric fields. Motion of charges also generates currents, and hence magnetic fields.These fields affect the motion of other charged particles far away.Let us consider the effect on each other of two slightly charged regions ofplasma separated by a distance r (Fig. 1.1).
The Coulomb force between A andB diminishes as 1/r2. However, for a given solid angle (that is, Δr/r ¼ constant), thevolume of plasma in B that can affect A increases as r3. Therefore, elements ofplasma exert a force on one another even at large distances. It is this long-rangedCoulomb force that gives the plasma a large repertoire of possible motions andenriches the field of study known as plasma physics. In fact, the most interestingresults concern so-called “collisionless” plasmas, in which the long-range electromagnetic forces are so much larger than the forces due to ordinary local collisionsthat the latter can be neglected altogether. By “collective behavior” we meanmotions that depend not only on local conditions but on the state of the plasmain remote regions as well.The word “plasma” seems to be a misnomer. It comes from the Greek πλάσμα,ατoς, τo, which means something molded or fabricated.
Because of collectivebehavior, a plasma does not tend to conform to external influences; rather, it oftenbehaves as if it had a mind of its own.41.31 IntroductionConcept of TemperatureBefore proceeding further, it is well to review and extend our physical notions of“temperature.” A gas in thermal equilibrium has particles of all velocities, and themost probable distribution of these velocities is known as the Maxwellian distribution. For simplicity, consider a gas in which the particles can move only in onedimension. (This is not entirely frivolous; a strong magnetic field, for instance, canconstrain electrons to move only along the field lines.) The one-dimensionalMaxwellian distribution is given byf ðuÞ ¼ A exp 12 mu2 =KTð1:2Þwhere f du is the number of particles per m3 with velocity between u and u + du,mu2 is the kinetic energy, and K is Boltzmann’s constant,12K ¼ 1:38 1023 J= KNote that a capital K is used here, since lower-case k is reserved for the propagationconstant of waves.
The density n, or number of particles per m3, is given by(see Fig. 1.2)n¼ð1f ðuÞduð1:3Þ1The constant A is related to the density n by (see Problem 1.2)A¼nFig. 1.2 A Maxwellian velocity distribution m 1=22πKTð1:4Þ1.3 Concept of Temperature5The width of the distribution is characterized by the constant T, which we callthe temperature. To see the exact meaning of T, we can compute the average kineticenergy of particles in this distribution:Eav ¼ð1122 mu f ðuÞdu1ð1ð1:5Þf ðuÞdu1Definingvth ¼ ð2KT=mÞ1=2andy ¼ u=vthð1:6Þwe can write Eq.
(1.2) asf ðuÞ ¼ A exp u2 =v2thand Eq. (1.5) as132 mAvthEav ¼ð1Avthð111exp y2 y2 dyexp y2 dyThe integral in the numerator is integrable by parts:ð11y exp y2 ydy ¼ 12½expðy2 Þy 1 1ð11¼2exp y2 dyð1112exp y2 dy1Canceling the integrals, we have1Eav ¼ 2mAv3th 12 1 2¼ 4 mvth ¼ 12 KTAvthð1:7ÞThus the average kinetic energy is 12 KT.It is easy to extend this result to three dimensions. Maxwell’s distribution is thenwheref ðu; v; wÞ ¼ A3 exp 12 mðu2 þ v2 þ w2 Þ=KTA3 ¼ n m 3=22πKTð1:8Þð1:9Þ61 IntroductionThe average kinetic energy isEav ¼ððð 1A3 12 mðu2 þ v2 þ w2 Þexp 12 mðu2 þ v2 þ w2 Þ=KT du dv dw1 ððð1A3 exp 12 mðu2 þ v2 þ w2 Þ=KT du dv dw1We note that this expression is symmetric in u, v, and w, since a Maxwelliandistribution is isotropic. Consequently, each of the three terms in the numerator isthe same as the others. We need only to evaluate the first term and multiply by three:Eav ¼ ÐÐÐ3A3 12 mu2 exp 12 mu2 =KT du exp 12 mðv2 þ w2 Þ=KT dv dwÐÐÐA3 exp 12 mu2 =KT du exp 12 mðv2 þ w2 Þ=KT dv dwUsing our previous result, we haveEav ¼ 32 KTð1:10ÞThe general result is that Eay equals 12KT per degree of freedom.Since T and Eav are so closely related, it is customary in plasma physics to givetemperatures in units of energy.
To avoid confusion on the number of dimensionsinvolved, it is not Eav but the energy corresponding to KT that is used to denote thetemperature. For KT ¼ 1 eV ¼ 1.6 1019 J, we haveT¼1:6 1019¼ 11, 6001:38 1023Thus the conversion factor is1 eV ¼ 11, 600 Kð1:11ÞBy a 2-eV plasma we mean that KT ¼ 2 eV, or Eav ¼ 3 eV in three dimensions.It is interesting that a plasma can have several temperatures at the same time.It often happens that the ions and the electrons have separate Maxwellian distributions with different temperatures Ti and Te. This can come about because thecollision rate among ions or among electrons themselves is larger than the rate ofcollisions between an ion and an electron.
Then each species can be in its ownthermal equilibrium, but the plasma may not last long enough for the two temperatures to equalize. When there is a magnetic field B, even a single species, say ions,can have two temperatures. This is because the forces acting on an ion along B aredifferent from those acting perpendicular to B (due to the Lorentz force). Thecomponents of velocity perpendicular to B and parallel to B may then belong todifferent Maxwellian distributions with temperatures T⊥ and T||.1.4 Debye Shielding7Before leaving our review of the notion of temperature, we should dispel thepopular misconception that high temperature necessarily means a lot of heat.People are usually amazed to learn that the electron temperature inside a fluorescentlight bulb is about 20,000 K.
“My, it doesn’t feel that hot!” Of course, the heatcapacity must also be taken into account. The density of electrons inside a fluorescent tube is much less than that of a gas at atmospheric pressure, and the totalamount of heat transferred to the wall by electrons striking it at their thermalvelocities is not that great.
Everyone has had the experience of a cigarette ashdropped innocuously on his hand. Although the temperature is high enough to causea burn, the total amount of heat involved is not. Many laboratory plasmas havetemperatures of the order of 1,000,000 K (100 eV), but at densities of only1018–1019 per m3, the heating of the walls is not a serious consideration.Problems1.1. Compute the density (in units of m3) of an ideal gas under the followingconditions:(a) At 0 C and 760 Torr pressure (1 Torr ¼ 1 mmHg).
This is called theLoschmidt number.(b) In a vacuum of 103 Torr at room temperature (20 C). This number is auseful one for the experimentalist to know by heart (103 Torr ¼ 1 μ).1.2. Derive the constant A for a normalized one-dimensional Maxwelliandistribution^f ðuÞ ¼ Aexp mu2 =2KTsuch thatð1^f ðuÞdu ¼ 11Hint: To save writing, replace (2KT/m)1/2 by vth (Eq. 1.6).1.2a. (Advanced problem). Find A for a two-dimensional distribution which integrates to unity. Extra credit for a solution in cylindrical coordinates.^f ðu; vÞ ¼ Aexp m u2 þ v2 =2KT1.4Debye ShieldingA fundamental characteristic of the behavior of plasma is its ability to shield outelectric potentials that are applied to it. Suppose we tried to put an electric fieldinside a plasma by inserting two charged balls connected to a battery (Fig.
1.3). Theballs would attract particles of the opposite charge, and almost immediately a cloudof ions would surround the negative ball and a cloud of electrons would surround81 IntroductionFig. 1.3 Debye shieldingFig. 1.4 Potentialdistribution near a gridin a plasmathe positive ball. (We assume that a layer of dielectric keeps the plasma fromactually recombining on the surface, or that the battery is large enough to maintainthe potential in spite of this.) If the plasma were cold and there were no thermalmotions, there would be just as many charges in the cloud as in the ball, theshielding would be perfect, and no electric field would be present in the body ofthe plasma outside of the clouds. On the other hand, if the temperature is finite,those particles that are at the edge of the cloud, where the electric field is weak, haveenough thermal energy to escape from the electrostatic potential well.
The “edge”of the cloud then occurs at the radius where the potential energy is approximatelyequal to the thermal energy KT of the particles, and the shielding is not complete.Potentials of the order of KT/e can leak into the plasma and cause finite electricfields to exist there.Let us compute the approximate thickness of such a charge cloud. Imagine thatthe potential ϕ on the plane x ¼ 0 is held at a value ϕ0 by a perfectly transparent grid(Fig. 1.4).
We wish to compute ϕ(x). For simplicity, we assume that the ion–electron mass ratio M/m is infinite, so that the ions do not move but form a uniformbackground of positive charge. To be more precise, we can say that M/m is large1.4 Debye Shielding9enough that the inertia of the ions prevents them from moving significantly on thetime scale of the experiment.
Poisson’s equation in one dimension isε 0 ∇2 ϕ ¼ ε 0d2 ϕ¼ eðni ne Þ ðZ ¼ 1Þdx2ð1:12ÞIf the density far away is n1, we haveni ¼ n1In the presence of a potential energy qϕ, the electron distribution function isf ðuÞ ¼ A exp 12 mu2 þ qϕ =KT eð1:13ÞIt would not be worthwhile to prove this here. What this equation says is intuitivelyobvious: There are fewer particles at places where the potential energy is large,since not all particles have enough energy to get there. Integrating f (u) over u,setting q ¼ e, and noting that ne(ϕ ! 0) ¼ n1, we findne ¼ n1 expðeϕ=KT e ÞThis equation will be derived with more physical insight in Sect.
3.5. Substitutingfor ni and ne in Eq. (1.12), we haveε0d2 ϕeϕ=KT e¼ene11dx2In the region where jeϕ/KTej 1, we can expand the exponential in a Taylor series: 2d2 ϕeϕeϕþ 12 KTþð1:14Þε0 2 ¼ en1edxKT eNo simplification is possible for the region near the grid, where jeϕ/KTej may belarge. Fortunately, this region does not contribute much to the thickness of the cloud(called a sheath), because the potential falls very rapidly there. Keeping only thelinear terms in Eq. (1.13), we haved 2 ϕ n1 e 2¼ϕdx2KT eð1:15Þε0 KT e 1=2λD ne2ð1:16Þε0Definingwhere n stands for n1, and KTe is in joules. KTe is often given in eV, in which case,we will write it also as TeV.101 IntroductionWe can write the solution of Eq. (1.14) asϕ ¼ ϕ0 expðjxj=λD Þð1:17ÞThe quantity λD, called the Debye length, is a measure of the shielding distance orthickness of the sheath.Note that as the density is increased, λD decreases, as one would expect, sinceeach layer of plasma contains more electrons.
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