Eur. Phys. J. D (2016) 70: 198 DOI: 10.1140/epjd/e2016-70425-9 THE EUROPEAN PHYSICAL JOURNAL D Regular Article
Drift velocity of charged particles in magnetic fields and its relation to the direction of the source current
Hanno Ess´ena and Arne B. Nordmark Department of Mechanics, Royal Institute of Technology (KTH), 10044 Stockholm, Sweden
Received 1 July 2016 / Received in final form 5 August 2016 Published online 13 October 2016 c The Author(s) 2016. This article is published with open access at Springerlink.com
Abstract. Integrable motion of charged particles in magnetic fields produced by stationary current dis- tributions is investigated. We treat motion in the magnetic field from an infinite flat current sheet, a Harris current sheath, an infinite rectilinear current, and a dipole in its equatorial plane. We find that positively charged particles as a rule will drift in the same direction as the current that is the source of the magnetic field in question. The conclusion is that charged particles moving under the influence of current distributions tend to enhance the current and that this indicates current self-amplification.
1 Introduction solutions start from the Lagrangian
1 2 q Finding the motion of classical charged particles in a given L = m v + v · A. (2) 2 c magnetic field is a problem of great practical interest in many areas of physics and engineering [1–3]. One reason where m is mass, q electric charge, and A the vector po- for investigating these problems is that they can throw tential. In the examples we study the motion is periodic some light on the much more difficult coupled problem in one of the coordinates, say x. We denote this period of particles and fields in mutual interaction, a problem with T . The drift velocity in another coordinate, say y,is that can be approached in fluid mechanics using magne- then, tohydrodynamics, in kinetic theory using the Vlasov equa- y(T ) − y(0) vy = , (3) tions, or using statistical mechanics based on the Darwin T Hamiltonian. The latter approach, with its long range ve- i.e. the mean velocity over one period. This is the defini- locity dependent interactions, takes the energy lowering tion used below. responsible for the attraction of parallel currents into ac- We investigate the problem of the drift of a classical count. Investigations by us [4–9] indicate that a plasma charged particle in a static inhomogeneous magnetic field will generate currents and magnetic fields in its approach produced by a given stationary current distribution. The to equilibrium. Here we will study the drift of charged following current distributions are treated: particles in the magnetic fields of given current distribu- – an infinite flat current sheet, tions and conclude that this drift will, as a rule, enhance – an infinite current sheath found as a solution to the the current producing the field, thereby supporting the Vlasov equations by Harris [12], conclusions based on the Darwin Hamiltonian. – an infinite straight current carrying wire [13], Drift velocities can be derived approximately for slowly – current due to a rigidly rotating spherical charged shell varying fields by Alfv´en’s guiding center theory [1,10]. An- i.e. the source of a dipole field. alytical solutions for drift velocities published by Head- land and Seymour [11] confirm the validity of the Alfv´en The infinite flat current sheet case is mathematically formula trivial. It thus makes the physical reason for the drift especially clear, but we are not aware that this has been mc 2 2 B ×∇|B| published before. vD = (v⊥ +2v ) . (1) 2q |B|3 Motion in the Harris sheath is integrable but the inte- grals require numerical solution (as far as we know). We The magnetic fields studied there are, however, difficult are not aware that this problem has been studied before. to realize in practice. Here we focus on analytical solution Analytical results can be found for the limiting cases of of drift velocities in magnetic fields that arise from ide- slow and fast particles. The case of fast particles reduce alized but reasonably realistic current distributions. Our to the flat sheet case. In addition to Hertweck [13] the rectilinear current a e-mail: [email protected] problem has been treated by M¨uller and Dietrich [14], Page 2 of 10 Eur. Phys. J. D (2016) 70: 198
x Yafaev [15], Ess´en [16], and Aguierre et al. [17]. Here we B focus on the drift velocity and how it is affected by the initial conditions. We point out that the maximum drift occurs for helical motion while zero angular momentum y motion results in minimum drift. We believe that this is a j new result. The problem of particles in a magnetic dipole field B is called the St¨ormer problem after Carl St¨ormer [18,19] who treated it in many publications from 1907 and on- Fig. 1. This figure illustrates how positively charged particles wards. There is an extensive literature (for a review see will drift in the direction of the current of a current sheet pro- Dragt [20]) on this non-integrable problem, which is of in- vided the initial conditions are such that the trajectory crosses terest as a model for the van Allen belts outside the Earth. the sheet. The current and drift are to the left in the figure. When one restricts the motion to the equatorial plane per- The magnetic field is into the plane of the figure for positive x pendicular to the dipole the problem is integrable [21,22]. and out of the plane for negative x. For certain initial conditions charged particles are bound between two radii and for these we calculate the angular drift velocity.
2 Charged particle near flat current sheet
In this section we calculate the motion of particles travers- ing a flat current sheet. We thus assume that there is a surface current density in the yz-plane k = σV yˆ where σ is the surface charge density of current carriers and V their velocity in the y-direction. From the relation, π Fig. 2. The drift velocity of a positively charged particle with 4 k nˆ × B − B , c = ( + −) (4) r = v = 1 moving in such a way that it crosses a current sheet. The velocity is shown as a function of ξ,thex-coordinate of where nˆ is normal to the surface of the current and B+ is the center of the circular trajectory. the magnetic field on the side that nˆ points to, while B− is the magnetic field on the opposite side, we find, using x ξ ≤ ξ
Fig. 4. Graph of the function − ln(cosh(x)) proportional to the vector potential Ay(x) of the Harris sheath (12).
as unit of length. The unit of time is then mcLD/(2θ)=1. The resulting Lagrangian is now 1 L = r˙ 2 − y˙ ln (cosh(x)) . (16) 2 One sees that the equation of motion in the z-direction (¨z =0,˙z = const.) decouples from the xy-motion and is Fig. 3. Graph of the number density n(x)(Eq.(9)), of the trivial. We will thus discuss only the xy-motion in what Harris current sheath, with a peak at x = 0 (red curve). This follows and use the Lagrangian charge density moves in the y-direction near the yz-plane. Also shown is the resulting magnetic field B(x), in the z-direction 1 2 2 L = (˙x +˙y ) − y˙ ln(cosh(x)). (17) (Eq. (13)). The field is zero at x = 0 (green curve). 2 There are two constants of the motion, the generalized where V is the speed of the particles in the y-direction, momentum, n0 the number density at x =0,c the speed of light, and LD is the Debye length, py ≡ κv =˙y − ln(cosh(x)), (18) θ and the energy, L , D = 2 (10) πn0q 4 1 1 1 2 E ≡ v2 = (˙x2 +˙y2)= x˙ 2 +[κv +ln(cosh(x))] , with θ = kB T , Boltzmann’s constant times the absolute 2 2 2 (19) temperature,while q is the charge of the particles, see Fig- so the problem is integrable, but the integrals are are not ure 3. The vector potential becomes expressible in terms of well known functions, it seems. A = Ay(x)eˆy, (11) To start with one notices that there is no magnetic field in the plane x =0(yz-plane). A particle in this plane where x θc xV with ˙ = 0 will therefore stay in this plane and move as 2 y Ay(x)=− ln cosh . (12) a free particle. The sign of ˙ depends only on the initial qV cLD condition. The magnetic field B corresponding to this vector po- Let us now focus of equation (18), i.e. tential is in the z-direction, B = B(x)eˆz,andthe z-component is y˙ = κv +ln(cosh(x)). (20) Vx We note that since |y˙|≤v,ln(cosh(x)) ≥ 0, and v ≥ 0, B(x)=− 16πn0θ tanh . (13) cLD we must have κ ≤ 1. (21) We now study the dynamics of a charged particle in this magnetic field. Note also that the positive sign in front of ln(cosh(x)) cor- The Lagrangian for a particle of charge q and mass responds to a positive charge. A graph of this function is m moving in the magnetic field is given by equation (2). shown in Figure 4. It is clear that for positive κv the parti- We first divide the Lagrangian with the particle mass. We cle will always have a positive velocity in the y-direction. then choose For negative κv the y-velocity is negative for values of 2θ |x| < arccosh(e−κv) but further away from the yz-plane = 1 (14) mV it will become positive. We now investigate the x-motion. as our unit of speed and We see from (19)that
cLD 1 2 = 1 (15) Veff (x)= [κv +ln(cosh(x))] , (22) V 2 Page 4 of 10 Eur. Phys. J. D (2016) 70: 198
1
0.8
0.6
0.4
0.2 v /
d 0 v
-0.2
-0.4
-0.6
-0.8
-1 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 κ
Fig. 6. Graphs of the drift velocity vd ≡y˙ in units of v as a function of κ. The top curve (red) is for large speed, v 1, and is given by equation (40). The lowest curve (yellow) is for small speed, v 1, and is given by equations (33)and(34). For intermediate speeds the drift velocities are between these V x v Fig. 5. Graphs of the function eff ( ), with =1,forfour two. The middle curve (blue) is for v = 1 and is calculated nu- κ κ / values of . The top curve is for =23 (green), the curve merically by doing the integrals (25)and(28). As a decreasing x x κ tangent to the -axis at = 0 (yellow) corresponds to =0, κ-value reaches −1 the trajectories bifurcate into two separate x and the curve with the small maximum at = 0 corresponds to trajectories outside the sheath. κ = −2/3 (blue). The curve with the large maximum (violet) has κ = −4/3. The horizontal line (red) is at E = v2/2= / 1 2 so the intersections of this line with the curves are at the The change in y-coordinate during one period of the x- x turning points in the -motion. motion is thus x + [κv +ln(cosh(x)]dx plays the role of an effective potential energy for the Δy =2 . (28) 2 x-motion. Graphs of this potential for different values x− v2 − [κv +ln(cosh(x))] of κv are shown in Figure 5. We therefore have from 2 2E =˙x − 2Veff(x)that The y-drift velocity can now be calculated as function of v and κ as vy = Δy/T but the integrals must be done dx 2 2 numerically (to the best extent of our knowledge) after = 2[E − Veff (x)] = v − [κv +ln(cosh(x))] . dt finding the relevant turning points x− and x+.Thisdrift (23) velocity as a function of κ for v = 1 is shown in Figure 6. From this one obtains dx dt = . (24) 2 3.1 Approximations for small and large v v2 − [κv +ln(cosh(x))] √ By introducing the scaled variablex ˆ = x/ v one can go to T x The period of the -motion is then given by the limit of small speeds in the integrals (25)and(28)and
x+ dx do them analytically. This then gives explicit expression T =2 . (25) for the drift velocity. 2 2 x− v2 − [κv +ln(cosh(x))] Assume that v 1. Using ln(cosh(√ x)) = ln(1+x /2+ ...)=x2/2+..., and putting x/ v =ˆx we get for (25) y dy/dt κv For the motion is we have from (20)that = + xˆ √ ln(cosh(x)) so that + dxˆ v T =2 . (29) 1 2 xˆ− v − κ x2 O v dy =[κv +ln(cosh(x)]dt, (26) 1 + 2 ˆ + ( ) or, using (24), that For v 1thisgives
[κv +ln(cosh(x)]dx 2 xˆ+ dxˆ dy = (27) T = √ . (30) 2 v 4 2 xˆ− 2 2 xˆ v − [κv +ln(cosh(x))] 1 − κ − κxˆ − 4 Eur. Phys. J. D (2016) 70: 198 Page 5 of 10
5 To the same approximation equation (28)gives 4.5 2 xˆ xˆ √ + κ + 2 dxˆ Δy =2 v . (31) 4 xˆ4 xˆ− − κ2 − κx2 − 1 ˆ 4 3.5
Theintegralscanthenbeexpressedintermsofthecom- 3 plete elliptic integrals K(k)andE(k)(seeRef.[23]). For 2.5 −1 <κ<1 we findx ˆ∓ = ∓ 2(1 − κ)and 2 √ T √4 K k , Δy v E k − K k = v ( ) and, =4 [2 ( ) ( )] (32) 1.5
1 where k = 2(1 − κ)/2. We thus find the drift velocity, 0.5 2E(k) − K(k) y˙ = v, (33) 0 K(k) -3-2-10123 Fig. 7. The trajectories of charged particles passing through for the case v small and −1 <κ<1. This drift velocity a Harris sheath with κ = −0.5. The trajectory with the becomes negative for κ-values less than κ ≈−0.65223 smaller amplitude is for large speed, v 1, (yellow curve). (see Fig. 6). For κ<−1 a similar calculation givesx ˆ∓ = The trajectory with the larger amplitude is for speed v =1 −2(κ ± 1) and (blue curve), and the one with intermediate amplitude is for small speed, v 1 (red curve). The drift is in the positive (k2 − 2)K(k)+2E(k) y-direction (upwards in the figure) for all three. Note that y˙ = v, (34) x y k2K(k) the -and -coordinates are scaled differently for the three trajectories, according to Section 3.1.Forv 1wethususe√ x, y /v x, y v x, y / v x, y where k =2/ 2(1 − κ). This drift is always in the nega- (˜ ˜)=(1 )( )andfor 1weuse(ˆ ˆ)=(1 )( ). tive y-direction. When v 1 it turns out to be advantageous to use the scaled variablex ˜ = x/v.Using 3.2 The nature of the sheath drift vx˜ −vx˜ e + e In summary we have found that a positively charged par- ln(cosh(vx˜)) = ln (35) 2 ticle will drift in the positive y-direction for all positive values of κ and also for most negative values of κ>−1. 1 v|x˜| −2v|x˜| =ln e 1+e Note that for κ<0 the function Veff (x) develops a bump 2 at x =0(seeFig.5) and this means that particles spend = v|x˜|−ln 2 + ln 1+e−2v|x˜| (36) more time inside the current sheath. For v = 1 drift in the negative y-direction starts at κ ≈−0.80747. For κ<−1 = v [|x˜| + O(1/v)] (37) the drift is always in the negative y-direction and the mo- tion is bound in a potential minimum on one side of the v we get for the period (25), in the limit of 1, sheath. The particles that drift in the positive y-direction are those that, not too slowly, pass through the sheath. x˜+ dx T ˜ . One notes that the guiding center formula (1) correctly =2 2 2 (38) x˜− 1 − κ − 2κ|x˜|−|x˜| predicts the drift in the negative y-direction for particles staying well outside the sheath i.e. in a direction opposite Similarly we find to the current in the sheath. In our following examples the corresponding drift will be found to be in the same x˜+ κ |x| dx Δy v ( + ˜ ) ˜ . direction as the current. The main reason for this differ- =2 2 2 (39) x˜− 1 − κ − 2κ|x˜|−|x˜| ence is that for a current localized near a point or near a line the magnetic field strength will fall off away from For −1 <κ<1 one finds,√ x ˜∓ = ∓(1 − κ), T =2π − the source, whereas for a current localized near a surface 4arcsin(κ) and, Δy =4v 1 − κ2. The drift velocity is it will increase away from the source. thus √ − κ2 y 1 v. ˙ = π − κ (40) 2 arcsin( ) 4 Particle outside line current This agrees with the results (7)and(8) for the for the flat current sheet with κ = −ξ/r and v = r.Forκ<−1the First some notation and kinematics. We will use cylin- drift is, of course, zero in this case. Some trajectories are drical coordinates, ρ, ϕ, z defined through, x = ρ cos ϕ, shown in Figure 7. y = ρ sin ϕ, z = z, in terms of cartesian coordinates. Page 6 of 10 Eur. Phys. J. D (2016) 70: 198
We also use the corresponding unit vectors eˆρ(ϕ)and Inserting ρ =1weseethat˙ρ = 0, as assumed. This means ˙ ρ ρ eˆϕ(ϕ) who’s time derivatives obey the relations eˆρ =˙ϕ eˆϕ, that = 1 is the inner turning point in the -motion. If ˙ we denote the outer turning point by ρ1 we find that the eˆϕ = −ϕ˙ eˆρ. A position vector is given by, r = ρ eˆρ +z eˆz, period of the ρ-motion is given by and the corresponding velocity is, v =˙ρ eˆρ + ρϕ˙ eˆϕ +˙z eˆz, so that vρ =˙ρ and vϕ = ρϕ˙. ρ1 dρ The general equations of motion can be found from the T =2 . (52) E − U ρ Lagrangian (2). For an (infinite) line current I along the 1 2 eff ( ) z-axis one can use, Use of equation (44)and(46)gives I A ρ −2 ρ eˆ , L2 ( )= c ln z (41) z ρρ z ˙ + ¨ = ρ2 (53) B r 2I eˆ with corresponding magnetic field ( )= cρ ϕ.The but d Lagrangian, divided by particle mass, is then ρρ ρ2 ρρ dt ˙ =˙ + ¨ (54) qI 1 L ρ, ρ, ϕ, z 1 ρ2 ρ2ϕ2 z2 − 2 z ρ. so ( ˙ ˙ ˙)= (˙ + ˙ +˙) 2 ˙ ln (42) L2 d m 2 mc z ρ2 z − ρρ . ˙ = ˙ + ρ2 dt ˙ (55) The quantity qI Integrating this over one period of the ρ-motion gives v ≡ 2 , 0 mc2 (43) T L2 Δz ≡ z T − z ρ2 z dt of dimension velocity, represents a characteristic velocity ( ) (0) = ˙ + 2 (56) 0 ρ of the system. Note that if q and I are of opposite sign this is actually a negative quantity. We use it as unit for The average (or drift) velocity in the z-direction is thus, velocity and put v0 =1below. Since the coordinates ϕ and z are cyclic one finds that Δz vz = (57) the corresponding generalized momenta are constants of T the motion, and it is positive if v0 of equation (43) is positive (q and 2 I ρ ϕ˙ = Lz, (44) of the same sign), otherwise negative. The same result z − ρ p , has been derived by Yafaev [15] and a similar one can be ˙ ln = z (45) found in reference [14]. Below we find explicit expressions for this drift velocity in the two special cases of helical and while the ρ-equation of motion is, plane motion. z ρ − ρϕ2 ˙ . ¨ ˙ + ρ =0 (46) 4.1 Helical trajectories In terms of the constants of the motion (44)-(45) the con- The equation for a helix of radius ρ = R can be written served (kinetic) energy (per mass) can be written (˙ϕ = ω =const.), L2 1 2 z 2 1 2 r(t)=R eˆρ(ωt)+vzt eˆz, (58) E = ρ˙ + +(pz +lnρ) = ρ˙ + Ueff (ρ) . 2 ρ2 2 (47) whereweusetimet as parameter. Introducing the pitch We now assume initial conditions angle θ through, v Rω v θ, v v θ, ρ(0) = 1, ρ˙(0) = 0, ϕ˙(0) = Lz. (48) ϕ = = sin z = cos (59)
or equivalently, vϕ/vz =tanθ,onegets, The energy is then v(t)=v[sin θ eˆϕ(ωt)+cosθ eˆz], (60) 1 2 2 E = Lz + pz . (49) 2 for the tangent (velocity) vector of this helix. Helical trajectories must be solutions of the equations Using (47) we find of motion (44)–(46)with¨ρ =˙ρ =0,andρ = R = 2 constant. These equations give ρ˙ =2E − Ueff(ρ) (50) 2 2 R ω = vz, (61) which using (49) is easily rearranged to and comparison with equation (59)thengives, 2 2 1 ρ L − − ρ p ρ . 2 2 ˙ = z 1 ρ2 ln (2 z +ln ) (51) v sin θ = v cos θ, (62) Eur. Phys. J. D (2016) 70: 198 Page 7 of 10 for the pitch angle θ of the helical trajectory. This can be written 2 2 v − vz = vz (63) 2 2 using sin θ =1− cos θ.Solvingforvz we find that 1 2 vz = 1+4v − 1 (64) 2 in units of v0 of equation (43), so that a negative v0 means that vz is in the opposite direction. One notes that for large v most of the velocity is in the z-direction.
4.2 Plane trajectories v ϕ Fig. 8. The drift velocity in the direction of the current, z We now consider plane trajectories with = constant and v v ϕ L ρ as a function of speed in units of 0 for the case of helical ˙ = z =0.Equation(51) at the outer turning point 1 motion equation (64), upper curve, and for the case of plane then gives, motion (Eq. (75)), lower curve. 1 pz = − ln ρ1, (65) 2 for the case of positive v0, which we assume in what fol- Now, returning to dimensional quantities, one finds that lows. Using this (51)gives Δz I1(v) vz = = vv0 (75) T I0 v |ρ˙| = ln ρ(ln ρ1 − ln ρ). (66) ( ) 1 where v = ln(ρ1/ρ0) is the dimensionless speed, ρ0 the Equation (56)withLz =0is 2 inner turning point, and v0 is given by equation (43). A
T T graph of this function is shown in Figure 8 where it is 2 Δz = ρ˙ dt = ρdρ˙ (t). (67) compared to the helix result (64). 0 0 Changing the integration variable to ρ we thus find 4.3 Results for all solutions ρ1 ρ1 Δz =2 |ρ˙| dρ =2 ln ρ(ln ρ1 − ln ρ) dρ. (68) Numerical results for arbitrary solutions show that the 1 1 ratio of drift velocity along the current to speed, vz/v, For the period we have from equation (52) is maximal for the helical trajectories and minimal for the plane ones. This means that all solutions will lie be- ρ1 dρ ρ1 dρ tween the two curves in Figure 8. We illustrate this in T . =2 =2 (69) Figure 9 where results for solutions between the two ex- 1 |ρ˙| 1 ln ρ(ln ρ1 − ln ρ) tremes are shown. On the horizontal axis is the (constant) 2 2 2 We now change variables in the integrals and put speed v = vρ + vϕ + vz and on the vertical axis the θ ln ρ ky (pitch) angle of equation (60) at the outer turning point ρ1 k, y ,dρ ke dy, ρ ln = = k = (70) 1. This angle is zero for the plane trajectories. It obeys v θ/ 2 θ =cos sin for the helical ones according to (62). so that ln ρ(ln ρ1 − ln ρ)=k y(1 − y) while ρ =1cor- This corresponds to the dashed curve at the top of the θ responds to y =0,andρ = ρ1 to y = 1. The integrals diagram. For the other trajectories the -value represents then become an initial condition. The numbers on the contours are the values of vz/v. The conclusion is that for any solution 1 v /v → v →∞ Δz k2 y − y eky dy of the equations of motion z 1for ,for =2 (1 ) (71) − 0 positive charge, otherwise to 1. and 1 eky dy T =2 . (72) 5 Charged particle in magnetic dipole field 0 y(1 − y) Here we study the non-relativistic motion of trapped par- These integrals can be expressed in terms of modified I z ticles in the equatorial plane of a magnetic dipole. We Bessel functions ν ( ), see Olver et al. [23], as follows calculate the angular drift velocity exactly and reach the Δz πkek/2I k/ , same conclusion as above: the drift of the trapped charged = 0( 2) (73) particles is in the same direction as the current producing k/2 T =2πe I1(k/2). (74) the field. Page 8 of 10 Eur. Phys. J. D (2016) 70: 198
π/2
0.05
0.1
π/3
0.15
0.25 0.2
0.3 0.35
0.05 θ
0.4 0.1 0.45
0.5 0.55
0.6 0.65 π/6 0.7
0.75
0.15 0.25 0.2 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0 0 0.5 1 1.5 2 f ρ v Fig. 10. Graphs illustrating the function ( )ofequation(81) for = 1 and the straight line 1/λ2 for λ =5.Theintersec- Fig. 9. The (constant) speed v of the particle is on the hori- tions of the line and the curve correspond to the three roots of zontal axis in units of v0. The pitch angle θ of the trajectory equation (84). at the outer turning point ρ1 is on the vertical axis. The con- tours represent constant values of vz/v and the value for each contour is given. The bottom line of the figure represents the the turning points in the ρ-motion are the solutions of plane solutions and the dashed upper end of the contours the 2 helical solutions. 1 − 1 − . λ2 ρ ρ2 =0 (83) λ> The vector potential of a dipole m = mzeˆz in the For 4 there are three positive roots of this equation z-direction is, given by: mz A(ρ, ϕ)= eˆϕ, (76) 1 2 ρ2 ρ1 = −λ + λ +4lλ , 2 when restricted to the equatorial plane, using cylindrical 1 2 B − m /ρ3 eˆ ρ2 = λ − λ − 4 lλ , (84) coordinates. The magnetic field is = ( z ) z.The 2 Lagrangian (2)is 1 2 ρ3 = λ + λ − 4 lλ , m q mz 2 L(ρ, ρ,˙ ϕ˙)= ρ˙2 + ρ2ϕ˙ 2 + ϕ.˙ (77) 2 c ρ and one negative (un-physical) root. From these one can show that, The angular momentum, ρ ρ ρ ρ ρ2 ρ2 q m 1 2( 1 + 2), λ 1 + 2 . L mρ2ϕ z = ρ2 ρ2 and = ρ − ρ (85) z = ˙ + c ρ (78) 1 + 2 2 1 Since on a circle with ρ = constant, the magnetic field is is conserved. The energy is, as always in magnetostatic constant, there must be circular solutions to the equations problems, just the kinetic energy and can be written of motion, as long as the speed is adapted the strength of 2 2 the field. These are easily found and turn out to have m 2 Lz 1 ρ λ E = ρ˙ + − , (79) radius c =2 and c =4. This means that they are 2 2m ρ ρ2 located at the maximum of the curve f(ρ)inFigure10. They are thus highly unstable. where qmz Orbits with λ>4 and with ρ-values between the two ≡ . (80) cLz smaller roots will remain there and will have an angular > drift velocity around the dipole, which we now calculate. For 0 the function Using (79)anddt = dρ/ρ˙, one finds that, 2 ρ2 2 f ρ 1 − |Lz| 1 ρ dρ ( )= 2 (81) T = , (86) ρ ρ E λ 2 2 ρ1 (ρ − λρ + λ )(ρ + λρ − λ ) has a minimum at ρ = where f =0andamaximumat ρ E m ρ2 ρ2ϕ2 2 is the period of the -motion. Using = 2 (˙ + ˙ )we ρ =2 where f =1/(16 )(seeFig.10). If we put also find
E m ρ2 (1 − /ρ)dρ 1 ≡ 2 , Δϕ =2λ . (87) 2 2 (82) 2 2 λ Lz ρ1 (ρ − λρ + λ )(ρ + λρ − λ ) Eur. Phys. J. D (2016) 70: 198 Page 9 of 10
6 Conclusions
We have calculated the drift velocity of charged parti- cles in the magnetic field of three different current distri- butions exactly: current located near a two-dimensional plane, current located near a one-dimensional line, and current in the neighborhood of a (zero-dimensional) point. In the two latter cases we find that charges will drift in a direction that enhances the current producing the field. In the dipole case this is valid for particles that remain near the dipole. In the case of current in the neighborhood of a plane the situation is complicated by the fact that the field increases away from the current but also in this case faster particles will drift to enhance the source current. In the Darwin statistical mechanical approach [4]cur- rents are predicted to be correlated in thermal equilib- rium. This however does not explain this behavior within the present model: charged test particles moving in time- independent external magnetic fields. The fact that there Fig. 11. The angular drift velocity (88) for particles trapped are exceptions to the rule for slow particles in the Harris by a dipole field in its equatorial plane. In the units used the sheath shows that a simple general explanation probably angular velocity increases with k where k = 0 corresponds to does not exist. For slowly varying fields one can find an zero velocity and k = 1 corresponds to the maximum trapped explanation in the derivation of the the Alfv´en formula (1) velocity. as given by e.g. Jackson [24]bynotinghowthefieldarises from the current distribution. For the straight wire case one has that the current is parallel to the z-axis so j zˆ. By doing these integrals one can calculate, The magnetic field is then B zˆ × ρˆ. Finally the gra- dient of |B| is ∇|B|−ρˆ. Putting this into (1)gives Δϕ ϕ˙ = , (88) vD (zˆ × ρˆ) × (−ρˆ)=ϕˆ × (−ρˆ)=zˆ j. For the dipole T equatorial plane case we similarly have j ϕˆ, B −zˆ, ∇|B|−ρˆ v −zˆ × −ρˆ ϕˆ j the angular drift velocity. After some hard work the inte- and ,so D ( ) ( )= .For grals turn out to give a current sheet on the other hand the magnetic field is not slowly varying. The explanation for this case and for |L | ρ ρ fast particles in a Harris sheath is instead found in Fig- T z 1 + 2 K k − E k , = E λk [ ( ) ( )] (89) ure 1. For slow particles outside the Harris sheath we get for the two regions outside the sheath, j yˆ, B ∓zˆ, and but now ∇|B|±xˆ,sotheAlfv´en drift formula gives ρ ρ vD (∓zˆ) × (±xˆ)=−yˆ −j. This explains the negative Δϕ 2( 1 + 2)K k − k Π −k, k = ρ ( ) 4( +1) ( ) (90) drift for slow particles outside the sheath. 1 The study was motivated by the much more difficult where K(k),E(k)andΠ(−k, k) denote complete elliptic problem of finding the behavior of systems where charged integrals, see Olver et al. [23], and where particles and fields move according to the coupled equa- tions of Lorentz and Maxwell. This difficult problem is at 2 2 ρ2 − ρ1 the heart of plasma physics. It seems that understanding k = . (91) how current is induced in a plasma in the neighborhood 2ρ1ρ2 of a given current distribution, due to the inhomogeneous Algebra shows that λ =2 (1/k + k). Also ρ1 and ρ2 can magnetic field from the source, should improve our quali- then be expressed as functions of k.Using|Lz|/E as unit tative understanding of this problem. of time we can now plot the angular drift velocity Δϕ/T as a function of 0
3. L.A. Artsimovich, S. Yu. Lukyanov, Motion of Charged 18. C. St¨ormer, Arch. Sci. Physic. Nat. Gen`eve 24, 350 (1907) Particles in Electric and Magnetic Fields (Mir, Moscow, 19. C. St¨ormer, The Polar Aurora (Oxford University Press, 1980) Oxford, 1955) 4. H. Ess´en, Phys. Rev. E 53, 5228 (1996) 20. A.J. Dragt, Rev. Geophys. 3, 255 (1965) 5. H. Ess´en, Phys. Rev. E 56, 5858 (1997) 21. C. Graef, S. Kusaka, Journal of Mathematics and Physics 6. H. Ess´en, J. Phys. A 32, 2297 (1999) 17, 43 (1938) 7. H. Ess´en, A.B. Nordmark, Phys. Rev. E 69, 036404–1 22. R. de Vogelaere, Canad. J. Math. 2, 440 (1950) (2004) 23. F.W.J. Olver, D.W. Lozier, R.F. Boisvert, Ch.W. Clark. 8. H. Ess´en, Eur. Phys. Lett. 79, 6002–1 (2007) NIST Handbook of Mathematical Functions (National 9. H. Ess´en, Eur. Phys. Lett. 84, 20011–1 (2008) Institute of Standards and Technology and Cambridge 10. H. Alfv´en, C.-G. F¨althammar, Cosmical Electrodynamics, University Press, Cambridge, 2010) 2nd edn. (Oxford University Press, Oxford, 1963) 24. J.D. Jackson, Classical Electrodynamics, 3rd edn. (John 11. M. Headland, P.W. Seymour, Aust. J. Phys. 28, 289 (1975) Wiley & Sons, New York, 1999) 12. E.G. Harris, Il Nuovo Cimento 23, 115 (1962) 13. F. Hertweck, Z. Naturforsch. 14, 47 (1959) 14. M. M¨uller, K. Dietrich, Z. Phys. D 33, 101 (1995) Open Access This is an open access article distributed 15. D. Yafaev, Math. Phys. Anal. Geometry 6, 219 (2003) under the terms of the Creative Commons Attribution 16. H. Ess´en, J. Phys. A 37, 9831 (2004) License (http://creativecommons.org/licenses/by/4.0), which 17. J. Aguirre, A. Luqueb, Da. Peralta-Salas, Physica D 239, permits unrestricted use, distribution, and reproduction in any 654 (2010) medium, provided the original work is properly cited.