Effective potential energy in St\ormer's problem for an inclined rotating magnetic dipole
EEffective potential energy in Størmer’s problem for aninclined rotating magnetic dipole
V. Epp • M. A. MasterovaAbstract
We discuss the dynamics of a charged non-relativistic particle in electromagnetic field of a rotatingmagnetized celestial body. The equations of motion ofthe particle are obtained and some particular solutionsare found. Effective potential energy is defined on thebase of the first constant of motion. Regions accessi-ble and inaccessible for a charged particle motion arestudied and depicted for different values of a constantof motion.
Keywords
Størmer’s problem: magnetic dipole: elec-tromagnetic field: inclined rotator: charge: equation ofmotion
The field of magnetic dipole and motion of chargedparticles in this field have large practical significancefor astrophysics. Magnetic field of many planets andstars can be thought of as a dipole field in good ap-proximation. Stationary case, when the magnetic mo-ment of a celestial body coincides with the rotationaxis, is well studied. Particularly, motion of a chargedparticle in the Earth magnetic field is studied in de-tail. Solution of equation of motion for charged particlein a dipolar magnetic field gives trapping regions forparticles, which have energy of limited range. Theseregions of planets are named radiation belts (Alfven1950; Holmes-Siedle & Adams 2002). The first theo-retical studies on the properties of the trajectories ofa charged particle in the dipolar field where done byStørmer (1907), DeVogelaere (1958) and Dragt (1965).
V. EppM. A. MasterovaTomsk State Pedagogical University, ul. Kievskaya 60, 634061Tomsk, Russia
Rather complicated mathematical methods were devel-oped in order to calculate the trajectory of charged par-ticle in the dipole magnetic field (see for example (Lan-zano 1968)). Solution of the Størmer’s problem was ex-tended to magnetic field that is a superposition of thefield of a dipole and a uniform magnetic field. Allowedand forbidden regions of the motion of charged parti-cles in such field was studied by Katsiaris & Psillakis(1986).There are also well known celestial bodies, which di-rection of magnetic moment differs from the direction ofaxis of rotation. In this case electric field is induced inthe neighbourhood of the body by magnetic field. Theneutron stars and pulsars are examples of such bod-ies. The first model of electric field which is generatedin the neighbourhood of a neutron star was developedby Deutsch (1955). Some other models were suggestedand studied by several authors (Michel 1991). All thesemodels are based on suggestion that the neutron staris a conducting body.Conducting body rotating in its own magnetic field,generates a potential difference between the poles andequator. This potential difference leads to developmentof a co-rotating magnetosphere. By assuming that themagnetic field is dipolar, and unaffected by the trappedparticles in the magnetosphere, and that the field dipoleaxis is parallel to the rotation axis, Goldreich, P. & Ju-lian (1969) determined many of the properties of themagnetosphere. Electromagnetic field and magneto-sphere of the oblique rotator and conducting body havebeen studied by many authors. See for example Cohen& Rosenblum (1972); Kaburaki (1981); Cohen & Kear-ney (1980).Electromagnetic field of conducting body differs es-sentially from the pure dipole field. But there are alsocelestial bodies, which consist of non-conducting mat-ter and their axis of rotation is inclined with respect tothe magnetic field axis. Magnetic field of such objects a r X i v : . [ a s t r o - ph . H E ] M a r in good approximation can be described as the field ofan inclined rotating magnetic moment or ”oblique ro-tator” (Babcock & Cowling 1953). Eelectromagneticfield in the near and intermediate zones of a magneticdipole rotating in free space has been studied recentlyby Sarychev (2009). He has also estimated the energy,which can be acquired by the particles during acceler-ation.In our previous paper (Epp & Masterova 2012) wehave calculated the field of rotating non-conductingbody with the dipole magnetic field, and the dipole axisinclined to the rotation axis. An integral of motion wasfound and appropriate effective potential energy wasstudied. It was shown that each stationary point of theeffective potential energy corresponds to particular so-lution of equations of motion for a charged particle. Inthis paper we develop investigation of the effective po-tential energy. We study here stability of the particlemotion in the vicinity of the stationary points. Specialattention is paid to investigation of regions accessibleand inaccessible for a charged particle motion. Theseregions are described and depicted for particles withdifferent values of the constant of motion. It is shownthat there are closed allowed regions at certain values ofthe constant of motion. These regions are co-rotatingwith the electromagnetic field of the magnetized body. Let us consider the field which is produced by precess-ing magnetic dipole. We define the law of motion of adipole moment µ in the Cartesian coordinate system asfollows: µ = µ (sin α cos ωt, sin α sin ωt, cos α ) , (1)where µ > ω > ≤ α ≤ π is theangle between the vector µ and rotational axis. Con-sider the general formulas for the field of an arbitrarychanging dipole (Feynman 1964): E = ( n × ˙ µ ) r c + ( n × ¨ µ ) rc , (2) H = ( n × ( n × ¨ µ )) rc + 3 n ( n ˙ µ ) − ˙ µ r c + 3 n ( nµ ) − µ r , (3)where c is the speed of light, n = r /r is the unit vector, r is the radius-vector. Field is calculated at time t , andall the quantities on the right side of these equationsare calculated at the retarded time t (cid:48) = t − rc . We use a spherical coordinate system ( r, θ, ϕ ), inwhich the components of electric field vector have theform: r E r = 0 , (4) r E θ = µρ sin α ( ρ sin τ − cos τ ) , (5) r E ϕ = − µρ cos θ sin α ( ρ cos τ + sin τ ) , (6)where τ = ωt (cid:48) − ϕ, ρ = rωc . (7)The magnetic field vector has components: r H r = 2 µ [sin α sin θ (cos τ − ρ sin τ )+ cos θ cos α ] , (8) r H θ = − µ (cid:2) cos θ sin α (cos τ − ρ sin τ − ρ cos τ ) − sin θ cos α ] , (9) r H φ = − µ sin α (sin τ + ρ cos τ − ρ sin τ ) . (10)The same formulas for α = π/ (cid:18) E θ E θ (cid:19) + (cid:18) E ϕ E φ (cid:19) = 1 , (11)where E θ = ρµ sin α (cid:112) ρ r , E ϕ = E θ cos θ . Hence,the vector E circumscribes an ellipse with semiaxises E θ and E ϕ in the plane orthogonal to the radius vec-tor. This ellipse degenerates to a circle in the directionof axis of rotation ( θ = 0 , π ), but in equatorial plane θ = π E oscillates in meridian plane (parallel to z -axis).It follows from Eqs (8) and (9) that the magneticfield has constant components (in contrast to the elec-tric field): H rc = 2 µr cos θ cos α, H θc = µr sin θ cos α. These components relate to the field of a static mag-netic dipole if the inclination angle α is equal to zero.The electric field tends to zero in this case.One can see that time dependence appears in theformulas for electric and magnetic fields as composi-tion ωt (cid:48) − ϕ . It means that change of ωt (cid:48) is equivalentto change of ϕ . In other words, geometry of the electricand magnetic field looks like as the field rotates with an-gular velocity ω around z -axis. This conclusion relatesalso to geometry of the field lines of electric and mag-netic fields. At first glance it would seem that we havea paradox: the linear velocity of motion is getting more then speed of light well away from z -axis. But motionof the field lines does not relate to transfer of matter orfield energy. The above equations state only that theelectromagnetic field at any point ( r, θ, ϕ ) of space isequal to value of the field at the point ( r, θ, ϕ − δϕ ) atthe moment t − δϕ/ω . Evidently, in the far field zoneonly radiation field remains, which moves radially withthe speed of light. L = − mc (cid:114) − v c + ec ( Av ) , (12)where v is the particle velocity vector, A is the vectorpotential. It is easy to prove that the vector potential A = − ( n × µ ) r − ( n × ˙ µ ) rc (13)gives the field (2) and (3). Substituting spherical com-ponents of vectors A and v A r = 0 , (14) A θ = µ sin αr (sin τ + ρ cos τ ) , (15) A ϕ = µr [cos α sin θ − sin α cos θ (cos τ − ρ sin τ )] , (16) v = ( ˙ r, ˙ θr, ˙ ϕr sin θ ) (17)into Lagrangian (12), we obtain L = − mc (cid:115) − ˙ r + ˙ θr + ˙ ϕ r sin θ c + eµcr (cid:110) ˙ θ sin α (sin τ + ρ cos τ ) + ˙ ϕ sin θ cos α − ˙ ϕ sin θ cos θ sin α (cos τ − ρ sin τ ) } . (18)This Lagrangian is a function of all spherical coordi-nates and time. The time dependence appears in thisformula as combination ωt (cid:48) − ϕ .Substituting this function into Lagrange’s equation ddt ∂L∂ ˙ q − ∂L∂q = 0 , (19)we obtain the relativistic equations of motion of acharged particle in the field of precessing magneticdipole (see Appendix A). Further we consider the charged particle to be a non-relativistic one. Let us introduce a new set of general-ized coordinates ρ, θ, ψ with ρ = rωc , ψ = ϕ − ωt. (20)Actually, this means that we use a co-rotating referencesystem. It is known that the rotating reference systemis applicable only inside the light cylinder, i.e. rω < c or ρ <
1. Then the nonrelativistic Lagrangian functiontakes the form L = mc ω (cid:104) ˙ ρ + ρ ˙ θ + ρ ( ˙ ψ + ω ) sin θ (cid:105) (21) − eµωc ρ sin α (cid:104) ( ˙ ψ + ω ) sin θ cos θ (cos ξ + ρ sin ξ )+ ˙ θ (sin ξ − ρ cos ξ ) (cid:105) + eµωc ρ cos α ( ˙ ψ + ω ) sin θ, where ξ = ψ + ρ . Now, the Lagrangian and Hamiltonianfunctions do not depend on time explicitly. In this casethe value of the Hamiltonian is an integral of motion H = q i ∂L∂ ˙ q i − L = mc ω ( ˙ ρ + ρ ˙ θ + ρ sin θ ˙ ψ ) − mc ρ sin θ − eµω c ρ cos α sin θ + eµω sin 2 θ c ρ sin α (cos ξ + ρ sin ξ ) . (22)The first term in this expression K = mc ω ( ˙ ρ + ρ ˙ θ + ρ sin θ ˙ ψ ) is always positive and it can be consideredas the kinetic energy. The rest terms are usually re-ferred to as effective potential energy. It can be writtenas: V ef = mc (cid:26) − ρ sin θ + N ⊥ sin 2 θρ (cos ξ + ρ sin ξ ) − N (cid:107) sin θρ (cid:41) , (23) N ⊥ = eµω sin αmc , N (cid:107) = eµω cos αmc . In this notation formula (22) can be represented as: K = H − V ef . (24)The inequality H − V ef ≥ particle can be in a stable, unstable or indifferent equi-librium. The aim of this investigation is the stabilityof a charged particle at the stationary points. As theLagrangian is written for a non-relativistic particle, weconsider that ωr (cid:28) c or ρ (cid:28)
1. This means that theparticles moving around the axis of precession with anangular velocity of about ω are non-relativistic. In thisapproximation ξ ≈ ψ and the effective potential energytakes the form V ef = mc (cid:26) − ρ sin θ + N ⊥ sin 2 θρ cos ψ − N (cid:107) sin θρ (cid:41) . (25)In order to obtain the stationary points of the function V ef we find solutions of the set of equations: ∂V ef ∂q i = 0 , (26)where q i = ρ, θ, ψ . Hence, we have a system of threeequations − ρ sin θ + 2 N (cid:107) sin θ − N ⊥ sin 2 θ cos ψ = 0 , (27) − ρ sin 2 θ − N (cid:107) sin 2 θ + 2 N ⊥ cos 2 θ cos ψ = 0 , (28) N ⊥ sin 2 θ sin ψ = 0 . (29)Equation (29) has next solutions: ψ = 0 , π, (30) θ = πn , ( n ∈ Z ) . (31) ψ = 0 , π Using Eq. (30) we can eliminate the variable ψ fromthe equations (27) and (28) by substitution cos ψ = ε , where ε = +1 corresponds to ψ = 0 and ε = − ψ = π . This results in a system of twoequations with two unknowns: − ρ + N (cid:107) − εN ⊥ cot θ = 0 (32) − ρ cot θ − N (cid:107) cot θ + εN ⊥ (cot θ −
1) = 0 (33)Expressing cot θ from equation (32) and substituting itinto equation (33), we obtain:2 ρ − N (cid:107) ρ − N = 0 , (34)where N = eµω mc . This equation has a solution: ρ = N (cid:104) cos α ± (cid:112) − sin α (cid:105) . (35) The sign before the square root is defined by the signof the charge which is hidden in N , and by condition ρ >
0. Therefore, Eq. (35) takes the form ρ = N (cid:104) cos α + q (cid:112) − sin α (cid:105) , (36)where q = e/ | e | = ± ρ into Eq. (32), we find:tan θ = − ε α (cid:104) α + q (cid:112) − sin α (cid:105) . (37)Thus, considering (30), (35), (37), for the positiveparticles in the case when sin 2 θ (cid:54) = 0 we get the follow-ing series of coordinates of stationary points for V ef : ρ = N (cid:104) cos α + (cid:112) − sin α (cid:105) , ψ = 0 , tan θ = −
12 sin α (cid:104) α + (cid:112) − sin α (cid:105) , (38) ρ = N (cid:104) cos α + (cid:112) − sin α (cid:105) , ψ = π, tan θ = 12 sin α (cid:104) α + (cid:112) − sin α (cid:105) . (39)For the negatively charged particles, we have next twostationary points: ρ = N (cid:104) cos α − (cid:112) − sin α (cid:105) , ψ = 0 , tan θ = 12 sin α (cid:104) − α + (cid:112) − sin α (cid:105) , (40) ρ = N (cid:104) cos α − (cid:112) − sin α (cid:105) , ψ = π, tan θ = 12 sin α (cid:104) α − (cid:112) − sin α (cid:105) . (41)Hence, the solution of equations (27) – (29) for thecase sin 2 θ (cid:54) = 0 gives two stationary points for a positivecharge and two points for a negative charge. θ = πn θ = 0 , π ) all the first derivatives from effectivepotential energy are equal to zero only in the planecos ψ = 0 and for any ρ . Which means that on the axis θ = 0 , π there are not stationary points.As to the equatorial plane θ = π ρ = N (cid:107) , cos ψ = 0 , N (cid:107) > . (42) This gives two solutions for ψ , because ψ ∈ (0 , π ).Thus, in case θ = π ρ = N (cid:107) , θ = π , ψ = π , (43) ρ = N (cid:107) , θ = π , ψ = 3 π . (44)It follows from (42) that e cos α >
0, which means thatthe two above mentioned stationary points correspondto a positive charge if α < π/ α > π/ ρ i , θ i , ψ i ) defined by Eqs (38)– (44) and initial zero velocity with respect to the ro-tating reference frame will be in equilibrium position.Substituting these coordinates and ˙ ρ = ˙ θ = ˙ ψ = 0 inequations of motion (A1) – (A3) and taking into ac-count that ϕ = ωt + ψ we obtain identical equalities.Hence, a particle being at rest at one of the stationarypoints in the co-rotating coordinate system will keepthis position. This means that in laboratory referenceframe the particle is moving along a circle with con-stant velocity v i = cρ i sin θ i . Thus, there are at leastsix particular solutions which describe circular motionof the particles in the field of inclined rotating dipole.Positions of the orbits defined by angle θ and their ra-dius are different for the positive and negative parti-cle. Two of the trajectories are laying in the equatorialplane z = 0.3.3 Labelling the stationary pointsIn order to define whether the stationary points (38)– (44) represent maximum or minimum of potentialenergy we apply Sylvester’s criterion: suppose that insome neighbourhood of a stationary point M i ( ρ i , θ i , ψ i )the second derivatives of the function V ef ( ρ, θ, ψ ) aredefined. The function has a local maximum if thequadratic form d V ef ( M ) is negative definite and alocal minimum if d V ef ( M i ) is positive definite. If theform d f ( M i ) is alternating then there is no extremumat the point M i (Merkin 1996). Let us find all sec-ond partial derivatives and define their values at the stationary points (38) – (41). a = 2 mc ∂ V∂ρ = cos α − − q cos α (cid:112) − sin α, (45) a = 2 mc ∂ V∂θ = 6 ρ , (46) a = 2 mc ∂ V∂ψ = − N sin α ρ (cos α − q (cid:112) − sin α ) , (47) a = 2 mc ∂ V∂ρ∂θ = 2 N cos ψ sin αρ , (48) a = 2 mc ∂ V∂θ∂ψ = 0 , (49) a = 2 mc ∂ V∂ρ∂ψ = 0 . (50)Using Eq. (37) we transform Eq. (45) into a = − θ and hence, it takes only negative values.For the first fundamental form, we have: D = (cid:12)(cid:12)(cid:12)(cid:12) a a a a (cid:12)(cid:12)(cid:12)(cid:12) = 2 Nρ (cid:2) − α − cos α + (cos α − q (cid:112) − sin α (cid:105) . (51)The last equation can be transformed to D = − N qρ sin θ (cid:112) − sin α, which makes obvious that D < (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a a a a a a a a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = D a (52)is definitely negative because a >
0. Thus, the form d f ( M i ) is alternating and there are no extremums ofthe function V ef at the four points with coordinates(38) – (41) .Calculation of the second derivatives in case of θ = π a = − , a = 6 N (cid:107) , a = 0 ,a = 0 , a = 0 , a = 2 N ⊥ N − (cid:107) sin ψ. The quadratic form combined of these derivatives is in-definite in its sign.Summing up, we state that the six stationary points(38) – (41), (43) and (44) are points of equilibrium for aparticle at rest, but the equilibrium is not stable. ”Par-ticle at rest” means that coordinates ρ, θ, ψ are fixed.
In the inertial reference frame the particle is moving ina circle with constant velocity.Stationary points that we considered above do notrepresent all solutions of the equations of motion forcircular orbits. If we substitute into equations of motion(A1) – (A3) ρ = const we find one more solution for e cos α > θ = π , ρ = 2 N, ϕ = Ω t + ϕ , (53)where Ω = (1 / ω , and ϕ is an arbitrary constant.This solution is valid for a positive charge if α < π/ α > π/ Let us return to the expression (24): K = H− V ef . Theintegral of motion H is defined by the initial coordinatesand velocity of the particle. If these are specified, theparticle can move only in space area where V ef < H .In this section we present plots of surfaces V ef = const.These surfaces restrict the permissible area for particlemotion.As we found in the previous section, the stationarypoints are at distance ρ of order | N | / from the co-ordinate origin. Hence, we introduce a variable ˜ ρ = ρ | N | − / in order to make ˜ ρ ∼
1, and a reduced poten-tial energy ˜ V = 2 V ef / ( mc N / )˜ V = − ˜ ρ sin θ + q sin α sin 2 θ ˜ ρ cos ψ − q cos α sin θ ˜ ρ , (54)where q = e/ | e | . The regions accessible to the particlemotion are defined by inequality˜ V ≤ C. (55)The constant C is related to the integral of motion H as C = 2 H / ( mc N / ).4.1 Aligned rotator, α = 0It is useful to start drawing plots from the simple caseof a stationary dipole α = 0 in order to show how theaxial symmetric equipotential surfaces transform intoantisymmetric ones. If α = 0, the equipotential surfacefor a given constant C is defined by equation ( θ (cid:54) = 0 , π ):˜ ρ + η ˜ ρ + 2 q = 0 , (56)where η = C/ sin θ . Let us consider a case of positively charged particle.Eq. (56) has not positive roots for ρ if C ≥ C ∈ ( −∞ ; −
3) Eq. (56) has two roots at any θ . The typical form of the equipotential surface in thiscase is given in Fig. 1. The inner surface has form of atorus. Inequality (55) is satisfied inside the torus – thisis a region accessible for motion. The region betweenthe torus and outer surface is inaccessible. Outside theexternal surface we have again accessible region. As theconstant C increases from −∞ to −
3, the inner surfaceof the torus is increasing, while the radius of outer sur-face is decreasing. At C = − ρ = 1. This is shownin Fig. 2. Fig. 1
Equipotential surfaces for α = 0, C = − . q = 1.C and A are accessible regions, B – inaccessible region Fig. 2
Equipotential surface for α = 0, C = − q = 1. Aand C are accessible regions, B – inaccessible region. In interval C ∈ ( −
3; 0) Eq. (56) has two positiveroots if sin θ < − C/
3, one root if sin θ = − C/ θ > − C/ C ∈ ( −
3; 0) is given inFig. 3. The space inside the surface is inaccessible formotion, while the outside space is accessible. The upperand lower parts of the surface in Fig. 3 have point-likeconnection at the coordinate origin ˜ ρ = 0. In the limit C → − Fig. 3
Equipotential surface for α = 0, C = − q = 1. Bis inaccessible region, A – accessible region For a negatively charged particle equation (56) hasone root for each C . In the case C ∈ (0 , + ∞ ) we havea surface in the form of a torus (Fig. 4). In other cases( C ∈ (0 , −∞ )), the equipotential surface has a formshown in Fig. 5. The radius of the surface increaseswith decreasing of C . Fig. 4
Equipotential surfaces at C = 3 , q = −
1, B - inac-cessible region, A - accessible region.
Fig. 5
Equipotential surface at C = − , q = −
1. A isaccessible region, B - inaccessible region. α (cid:54) = 0Let us consider the equipotential surfaces in the generalcase α (cid:54) = 0. One can see from equation (54) that the equipotential surfaces are symmetric with respect tosubstitution θ → π − θ , ψ → ψ + π , i.e. they aresymmetric with respect to reflection in the plane z = 0and rotation around axis OZ through π .Instead of (56) we have next equation˜ ρ + η ˜ ρ + 2 χ = 0 , χ = q (cos α − sin α cot θ cos ψ ) . (57)In case χ < ρ , hence,there is one equipotential surface which divides all spaceinto accessible and inaccessible regions. If χ > C > χ >
C <
0, thenumber of solutions depends on the sign of expression b = χ − ( − η/ / . Namely, equation (57) has no so-lutions for ˜ ρ if b >
0, one solution for b = 0 and twosolutions if b < b is the same as the sign of expression q sin θ (cos α sin θ − sin α cos θ cos ψ ) − (cid:18) − C (cid:19) − / . Evidently, the sum of first two terms lies in the range[-1, 1]. Hence, if
C < − b is definitely negative. Inother cases sign of b depends on θ and α . Fig. 6
Equipotential surfaces at C = − , q = 1. A and Care accessible regions, B - inaccessible region. Thus, if C ∈ ( −∞ ; − θ . The typical form of the equipotential surfacein this case is given in Fig. 6. But unlike the case α = 0(Fig. 1), there are three equipotential surfaces.Equipotential surfaces for other values of C and q are shown in Figs 7-9. All pictures 6-9 are plotted for α = π/ α → θ = π/ ρ = N / , ψ = 0 , π, π/ , π/
2. Aparticle being at rest at these points in the rotating ref-erence frame, is moving along a circle in the laboratory
Fig. 7
Equipotential surface at C = − , q = 1. A and Care accessible regions, B - inaccessible region. Fig. 8
Equipotential surface at C = 3 , q = −
1. A isaccessible region, B and C - inaccessible region.
Fig. 9
Equipotential surfaces at C = − , q = −
1. A andC are accessible regions, B - inaccessible region. reference frame. Substituting the values of coordinatesand velocity in Eq. (22) we find that C = −
3. Hence,the particle moves along the border between two regionsA and C of Fig.2. This orbit corresponds to unstablemotion.Let us find the constant C for stationary points incase α (cid:54) = 0. Substituting (36) and (37) in equation (54)we obtain: C = − (cid:16) α + q (cid:112) − sin α (cid:17) / (cid:16) cos α + q (cid:112) − sin α (cid:17) / . (58) Then, for the stationary points (38), (39) we have C ≈− . C is shown in Fig. 10. The stationary points are pointsof contiguity of regions A and C . Again, a particle atthese two positions is in unstable state of motion.For the solutions (40), (41) we have C ≈ − . Fig. 10
Equipotential surfaces at C = − . , q = 1 . Aand C are accessible regions, B - inaccessible region.
Fig. 11
Equipotential surfaces at C = − . , q = − . Aand C are accessible regions, B - inaccessible region.
We have calculated the components of the magneticand electric fields of a rotating non-conducting bodywhen the axis of the dipole magnetic field of the bodyis inclined to the rotation axis.In order to find whether the ”radiation belts” arepossible in such system, we have investigated the ef-fective potential energy in the field of inclined rotat-ing magnetic dipole. It is shown that there are closedequipotential surfaces which are co-rotating with thedipole field. The particles with definite initial energyare confined within such surfaces. The circular orbitsof charged particles correspond to the points at whichthe two allowed for motion regions are contiguous toeach other. It is shown that these orbits are unstable.
Equipotential surfaces are plotted for different val-ues of the constant of motion for positive and negativecharge of a particle. The results can be used for descrip-tion of the radiation belts around some specific celestialbodies possessing inclined dipolar magnetic field.This research has been supported by the grant forLRSS, project No 224.2012.2 A Relativistic equations of motion
Here we record relativistic equations of motion of a charged particle in the field of rotating magnetic dipole.Lagrangian (18) leads to the following equations of motion in a spherical coordinate system. m (1 − β ) / (cid:26) ¨ rc ( c − r ˙ θ − r ˙ ϕ sin θ ) + r ˙ rc ( ˙ r ˙ θ + r ˙ θ ¨ θ + ˙ ϕ sin θ ( ˙ r ˙ ϕ + r ¨ ϕ )+ 12 r ˙ ϕ ˙ θ sin 2 θ ) (cid:27) − mr ( ˙ θ + ˙ ϕ sin θ ) (cid:112) − β + eµ sin α cr (cid:110) θ (sin τ + ρ cos τ − ρ sin τ ) − ˙ ϕ sin 2 θ (cos τ − ρ sin τ − ρ cos τ ) (cid:9) + eµ cos αcr ˙ ϕ sin θ = 0 , (A1) mc (1 − β ) / (cid:110) r ¨ θ ( c − ˙ r − r ˙ ϕ sin θ ) + r ˙ r ˙ θ (2 c − r − r ˙ θ − r ˙ ϕ sin θ )+ r ˙ θ ( ˙ r ¨ r + r ˙ ϕ ¨ ϕ sin θ + 12 r ˙ ϕ ˙ θ sin 2 θ ) (cid:27) − m ˙ ϕ r cos θ sin θ (cid:112) − β − eµ sin αr ω (cid:8) ρ ( ω − ˙ ρ − ϕ sin θ )(cos τ − ρ sin τ ) − ˙ ρ sin τ (cid:9) − eµ ˙ ϕ cos α sin 2 θcr = 0 , (A2) m (1 − β ) / (cid:26) sin θ (2 r ˙ r ˙ ϕ + r ¨ ϕ ) + r ˙ ϕ ˙ θ sin 2 θ − r sin θc (cid:104) ˙ r ˙ ϕ (2 ˙ r + r ˙ θ + r ˙ ϕ sin θ )+ r ¨ ϕ ( ˙ r + r ˙ θ ) − r ˙ ϕ ( ˙ r ¨ r + r ˙ θ ¨ θ + 12 r ˙ ϕ ˙ θ sin 2 θ ) (cid:21) − r ˙ ϕ ˙ θc sin 2 θ ( ˙ r + r ˙ θ + r ˙ ϕ sin θ ) (cid:41) + µe sin αr (cid:40) r ˙ θ sin θc (cos τ − ρ sin τ ) + ˙ r sin 2 θ c (cos τ − ρ sin τ − ρ cos τ )+ ρ sin 2 θ τ + ρ cos τ ) (cid:27) + µe cos αcr ( ˙ θr sin 2 θ − ˙ r sin θ ) = 0 . (A3) B Stability of the orbit in the equatorial plane
It is easy to check that equations of motion (A1) – (A3) have partial solution r = R, θ = π , ϕ = ωt + ϕ (B1)with restiction on the initial phase ϕ :cos( ρ + ϕ ) + ρ sin( ρ + ϕ ) = 0 , ρ = Rω/c.
We assume now that the particle is a non relativistic one. Then ρ (cid:28) ϕ = 0 . This solution corresponds to motion of a particle being at rest at the stationary points (43) and (44). Let us verifystability of this solution. We develop equations of motion as series in increments of coordinates in a neighbourhoodof the solution. Let us introduce the increments of coordinates as follows θ = π δθ, ϕ = ωt + ϕ + δϕ, r = R + δr. (B2)We substitute these coordinates in equations (A1) – (A3) and expand the equations up to first order of coordinatesincrements. Taking into account that for a non reativistic particle ρ (cid:28) δ ¨ r − δrω − ωRδ ˙ ϕ + εωR tg αδ ˙ θ = 0 ,δ ¨ θ − ε δ ˙ rω tg αR + 3 ω δθ − εω tg αδϕ = 0 , (B3) δ ¨ ϕ + δ ˙ rωR − εω tg αδθ = 0 , where ε = sin ϕ = ±
1. We investigate the trivial solution of this system for stability. By Lyapunov’s theorem thetrivial solution is stable when all roots of characteristic first-approximation equation of the system have negativereal parts (Merkin 1996). We write Eq. (B3) in the form: x (cid:48)(cid:48) − x − x (cid:48) + T x (cid:48) = 0 ,x (cid:48)(cid:48) − T x (cid:48) + 3 x − T x = 0 , (B4) x (cid:48)(cid:48) + x (cid:48) − T x = 0 . where x = δrR , x = δθ, x = δϕ, T = (cid:15) tg α . Prime denotes ξ derivative , where ξ = ωt . Reduce the system (B4)to a normal form: x (cid:48) − d = 0 x (cid:48) − d = 0 x (cid:48) − d = 0 d (cid:48) − x − d + T d = 0 d (cid:48) − T d + 3 x − T x = 0 d (cid:48) + d − T x = 0 (B5)A set of solutions of the system has the form d = a e iνξ , d = a e iνξ ....x = a e iνξ . This gives the characteristicequation: k − k (1 + T ) − k (1 + T ) − T = 0 , (B6)where k = ν .Important criteria that give necessary and sufficient conditions for all the roots of the characteristic polynomial(with real coefficients) to lie in the left half of the complex plane are known as Routh-Hurwitz criteria (Gantmacher1959).Given a polynomial P ( λ ) = λ n + a λ n − + ... + a n − λ + a n , (B7)where the coefficients a i are real constants, i = 1 , , ..., n . Define the n Hurwitz matrices using the coefficients a i of the characteristic polynomial: H = ( a ) , H = (cid:18) a a a (cid:19) , H = a a a a a a a . . . (B8)All of the roots of the polynomial P ( λ ) are negative or have negative real part if the determinants of all Hurwitzmatrices are positive (Det H j ≥ j = 1 , , . . . , n ).For the polynomials of degree n=3, the Routh-Hurwitz criteria simplify to a > , a a − a > , a ( a a − a ) > . (B9)From (B6) we get: a = − (1 + T ) , a = − T ) , a = − T . It is easy to see that conditions (B9) ofRouth-Hurwitz criteria are not satisfied. So, the trivial solution of system (B3) is not stable. References
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