qb 3 B ( B) r 3 e r = 3 B r e r B = B/ r e r = 3 B ER 3 E r 4

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1 Magnetospheric Physics - Homework solution, /8/14 18 Gradient and curvature drift (a) A single proton has a parallel and perpendicular energy of 1 kev. Compute (B B) /B 3 and determine the instantaneous curvature and gradient drift velocity for the Earth s dipole field in the magnetic euator at a radial distance of 5 R E (b) Consider an isotropic Maxwell plasma distribution for protons with a temperature T = 1 8 K in the euatorial plane. Evaluate the instantaneous bulk (average) velocity of the distribution at a radial distance of 5 R E based on the gradient and curvature drift. Solution: (a) Gradient and curvature drifts in the absence of electric currents (dipole field): With the magnetic field (B E = T) v = mv (B B) B3 v C = mv B 3 B ( B) B r = 3 cos θ B θ = B E R 3 E r 3 sin θ r 3 B = B ER 3 E r 3 ( cos θ ) 1/ In the euatorial plane B has only a radial component: B = B/ r e r = 3 B ER 3 E r 4 e r = 3 B r e r B θ 1 B (B B) 3 θ=9 = 3 B r B (e 3 θ e r ) = 3 B E R 3 ( ) E r 3 e r r 3 3 φ = 3r e 3 φ Gradient drift of an individual particle with mv / = 1keV : v = mv B (B B) = 3 mv 3r e 3 φ = 14 e 3 5 e e 6 φ ms 1 = ms 1 = ms 1 = 3.77 kms 1

2 Curvature drift of an individual particle with mv / = 1keV : v C = mv B (B B) = 3 mv 3r e 3 φ = 7.54 kms 1 such that the combined gradient and curvature drift is v D,1keV = 11.3 kms 1. (b) Gradient and curvature drift for a Maxwell distribution with T = 1 8 K: Gradient drift (using z along the parallel direction): u = 1 n ( ) mv B B B fd 3 v = 1 B B ( m v 3 n B 3 x + vy) fd 3 v From homework problem 8 we have for a Maxwell distribution m v xfd 3 v = m v yfd 3 v = nk B T such that Curvature drift: u C = 1 n u = 1 n = k BT = k BT B B nk B 3 B T = k BT 3L B E R E B B B 3 ( mv ) B B B fd 3 v = 1 B B mv 3 n B zfd 3 v 3 B B B 3 = k BT 3L B E R E Such that the drifts are identical for the same parallel and perpendicular temperatures: u = u C = k BT 3L B E R E = L ms 1 = 13 L ms 1 = 3.5 kms 1 and for the combined drift u D = u + u C = 6.5 kms 1

3 19. Loss cone distribution Consider an initial Maxwell distribution function f(v) = n ( n and temperature T in the euatorial plane. m πk B T ) 3/ ( ) exp mv with density (a) For a given loss cone with angle α determine the fraction of particles lost from the isotropic distribution function. Determine the number density ñ for the new distribution function f (distribution without particles in the loss cone; hint: Represent the distribution in velocity space in spherical velocity coordinates.). (b) Compute the parallel and perpendicular energy density for the distribution function with the loss cone as a function of temperature T, density n, and angle α. (c) What is the angle α for the loss cone if the energy ratio is W /W = 1/4? Solution: (a) For a given loss cone with angle α determine the fraction of particles lost from the originally isotropic distribution function. Determine the number density ñ for the new distribution function f.with B v ll k B T x exp ( a x ) dx = π 4a 3 and using sperical coordinates for the velocity integration the number of particles in the loss cone is n L π α = dϕ sin ϑdϑ v f(v)dv v^1 ( ) m 3/ ( ) = 4π [ cos α] α n dvv exp mv πk B T k B T ( ) m 3/ ( ) 3/ π kb T = 4π (1 cos α) n πk B T 4 m where n is the original number of particles. Total Number of Particles in the loss cone: ϕ θ a v v^ n L = n (1 cos α) Particles remaining in the distribution: ñ = n n L = n cos α Fraction of particles lost: r = n L /n = 1 cos α The loss cone distribution function (= distribution function with the loss cone) does actually not change outside of the loss cone but should be re-written in the form f(ϑ, v) = ñ ( ) ) m 3/ exp ( mv S(ϑ, α) cos α πk B T k B T to account for the normalization to ñ. Here S(ϑ, α) is a step function with satisfies S(ϑ, α) = 1 for α ϑ π α.

4 (b) Parallel and perpendicular energy: With x 4 exp ( a x ) dx = 1 3x exp ( a x ) π dx = 3 a 8a 5 and v = v cos ϑ and v = v sin ϑ the energy densities are w = π = π m π α dϕ sin ϑdϑ m f(v)v α v dv ( ) ñ m 3/ π α ( sin ϑ cos ϑdϑ v 4 exp cos α πk B T α ) 3/ cos 3 ϑ π α 3 ( ) 5/ π kb T 3 8 m = πm ñ ( m cos α πk B T = m ñ cos α cos 3 α k B T m α = 1 ñk BT cos α mv k B T ) dv π π α dϕ sin ϑdϑ m w = f(v)dv α v = π m ( ) ñ m 3/ π α ( ) sin 3 ϑdϑ v 4 exp mv dv cos α πk B T α k B T = πm ñ ( ) [ ] m 3/ π α cos ϑ + cos3 ϑ 3 ( ) 5/ π kb T cos α πk B T 3 8 m α = m ñ [ ] cos α cos α cos3 α 3 k B T 3 8 m = 3 ( ñk BT 1 1 ) 3 cos α The average energies per particle (density for the losscone distribution is ñ) are W = 1 k BT cos α W = 3 ( k BT 1 1 ) 3 cos α Note anothe way to express this would be through parallel and perpendiculat effective temperature defined through p = ñk B T and p = ñk B T with p = w and p = w such that T = 3 T ( cos α ) and T = T cos α. (c) Angle α for the loss cone if the energy ratio is W /W = 1/4? such that cos α = 3/5 or α = 39.. W /W = cos α 3 cos α = 1 4

5 Particle drifts (a) What perpendicular particle energy is reuired to compensate the co-rotational drift in the Earth s magnetosphere through the gradient B drift at the magnetic euator? Sketch or plot the reuired perpendicular velocity as a function of r. What is this energy for particles at L = 4, 6, and 8? (b) Express the perpendicular energy through the magnetic moment. Assume that the particle is on field lines with L = 8 and is mirrored just above the ionosphere. What are the latitude of the mirror point and what is the parallel energy of the particle? What is wrong with this problem? Solution: (a) Gradient drift and angular velocity: v = mv B (B B) = 3 mv 3r ω = v r = mv Corotational angular velocity with T d = 1 day: such that B E R 3 E ω cor = π T d = mv 4π 3r 3 3L mv 4π 3L B E R E w = mv = π T d = π T d 3L π = e L = e π L = = 3.9keV L which corresponds to 7.7 kev, 5.1 kev, and 3.9 kev for L = 4, 6, and 8. (b) Magnetic moment in the euatorial plane and at the mirror point: µ = mv e B e = mv m B m such that mv m = mv e B m B e

6 with and B = B E R 3 E r 3 r = r e cos λ ( sin λ ) 1/ = BE R 3 E r 3 ( 4 3 cos λ ) 1/ such that the mirror latitude is λ m = arccos r m /r e = arccos 1/L = The magnetic field magnitude along the field line with the euatorial crossing point r e = LR E is R 3 ( E B = B E 4 3 r r 3 r e such that 1 B e = B E L 3 ( B m,1re = B E ) 1/ L ) 1/ and the perpendicular energy at the mirror point for L = 8 is mv m = mv e B m = mv e B e ( L ) 1/ L = 974 mv e = 3.797MeV Energy conservation implies that the total energy in the euatorial plane is the same as at the mirror point such that the parallel energy is w = mv m mv e = 973 mv e = 3.794MeV The problem assumed a particle for which the gradient drift in the euator compensates corotation. However, a particle that mirrors just above the ionosphere has a parallel energy that is almost a 1 times the perpendicular energy in the euatorial plane. For such a particle the curvature drift is about a 1 times faster than the gradient drift in the euator such that the particle actually does not compensate corotation.

Magnetospheric Physics - Homework, 02/21/ Local expansion of B: Consider the matrix B = (1)

Magnetospheric Physics - Homework, //4 4. Local expansion of B: Consider the matrix B α x γ δ x γ α y δ y β x β y α z () with constant parameters α i, β i, δ i, γ, and γ for the expansion of the magnetic