( 1 jj) = p + j B (1) j = en (u i u e ) (2) ρu = n (m i u i + m e u e ) (3)

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1 Magnetospheric Physics - Homework, 2/14/ MHD Equations: a Consider a two component electrons and ions charge neutral ρ c = 0 plasma where the total bulk velocity is defined by ρu = n m i u i + m e u e with ρ = n m i + m e. Compute u i and u e as a function of u and the current density j only. b Use the two fluid momentum equations for this plasma eq in the manuscript for isotropic pressure with p = p i + p e to derive the single fluid momentum equation. + ρuu + m em i 1 jj = p + j B 1 e 2 ρ c Demonstrate that this equation conserves momentum, i.e., that it can be brought into the form / = T. Hint: Use Ampere s law and appropriate vector identities to modify the j B term. Solution: a Using j = en u i u e 2 ρu = n m i u i + m e u e 3 we can multiply equation 2 with m e /e and take the sum to obtain jm e /e + ρu = m i + m e nu i or u i = u + m e enm j We can now either use this result in equation 2 or 3 to obtain b Isotropic two fluid equations u e = u m i enm j ρ e u e ρ i u i = ρ e u e u e p e ene + u e B = ρ i u i u i p i + ene + u i B Taking the sum of the two equations with ρu = ρ e u e + ρ i u i yields for the time derivative term with ρu = n m i u i + m e u e : ρ e u e + ρ iu i = First term on the rhs using the expressions for u i and u e : ρ e u e u e + ρ i u i u i = m e n u m i enm j u 1 m i enm j + m i n u + m e enm j u + m e enm j

2 = m e n +m i n uu uu + m i enm uj m i enm ju + mi 2 jj enm m e enm uj + m e enm ju + me enm 2 jj = Mnuu + m em i m e + m i jj = ρuu + m em i e 2 nm 2 e 2 ρ jj For the pressure terms we can simply use the definition p = p i + p e. Finally, for the last terms we obtain for the sum ene + u i B ene + u e B = j i B + j e B = j B such that collecting all terms we recover the equation 1. c With B = µ 0 j we obtain Using the vector identities we obtain = ρuu p + j B = ρuu p + 1 µ 0 B B A B = A B + B A + A B + B A AB = A B + A B B B = 1 B B B B 2 = 1 2 B2 BB + B B = 1 2 B2 BB because B = 0. Substitution in the momentum equation yields = ρuu p 1 B BB 2µ 0 µ 0 = ρuu p + 1 B BB 2µ 0 µ 0 or combining all terms [ = ρuu + p + 1 B ] BB 2µ 0 µ 0 which illustrates the concept of momentum conservation! 2

3 12. Dipole field in cylindrical coordinates: a Demonstrate that the transformation between spherical coordinate using latitude λ and cylindrical coordinates z, R, ϕ is given by e r = e z sin λ + e R cos λ e λ = e z cos λ e R sin λ b Using this transformation, compute the Earth s dipole field in cylindrical coordinates. Show that this field can also be derived from the φ component of the vectorpotential A ϕ z, R = κr/ z 2 + R 2 3/2 with κ = µ 0 M E / 4π. c Demonstrate that f z, R = RA ϕ z, R satisfies B f = 0, i.e., such that f = const represents magnetic field lines in cylindrical coordinates. Derive the field line equation and demonstrate that it is identical to the equation we have derived in class for spherical coordinates. Solution: a Cylindrical coordinates are determined by z = r sin λ R = r cos λ such that the inverse transformation is sin λ = z/r cos λ = R/r with r 2 = z 2 + R 2 The unit vector along the r direction has a z component of sin λ and an R component of cos λ such that it is given by e z sin λ + e R cos λ see Figure. The unit vector along the λ direction has to have a negative value along the R direction and must be perpendicular to unit vector along the r direction. Therfore, e r = e z sin λ + e R cos λ e λ = e z cos λ e R sin λ b In sperical coordinates the dipole magnetic field is given by B r = 2κ sin λ r 3 B λ = κ cos λ r 3 The magnetic field in cylindrical coordinates is then determined by B = B r e r + B λ e λ = 2κ sin λ e r 3 z sin λ + e R cos λ + κ cos λ e r 3 z cos λ e R sin λ = sin λ cos λ 3κ e r 3 R + κ 1 cos 2 λ 2 sin 2 λ e r 3 z = 3κ zr r e 5 R + κ 1 R 2 2z 2 e r 5 z 3

4 such that B z = κ R2 2z 2 zr B z 2 + R 2 5/2 R = 3κ z 2 + R 2 5/2 Derivation from the vectorpotential A ϕ z, R = κr/ z 2 + R 2 3/2 using B = A φ e φ B R = A φ z = 3κ zr z 2 + R 2 5/2 [ B z = 1 RA φ R R = κ R 2R z 2 + R 2 3R 2 R 3/2 z 2 + R 2 5/2 = κ 2 z2 + R 2 3R 2 = κ 2z2 R 2 z 2 + R 2 5/2 z 2 + R 2 5/2 which is identical to the magnetic field from the coordinate transformation. c The gradient operator in cylindrical coordinates is f = f R e R + 1 f R φ e φ + f z e z ] such that 1 R RA φ z B f = A φ z implies f R = 1 R f R + 1 R RA φ f R z RA φ f R z = 0 This is solved by the function f = RA φ or any power of RA φ such that magnetic field lines are given by κr 2 RA φ = = const 4 z 2 + R 2 3/2 Using a constant κc 0 with κc 0 > 0 the field line equations are R 2 = c 0 z 2 + R 2 3/2 or z 2 = c 2/3 0 R 4/3 R 2 Note that that this equation has a maximum value of R which is determined by R = c 1 0. We can substitute R = r cos λ into 4 and the field line equations become cos2 λ r = const which is the same as in spherical coordinates! 4

5 13. Dipole magnetic field: a Assume the magnetic field of the Earth to be dipolar. Two magnetic field lines are radially separated by 1000 km in the magnetic equator at a distance of 5 Earth radii 5 R E. What is the separation at the Earth s surface? b Consider the superposition of a constant IMF B IMF = e z to the dipole field of the Earth. Compute the radial distance of the X-line that separates open and closed field for = T and = T. Determine the latitude on the Earth s surface of the boundary between closed and open magnetic field polar cap for the two IMF values. Assume as illustrated in class that the IMF and the Earth s dipole field can be linearly superimposed. Hint: Use the field line equation to map the magnetic field lines connected to the X-line to the Earth s surface. Solution: a Fieldline equation: r = L cos 2 λ Latitude at the Earth s surface: λ = arccos For L = 5: For L = /6400 = 5.156: /L This corresponds to a distance of about 48.8 km or 30 miles Note 1 corresponds to about km. b Estimate the latitude on the Earth s surface from which the magnetic field is open into space for B IMF = e z with > 0. Equatorial magnetic field: B eq = + B E R 3 E/r 3 with B E = T X line, B eq = 0 => r xl = B E / 1/3 R E or for L-shell: L xl = B E / 1/3 For = T : For = T : Field line equation: r 2 cos 2 λ r xl,1 = 21.8 R E r xl,1 = 10.1 R E 2 B ERE 3 r 3 = const = r 2 xl 2 B ER 3 E r 3 xl On Earth s surface r = 1 R E : cos 2 B0 λ E 2 + B E = L 2 xl = BE B0 2 + B E L 3 xl 2/3 B0 2 + or For B E : cos 2 λ E = BE cos 2 λ E 3 2/ B E = 3 BE 2/3 BE 1 2/3 = 3 1/3 B0 2B E 2 B E 1 + 2B E / 5

6 and For = T : For = T : λ E arccos λ e = λ e = [ 3 1/3 ] 1/2 [ ] B0 3 1/2 = arccos 2 B E 2L xl 6