Magnetospheric Physics - Final Exam - Solutions 05/07/2008

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1 Magnetospheric Physics - Final Exam - Solutions 5/7/8. Dipole magnetic field a Assume the magnetic field of the Earth to be dipolar. Consider a flux tube with a small quadratic cross section in the equatorial plane where the two inner corners are at the same radial distance r and the two outer corners are at the distance r + ε with ε. What is the latitudinal and azimuthal separation of the corner field lines at the Earth s surface. For the sake of numbers assume r = 4R E and ε = 3. b Is the cross section of the flux tube still a square and what is the aspect ratio at the Earth s surface? Does the aspect ratio depend on the L value radial distance? c Show explicitly that the magnetic flux through the crossection in the equatorial plane is the same as on the Earth s surface, i.e., that B A = const for your result for the small value of ε you can assume B approximately constant for each crossection. Solution: a Flux tube mapping to Earth: Fieldline equation: r = R eq cos λ Latitude of outer field line at the Earth s surface: λ E = arccos R E /r = arccos /4 = 6 R E /r + ε = arccos Latitude of inner field line at the Earth s surface: λ E = arccos 6.6 Latitudinal difference: λ =.64 =.86 3 rad or.83km Taylor expansion to lowest order in ε = r /r : /4.4 = RE λ E = arccos + ε d RE arccos = arccos r dε r + ε L + ε d arccos dε L + ε = arccos L + ε /L /L + ε 3/ = arccos L + ε L such that the latitudinal separation is λ = ε = 3 L = rad or Latitudinal distance: x = λr E = ε R L E =.85km. Azimuth Longitude: Since dipole field lines are in planes with φ = const the azimuthal separation in rad at the Earth s surface is φ = εr r = ε = 3 rad The physical distance in longitude depends on latitude as well: b Aspect ratio: Thus the aspect ratio on the Earth s surface is y = εr E cos λ E = εr E /L = 3. km δ = y = εr E L = L = /L x L / εr E L /

2 The aspect ratio depends on the L shell or on latitude of the foot points on the Earth. For high latitudes L the aspect ratio approaches. c Crossections and magnetic flux Crossection in the equatorial plane: A eq = ε r = ε L R E Crossection on the Earth s surface: A E = y x = εr E L / ε L R E = ε R E L / L Magnetic field in the equatorial plane: B eq = B E /L 3 normal to the equator such that the flux is Φ eq = B eq A eq = ε RE L B E On the Earth s surface we could use the magnitude of B at λ E and then determine the radial component through the orientation of B on the surface. It is easier to directly use the radial component of B for the dipole field Magnetic flux on the Earth s surface: B r r = R E = B E RE 3 sin λ E = B r 3 E sin λ E = B E cos λ E = B E /L ε Φ E = B r,e A E = B E RE /L L / L = ε RE L B E Note that using the field line equation with r = R E implies = L cos λ E such that all quantities could be expressed as functions of λ E instead of L. Note also that, instead of computing λ through mapping of r + ε, we could have computed λ through λ E and φ using magnetic flux conservation.

3 . Anisotropic distribution function An anisotropic bi-maxwellian distribution function for a two-dimensional system / y = can be written as F H, µ = GA exp βah αaµb with the Hamiltonian H = m/ v + v + qφ, the adiabatic moment µ = m/v /B, the magnetic field strength B, and the y component of the vector potential A, where GA, βa, and αa are arbitrary functions. a Relate α and β to the temperatures T and T for a bi-maxwellian distribution. b Calculate the number density n by integrating F over velocity space to show na, B = πkb T m πk B T GA exp βqφ m c Show that the parallel and perpendicular pressures are p = na, Bk B T and p = na, Bk B T d Determine the pressure anisotropy D = p p p as a function of B, B, β, and α. Solution: a Distribution function: F H, µ = GA exp βah αaµb = GA exp β m v + v B v βqφ α B = GA exp β m v βqφ α B m B + β v Such that kt = /β and kt = /λ with λ = α B + β B b Calculate the number density n by integrating F over velocity space. na = GA exp βqφ = πga exp βqφ π = πga exp βqφ βm π π = GA exp βqφ βm λm d 3 v exp β m v λ m v dv exp β m v λm exp λ m v dv v exp λ m v c Show that the parallel and perpendicular pressures are p = na, Bk B T and p = na, Bk B T p = mga exp βqφ = m π λm GA exp βqφ = m π λm π GA exp βqφ = π GA exp βqφ λm = nakt π βm kt d 3 vv exp β m v λ m v dv v exp β m v 3/ βm

4 p = mga exp βqφ = π πm βm GA exp βqφ π = πm = π π GA exp βqφ βm d 3 vv exp β m v λ m v [ dv v3 exp v λm λ m GA exp βqφ βm λm kt = nakt λ m v exp λ m ] v d Determine the pressure anisotropy D = p p p as a function of B, B, β, and η. Assuming B and B > this particularly implies: Note also that this requires αb βb T T = D = T T T /β /λ = = β /β λ = αb + βb αb + = βb { < if α/β > > if α/β < > otherwise T or T is negative.

5 3. Whistler waves In the electron magnetohydrodynamic EMHD limit u Ohm s law can be written as E = ne j B + ηj Assume a uniform plasma with a magnetic field B = B e z and perturbations δb x and δb y exp ik z z iωt. a Use the continuity equation to show that δn = in the EMHD u approximation. b Use the the above form of Ohm s law with together with the induction equation and Ampere s law to derive equations for linear waves in δb x and δb y. Show that the dispersion relation is ω = ±kzb / ne ikzη/µ = ±k z V A k z c/ω pi ikzη/µ. These waves are the so-called whistler waves. c Show that the group and phase velocities are V A k z c/ω pi for this wave for η =? What is the wave length when the phase velocity is V A. Discuss the group and phase velocities in the limit of k z to infinity. Solution: a Show that δn = in the EMHD u approximation. Continuity equation: δn = nδu = t b Derivation of the dispersion relation: The linearized induction equation becomes or δb t = δe = ne δj B + ηδj iωδb x = ik z B ne δj x + ηδj y B iωδb y = ik z ne δj y + ηδj x With Ampere s law B = j the current density is given by Inserting δj yields Re-arranging: ωδb x = k z ωδb y = k z δj x = z δb y / = ik z δb y / δj y = z δb x / = ik z δb x / i B ne δb y + iηδb x i B ne δb x iηδb y = ikz B ne δb y iη k z δb x = ikz B ne δb x iη k z δb x B ω + i ηk z δb x = ikz ne δb y ω + i ηk z B δb y = ikz ne δb x

6 Multiplication of the two equations yields With the dispersion relation is ω + i ηk z = k z B ne / B ne = B mp ɛ m / p c = V m p n / ne A c = VA ne ω pi ω = ±kzv c A ik η z ω pi with the damping rate γ = k zη/. Here the resistive diffusion time for a wve length λ z is τ diff,λ = λ z/η such that the exponential damping rate is γ = 4π /τ diff,λ. Here c/ω pi = λ i the ion inertia length. c Phase velocity along z and group velocity dω/dk z : ω k z = ±k z V A λ i dω dk z = ±k z V A λ i Discussion of the wave propagation: For a phase velocity of ω/k z = ±V A the k vector satisfies k z = π/λ z = /λ i or the wave length is λ z = πλ i. For any wavelength smaller than this the phase velocity is faster. In particular for k z or λ z both the group and the phase velocities approach infinity. Note, however, that for λ z also the damping becomes infinitely fast. The time for the wave to travel one wave length is τ λ = λ z / ω/k z = λ z / k z V A λ i = λ z/ πv A λ i. The amount of resistive damping during this time is τ λ γ = λ z 4π η πv A λ i λ z = πλ i V A η λ i = τ Ai τ diff,i Here τ Ai = πλ i /V A i.e., the travel time of an Aflven wave over the distance πλ i and τ diff,i = λ i /η, i.e., the diffusion time for the ion inertia length. Note that τ Ai /τ diff,i = πr with R being the Lundquistnumber based on the ion inertia length.

7 4. Shocks a What are the conditions in terms of the upstream velocity in the rest frame of the discontinuity for the formation of a slow shock, a fast shock, and a rotational discontinuity? b Summarize the typical properties of slow shocks, fast shocks, and rotational discontinuities. What are switch-off and switch-on shocks and how do they relate to slow and fast shocks? c Assume a fixed upstream tangential field of B. Can the sheet current of a slow shock and a fast shock be greater than that of a rotational discontinuity? The sheet current is B u B d / for x pointing upstream and y along the upstream tangential field.. Are the directions of the slow shock current and the fast shock current the same as the direction of the current for the rotational discontinuity? d Why is the Earth s bow shock almost always a fast shock? What is the physical reason for the strong heating of plasma behind the bow shock? e What are the asymptotic values for the downstream density, magnetic field, and pressure if the upstream velocity goes to infinity? In this limit the downstream velocity also goes to infinity. How can this be if the downstream velocity has to satisfy the condition that it needs to be slower than the fast mode speed? Solution: a Upstream conditions: Fast shock: Upstream velocity faster than the fast mode speed Rotational discontinuity: Upstream velocity Alfven speed and parallel to the magnetic field in the upstream region Slow shock: Upstream velocity smaller or equal to the Alfven speed but faster than slow mode speed. b Properties: Fast shock: Compression of density, pressure required for all shocks, compression of magnetic field such that downstream field is bent away from shock normal. Rotational discontinuity: Not a shock. Downstream plasma density and pressure equal to upstream density and pressure. Downstream tangential magnetic field is equal in magnitude and opposite in direction to upstream field. Normal magnetic field required. Slow shock: Upstream velocity smaller or equal to the Alfven speed but faster than slow mode speed. Density and pressure increase; down stream magnetic field smaller than upstream field such that magnetic field is bent toward shock normal. Switch-off shock: Slow shock for which downstream tangential magnetic field is. Switch-on shock: Fast shock for which the upstream tangential magnetic field is. c Sheet current: Fast shock: The maximum downstream tangential magnetic field is 4B. Thus the maximum sheet current is I fs = 3B / Rotational discontinuity: The downstream tangential magnetic field is B. Thus the sheet current is I rd = B /.

8 Slow shock: The minimum downstream magnetic field is, such that the maximum sheet current density is B /. d Bow shock: The solar wind speed relative to Earth is almost always faster than the fast mode speed. Therefore the bow shock is almost always a fast shock. The strong heating is due to the conversion of flow energy into magnetic and thermal energy. Since the amount of magnetic compression is limited for a strong shock, a large often dominant amount of energy is going into plasma heating. e Asymptotic values: The maximum plasma and magnetic field compression is a factor of 4 of the upstream values. The downstream pressure goes to infinity in the limit of infinite upstream plasma velocity. Although the downstream velocity goes to infinity in this limit the velocity remains sub-fast because the fast mode speed for pressure also going to infinity approaches infinity faster than the downstream flow velocity.

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