Magnetospheric Physics - Homework, 4/04/2014

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1 Magnetosheric hysics - Homework, // 7. Fast wave, fast shock, erendicular shock, entroy, jum relation. Consider the fast erendicular shock. a) Determine the ositive root of the solution for the comression ratio. How does the comression ratio behave for very large and very small ustream lasma β? b) Derive the downstream Mach number as a function of ustream Mach number and for small and large lasma β. How does the downstream Mach number change as a function of β for fied ustream Mach number? c) An imortant quantity for sace lasma rocesses is entroy. A measure of entroy is s = /n γ. Determine the downstream to ustream ratio of s as a function of ustream Machnumber and lasma β. lot this ratio and the comression ratio as a function of lasma β for a fied ustream Machnumber of. What is the value of the minimum ustream lasma β to sustain a shock? Why? Solution: a) Relations: M = u u c s = c s γ ρ β = thu Bu = µ u A = B u µ ρ u B u = γ c s u A The equation for the comression ratio X = ρ d /ρ u is X = γ) X + [ β + γ ) βm + ] γx γ γ + ) βm = Substitute of the sonic Machnumber the fast Machnumber: yields M = u c s = u c s + va c s + va c s = M f ) + B ρ µ ργ = M f ) + B µ γ = M f + γβ γβ X = γ) X + [ γβ + γ ) + γβ) M f + γ ] X γ + ) + γβ) M f = Formal eact) solution: Division γ) X + ax b = with a = γβ + γ ) + γβ) M f + γ γ) b = γ + ) + γβ) M f γ) with the solution : X = a ± b + a

2 Note not necessary) eansion into Taylor series valid for large M f ): X = a + a + ba = a + a + b ) a b 8a = b a b 8a 3 Here only the ositive solution is hysical. For both β and β the comression goes to for M f such that we consider M f here: For β : X = γ) X + [ γ ) M f + γ ] X γ + ) M f = a = γ ) M f + γ γ) b = γ + ) M f γ) X = γ ) M f + γ γ) + γ) γ + ) Mf γ) + [ γ ) Mf + γ] γ + ) M f γ ) M f + γ for M f For β : X = γ) X + γβ [ + γ ) M f ] X γβ γ + ) M f = a = γβ + γ ) M f γ) b = γβ γ + ) M f γ) such that X = γ) X + [ γβ + γ ) + γβ) M f + γ ] X γ + ) + γβ) M f γβ [ + γ ) M f ] X γβ γ + ) M f = with X = γ + ) M f + γ ) M f Note that this is the solution for the gasdynamic shock because the magnetic field is negligible for β. b) Derive the downstream Mach number as a function of ustream Mach number and for small and large lasma β. How does the downstream Mach number change as a function of β for fied ustream Mach number? This illustrates the comutation for the fast mode Machnumber but a solution for the sonic Machnumber is also accetable since this was not secified in the roblem. ressure ratio: = + β X ) + γm X ) = + β X ) + β + γβ) M ) f X

3 Machnumber ratio: M d M u = u nd vau + c su u nu vad + c sd = X 3 + γβ u/ + γβ d / = u nd + c su/vau vau u nu + c sd /v Ad vad so we need to comute β d /β u : Substituting: β d = µ Bu = Bu β u Bd µ Bd = X [ + β X ) + β + γβ) M f X )] M d M u = X 3 + γβ u + γβ d = X 3 + γβ u [ + γβ u X + β u X ) + βu + γβ u ) Mf X ) ] = X + γβ u X + γβ u + γ X ) + γ + γβ u ) M f X ) Note that X = yields M d = M u. Also the derivative of the denominator is ositive for X such that the denominator is only increasing for X while the numerator is constant. Therfore M d /M u is monotonically decreasing for X. For large β and large M f : M d M u = X β u β u + γβ u M f X ) For small β: M d M u = X X + γ X ) + γm f X ) c) Entroy change: H = /ρ γ Relations: and Entroy: X = ρ d ρ u u nd u nu = X B yd B yu = X = +β X ) +γm X ) = +β X ) +β + γβ) M ) f X S d S u = ) γ ρu ρ d = [ + β X ) + β + γβ) M f X )] X γ

4 Note: The simulation treats the energy equation usually as /γ / t = /γ u ) which use a one-dimensional discontinuity imlies /γ u = const or d ) /γ = u nu u nd or = X γ In this case S d /S u =. For the actual solution of the comression ratio, the down- to ustream ressure ratio, and the ratio of the entroy function S the following lots show these quantities as functions of lasma β and of fast mode Machnumber for a small and a large value of lasma β and Mach Number. Mach Number:. Mach Number: lasma :. lasma :. Machnumber Machnumber 8. Simulation of a hydrodynamic shock. The initial condition 8 in the simulation code is an eamle for a hydrodynamic shock. Note, in order to run this stable and without large oscillation you need to increase the viscosity arameters significantly. a) For what values of the viscosities do you get a reasonably stable solution? b) Run the code for two different ustream Machnumbers and use methods,, and for the treatment of the energy equation arameter intu). Describe any differences in the results. Comare the analytic jum conditions for a hydrodynamic shock with your results for the two Machnumbers. Discuss these results. Note, that for larger Machnumbers you will need higher viscosity and ossibly also a wider transition region arameterized with l in the rogram. Solution: a) The required viscosity deends somewhat on the ustream Machnumber. For the new version of the code smaller Machnumbers u to 3 or require diffusion and viscosity of about. and they should

5 be increased to about. for stronger shocks Mach numbers as high as ). Of equal or even higher imortance is the time ste. Theoretically the maimum time ste is limited t / ma v). The grid sacing is. for the default values. The maimum velocity is the maimum seed of information transort which in the case of a flowing lasma is the bulk velocity lus the fastest wave mode. In this case it is the ustream velocity lus the ustream seed of sound which for the current arameters is ma v) M u c s + c s. Here we have chosen ustream Machnumbers of and 8 and the sound seed is γ/ρ.9 such that for M u = the limiting time ste is aroimately.3 and for M u = 8 c s = it is.. Indeed the simulation for M u = runs stable for t =.35 but is unstable for t =.. For M u = runs stable for t =. but is unstable for t =.5 consistent with the teoretical eectation. b) The following lots show the results for the cases. The first two lots show the cases with M u = and the net two lots show the results for M u = 8. There are two aarent differences although small for M u =. The density comression is sightly larger and the shock moves in the ositive direction with a velocity of.7) for intu =. The theoretical rediction for the shock are X = γ + ) M u + γ ) M u = γm u γ ) γ + which yields X =.3 and / =.75 which is almost eactly satisfied the nd method energy conservation). However, for small Machnumbers the error usinf the st method is only about %. Note that the fact that the shock move in this case imlies that the actual Machnumber is slightly less the shock velocity) than in the initial conditions time = time = M u =, energy equation using /γ M u =, energy conservation

6 time = time =. M u = 8, energy equation using /γ M u = 8, energy conservation For the nd case with M u = 8 the differences are much larger. The theoretical redictions are X = 3.9 and / = Again the energy consevation method is highly accurate. The method for intu = achieves almost the same results for the ressure and velocity change, but it fails strongly regarding the density change which is almost 3 for the comression here. Also, the shock moves now with considerable velocity about.) into the downstream direction, thus altering the actually alicable Machnumber. 7. roject Decide a roject toic, collect the material, and make a first attemt to understand the basic motivation, methodology, and imact of the results of the article. In case you choose a simulation toic, outline your goals for this toic and formulate an aroach how to achieve those. Solution: Individual