The Virial Theorem, MHD Equilibria, and Force-Free Fields

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1 The Virial Theorem, MHD Equilibria, and Force-Free Fields Nick Murphy Harvard-Smithsonian Center for Astrophysics Astronomy 253: Plasma Astrophysics February 10 12, 2014 These lecture notes are largely based on Plasma Physics for Astrophysics by Russell Kulsrud, Lectures in Magnetohydrodynamics by the late Dalton Schnack, Ideal Magnetohydrodynamics by Jeffrey Freidberg, Hydrodynamic and Hydromagnetic Stability by S. Chandrasekhar, Classical Electrodynamics by J. Jackson, and examples by A. Savcheva and A. Spitkovsky.

2 Outline We will look at the properties and key characteristics of MHD equilibria. Our discussion will focus on: The virial theorem 0 = 2E V + 3 (γ 1) E p + E B + E g MHD equilibria J B Force-free fields c = p J B = 0 B = αb

3 The Virial Theorem for MHD (following Kulsrud 4.6) The Virial Theorem allows us to understand broadly the equilibrium properties of a system in terms of energies Suppose there exists a magnetized plasma within a finite volume. The scalar moment of inertia is I = 1 ρr 2 dv (1) 2 where r is the position vector about some arbitrary origin Our strategy: Calculate di/dt and d 2 I/dt 2 Ignore surface integrals by assuming the volume is large Put the result in terms of energies Set d 2 I/dt 2 = 0 for an equilibrium Determine the conditions under which the resulting equation can be satisfied V

4 Take the first time derivative of I Use the continuity equation, the radial form r 2 = 2r, the identity (f A) f A + A f, and Gauss theorem. di = 1 ρ dt 2 V t r 2 dv = 1 (ρv) r 2 dv 2 V = 1 (ρvr 2) dv + 1 ρv r 2 dv 2 V 2 = 1 r 2 ρv ds + ρv r dv (2) 2 S The moment of inertia changes by mass entering the system, or mass moving toward or away from the origin Consider a volume large enough so that no mass enters or leaves. The surface integral then vanishes: di dt = ρv r dv (3) V V

5 Take the second time derivative of I Use the momentum equation, r = I, the tensor identity A T = (A T) + T : A, and Gauss theorem d 2 I dt 2 = r (ρv) dv = ( T) r dv V t V = (T r) dv + T : r dv V V = ds T r + trace (T) dv (4) S The first term represents surface stresses. If we assume that surface stresses are negligible, then we are left with V d 2 I dt 2 = trace (T) dv (5) V

6 Let s look again at the stress tensor T (with gravity) We need to take T = ρvv + pi + B2 8π I BB 4π + ( φ)2 I 8πG ( φ)( φ) 4πG (6) The Reynolds stress is ρvv = V x V x V x V y V x V z V x V y V y V y V y V z V x V z V y V z V z V z (7) Then take its trace by adding up the diagonal elements: trace (ρvv) = ρ ( Vx 2 + Vy 2 + Vz 2 ) = ρv 2 (8) It s just twice the kinetic energy density!

7 Now evaluate the traces of the other terms in T The traces of the other stress tensors yield energy densities times constants! trace ( B 2 8π I BB trace ( ( φ) 2 I 8πG ( φ)( φ) 4πG trace (pi) = 3p (9) ) = B2 (10) 4π 8π ) = ( φ)2 (11) 8πG Recall that the internal energy density is given by p/(γ 1)

8 Now let s put these back into the volume integral By replacing trace (T) in Eq. 5 we arrive at d 2 ( ) I dt 2 = ρv 2 + 3p + B2 8π ( φ)2 dv 8πG where V = 2E V + 3 (γ 1) E p + E B + E g (12) Kinetic energy: EV 0 Internal energy: Ep 0 Magnetic energy: EB 0 Gravitational energy: E g 0 (only possible negative term!) In an equilibrium, this expression must equal zero: 0 = 2E V + 3 (γ 1) E p + E B + E g (13) We often consider averages of confined systems over a long time.

9 What happens when we neglect magnetic and internal energy? If E B = E g = 0, then we recover E V = 1 2 E g (14) The kinetic energy must equal half the magnitude of the gravitational energy. This is a well-known result in self-gravitating systems such as star clusters, galaxies, and galaxy clusters This result has been used to infer the presence of dark matter

10 What happens when we drop gravity in a static system? In the absence of gravity and bulk motions, we are left with 0 = 3 (γ 1) E p + E B (15) But E p 0 and E B 0! We have a contradiction! A magnetized plasma cannot be in MHD equilibrium under forces generated solely by its own internal currents. Equilibria are possible if there are external currents as in laboratory plasmas Accounted for from the surface integrals we dropped In astrophysics, this might not be satisfied

11 What limits on the magnetic energy does the Virial Theorem imply? If E V = E p = 0, then 2E V + 3 (γ 1) E p + E B + E g = 0 (16) For a stable equilibrium, the magnetic energy must not exceed the magnitude of the gravitational energy. The virial theorem provides broad insight into the equilibrium properties of a system without having to worry about the details

12 MHD Equilibria We often care about systems that are in equilibrium Let s look at the momentum equation (neglecting gravity) ( ) ρ t + V V = J B p (17) c For a static equilibrium, the configuration must have J B c in the absence of other forces = p (18)

13 Properties of MHD equilibria Take B of the equilibrium equation: B ( p) = ( ) J B B c (19) B p = 0 (20) where we use that B is orthogonal to J B. B p is the directional derivative of p in the direction of B Plasma pressure is constant along magnetic field lines Similiarly, if we take J the equilibrium equation then J p = 0 (21) since J is orthogonal to J B also.

14 Effects of fast thermal conduction on equilibria Ideal MHD does not include thermal conduction However, thermal conduction is very fast along field lines! If temperature is constant along field lines, then For p = nkt, then we also have B T = 0 (22) B n = 0 (23) Eqs. 22 and 23 are not exact results, but rather arise from the approximation of infinitely fast parallel thermal conduction.

15 Example: equilibria with a unidirectional magnetic field Consider a configuration where the magnetic field is purely in the ẑ direction The equilibrium condition is then p + B z 8π = p tot (24) where the total pressure, p tot, is a constant The tension forces disappear because the field lines are straight

16 Example: consider 1D cylindrical equilibria A Z-pinch (above) has current flowing in the ẑ direction so that B is purely azimuthal A θ-pinch has current flowing in the ˆθ direction so B is purely axial A screw pinch has components of J and B in both the axial and azimuthal directions For these configurations, we look for solutions of the form p = p(r) ; J = J θ (r) ˆθ + J z (r)ẑ ; B = B θ (r) ˆθ + B z (r)ẑ (25) for which B = 0 is trivially satisfied

17 Finding a Z-pinch 1D equilibrium Set J θ = 0 and B z = 0 since current is purely axial Ampere s law becomes J z (r) = c 1 d 4π r dr (rb θ) (26) The ˆr component of the momentum equation is We then apply Eq. 26 J z B θ = dp dr (27) dp dr + c B θ d 4π r dr (rb θ) = 0 (28)

18 Finding a Z-pinch 1D equilibrium This can be rearranged to ( ) d p + B2 θ + B2 θ = 0 (29) dr 8π 4πr }{{}}{{} total pressure tension Or, putting this in terms of the curvature vector, ( ) p + B2 B2 8π 4π κ = 0 (30)

19 A Z-pinch equilibrium can be found by specifying B θ (r) and then solving for p(r) Shown above is the Bennett pinch with B θ r r 2 + r 2 0 ; p, J z r 2 0 (r 2 + r 2 0 )2 (31) If the domain is r [0, ] then the magnetic energy diverges! Need an outer wall, which is not present in astrophysics.

20 Axisymmetric equilibria are found by solving the Grad-Shafranov equation Astrophysical applications include flux ropes in the corona/solar wind, compact object magnetospheres, etc. Fundamentally important for fusion devices like tokamaks

21 To derive the Grad-Shafranov equation, we first define a flux function We introduce a flux function ψ such that B r = 1 ψ r z 1 ψ B z = r t which satisfies the divergence constraint for any B θ (r, z). Contours of constant ψ represent the projection of magnetic field lines into the poloidal (r-z) plane (32) (33)

22 We apply Ampere s law with this magnetic field configuration The current density for this configuration is where the operator is J r = c 1 4π r z (rb θ) (34) c 1 J z = 4π r r (rb θ) (35) c ψ J θ = (36) 4π r r r ( 1 r r ) + 2 z 2 (37)

23 Now look again at the equilibrium relation Equilibrium requires J B c By symmetry, p/ θ = 0 so that This then becomes 1 r 2 = p (38) J z B r J r B z = 0 (39) [ r (rb θ) ψ z z (rb θ) ψ r ] = 0 (40) To satisfy this, rb θ must be a function of ψ alone, so we define rb θ F (ψ) (41)

24 Deriving the Grad-Shafranov equation Likewise, p is constant along field lines and is also a function of ψ: p = p (ψ) (42) The r component of the equilibrium relation is j θ B z j z B θ = p r Combining these results we arrive at ψ 4πr 1 ψ r r 1 df ψ F πr dψ r r = dp ψ dψ r (43) (44) which simplifies to the Grad-Shafranov equation ψ + F df dψ = 4πr 2 dp dψ (45)

25 How do we find a solution of the Grad-Shafranov equation? The Grad-Shafranov equation is given by where ψ + F df dψ = 4πr 2 dp dψ, (46) r ( ) r r r z 2 (47) F (ψ) rb θ (48) To solve the Grad-Shafranov equation, we need to Specify p(ψ) Specify F (ψ) Solve for ψ The Grad-Shafranov equation is usually solved numerically

26 Equilibria and the Virial Theorem The natural state of a flux rope is to try to expand to infinity In laboratory plasmas, conducting wall outer boundaries and externally applied magnetic fields prevent this These show up as surface integrals in the Virial Theorem In astrophysical plasmas, a flux rope can be held in place from J B and p forces from the surrounding medium The surrounding medium, in turn, can be held in place by gravitational forces Example: the solar corona

27 How does gravity change things? The equilibrium condition becomes 0 = J B c p + ρg (49) B and J are no longer necessarily orthogonal to p B p = ρb g (50) J p = ρj g (51) A radially outward flux tube reduces to the case of hydrostatic equilibrium: p = ρg The Grad-Shafranov equation can be generalized to include gravitational forces I decided against assigning this as a homework problem.

28 Force-free fields When pressure is constant or in the limit of β 0, the pressure gradient force vanishes. Equilibria then have J B = 0 (52) Such configurations are called force-free because there is no Lorentz force and no plasma pressure gradient force Using Ampere s law, this reduces to where α is constant along field lines B = αb (53) Vector fields parallel to their own curl are called Beltrami fields

29 Linear force-free fields have constant α Start with the condition for a linear force-free field: Take its curl: B = αb (54) ( B) = α B (55) Use Eq. 54 for the RHS and vector identities ( B) 2 B = α 2 B (56) Linear force-free fields obey the Helmholtz equation 2 B + α 2 B = 0 (57) which can be solved using separation of variables, Green s functions, Fourier series, or numerically.

30 Nonlinear force-free fields (NLFFF) have non-constant α Finding nonlinear force-free fields requires solving B = α (x) B (58) where α must still remain constant along field lines Some analytical examples exist in 2D 3D solutions must generally be found numerically

31 Example: pulsar magnetospheres I Force-free field models are able to capture features that do not exist in a vacuum solution I Figure of oblique rotator by Anatoly Spitkovsky. Color marks direction of toroidal field.

32 Example: NLFFF modeling of sigmoidal solar active regions Hinode/XRT NLFFF model/photospheric magnetic flux contours Significant free energy is stored in the sheared/twisted magnetic field of sigmoids (S-shaped structures in the corona) The modeling strategy is to introduce a flux rope into a potential field model, let it relax, and see if it matches the observed X-ray emission (Savcheva et al. 2012)

33 Potential magnetic fields represent the minimum energy condition for a configuration s boundary conditions These are a special case of force-free fields with α = 0 Potential fields have J = 0, so Ampere s law becomes 0 = B (59) The curl of a gradient is zero, so this is satisfied if where ζ is a scalar magnetic potential B = ζ (60) Taking the divergence of both sides shows that ζ can be found by solving Laplace s equation B = ζ (61) 0 = 2 ζ (62)

34 Potential magnetic fields can be calcuated using Laplace s equation for the vector potential From Ampere s law with no currents we know that From these two expressions we arrive at B = A (63) 0 = c 4π B (64) ( A) = 0 (65) Using vector identities and choosing the Coulomb gauge ( A = 0) we arrive at 2 A = 0 (66)

35 Determining the magnetic free energy of a system The magnetic free energy of a system is given by ( ) B 2 E M 8π B2 p dv (67) 8π V where B p is the potential field solution for the system s boundary conditions However, the quantity B 2 8π B2 p 8π (68) should not be considered a free energy density since free energy is an integral quantity. At times this could be negative! Current density is a good proxy for how much stress exists in a magnetic field

36 Summary The Virial Theorem allows us to understand key requirements for equilibria even before we know the details A plasma cannot be in an MHD equilibrium generated solely by its internal currents 2D MHD equilibria are found using the Grad-Shafranov equation However, many equilibria are unstable 3D MHD equilibria are difficult to find and/or generalize Force-free fields have J B = 0 so that B = αb Linear force-free fields have constant α and can be solved for using the Helmholtz equation Nonlinear force-free fields have nonconstant α and usually are solved numerically Potential fields are the minimum energy state for given boundary conditions