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. Outline I We will look at the properties and key characteristics of MHD equilibria. Our discussion will focus on: I The virial theorem 0 = 2EV + 3 (γ − 1) Ep + EB + Eg I MHD equilibria J × B = rp c I Force-free fields J × B = 0 ) r × B = αB The Virial Theorem for MHD (following Kulsrud §4.6) I The Virial Theorem allows us to understand broadly the equilibrium properties of a system in terms of energies I Suppose there exists a magnetized plasma within a finite volume. The scalar moment of inertia is 1 Z I = ρr 2 dV (1) 2 V where r is the position vector about some arbitrary origin I Our strategy: 2 2 I Calculate dI=dt and d I=dt I Ignore surface integrals by assuming the volume is large I Put the result in terms of energies 2 2 I Set d I=dt = 0 for an equilibrium I Determine the conditions under which the resulting equation can be satisfied Take the first time derivative of I 2 I Use the continuity equation, the radial form rr = 2r, the identity r · (f A) ≡ f r · A + A · rf , and Gauss' theorem. dI 1 Z @ρ 1 Z = r 2 dV = − r · (ρV) r 2 dV dt 2 V @t 2 V 1 Z 1 Z = − r · ρVr 2 dV + ρV · rr 2 dV 2 V 2 1 I Z = − r 2ρV · dS + ρV · r dV (2) 2 S V I The moment of inertia changes by mass entering the system, or mass moving toward or away from the origin I Consider a volume large enough so that no mass enters or leaves. The surface integral then vanishes: dI Z = ρV · r dV (3) dt V Take the second time derivative of I I Use the momentum equation, rr = I, the tensor identity A · r · T = r · (A · T) + T : rA, and Gauss' theorem d2I Z @ Z 2 = r · (ρV) dV = − (r · T) · r dV dt V @t V Z Z = − r · (T · r) dV + T : rr dV V V I Z = − dS · T · r + trace (T) dV (4) S V The first term represents surface stresses. If we assume that surface stresses are negligible, then we are left with d2I Z 2 = trace (T) dV (5) dt V Let's look again at the stress tensor T (with gravity) I We need to take B2 BB (rφ)2 I (rφ)(rφ) T = ρVV + pI + I − + − (6) 8π 4π 8πG 4πG I The Reynolds stress is 0 1 Vx Vx Vx Vy Vx Vz ρVV = @ Vx Vy Vy Vy Vy Vz A (7) Vx Vz Vy Vz Vz Vz Then take its trace by adding up the diagonal elements: 2 2 2 trace (ρVV) = ρ Vx + Vy + Vz = ρV 2 (8) It's just twice the kinetic energy density! Now evaluate the traces of the other terms in T I The traces of the other stress tensors yield energy densities times constants! trace (pI) = 3p (9) B2 BB B2 trace I − = (10) 8π 4π 8π ! (rφ)2 I (rφ)(rφ) (rφ)2 trace − = − (11) 8πG 4πG 8πG I Recall that the internal energy density is given by p=(γ − 1) Now let's put these back into the volume integral I By replacing trace (T) in Eq. 5 we arrive at 2 Z 2 2 ! d I 2 B (rφ) 2 = ρV + 3p + − dV dt V 8π 8πG = 2EV + 3 (γ − 1) Ep + EB + Eg (12) where I Kinetic energy: EV ≥ 0 I Internal energy: Ep ≥ 0 I Magnetic energy: EB ≥ 0 I Gravitational energy: Eg ≤ 0 (only possible negative term!) I In an equilibrium, this expression must equal zero: 0 = 2EV + 3 (γ − 1) Ep + EB + Eg (13) We often consider averages of confined systems over a long time. What happens when we neglect magnetic and internal energy? I If EB = Eg = 0, then we recover 1 E = − E (14) V 2 g The kinetic energy must equal half the magnitude of the gravitational energy. I This is a well-known result in self-gravitating systems such as star clusters, galaxies, and galaxy clusters I This result has been used to infer the presence of dark matter What happens when we drop gravity in a static system? I In the absence of gravity and bulk motions, we are left with 0 = 3 (γ − 1) Ep + EB (15) But Ep ≥ 0 and EB ≥ 0! We have a contradiction! I A magnetized plasma cannot be in MHD equilibrium under forces generated solely by its own internal currents. I Equilibria are possible if there are external currents as in laboratory plasmas I Accounted for from the surface integrals we dropped I In astrophysics, this might not be satisfied What limits on the magnetic energy does the Virial Theorem imply? I If EV = Ep = 0, then 2EV + 3 (γ − 1) Ep + EB + Eg = 0 (16) For a stable equilibrium, the magnetic energy must not exceed the magnitude of the gravitational energy. I The virial theorem provides broad insight into the equilibrium properties of a system without having to worry about the details MHD Equilibria I We often care about systems that are in equilibrium I Let's look at the momentum equation (neglecting gravity) @ J × B ρ + V · r V = − rp (17) @t c I For a static equilibrium, the configuration must have J × B = rp (18) c in the absence of other forces Properties of MHD equilibria I Take B· of the equilibrium equation: J × B B · (rp) = B · (19) c B · rp = 0 (20) where we use that B is orthogonal to J × B. I B · rp is the directional derivative of p in the direction of B I Plasma pressure is constant along magnetic field lines I Similiarly, if we take J· the equilibrium equation then J · rp = 0 (21) since J is orthogonal to J × B also. Effects of fast thermal conduction on equilibria I Ideal MHD does not include thermal conduction I However, thermal conduction is very fast along field lines! I If temperature is constant along field lines, then B · rT = 0 (22) For p = nkT , then we also have B · rn = 0 (23) Eqs. 22 and 23 are not exact results, but rather arise from the approximation of infinitely fast parallel thermal conduction. Example: equilibria with a unidirectional magnetic field I Consider a configuration where the magnetic field is purely in the ^z direction I The equilibrium condition is then B p + z = p (24) 8π tot where the total pressure, ptot , is a constant I The tension forces disappear because the field lines are straight Example: consider 1D cylindrical equilibria I A `Z-pinch' (above) has current flowing in the ^z direction so that B is purely azimuthal I A `θ-pinch' has current flowing in the θ^ direction so B is purely axial I A `screw pinch' has components of J and B in both the axial and azimuthal directions I For these configurations, we look for solutions of the form p = p(r); J = Jθ(r)θ^ + Jz (r)^z ; B = Bθ(r)θ^ + Bz (r)^z (25) for which r · B = 0 is trivially satisfied Finding a Z-pinch 1D equilibrium I Set Jθ = 0 and Bz = 0 since current is purely axial I Ampere's law becomes c 1 d J (r) = (rB ) (26) z 4π r dr θ The ^r component of the momentum equation is dp J B = (27) z θ dr We then apply Eq. 26 dp c B d + θ (rB ) = 0 (28) dr 4π r dr θ Finding a Z-pinch 1D equilibrium I This can be rearranged to d B2 B2 p + θ + θ = 0 (29) dr 8π 4πr | {z } |{z} total pressure tension Or, putting this in terms of the curvature vector, B2 B2 r p + − κ = 0 (30) ? 8π 4π A Z-pinch equilibrium can be found by specifying Bθ(r) and then solving for p(r) I Shown above is the `Bennett pinch' with r r 2 B / ; p; J / 0 (31) θ 2 2 z 2 2 2 r + r0 (r + r0 ) I If the domain is r 2 [0; 1] then the magnetic energy diverges! Need an outer wall, which is not present in astrophysics. Axisymmetric equilibria are found by solving the Grad-Shafranov equation I Astrophysical applications include flux ropes in the corona/solar wind, compact object magnetospheres, etc. I Fundamentally important for fusion devices like tokamaks To derive the Grad-Shafranov equation, we first define a flux function I We introduce a flux function such that 1 @ B = − (32) r r @z 1 @ B = (33) z r @t which satisfies the divergence constraint for any Bθ(r; z). I Contours of constant represent the projection of magnetic field lines into the poloidal (r-z) plane We apply Ampere's law with this magnetic field configuration I The current density for this configuration is c 1 @ J = − (rB ) (34) r 4π r @z θ c 1 @ J = (rB ) (35) z 4π r @r θ c ∆∗ J = (36) θ 4π r where the operator ∆∗ is @ 1 @ @2 ∆∗ ≡ r + (37) @r r @r @z2 Now look again at the equilibrium relation I Equilibrium requires J × B = rp (38) c I By symmetry, @p=@θ = 0 so that Jz Br − Jr Bz = 0 (39) This then becomes 1 @ @ @ @ (rB ) − (rB ) = 0 (40) r 2 @r θ @z @z θ @r To satisfy this, rBθ must be a function of alone, so we define rBθ ≡ F ( ) (41) Deriving the Grad-Shafranov equation I Likewise, p is constant along field lines and is also a function of : p = p ( ) (42) I The r component of the equilibrium relation is @p j B − j B = (43) θ z z θ @r I Combining these results we arrive at ∆∗ 1 @ 1 dF @ F dp @ − − = (44) 4πr r @r πr d @r r d @r which simplifies to the Grad-Shafranov equation dF dp ∆∗ + F = −4πr 2 (45) d d How do we find a solution of the Grad-Shafranov equation? I The Grad-Shafranov equation is given by dF dp ∆∗ + F = −4πr 2 ; (46) d d where @ 1 @ @2 ∆∗ ≡ r + (47) @r r @r @z2 F ( ) ≡ rBθ (48) I To solve the Grad-Shafranov equation, we need to I Specify p( ) I Specify F ( ) I Solve for I The Grad-Shafranov equation is usually solved numerically Equilibria and the Virial Theorem I The natural state of a flux rope is to try to expand to infinity I In laboratory plasmas, conducting wall outer boundaries and externally applied magnetic fields prevent this I These show up as surface integrals in the Virial Theorem I In astrophysical plasmas, a flux rope can be held in place from J × B and −∇p forces from the surrounding medium I The surrounding medium, in turn, can be held in place by gravitational forces I Example: the solar corona How does gravity change things? I The equilibrium condition becomes J × B 0 = − rp + ρg (49) c I B and J are no longer necessarily orthogonal to rp B · rp = ρB · g (50) J · rp = ρJ · g (51) I A radially outward flux tube reduces to the case of hydrostatic equilibrium: rp = ρg I The Grad-Shafranov equation can be generalized to include gravitational forces I I decided against assigning this as a homework problem.