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 = ∇p c
I Force-free fields
J × B = 0 ⇒ ∇ × 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 ∇r = 2r, the identity ∇ · (f A) ≡ f ∇ · A + A · ∇f , and Gauss’ theorem.
dI 1 Z ∂ρ 1 Z = r 2 dV = − ∇ · (ρV) r 2 dV dt 2 V ∂t 2 V 1 Z 1 Z = − ∇ · ρVr 2 dV + ρV · ∇r 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, ∇r = I, the tensor identity A · ∇ · T = ∇ · (A · T) + T : ∇A, and Gauss’ theorem
d2I Z ∂ Z 2 = r · (ρV) dV = − (∇ · T) · r dV dt V ∂t V Z Z = − ∇ · (T · r) dV + T : ∇r 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 (∇φ)2 I (∇φ)(∇φ) T = ρVV + pI + I − + − (6) 8π 4π 8πG 4πG
I The Reynolds stress is Vx Vx Vx Vy Vx Vz ρVV = Vx Vy Vy Vy Vy Vz (7) Vx Vz Vy Vz Vz Vz
Then take its trace by adding up the diagonal elements: