Conservation Laws in Ideal MHD Nick Murphy Harvard-Smithsonian Center for Astrophysics Astronomy 253: Plasma Astrophysics February 3, 2016 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, Magnetic Reconnection by Eric Priest and Terry Forbes, and various papers by M. Berger & J. Perez. The description of Stokes' Theorem and the Divergence Theorem come from Stewart's Multivariable Calculus with figures adapted from Wikimedia Commons and R. Fitzpatrick's online notes. Outline I Mathematical interlude: I Stokes' Theorem I Divergence Theorem I Dyadic tensors I Conservation of mass, momentum, and energy I The frozen-in condition I Helicity and cross-helicity Stokes' Theorem I Let S be an oriented piecewise-smooth surface that is bounded by a simple, close, piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open 3 region in R that contains S. Then I Z F · dr = r × F · dS (1) C S where dS = n^ dS and n^ is a unit normal vector to S The Divergence Theorem (also called Gauss's Theorem) I Let V be a simple solid region and let S be the boundary surface of V, given with positive (outward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains V. Then I Z F · dS = r · F dV (2) S V I The interior contributions to the divergence integral cancel, so only the exterior contributions remain While somewhat horrible, dyadic tensors1 allow the equations of MHD to be written in a compact/useful form I A dyadic tensor is a second rank tensor that can be written as the tensor product of two vectors: Aij = Bi Cj A = BC = (B1e^1 + B2e^2 + B3e^3)(C1e^1 + C2e^2 + C3e^3) = B1C1e^1e^1 + B1C2e^1e^2 + B1C3e^1e^3 + B2C1e^2e^1 + B2C2e^2e^2 + B2C3e^2e^3 + B3C1e^3e^1 + B3C2e^3e^2 + B3C3e^3e^3 (3) where e^1. are unit vectors and e^1e^3. are unit dyads, e.g.: 0 0 0 1 1 e^1e^3 = @ 0 0 0 A (4) 0 0 0 1Good resources: Lectures in Magnetohydrodynamics by Dalton Schnack and http://people.rit.edu/pnveme/EMEM851n/constitutive/tensors rect.html Dot products and double dot products with dyadic tensors I In general, BC 6= CB I Dot products from the left and right are different: e^1 · A = B1C1e^1 + B1C2e^2 + B1C3e^3 (5) A · e^1 = B1C1e^1 + B2C1e^2 + B3C1e^3 (6) I Double-dot notation is ambiguous but we use: A : T ≡ Aij Tij where we sum over repeated indices The divergence of a dyadic tensor is a vector I The divergence of a dyadic tensor is r · T = @i Tij (7) This is a vector whose jth component is @T1j @T2j @T3j (r · T)j = + + (8) @x1 @x2 @x3 I We can even generalize Gauss' theorem! Z I r · T dV = dS · T (9) S What about the dyad T ≡ rV? I The gradient operator is given by @ @ @ @ r ≡ e^i ≡ e^1 + e^2 + e^3 (10) @xi @x1 @x2 @x3 I Then T ≡ rV is given in Cartesian coordinates by @Vj (rV)ij = (11) @xi @Vj rV = e^i e^j (12) @xi 0 1 @x Vx @x Vy @x Vz = @ @y Vx @y Vy @y Vz A (13) @z Vx @z Vy @z Vz where i is the row index and j is the column index Conservative form I As mentioned previously, conservative form is usually given by @ (stuff) + r · (flux of stuff) = 0 (14) @t where source and sink terms go on the RHS I If ‘stuff' is a scalar, then ‘flux of stuff' is a vector I If ‘stuff' is a vector, then ‘flux of stuff' is a dyadic tensor I If Eq. 14 is satisfied, then ‘stuff' is locally conserved The continuity equation is already in conservative form 2 I The continuity equation is @ρ + r · (ρV) = 0 (15) @t where ρV is the mass flux I Mass is locally conserved ) the universe is safe once again! 2This expression is similar to the Law of Conservation of Space Wombats Is mass conserved globally? I Integrate the continuity equation over a closed volume V bounded by a surface S Z @ρ + r · (ρV) dV = 0 (16) V @t @ Z Z ρ dV + r · (ρV) dV = 0 (17) @t V V I Now define M as the mass within V and use Gauss' theorem dM I + dS · (ρV) = 0 (18) dt S |{z} | {z } change of mass in V mass flow through S I Yep, we're still good. Mass is conserved globally. I Recall that this was how we derived the continuity equation in the first place! The momentum equation in conservative form I The momentum equation can be written as @ρV + r · T = 0 (19) @t where the ideal MHD stress tensor T is B2 BB T = ρVV + p + I − (20) 8π 4π and I ≡ e^1e^1 + e^2e^2 + e^3e^3 is the identity dyadic tensor 2 I The quantity p + B =8π is the total pressure I One can also include the viscous and gravitational stress tensors, or use a pressure tensor rather than a scalar pressure Physical meaning of the stress tensor I Imagine you have an infinitesimal box I Forces are exerted on each face by the outside volume I The forces on each side of the box could have components in three directions I The stress tensor includes the nine quantities needed to describe the total force exerted on this box The Reynolds stress tensor I The Reynolds stress is given by TV ≡ ρVV (21) I TV represents the flux of momentum. Think: TV ≡ ρV V (22) |{z} |{z} Momentum density times a velocity I (ρVx )Vy is the rate at which the x component of momentum is carried in the y direction (and vice versa) The Maxwell stress tensor I The Maxwell stress is B2 BB T ≡ I − (23) B 8π 4π I Momentum is transported by the magnetic field BB I The quantity 4π is the hoop stress which resists shearing motions We can't always drop gravity! (Kulsrud, p. 77) I Gravity is often treated as a body force where g is constant I More generally, g is found using the gradient of a scalar potential, φ, g = −∇φ (24) where Poisson's equation is used to solve for the potential, r2φ = 4πGρ (25) I The gravitational stress tensor is (rφ)2 I rφrφ T = − (26) g 8πG 4πG where rφrφ is a dyadic tensor I The usual form of the force may be recovering using ρg = −∇ · Tg (27) CGL theory allows for pk 6= p? I For a scalar pressure, the corresponding part of the stress tensor is symmetric 0 p 0 0 1 Tp = pI = @ 0 p 0 A (28) 0 0 p I Chew, Goldberger, & Low (1956) allow for different perpendicular and parallel pressures in the `double adiabatic' approximation 0 1 p? 0 0 TCGL = @ 0 p? 0 A (29) 0 0 pk I Used more frequently in astrophysics than in laboratory or space plasma physics Conservation of energy I The energy equation can be written as @w + r · s = 0 (30) @t I The energy density is ρV 2 B2 p w = + + (31) 2 8π γ − 1 |{z} |{z} | {z } kinetic magnetic internal I The energy flux is ρV 2 p cE × B s = V + V + pV + (32) 2 γ − 1 |{z} 4π | {z } | {z } work | {z } kinetic internal Poynting I pV represents the work done on the plasma from −∇p forces We now have a set of three equations describing conservation of mass, momentum, and energy I Conservation of mass, momentum, & energy are given by @ρ + r · (ρV) = 0 (33) @t @ρV + r · T = 0 (34) @t @w + r · s = 0 (35) @t where the stress tensor, energy density, and energy flux are B2 BB T = ρVV + p + I − (36) 8π 4π ρV 2 B2 p w = + + (37) 2 8π γ − 1 ρV 2 γ E × B s = + p V + (38) 2 γ − 1 4π The most important property of ideal MHD is the frozen-in condition I Define S as a surface bounded by a closed curve C that is co-moving with the plasma I The magnetic flux through S is Z Ψ = B · dS (39) S dΨ I There are contributions to dt from: dΨ1 I Changes in B with S held fixed, dt dΨ2 I The flux swept out by C as it moves with the plasma, dt How does the flux change due to B changing? I Take the time derivative of the magnetic flux with S fixed: dΨ Z @B 1 = · dS dt S @t Z = −c (r × E) · dS S I = −c E · dl (40) C where we used Faraday's law and Stokes' theorem. How does the area of S change as it is swept with the plasma? V dS C2 dl C1 I The lateral area swept out by C as it flows with the plasma is dS = Vdt × dl (41) where the line element dl is tangent to C I The flux through this area is B · dS = B · Vdt × dl, so that dΨ I 2 = B · V × dl dt C I = − V × B · dl (42) C Putting it all together I The total change in magnetic flux through S is then dΨ dΨ dΨ = 1 + 2 dt dt dt I I = − c E · dl − V × B · dl C C | {z } | {z } changes in B changes in S I V × B = −c E + · dl (43) C c V×B I But E + c = 0 in ideal MHD, so that dΨ = 0: (44) dt The magnetic flux through any co-moving fluid element remains constant in ideal MHD.