1 Lorentz Force in MHD and Its Relation to the Maxwell Stress Tensor

1 Lorentz force in MHD and its relation to the Maxwell stress tensor

One of the main simplifications of the MHD approximation is the fact that the fact that the electro- magnetic force on the plasma (Lorentz force) can be expressed entirely in terms of the magnetic field, with no reference to the electric field E or the current density J.

In its ground form the Lorentz force density is

J × B f = , (1.1) L c where Faraday’s law for slow phenomena allows the elimination of the current density J through

c J = (∇ × B) . (1.2) 4π

In addition, the magnetic ﬁeld is divergence-free:

∇ · B = 0 . (1.3)

In what follows, the following vector/tensor identities may prove useful:

∇ (A · B) = A × (∇ × B) + B × (∇ × A) + (A · ∇) B + (B · ∇) A (1.4) ∇ · (A ⊗ B) = (∇ · A) B + (A · ∇) B .

A and B are two arbitrary vectors with components Ai and Bi respectively , and A ⊗ B is the dyadic ’direct product” tensor of the vectors A and B, with tensor components (A ⊗ B)ij = AiBj.

1 a. Show, by whatever means you see ﬁt (e.g. vector manipulation, writing things out in terms of components ···) and using the above equations, that the Lorentz force density can be written as

f L = −∇ · M , (1.5)

with

B2 B ⊗ B M ≡ I − (1.6) 8π 4π

the Maxwell stress tensor. The tensor I ≡ diag(1 , 1 , 1) is the 3 × 3 unit tensor.

For those who forgot: if we put x1 = x, x2 = y and x3 = z the cartesian components of the vector V ≡ ∇ · M are:

∂Mji Vi = , (1.7) ∂xj

where summation over repeated indices is implied (i.e. the Einstein summation convention). b. Consider a magnetic field that has a fixed direction, chosen along the x-axis, but varies in strength in the direction perpendicular to the field (i.e. in the z-direction):

B(x) = B(z) xˆ . (1.8)

Calculate the Maxwell stress tensor for this case, and show explicitly that the resulting Lorentz force is perpendicular to B, as the ground form (1.1) obviously demands.

2 2 A simple two-dimensional MHD equilibrium (Exam question in 2014!)

Consider a static (V = 0) and time-independent (∂/∂t = 0) MHD equilibrium with gravity. The gravitational acceleration is g = −∇Φ, with Φ(x) the gravitational potential. The magnetic ﬁeld B(x) lies in the x − z plane. The plasma has a density ρ and a pressure P . We will assume that the plasma has a constant temperature T0, so that pressure and density are related through

RT P (x) = 0 ρ(x) , (2.1) µ with R the universal gas constant and µ = m/mH the mass of the atoms in units of the hydrogen mass. The force balance for this simple equilibrium reads:

J × B ∇P = + ρ g . (2.2) c

a. Assume that gravity acts along the negative z-axis,

g = −g zˆ (2.3)

with g > 0. Also assume that locally the magnetic field is inclined at an angle θ(s) with respect to the vertical direction, see Figure 1. Here s measures the distance along a given magnetic field line. Give the component of the force-balance equation (2.2) along the magnetic field, and show that it can be written as

dP = −ρg (2.4) dz

regardless the local direction of the magnetic ﬁeld. b. Show that the density and pressure along a ﬁeld line vary as

ρ(s) = ρ0 exp(−z(s)/H) ,P (s) = P0 exp(−z(s)/H) (2.5)

and determine the constant scale height H. Here s is the length measured along the magnetic ﬁeld.

3 z

g q(s)

s z(s) B

Figure 1: The geometry as it applies to question 2: a magnetic ﬁeld B = (Bx , 0 ,Bz) in the x − z-plane is locally inclined at an angle θ(s) with respect to the vertical (z-)direction. The quantity s parametrizes the length along a given magnetic ﬁeld line. Gravity acts in the negative z-direction: g = −g zˆ.

c. The magnetic ﬁeld is divergence-free: ∇ · B = 0. Show that the magnetic ﬁeld that is generated by a scalar function A(x , z) through the relations

∂A ∂A B (x , z) = ,B (x , z) = − (2.6) x ∂z z ∂x

is always divergence-free. The function A(x , z) is known as the ﬂux function. d. Show that such a magnetic ﬁeld has a current density J equal to J(x) = J(x , z) yˆ, where

c ∂2 ∂2 J(x , z) = (∆ A) with ∆ ≡ + . (2.7) 4π 2 2 ∂x2 ∂z2

4 e. Now show the following: – The ﬂux function A(x , y) is constant along the magnetic ﬁeld; – Any solution of the force balance equation must have

∇P + ρ ∇Φ = F (A) ∇A, (2.8)

where F (A) satisﬁes

∆2A + 4π F (A) = 0 . (2.9)

5 3 A buoyant slender ﬂux tube in an isothermal atmosphere

Consider a thin strand of magnetic field lines (thin flux tube or flux rope) with a length much larger than its width. This quasi-cylindrical flux rope will serve as a simple model of a Solar Prominence, see Figure 2, a magnetically confined strand of gas suspended above the Solar surface, or of a magnetic loop connecting two small Sunspots (see Figure).

The magnetic ﬁeld inside the rope is

B = B `ˆ. (3.1)

with `ˆ a unit vector tangent to the rope’s axis. If the rope traces its trajectory in the x − z plane we can always write

`ˆ= (cos θ , 0 , sin θ) (3.2)

with θ the angle between `ˆ and the x-axis. We take the cross section to be circular, with radius a and cross sectional area πa2. The rope hangs in an atmosphere with a constant temperature T . This atmosphere is stratiﬁed in the vertical direction due to gravity with a gravitational acceleration equal to g = −g zˆ. As a result, the atmospheric pressure Pe and density ρe varies with height z as

Pe(z) = Pe0 exp(−z/H) , ρe(z) = ρe0 exp(−z/H) . (3.3)

Here

RT H = (3.4) µg

is the pressure scale height. There is no magnetic ﬁeld in the atmosphere that surrounds the ﬂux rope. As always, pressure and density are related by the ideal gas law:

ρRT P = nk T = . (3.5) b µ

In this assignment you will calculate the trajectory of this ﬂux rope. The trajectory is curved in general, but can be shown to lie in a plane, which we take to be the x − z plane. You are also asked to calculate how quantities vary along the length of the rope under the assumption that the local radius of the quasi-cylindrical rope is small and satisﬁes a H.

6 Figure 2: A large ﬂux rope above the Solar surface, as seen by the Solar Dynamics Observatory on March 30, 2010. This rope shows some twist, an eﬀect that is neglected in this calculation.

7 Figure 3: The geometry of a slender rope of magnetic ﬂux that extends from the Solar photosphere into the corona. The rope runs from foot point to foot point, and reaches a maximum height above the photosphere equal to zmax. The inclination angle of the rope trajectory (deﬁned by the tangent vector `ˆ) with respect to the horizontal is θ(z). The unit vector nˆ points towards the center of curvature, and is perpendicular to the rope’s axis. The length along the rope is measured by the parameter `, with ` = 0 at the left foot point.

The shape of the rope is determined by the balance between the buoyancy force f buoy due to the density diﬀerence between the rope and its surroundings, and the magnetic tension force f tens due to the magnetic ﬁeld that runs through the rope.

8 a. If the flux rope is sufficiently thin, the temperature of the material in the rope (internal temperature Ti) and that of the atmosphere ( external temperature T ) will quickly equilibrate. At the edge of the flux rope the total pressure (magnetic + gas pressure) must be the same on both sides. Show that this implies that the mass density inside the rope satisfies the following relation:

µB2 ρ = ρ − . (3.6) i e 8πRT

You may neglect the variation of the atmospheric pressure around the axis of the tube as a H.

Along the magnetic ﬁeld (in the direction of the rope’s axis) the Lorentz force (J × B)/c vanishes. Therefore the force balance in this direction only involves the gradient of the internal gas pressure Pi and gravity:

∂P `ˆ· ∇ P ≡ i = −ρ g = −ρ g sin θ . (3.7) i ∂` i k i

Here θ is the angle between the tangent to the tube and the horizontal direction (the x-direction), ˆ gk = g · ` is the component of the gravitational acceleration along the rope’s axis and ` is the length measured along the rope trajectory.

b. Use the geometry of the problem, the ideal gas law and the isothermal assumption Ti = T = constant to show that this equation is solved by choosing the density inside the flux rope to have the same exponential profile as quantities outside the flux rope: for a section of rope at height z one has

ρi(z) = ρi0 exp(−z/H) (3.8)

(Hint: Use that every position on the rope’s axes can be uniquely parametrized by `, but also by z: there is a one-to-one relationship between the two!)

9 c. The total magnetic ﬂux through the rope,

2 Φm = πa B (3.9)

is a conserved quantity that can not vary along the length of the (curved) tube. Show that results a and b imply that the magnetic ﬁeld and the tube radius are functions of the height z, and vary as:

B(z) = B0 exp(−z/2H) , a(z) = a0 exp(z/4H) . (3.10)

The material inside the ﬂux rope is lighter than the atmosphere surrounding the tube (result a). That leads (by Archimedes’s law) to a buoyant force per unit length in the direction perpendicular to the rope’s axis equal to

2 f buoy = πa (ρe − ρi) g⊥ . (3.11)

Here g⊥ is the component of the gravitational acceleration perpendicular to the tubes axis, formally deﬁned as

ˆ ˆ g⊥ = g − (g · `) ` = g cos θ nˆ . (3.12)

The unit vector nˆ is perpendicular to the rope’s axis, and is formally deﬁned by the relation

∂`ˆ dθ (`ˆ· ∇)`ˆ= = nˆ ≡ κ nˆ , (3.13) ∂` d` with 1/κ the radius of curvature of the rope (see the Figure again, and Chapter 3.1, Eqn. 3.1.13). This vector equals

nˆ = (sin θ , 0 , − cos θ) . (3.14)

Without a compensating force, the tube would ﬂoat up.

10 However, in this case, the buoyancy force is balanced by the magnetic tension force/unit length, which equals

B2 f = πa2 κ nˆ . (3.15) tens 4π

This force results from the fact that the rope’s trajectory is curved. The perpendicular force balance f buoy + f tens = 0 can be written as:

B2 πa2 κ + πa2 (ρ − ρ ) g cos θ = 0 . 4π e i

(3.16)

d. Use the geometry of the problem to show that

dθ dθ κ = = sin θ (3.17) d` dz

e. Use the result of question (a) and the force balance to show that the inclination angle θ satisﬁes the following diﬀerential equation:

dθ cot θ = − . (3.18) dz 2H

This equation determines the trajectory of the ﬂux rope through the Corona. f. Show that the solution of (3.18) can be represented as

cos θ(z) = (cos θ0) exp(z/2H) , (3.19)

with θ0 = θ(z = 0). g. Physically one must have cos θ ≤ 1. What does that requirement give for the maximum height zmax of the ﬂux rope?

11 h. Again using the geometry of the problem, show that a loop that leaves the Solar surface at z = 0 with an inclination angle θ0 with respect to the horizontal direction re-enters the solar surface at a distance

∆x = 4Hθ0 (3.20)

from the point-of-origin. This is the distance between the so-called foot points of the loop in the Solar photosphere.

(Hint: for symmetry reasons the inclination angle runs through the range −θ0 ≤ θ ≤ θ0 (from right to left) over the entire loop. Try to ﬁnd an equation for dx/dθ.) i. Now calculate the height zmax and the distance ∆x between the foot points of the rope for a o starting value of the inclination angle equal to θ0 = 45 .

12 4 HAND-IN ASSIGNMENT: A two-dimensional MHD equilibrium with translation symmetry in the z-direction

An example where the symmetries of the problem can be used to simplify the mathematics.

Consider a static equilibrium (ﬂow velocity vanishes everywhere) and without gravity. Then the MHD Lorentz force must be balanced by a pressure gradient. This balance can be written in various ways:

J × B (∇ × B) × B B2 (B · ∇)B ∇P = = = −∇ + , (4.1) c 4π 8π 4π

We assume that the equilibrium is invariant in the z-direction:

∂ (any physical quantity) = 0 . (4.2) ∂z

We try a solution for the magnetic ﬁeld of the form:

∂A ∂A B = ∇A × zˆ + B zˆ = , − ,B . (4.3) z ∂y ∂x z

a. Show that the trial solution satisﬁes the constraint of a divergence-free magnetic ﬁeld, ∇ · B = 0.

b. Show that the z-component of the force balance equation, together with the assumption (4.3) for the form of B and the assumed symmetry, imply that

∂A ∂B ∂A ∂B z − z = 0 . (4.4) ∂y ∂x ∂x ∂y

Hint: choose the most convenient form of the equation!

c. Show that this equation is solved if Bz is a function of A(x , y):

Bz = Bz(A) . (4.5)

Hint: think carefully about the diﬀerentiation rules for Bz in the case that Bz is a function of A alone, while A is a function of x and y!

13 d. Show that the force balance (4.1) implies that

∂P ∂P B · ∇P = B + B = 0 , (4.6) x ∂x y ∂y

and that this equation is solved if the pressure is a function of A alone:

P = P (A) . (4.7) e. Show that the force balance equations (4.1) in the x and y-direction can be satisﬁed simultane- ously if A(x , y) is the solution of

∂2 ∂2 d B2 + A(x , y) + 4π P + z = 0 . ∂x2 ∂y2 dA 8π

(4.8)