solar theory mt45100.4. integral theorems 9 0.4 integral theorems the two theorems that are commonly...

1

(Revision 2)

Solar TheoryMT4510

Dr Clare Parnell2010

Contents

0 Review of Vector Calculus. 50.1 Operators in Various Coordinate Systems. . . . . . . . . . . . .. . . . . . . . 60.2 Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80.3 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 80.4 Integral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 9

1 Maxwell’s Equations and Magnetic Fields 111.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 111.2 Electromagnetic Waves in a Vacuum . . . . . . . . . . . . . . . . . . .. . . . 121.3 Magnetic Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 13

2 MHD Equations 172.1 Electromagnetic Equations . . . . . . . . . . . . . . . . . . . . . . . .. . . . 17

2.1.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.1.2 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2 Fluid Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .182.2.1 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2.2 The Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . 192.2.3 The Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2.4 Summary of the MHD Equations: Important to know these .. . . . . . 202.2.5 General Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Magnetic Induction 233.1 The Induction Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 233.2 Induction Equation - The Diffusion Limit . . . . . . . . . . . . .. . . . . . . 25

3.2.1 Diffusion in a current sheet . . . . . . . . . . . . . . . . . . . . . .. . 253.3 Induction Equation - Frozen-in-Flux Theorem . . . . . . . . .. . . . . . . . . 29

3.3.1 Frozen-in-Flux Theorem (Alfven’s Theorem) . . . . . . .. . . . . . . 303.4 Steady State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4 Magnetic Forces 374.1 The Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37

4.1.1 Magnetic Tension Force . . . . . . . . . . . . . . . . . . . . . . . . . 384.1.2 Magnetic Pressure Force . . . . . . . . . . . . . . . . . . . . . . . . .394.1.3 Magnetic Force Balance . . . . . . . . . . . . . . . . . . . . . . . . . 40

3

4 CONTENTS

5 Magnetohydrostatic Equilibria 415.1 Hydrostatic Pressure Balance . . . . . . . . . . . . . . . . . . . . . .. . . . . 415.2 The Plasma Beta -β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.3 The Force-Free Approximation . . . . . . . . . . . . . . . . . . . . . .. . . . 435.4 Potential Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 45

5.4.1 Uniqueness of Potential Fields . . . . . . . . . . . . . . . . . . .. . . 485.5 Force-Free Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 49

5.5.1 Property ofα . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.5.2 Constantα . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.5.3 Non-constantα . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5.6 The Vector Magnetic Potential . . . . . . . . . . . . . . . . . . . . . .. . . . 515.7 Grad-Shafranov Equation for 2D MHS Equilibria . . . . . . . .. . . . . . . . 54

5.7.1 Sheared Arcades . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.8 Prominence Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 59

5.8.1 Kippenhahn and Schluter Prominence Model . . . . . . . . .. . . . . 61

6 The Solar Wind 656.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .656.2 Parker’s Solar Wind Model . . . . . . . . . . . . . . . . . . . . . . . . . .. . 656.3 Super-radial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 696.4 Polytropic Solar Wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 71

6.4.1 Nature of the Polytropic Solar Wind Solutions . . . . . . .. . . . . . 736.5 The Heliopause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

7 Magnetohydrodynamic Waves 777.1 Linearised MHD Equations . . . . . . . . . . . . . . . . . . . . . . . . . .. . 777.2 Acoustic (Sound) Waves - Basic Waves Properties . . . . . . .. . . . . . . . . 78

7.2.1 Dispersion Relations . . . . . . . . . . . . . . . . . . . . . . . . . . .817.3 Alfven Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

7.3.1 Phase and Group Speed. . . . . . . . . . . . . . . . . . . . . . . . . . 85

Chapter 0

Review of Vector Calculus.

Vector Calculus is the fundamental tool used in the formulation and analysis of problems inelectromagnetism, fluid dynmaics & magnetohydrodynamics.It is, therefore, important to re-view the basics of vectors and vector calculus. The basic definitions of quantities are givenbelow:

Scalar function - e.g.,p, ρ may be dependent on space and/or time -p(x, y, z, t), with only amagnitude (e.g., pressure or density) and written in light type in books.

Vector function - e.g.,B, v with a magnitude and direction (e.g. magnetic field, velocity)that may be dependent on space and/or time, written asB or B in books (B = Bxx + Byy +Bzz). You must identify vectors by underlining them. This is important.

Sum of vectors : Assuming thatA = (Ax, Ay, Az) = Axx + Ayy + Azz (and similarly forthe vectorB), we have

A ±B = C, (Cx = Ax ± Bx, Cy = Ay ± By, Cz = Az ± Bz).

Scalar Product :

A · B = AxBx + AyBy + AzBz =| A || B | cos θ = B · A,

whereθ is the acute angle between the two vectors. Note that this works for any orthogonalcoordinate system. Thus, in cylindrical coordinates, ifA = (AR, Aφ, Az) = ARR+Aφφ+Azz

(and similarly for the vectorB), then

A ·B = ARBR + AφBφ + AzBz.

Vector Product :A× B =| A || B | sin θ n = −B ×A

whereθ is the acute angle between the two vectors.n unit vector pointing perpendicular tothe plane containingA andB. The standard way to calculateA × B is to use the determinantmethod. Note this approach works in all orthogonal coordinate systems.

A ×B =

∣

∣

∣

∣

∣

∣

x y z

Ax Ay Az

Bx By Bz

∣

∣

∣

∣

∣

∣

= (AyBz − AzBy) x + (AzBx − AxBz) y + (AxBy − AyBx) z

5

6 CHAPTER 0. REVIEW OF VECTOR CALCULUS.

0.1 Operators in Various Coordinate Systems.

Three coordinate systems will be used throughout this course.

1. Cartesian Coordinates (x,y,z).

(a) Increment and Volume Element.

dl = dxx + dyy + dzz, (0.1)

dV = dxdydz. (0.2)

(b) Grad. (∇ acts on a scalar function to produce a vector.)This operator gives the rate of change of the scalar functionit acts on. The directionis the direction of the maximum increase of the function. Itsmagnitude gives theslope or rate of increase along this direction.

∇f =∂f

∂xx +

∂f

∂yy +

∂f

∂zz. (0.3)

(c) Div. (∇· acts on a vector function to produce a scalar.)The divergence of a vector functionA = (Ax(x, y, z), Ay(x, y, z), Az(x, y, z)) at apoint is a measure of how much the function spreads out or diverges at that point.

∇ · A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂z. (0.4)

In 1D, if A = (Ax(x), 0, 0), and the functionAx(x) is increasing, then∂Ax(x)/∂x >0, and∇ · A > 0.

(d) Curl. (∇× acts on a vector function to produce a vector.)The curl of a function is a measure of how much the function curls, rotates or twistsat the point in question.

∇ × A =

(

∂Az

∂y− ∂Ay

∂z

)

x +

(

∂Ax

∂z− ∂Az

∂x

)

y +

(

∂Ay

∂x− ∂Ax

∂y

)

z. (0.5)

(e) Laplacian of a Scalar function,f produces a scalar.

∇2f = ∇ · ∇f =∂2f

∂x2+

∂2f

∂y2+

∂2f

∂z2. (0.6)

(f) Laplacian of a Vector functionA produces a vector.

∇2A = ∇2Axx + ∇2Ayy + ∇2Azz = ∇(∇ · A) − ∇ × (∇ × A). (0.7)

2. Cylindrical Coordinates (R,φ,z).


dl = dRR + Rdφφ + dzz, (0.8)

dV = RdRdφdz. (0.9)

0.1. OPERATORS IN VARIOUS COORDINATE SYSTEMS. 7

(b) Grad.

∇f =∂f

∂RR +

1

R

∂f

∂φφ +

∂f

∂zz. (0.10)

(c) Div.

∇ · A =1

R

∂

∂R(RAR) +

1

R

∂Aφ

∂φ+

∂Az

∂z. (0.11)

(d) Curl.

∇ ×A =

(

1

R

∂Az

∂φ− ∂Aφ

∂z

)

R +

(

∂AR

∂z− ∂Az

∂R

)

φ +

(

1

R

∂

∂R(RAφ) −

1

R

∂AR

∂φ

)

z. (0.12)

(e) Laplacian of a Scalar.

∇2f =1

R

∂

∂R

(

R∂f

∂R

)

+1

R2

∂2f

∂φ2+

∂2f

∂z2. (0.13)

3. Spherical Coordinates (r,θ,φ)


dl = drr + rdθθ + r sin θdφφ, (0.14)

dV = r2 sin θdrdθdφ. (0.15)

(b) Grad.

∇f =∂f

∂rr +

1

r

∂f

∂θθ +

1

r sin θ

∂f

∂φφ. (0.16)

(c) Div

∇ · A =1

r2

∂

∂r(r2Ar) +

1

r sin θ

∂

∂θ(Aθ sin θ) +

1

r sin θ

∂Aφ

∂φ. (0.17)

(d) Curl

∇× A =1

r sin θ

(

∂

∂θ(Aφ sin θ) − ∂Aθ

∂φ

)

r +

1

r

(

1

sin θ

∂Ar

∂φ− ∂

∂r(rAφ)

)

θ +1

r

(

∂

∂r(rAθ) −

∂Ar

∂θ

)

φ. (0.18)

(e) Laplacian of a Scalar.

∇2f =1

r2

∂

∂r

(

r2∂f

∂r

)

+1

r2 sin θ

∂

∂θ

(

sin θ∂f

∂θ

)

+1

r2 sin2 θ

∂2f

∂φ2. (0.19)


0.2 Flux

The fluxF of a vector quantityB through a surfaceS is given by

F =

∫

S

B · dS (0.20)

wheredS = dSn is a vector normal to the surfaceS, ( B ·dS = BndS whereBn is the normalcomponent at the surface). By conventiondSpoints outwards from the surface when the surfaceis closed.

0.3 Vector Identities

You are expected to know and memorize the following vector identities

1. (A×B) ·C = A · (B×C) = B · (C×A) = (B×C) ·A = C · (A×B) = (C×A) ·B

2. A × (B× C) = (C × B) ×A = (A · C)B− (A ·B)C

3. ∇(fg) = f∇g + g∇f

4. ∇ · ∇f = ∇2f

5. ∇ × (∇f) = 0

6. ∇ · (∇ × A) = 0

7. ∇2A = ∇(∇ ·A) − ∇ × ∇ ×A

The following are given as a useful reference and will be usedin subsequent examples. Youwould not be expected to memorize them.

1. ∇(a/b) = (1/b)∇a − (a/b2)∇b

2. ∇ × (fA) = (∇f) × A + f(∇ ×A)

3. ∇ · (fA) = (∇f) · A + f(∇ · A)

4. ∇ × (A× B) = (B · ∇)A− (A · ∇)B + (∇ · B)A − (∇ · A)B

5. ∇ · (A× B) = B · (∇ × A) − A · (∇ × B)

6. A × (∇ × B) = (∇B) · A− (A · ∇)B

7. ∇(A · B) = (B · ∇)A + (A · ∇)B + B× (∇ ×A) + A × (∇ × B)

Example 0.3.1Simplifyk × (v × B).Using the vector identity no 2 from the list of ones you shouldmemorize, we have

k × (v × B) = (k ·B)v − (k · v)B.

This can be used repeatedly to simplify more complicated expressions such as

{k × [k × (v × B)]} × B.

0.4. INTEGRAL THEOREMS 9

0.4 Integral Theorems

The two theorems that are commonly used are :

1. Divergence Theorem∫ ∫ ∫

V

(∇ · B)dV =

∫ ∫

S

B · dS, (0.21)

wheredS = ndS andn is the outward pointing normal to the surface. The integral on theright hand side involves the normal component ofB integrated over some closed surface(S) and is therefore the flux ofB through that surface. The left hand side calculates howmuch the entire function “diverges” or spreads out within the volume (V) bounded by theclosed surface (S). The spreading out of the function withinthe volume is equal to the fluxthrough the surface.

2. Stokes Theorem∫ ∫

S

(∇ ×A) · dS =

∮

C

A · dl (0.22)

The left hand side calculates the flux of the curl ofA passing through a surface (S). Thisjust equals the line integral of function around the boundary line (C) of the surface (S).Remember the direction of the line integral is taken so that the surface is on the left.

Chapter 1

Maxwell’s Equations and Magnetic Fields

1.1 Maxwell’s Equations

Maxwell’s equations are the set of electromagnetic equations and they are given in mks units.In the followingE, B are vector functions of space and time,ρ∗ is a scalar function of spaceand time andµ, c, e andǫ are constants with specific values.Ampere’s Law

∇ × B = µj +1

c2

∂E

∂t, (1.1)

[the last term on rhs is called the displacement current.]- gradients in magnetic field create electric currents.Solenoidal constraint

∇ · B = 0, (1.2)

- indicates that there are no magnetic monopoles, i.e., no sources or sinks of magnetic field.Faraday’s law

∇ ×E = −∂B

∂t, (1.3)

- says that spatially varying electric field can induce a magnetic field.Gauss’ Law

∇ · E =1

ǫρ∗. (1.4)

- implies charge conservation.

• B - magnetic induction, but usually referred to as magnetic field. Frequently quoted inGauss (cgs unit), where 1 Tesla (mks unit) =104 Gauss.

• j - current density,

• E - electric field,

• ρ∗ = e(z+n+ − n−) - charge density, wheree - electron charge,z+ - number of chargeson the ions,n+ - number density of ions andn− - number density of electrons. [Note, in aplasma you can only get positive ions since the plasma is too hot for molecules to form -can only strip electrons.] On the Sun, most of plasma is hydrogen soz+ = 1. The numberdensities are the number of particles per unit volume and they can be functions of positionand time.

11

12 CHAPTER 1. MAXWELL’S EQUATIONS AND MAGNETIC FIELDS

• µ - magnetic permeability in a vacuum (4π × 10−7 H m−1)

• Speed of light in a vacuum -c = (µǫ)−1/2 ≈ 3 × 108 m s−1

• ǫ - permittivity of free space.

In textbooks you will also see references to the Electric DisplacementD = ǫE, and the Mag-netic FieldH = B/µ. We will not useD andH.

Maxwell’s equations are the key equations of electromagnetism with E andj the primaryvariables. However, on the Sun, they are part of a larger set of equations called themagneto-hydrodynamic equations, which describe the behaviour of the plasma and in this situation themagnetic fieldB is a primary variable.

1.2 Electromagnetic Waves in a Vacuum

In a vacuum, there are no charges and no currents so thatρ∗ = 0 andj = 0. Hence, (1.1) and(1.3) give

∇ × B =1

c2

∂E

∂t

∇ × E = −∂B

∂t

Taking the curl of the first equation and the time derivative of the second equation gives

∇ × ∇ × B =1

c2

∂

∂t(∇ ×E)

∂

∂t(∇ × E) = −∂2B

∂t2

Eliminating the electric field,E, leaves a single vector equation for the magnetic field as

∇ × ∇ × B = −∂2B

∂t2

Finally, using the vector identity in the previous section and using (1.2), we have

∇ × ∇ ×B = ∇(∇ · B) −∇2B = −∇2B,

and so

∇2B =1

c2

∂2B

∂t2. (1.5)

This is a wave equation for the magnetic induction. These waves travel at the speed of light,c.If the wave equation seems frightening, look back over your notes on MT2003 and Funda-

mentals of Applied Mathematics.

1.3. MAGNETIC FIELD LINES 13

Example 1.2.1Since the wave equation (1.5) has constant coefficients in both space and timewe can look for solutions of the form

B = B0ei(kx+ly+mz−ωt),

whereB0 is a constant vector,k, l andm are constant wavenumbers andω is a constant fre-quency. Substituting into (1.5) we have

−(k2 + l2 + m2) = −ω2

c2,

or, equivalentlyω2 = c2

(

k2 + l2 + m2)

.

This is called a dispersion relation that gives the frequency, ω, in terms of the given wavenum-bersk, l andm.

1.3 Magnetic Field Lines

On the Sun, the magnetic field outlines the structure of different phenomena and so it is impor-tant to know how to draw magnetic field lines.

If the magnetic fieldB = (Bx, By, Bz) is known [as a function of position], then the mag-netic lines of force, (magnetic field lines) are defined by

dx

Bx=

dy

By=

dz

Bz=

ds

B. (1.6)

whereB = (B2x+B2

y+B2z)

1/2. The solution to (1.6), a system of ordinary differential equations,defines a curve in three dimensional space that is a field line.In parametric form, the field linesare curves in 3D satisfying

dx

ds=

Bx

B,

dy

ds=

By

B,

dz

ds=

Bz

B, (1.7)

where the parameters is the distance along the field line.Furthermore, the spacing of the fieldlines corresponds to the magnitudeB of the field. The

closer the fieldlines the stronger the magnetic field. Also these lines must be given arrows toindicate the direction of the field.

Example 1.3.1Given magnetic fieldB = B0(y/a, x/a, 0), whereB0 and a are constants,calculate the equations of the field lines and sketch the magnetic field.

CheckB satisfies solenoidal constraint Eqn (1.2),

∇ · B =∂

∂x(B0y

a) +

∂

∂y(B0x

a) = 0

Using (1.6), the field lines are given by

dx

(B0y/a)=

dy

(B0x/a), ⇒

∫

xdx =

∫

ydy, ⇒ 1

2x2 =

1

2y2 +

1

2C


⇒ x2 − y2 = C = constant.

Therefore field lines are hyperbolae (See Fig 1.1).Plottingx2 − y2 = C with C increasing in fixed increments gives a field where the spacing

between field lines is representative of the magnitude ofB, e.g., along the liney = 0 themagnitude of the fieldB increases as|x| increases and along the linex = 0 B increases as|y|increases.

SinceB is a vector the field lines have a particular direction and so they need arrows.Consider the magnetic field alongy = 0. B = B0(0, x/a, 0) so the field is perpendicular toy = 0 and whenx > 0 the field lines point upwards and whenx < 0 they point downwards.Similarly alongx = 0, B = B0(y/a, 0, 0) so the field is perpendicular tox = 0 and wheny > 0 the field lines point to the right and whenx < 0 they point to the left.

Figure 1.1: Magnetic field lines for the fieldB(x, y, z) = (y, x, 0).

Example 1.3.2Sketch the field lines for the magnetic fieldB = B0(0, x2/a2, 0), whereB0 and

a are constants.Again, checkB satisfies solenoidal constraint,

∇ · B =∂

∂y(B0x

2

a2) = 0

Using (1.6), the field lines are given by

dx

dy=

0

(B20x

2/a2)= 0 ⇒ x = constant.

Therefore field lines are vertical lines given byx = constant.PlottingB = B2

0x2/a2 vsx (Fig 1.2a) shows us that the magnitude of the field increases as

|x| increases, hence the field lines get closer the further they are from the origin (Fig 1.2b).Finally, all field lines point upwards sinceBy > 0 for all x.

1.3. MAGNETIC FIELD LINES 15

Figure 1.2: (a) A plot ofB vsx and (b) the magnetic field lines for Example 1.3.2.

Chapter 2

MHD Equations

2.1 Electromagnetic Equations

2.1.1 Maxwell’s Equations

Ampere’s Law

∇ × B = µj +1

c2

∂E

∂t, (2.1)

Solenoidal constraint∇ · B = 0, (2.2)

Faraday’s law∂B

∂t= −∇ × E, (2.3)

Gauss’ Law

∇ · E =1

ǫρ∗. (2.4)

The MHD ApproximationThe displacement current (2nd term on right hand side) in Ampere’s law may be neglected

if the typical plasma velocities are much less than the speedof light, c.Proof

• Typical lengthscale -l0 - distance over which quantities vary - [Ify = e−x/l0 , theny variesappreciably (namely by a factor of1/e) over a distancex = l0]

• Typical timescales -t0• Typical plasma velocity -v0 = l0/t0

(2.3) =⇒|∇ × E| ≈ E

l0and

∂B

∂t≈ B

t0.

Therefore,

E =l0t0

B = v0B.

(2.1) =⇒L.H.S. : ∇ ×B ≈ B

l0

17

18 CHAPTER 2. MHD EQUATIONS

R.H.S. 2nd term :1

c2

∂E

∂t≈ 1

c2

E

t0=

v0

c2

B

t0=

v0

c2

l0t0

B

l0=

v20

c2

B

l0.

Hence, ifv20 ≪ c2 then

|∇ × B| ≫ 1

c2

∂|E|∂t

.

This is theMHD approximation, so that Ampere’s law becomes

∇ × B = µj. (2.5)

Charge NeutralityBy assuming charge neutrality, i.e.,z+n+ − n− ≪ n, wheren = n+ + n−,it can be shown that the charge densityρ∗ in (2.4) maybe neglected if the total number densityn satisfies

n ≫ ǫBv0

eL.

z+ - charge number of ions,n+ - ion number density andn− - electron number density(Proof is a homework question.)

Charge neutrality is important to ensure that the dominant electric fields in the plasma aregenerated by temporal changes in the magnetic field (2.3) rather than from a multitude of ionsand electrons (i.e., electric sinks and sources).

2.1.2 Ohm’s Law

The final electromagnetic equation is Ohm’s law,

j = σ(E + v × B), (2.6)

σ – electrical conductivity (siemens m−1).v – plasma velocity

This is a generalisation ofV = IR (voltage = currenttimesresistance) to a moving conductor.Note: it couples the electromagnetic equations to the plasma fluid equations throughv, the

plasma velocity.

2.2 Fluid Equations

Mass Continuity∂ρ

∂t+ ∇ · (ρv) = 0, (2.7)

ρ – densityStates that matter is neither created nor destroyed.Equation of Motion

ρDv

Dt= −∇p + F. (2.8)

p – pressureF – external forceNewton’s second law of motion,mass× acceleration = applied force

2.2. FLUID EQUATIONS 19

D/Dt is theconvective time derivative

D

Dt=

∂

∂t+ v · ∇,

is the time derivative as a quantity moves with the plasma.Energy Equation

D

Dt

(

p

ργ

)

= −L. (2.9)

L - total energy loss functionγ - ratio of specific heats,cp/cv (normally taken as 5/3) [cp- specific heat at constant pressure,

cv- specific heat at constant volume]Ideal Gas Law

p =ρRT

µ.

T - plasma temperatureR = 8.3 × 103J K−1kg−1 - the gas constantµ - mean atomic weight (average mass per particle in units ofmp (the mass of a proton))µ = 0.5 - for a fully ionised hydrogen plasmaµ = 1 - for a neutral plasmaHere, we will useµ = 1 and so

p = ρRT. (2.10)

(2.10) can be written in terms of the number density sinceρ = nmp so

p = nkBT, (2.11)

kB - Boltzmann’s constant

2.2.1 Continuity Equation

(2.7) can be expanded to give

∂ρ

∂t+ (v · ∇)ρ + ρ∇ · v = 0.

If plasma isincompressible[Dρ/Dt = 0] then (2.7) reduces to

∇ · v = 0.

i.e., there are no sources or sinks inv.

2.2.2 The Equation of Motion

In (2.8) the external force,F, is made up of:

• magnetic force j ×B - theLorentz force

• gravitational force ρg


• viscous force (most situations we will come across are inviscid so can ignore viscousforces)

Thus, (2.8) becomes

ρDv

Dt= −∇p + j× B + ρg (2.12)

Note: the Lorentz force couples the fluid equations to the electromagnetic equations.

2.2.3 The Energy Equation

In this course, we will only consider the adiabatic evolution of the plasma, i.e., we will assumeL = 0 (since energy losses greatly complicate the equations but tend to influence the solutionsmuch more slowly than the adiabatic expansion/compression) and, hence, (2.9) becomes

D

Dt

(

p

ργ

)

= 0. (2.13)

Adiabatic (or isentropic) means without energy loss, at constant energy.Using (2.7), (2.13) can be written as (proof is a tutorial question)

∂p

∂t+ v · ∇p = −γp∇ · v.

2.2.4 Summary of the MHD Equations: Important to know these

For this course we will use the inviscidand adiabaticform of the MHD equations.

∂ρ

∂t+ ∇ · (ρv) = 0, (2.14)

ρ∂v

∂t+ ρ(v · ∇)v = −∇p + j× B + ρg, (2.15)

∂p

∂t+ v · ∇p = −γp∇ · v, (2.16)

p = ρRT, (2.17)

j = ∇ × B/µ, (2.18)

∇ · B = 0, (2.19)

∂B

∂t= −∇ × E, (2.20)

1

σj = E + v × B. (2.21)

2.2. FLUID EQUATIONS 21

In these equations it is important to realise which quantities are constants and which arevariables. The constants (and hence are not differentiatedin general) are

g gravitational acceleration, constant for most cases, g = −gz

but follows inverse square law for the large distances in thesolar wind

γ =5

3is the ratio of specific heats and is equal to 5/3 for a fully ionised hydrogen plasma

R = 8 × 103JK−1 the gas constant

µ = 4π × 10−7Hm−1 the magnetic permeability of a vacuum

σ siemens/m the electrical conductivity, constant but the value depends on situation

The dependent variables that can depend on space and time, and hence the quantities we arenormally trying to obtain by solving the differential equations, are (for cartesian coordinates)

ρ the mass density

v = (vx, vy, vz) , the fluid velocity

p the gas pressure

j = (jx, jy, jz) , the current density

B = (Bx, By, Bz) , the magnetic induction but normally called the magnetic field

T the plasma temperature

E = (Ex, Ey, Ez) , the electric field

2.2.5 General Remarks

Different from usual fluid dynamics equations because the magnetic field introduces severaleffects:

1. it exerts aforcewhich may either move the plasma (flare eruptions, coronal transients) orguide the plasma motions (spicules and surges)

2. it providessupportandstability (coronal loops, prominences and sunspots). Many solarphenomena are observed for days and months whereas the magnetic field in lab plasmadevices becomes unstable in less than 1 second.

3. it is elasticand so can supportwaves.

4. it storesenergywhich may be released (flares, transients and coronal heating).

Chapter 3

Magnetic Induction

One of the key processes on the Sun, indeed in general plasma physics, is ‘magnetic reconnec-tion’. It is a mechanism which allows stresses within a magnetic field to dissipate and thereforeallows magnetic energy to be released. It is important, for example, in heating the Sun’s corona,powering solar flares and CMEs, and driving the solar wind.

In this chapter, we will look at the two key elements of magnetic reconnection, namelymagnetic diffusionandmagnetic advection, and determine the energy released from magneticdiffusion.

3.1 The Induction Equation

In solar MHD, we can eliminate the electric field,E, and the electric current density,j and workwith the primary variable,B. To eliminateE, we use [Ohm’s Law] (2.21) and [Ampere’s Law](2.18) to get

E = −v ×B +1

µσ∇ × B,

and so [Faraday’s Law] (2.20) becomes

∂B

∂t= ∇ × (v ×B) − ∇ × (η∇ × B), (3.1)

where

η =1

µσ. (3.2)

Assumingη is constant (it really depends onT but we treat it as constant), (3.1) becomes

∂B

∂t= ∇ × (v ×B) − η∇ × (∇ × B),

= ∇ × (v ×B) + η[∇2B − ∇(∇ · B)].

[vector identity, remember one of the ones you were going to memorise!].Using [the solenoidal constraint] (2.19) we get theinduction equation,

∂B

∂t= ∇ × (v × B) + η∇2B, (3.3)

23

24 CHAPTER 3. MAGNETIC INDUCTION

Example 3.1.1What is the dimension ofη?

L.H.S. (3.3)≈ B

t02nd term R.H.S.(3.3)≈ η

B

l20

⇒ B

t0≈ η

B

l20⇒ η ≈ l20t

−10 (m2s−1)

η = 109T−3/2 m2s−1 ≈ 1. in the solar corona(T ≈ 106 K)

Because the dimensions are of a diffusivity, we sometimes call η the magnetic diffusivity.

Magnetic Reynolds Number: An importantdimensionless parameter.Define,Rm, themagnetic Reynolds number, as the ratio of the advection and diffusion terms

in the induction equation.

Rm =∇ × (v ×B)

η∇2B,

=v0B/l0ηB/l20

,

=l0v0

η=

l20ηt0

, (3.4)

Rm is an important parameter since it is dimensionless, as can be seen from above. IfRm

is large, we can neglect the magnetic diffusion term and ifRm is small we can neglect theadvection term. This is illustrated below. On the Sun,Rm is normally very large becausel0 andv0 are typically large.

Example 3.1.2What is the magnetic Reynolds number in a sunspot?Sunspots are cooler than the coronal plasma by about a factorof 100 andη = 103 m2s−1,

l0 = 104 km, v0 = 1 kms−1 = 103 ms−1.Therefore,

Rm =107103

103= 107 ≫ 1.

If Rm ≫ 1 then the induction equation, (3.3), is approximated by

∂B

∂t= ∇ × (v × B). (3.5)

However, ifRm ≪ 1 then (3.3) reduces to

∂B

∂t= η∇2B. (3.6)

It can be helpful to think in terms of the timescales that describe the evolution of two differentphysical processes with(1) changes due to fluid motions from (3.5)

B

τmotion≈ v0

l0B ⇒ τmotion ≈ l0

v0,

3.2. INDUCTION EQUATION - THE DIFFUSION LIMIT 25

and(2) changes due to diffusion of the field from (3.6)

B

τdiffusion≈ η

B0

l20⇒ τdiffusion ≈ l20

η.

[Compare with e.g. the motion of a block of ice: changes due tomoving and changes due tomelting]

In general solar situationsτmotion ≪ τdiffusion

=⇒ τdiffusion

τmotion=

l20η

v0

l0=

l0v0

η= Rm ≫ 1

and the diffusion term is negligible. However, there are several important exceptions, for ex-ample, where length scales become small (e.g., in current sheets, during magnetic reconnectionand during solar flares).

3.2 Induction Equation - The Diffusion Limit

In this section, we takeRm ≪ 1 and study the process of pure magnetic diffusion on its own.Hence, the induction equation reduces to

∂B

∂t= η∇2B. (3.7)

This is a diffusionequation (also categorised as a parabolic partial differential equation) and soindicates that any irregularities in an initial magnetic field will diffuse away and be smoothedout. This process occurs on the diffusion timescale, namely,

τdiffusion ≈ l20/η

.

Example 3.2.1What is the diffusion time of a sunspot’s magnetic field, if the radius of thesunspot isl0 = 107 m, T = 104 K andη = 103 m2s−1?

τdiffusion ≈ 1011 seconds≈ 3, 000 years!

3.2.1 Diffusion in a current sheet

Consider the diffusion of a current sheet, where there is a vertical magnetic field of the form

B = Bz(x, t)z

Suppose there is an initial anti-parallel magnetic field with

B = Bz(x, 0)z

where

Bz(x, 0) =

{

B0, x > 0,−B0, x < 0,

(3.8)


Figure 3.1: Plot of the initial magnetic field with current lying along they-coordinate. Solid field linesare directed upwards and dashed field lines are directed downwards.

andBz(±l, t) = ±B0. (3.9)

At t = 0 this field has current equal to

j =1

µ(∇× B) =

1

µ(0,−∂Bz

∂x, 0) .

Hence, att = 0, j = 0 for all x, exceptx = 0 where it is infinite! How does this initialdiscontinuity inB atx = 0 diffuse away?

If B = Bz(x, t)z, then Eqn (3.7) only has one non-zero component, namely thez compo-nent, and the diffusion equation becomes

∂Bz

∂t= η

∂2Bz

∂x2. (3.10)

subject to the initial conditionBz = Bz(x, 0) (Equation (3.8)) and the boundary conditions,Bz(l, t) = B0 andBz(−l, t) = −B0 (Equation (3.9)).

We look for separable solutions of the formBz = X(x)T (t). Substituting into (3.10) gives

X(x)dT

dt= ηT (t)

d2X

dx2.

Dividing by X(x)T (t) gives1

T

dT

dt= η

1

X

d2X

dx2.

Since the left hand side is a function oft and the right hand side is a function ofx, this can onlybe true if they are both equal to a constant, sayλ. Thus, we generate two equations

1

T

dT

dt= λ, (3.11)

1

X

d2X

dx2=

λ

η. (3.12)

The separation constant can be either positive, negative orzero. All the possible solutions areadded together to form the final solution.

3.2. INDUCTION EQUATION - THE DIFFUSION LIMIT 27

Positiveλ would give rise to exponentially growing solutions, which are physically unreal-istic. λ = 0 is possible and generates the steady state solution. Thus, for λ = 0,

1

T

dT

dt= 0, ⇒ dT

dt= 0, ⇒ T (t) = 1

1

X

d2X

dx2= 0, ⇒ d2X

dx2= 0, ⇒ X(x) = ax + b,

wherea andb are constants.Now we add on the separable solutions for negativeλ, λ = −p2. Hence,

Bz(x, t) = ax + b +∑

λ<0

X(x)T (t).

1

T

dT

dt= −p2, (3.13)

1

X

d2X

dx2= −p2

η. (3.14)

Solving forX(x) first we have

d2X

dx2+

p2

ηX = 0, ⇒ X(x) = A cos(

p√ηx) + B sin(

p√ηx).

Solving forT (t) we have

dT

dt= −p2T, ⇒ T (t) = C exp(−p2t).

Thus the solution is

Bz(x, t) = ax + b +∞∑

p=1

(

A cos(p√ηx) + B sin(

p√ηx)

)

e−p2t.

We now determine the boundary conditions to determine the constantsa andb and values ofthe separation constants.

The boundary conditions areBz(l, t) = B0 andBz(−l, t) = −B0. Clearly, these do notdepend on time and so the steady state solution is chosen to satisfy the boundary conditionsand the values of the separation constants are chosen such that the time dependent part of thesolution is zero at the boundaries. Thus

B0 = al + b and − B0 = −al + b.

Hence,b = 0 anda = B0/l and the steady state solution is simply

B0x

l.

Since the initial state, steady state and boundary conditions are all odd functions we need an oddsolution for the time dependent solution, so the cosine terms are neglected. Thus,A = 0 and


Figure 3.2: Plot of the field lines during diffusion at timest = 0, t1, t2, t3. Solid field lines are directedupwards and dashed field lines are directed downwards.

p = nπ√

η/l wheren is an integer. [Note, one could chooseB = 0 andp = (n + 1/2)π√

η/lbut we would eventually find that all the coefficientsAn would be identically zero.]

Hence, the final solution can be expressed as

Bz(x, t) = B0x

l+

∞∑

n=1

Bn sin(nπ

lx)e−n2π2ηt/l2 .

The last stage is to determine the constantsBn from the initial conditions. Hence,

Bz(x, 0) = B0x

l+

∞∑

n=1

Bn sin(nπ

lx).

Thus,∞∑

n=1

Bn sin(nπ

lx) = Bz(x, 0) − B0

x

l.

To determineBn, we multiply bysin(mπx/l) and integrate fromx = −l to x = l. Since∫ l

−l

sin(nπx/l) sin(mπx/l)dx =

{

0, n 6= m,l, n = m,

only one term remains on the left hand side, namely the term with Bm,

Bm =1

l

∫ 0

x=−l

−B0

(

1 +x

l

)

sin(mπ

lx)

dx +1

l

∫ l

x=0

B0

(

1 − x

l

)

sin(mπ

lx)

dx.

3.3. INDUCTION EQUATION - FROZEN-IN-FLUX THEOREM 29

Using integration by parts, we find

Bm =2B0

mπ(3.15)

Thus, our final solution is

Bz(x, t) = B0x

l+

∞∑

n=1

2B0

nπsin(nπ

lx)

e−n2π2ηt/l2 (3.16)

Figure 3.2 shows plots of the magnetic fieldlines at four different times showing how diffusionsmoothes out the discontinuity atx = 0.

Figure 3.3: Plots ofBz(x, t) andj(x, t) againstx during diffusion at timest = 0, t1, t2, t3.

Plots ofBz(x, t) againstx at four different times (Figure 3.3a) also illustrates the smoothingof the discontinuity. Att = 0 there is a jump atx = 0, whereas att → ∞ (t = t3) the magneticfield is linearly dependent onx, Bz(x, t) = B0x/l.

The evolution of the current density can be calculated from (3.16) and it equals

j(x, t) =1

µ

(

0,−∂Bz

∂x, 0

)

=

(

0,−B0

l−

∞∑

n=1

2B0l cos(nπ

lx)

e−n2π2ηt/l2 , 0

)

(3.17)

Figure 3.3b shows how the infinitely thin current sheet atx = 0 andt = 0 smoothes out so thatfinally at t = t3 the current is constant across the region isjy(x, t3) = B0/l.

3.3 Induction Equation - Frozen-in-Flux Theorem

Now consider the usual solar limit,Rm ≫ 1, so that (3.3) simplifies to (3.5), namely

∂B

∂t= ∇ × (v × B).


3.3.1 Frozen-in-Flux Theorem (Alfven’s Theorem)

(Alfven, 1943) “ In a perfectly conducting fluid (Rm → ∞), magnetic field lines move with thefluid: the field lines are ‘frozen’ into the plasma. ”

Motions along the field lines do not change them, but motions transverse to the field carrythe field with them.

Before showing a general proof, consider the simpler situation where the velocity is actuallyjust a uniform flow,v = (vx, vy, vz) = constant vector. Then, we can use a vector identity tosimplify the right hand side

∇× (v × B) = (B · ∇)v − (v · ∇)B + (∇ · B)v − (∇ · v)B.

Sincev is a constant vector, the first and fourth terms on the right hand side are automaticallyzero. The third term is also zero (one of our basic equations in MHD). Hence,

∇× (v × B) = −(v · ∇)B.

Therefore, the induction equation can be expressed as

∂B

∂t+ (v · ∇)B = 0 ⇒ DB

Dt= 0.

Thus, since this is the time derivative moving with the fluid,B does not change in time as itmoves with the fluid. It is frozen to the plasma and moves with it. Now we generalise this resultto a non constant velocity.Proof

What we need to show is that the magnetic flux through a closed curve that moves with thefluid does now change in time. The flux that is initially confined by the closed curve remainsthe same even if the curve has moved.

The magnetic flux through a closed curveC bounding an open surfaceS is equal to

F =

∫

S

B · dS.

This magnetic flux can change in time in two ways such that

dF

dt=

d

dt

∫

S

B · dS = G1 + G2.

(i) Changes ofB in time produce the term

G1 =

∫

S

∂B

∂t· dS.

(ii) Also through motions ofC, the surface’s boundary, in time. Consider what happens tothe closed curveC as it moves with the plasma (See Fig 3.4). During a timeδt, an elementδlof C sweeps out an element of area

vδt × δl.

A magnetic fluxB · (vδt × δl)


C

δlδt v

δt v lx δS

Figure 3.4: An elementδl of the curveC moves with the plasma velocityv for a timeδt and sweeps outan areavδt × δl.

passes through this area, which can be rewritten, using the properties of a triple scalar product,to give

(−δtv ×B) · δl.Hence, the change in magnetic flux due to motions of the boundary is

G2 = −∮

C

v ×B · dl.

Thus, the total rate of change of F is

dF

dt=

∫

S

∂B

∂t· dS−

∮

C

v × B · dl.

Invoking Stoke’s theorem and then applying (3.5) we find

dF

dt=

∫

S

(

∂B

∂t− ∇ × (v × B)

)

· dS = 0 . (3.18)

Hence,F does not change in time, i.e.,

dF

dt=

d

dt

{∫

S

B · dS}

= 0, (3.19)

whereS is an open surface bounded byC any closed contour moving with the fluid. The mag-netic lines of force arefrozen into the fluid, in other words they areadvected with the plasma.

Example 3.3.1How does the flux within a flux tube vary when the flux tube’s areachanges?Alfven’s theorem says that the identity of a flux tube is preserved by plasma motions. Hence,

the same flux occupies the interior of the flux tube at timet2 as it did at timet1.If the area of flux tube is small, then the flux in the tube can be approximated by

B × cross sectional area= F = constant,

whereB = |B|, so thatBA = F . (3.20)

Therefore, if the area,A, is reduced by fluid motion then magnetic fieldB becomes strongerand ifA is increased thenB becomes weaker.


flux tube at time t >tflux tube defined byfield passing through C

at time t1

Figure 3.5: A flux tube at timet1 is deformed by the fluid motion at the later timet2.

Example 3.3.2Advection of a magnetic field.Consider what happens to a magnetic field of the formB = (0, 0, Bz(x, t)) that is moved

by the flowv = (−x, 0, z) in a plasma whereRm ≫ 1, assuming the initial condition,B =(0, 0, B0/(1 + x2)) at t = 0.

The magnetic field must satisfy the ideal induction equation, i.e., (3.5), namely,

∂B

∂t= ∇× (v × B).

So R.H.S.:v ×B = (0, xBz, 0)

and

∇× (v ×B) = (0, 0,∂xBz

∂x)

L.H.S.∂B

∂t= (0, 0,

∂Bz

∂t)

=⇒ ∂Bz

∂t=

∂(xBz)

∂x

=⇒ ∂Bz

∂t= Bz + x

∂Bz

∂x(3.21)

(i) To solve (3.21) use separation of variables, i.e. assumeBz(x, t) = X(x)T (t), so (3.21)becomes

XdT

dt= XT + x

dX

dxT

1

T

dT

dt= 1 +

x

X

dX

dx(3.22)

The L.H.S. of (3.22) can only equal R.H.S. if they both equal aconstant, sayλ. Therefore,

1

T

dT

dt= λ (3.23)

1 +x

X

dX

dx= λ (3.24)


(ii) First solve (3.23):∫

dT

T=

∫

λdt =⇒ T = Aeλt (3.25)

whereA is an arbitrary constant.(iii) Then solve (3.24):

∫

dX

X=

∫

λ − 1

xdx =⇒ X = Cxλ−1 (3.26)

whereC is an arbitrary constant.(iv) Therefore,

Bz(x, t) = XT = Aλxλ−1eλt.

Note,λ does not have to be single valued, but can take many values.(v) Putting in the initial conditionBz(x, 0) = B0/(1 + x2) implies that

Bz(x, 0) = B01

1 + x2=∑

λ

Aλxλ−1.

Thus, we need to expand our initial condition in terms of powers ofx,

1

1 + x2= (1 + x2)−1 = 1 − x2 + x4 − x6 + ....

Strictly speaking, this expansion is only valid forx < 1 but, in fact, it turns out this will generatethe solution to the induction equation for allx. Hence, considering different values ofλ we get:

λ = 0 =⇒ A0x0−1 = 0 A0 = 0

λ = 1 =⇒ A1x1−1 = B0 A1 = B0

λ = 2 =⇒ A2x2−1 = 0 A2 = 0

λ = 3 =⇒ A3x3−1 = −B0x

2 A3 = −B0

λ = 4 =⇒ A4x4−1 = 0 A4 = 0

λ = 5 =⇒ A5x5−1 = B0x

4 A5 = B0

Thus

Bz(x, 0) = B0

∞∑

n=0

(−1)nx(2n+1)−1

whereλ = 2n + 1. Hence,

B(x, t) = (0, 0, B0

∞∑

n=0

(−1)nx2ne(2n+1)t)

or, by comparison with the expansion for(1 + x2)−1,

B(x, t) =

(

0, 0, B0et 1

1 + (xet)2

)

.

It is straightforward to confirm thatBz(x, t) does indeed satisfy the induction equation and theinitial condition and so it is the unique solution.Let us now sketch whatB(x, t) againstx looks like at various different times.


The field atx = 0 grows exponentially,

B(0, t) = (0, 0, B0et)

but the field atx = ±1/2 decreases exponentially

B(x, t) =

(

0, 0, B0et 1

1 + e2t/4

)

≈ (0, 0, 4B0e−t)

This happens because the field is being advected towards thex-axis where it is piling up.

Figure 3.6: Magnetic advection. Plot of (a)Bz and (b)jy againstx at timest = 0, 0.25, 1 and2 (solid,dashed, dotted and dot-dashed lines).

During the process of advection the current density in the domain varies,

j(x, t) =1

µ∇ × B = (0,−∂Bz(x, t)

∂x, 0) = (0,

2B0xe3t

µ(1 + (xet)2)2, 0).

At x = 0, jy(0, t) = 0 and is held fixed at this value for all time.Initially, whent = 0 we have

jy ≈ 2B0x

µif x ≪ 1,

jy ≈ B0

2µif x = 1,

jy ≈ 0 if x ≫ 1.

and sojy has two turning points either side of zero.These turning points occur when

∂jy

∂x=

2B0e3t

µ

(

1

(1 + (xet)2)2+

(−2x)(2xe2t)

(1 + (xet)2)3= 0

)

,

or1 + x2e2t − 4x2e2t = 1 − 3x2e2t = 0,

so they occur at

x = ±xtp = ± 1√3et

.

3.4. STEADY STATE 35

Hence, ast → ∞ thenxtp → 0 and

jy(xtp) =3√

3B0e2t

8µ→ ∞ .

Finally, the total magnetic flux, at timet, is given by∫ ∞

−∞

B0et

1 + x2e2tdx = B0e

t

(

1

et

[

tan−1(xet)]∞

−∞

)

= B0π.

Note that the total flux is independent of time (just as Alfven’s theorem states).

3.4 Steady State

In the previous section, the last example illustrated how the advection caused the magnetic fieldatx = 0 to continually build up. The current also builds up in time and eventually the neglecteddiffusion term,∇2B will become important. The advection and diffusion will continue tocompete with each other as the system tries to settle down to asteady state. If there is a steadystate, it will be reached once the time derivative term is zero. Hence, for the imposed flowv = (−x, 0, z), the steady state for a vertical magnetic field satisfies

0 =∂

∂x(xBz) + η

∂2Bz

∂x2(3.27)

In actual fact the derivatives should be straight ”d”s because the steady state only depends onx. Thus,

ηd2Bz

dx2+ x

dBz

dx+ Bz = 0.

The simplest way to find a solution is to look for a series solution (in this case in powers ofx2).The series solution can be recognised as simply

Bz = CB0e−x2/2η = CB0 exp

(

−x2

2η

)

,

whereC is an arbitrary constant.C must be chosen so that the flux is the same as that evaluatedearlier. Hence,

∫ ∞

−∞

Bzdx = B0π = CB0

∫ ∞

−∞

exp

(

−x2

2η

)

= CB0

√

2ηπ.

Therefore,C =

π√2ηπ

.

Hence, the steady state solution is

Bz =

√

π

2ηB0 exp

(

−x2

2η

)

.

Note that the value atx = 0 is√

π2η

B0. Since,η is small, this is large compared to the initial

value ofB0.


(You can check the solutions satisfies the steady state equation by differentiating and substi-tuting into (3.27)).

This short section has illustrated how the three terms in theinduction equation can interactwith each other. Advection brings the magnetic field closer together, diffusion tries to spread itout and eventually a steady state can be reached.

Chapter 4

Magnetic Forces

4.1 The Lorentz Force

The Lorentz force,j × B, in the equation of motion (2.15) provides a link between thefluidequations and the electromagnetic equations. Given a prescribed flow,v, the induction equation(3.3) tells us how the magnetic field will evolve in time. AsB changes, the Lorentz forceprovides a back reaction on the plasma producing - a force that modifies the velocity.

Here, we analyse the properties of the Lorentz force and giveit some physical meaning.From [Ampere’s Law] (2.18)

j ×B =1

µ(∇ × B) ×B.

Using the vector identity (no. 14),

∇ (B ·B) = 2B × (∇ × B) + 2(B · ∇)B,

it becomes

j ×B =1

µ(B · ∇)B − ∇

(

B2

2µ

)

. (4.1)

The first term,(B·∇)B/µ, represents amagnetic tension forceand the second term ,−∇ (B2/2µ),represents amagnetic pressure force.

Note, that the Lorentz force has no component parallel to themagnetic field,

B · (j ×B) = 0 .

This implies that the components of the magnetic pressure and tension forces parallel to themagnetic field must balance,

B ·(

1

µ(B · ∇)B

)

= B ·(

∇

(

B2

2µ

))

,

and so the resulting Lorentz force must be perpendicular to the magnetic fieldB.

37

38 CHAPTER 4. MAGNETIC FORCES

4.1.1 Magnetic Tension Force

The magnetic tension force,(B · ∇)B/µ,

has magnitudeB2/µ. It appears whenever the magnetic field lines are curved.

Example 4.1.1Magnetic field under magnetic tensionConsider the magnetic field given byB(R, φ, z) = (0, B0, 0).This magnetic field consists of nested circular field lines (Fig 4.1) and its magnitude is

constant. Hence, we would expect a magnetic tension force, but no magnetic pressure force.

Figure 4.1: Magnetic fieldB(R,φ, z) = (0, B0, 0) with tension forces illustrated.

Magnetic tension force:

1

µ(B · ∇)B =

1

µ

(

BR∂

∂R+ Bφ

1

R

∂

∂φ+ Bz

∂

∂z

)

B =B0

µ

1

R

∂(B0φ)

∂φ= −B2

0

µRR

So there is a tension force acting radially inwards, as one would expect.Magnetic pressure force:

−∇

(

B2

2µ

)

= −(

∂

∂R

(

B20

2µ

)

,1

R

∂

∂φ

(

B20

2µ

)

,∂

∂z

(

B20

2µ

))

= 0

The magnetic pressure force is zero since the magnitude ofB does not vary.The Lorentz force,j× B:

B(R, φ, z) = (0, B0, 0) and j(R, φ, z) =1

µ(0, 0,

B0

R)

so

j ×B = −B20

µRR,

which confirms that the overall force is simply due to the tension of the field lines and is purelyradial.

4.1. THE LORENTZ FORCE 39

4.1.2 Magnetic Pressure Force

The magnetic pressure force,−∇

(

B2/2µ)

,

also has magnitudeB2/2µ per unit area. It occurs when the field strength,B = |B|, varies withposition. Note, from the equation of motion (2.15), that themagnetic pressure force has thesame form as the plasma pressure force,−∇p.

Example 4.1.2Magnetic field under magnetic pressureConsider the magnetic field given byB = B0(0, x/a, 0).The magnetic field lines are straight, as shown in Figure 4.2,but the magnitude, indicated

by the closeness of the lines of force, varies withx.

Figure 4.2: Magnetic fieldB = B0(0, x/a, 0) with pressure forces illustrated.

Magnetic tension force:

1

µ(B · ∇)B =

1

µ

(

Bx∂

∂x+ By

∂

∂y+ Bz

∂

∂z

)

B =B0

µ

x

a

∂

∂y(B) = 0.

As expected since the field lines are all straight.Magnetic pressure force:

−∇

(

B2

2µ

)

= −∇

(

B20x

2

2µa2

)

= −B20x

µa2x.

The magnitude ofB is strong for large|x| and gets weaker as|x| → 0 therefore the magneticpressure force is directed in towards they-axis.Lorentz force,j× B:

B = (0, B0x/a, 0) and j = (0, 0, B0/µa)

so

j ×B = −B20x

µa2x,

which shows that the overall force is purely a pressure forceparallel to thex-axis and directedin towards they-axis.

40 CHAPTER 4. MAGNETIC FORCES

4.1.3 Magnetic Force Balance

Example 4.1.3Magnetic field in equilibrium.When the magnetic tension and pressure forces balance the magnetic field is said to be in

magnetic force balance.ConsiderB = B0(y/a, x/a, 0). The field lines for this field are shown in Figure 4.3.

Figure 4.3: Magnetic fieldB = B0(y/a, x/a, 0) with the expected pressure and tension forces shown.

The magnetic tension force is given by

(B · ∇)B =B2

0

a2xx +

B20

a2yy.

Thus, ony = 0 the tension force is outwards along thex-axis and onx = 0 it is outwards alongthey-axis.

The magnetic pressure force is given by

−∇

(

B2

2

)

= −B20

a2xx − B2

0

a2yy.

Thus, ony = 0 the pressure force is inwards along thex-axis and onx = 0 it is inwards alongthey-axis.

The pressure force balances the tension force so the Lorentzforce is identically zero. This isclear fromj ×B sincej = 0.

Each small flux tube (or field line) is like an elastic band under tension. Neighbouring fluxtubes expand against each other with a pressureB2/2µ. To achieve an equilibrium either theremust be a balance between magnetic tension and magnetic pressure (magnetic force balance) orother terms in the equation of motion (2.15) must identically cancel with the magnetic forces.For instance, a plasma pressure can give a magnetohydrostatic equilibrium.

Chapter 5

Magnetohydrostatic Equilibria

Many solar phenomena are observed to remain in a steady, essentially static state for longperiods of time. In these circumstances they may be approximated by a magnetohydrostaticequilibrium solution to the MHD equations. Examples of magnetohydrostatic equilibria includecoronal loops, coronal arcades as models of active regions,sunspots and prominences.

To derive the magnetohydrostatic equations we neglect flows, so thatv ≡ 0, and assumethere is no time variation, so that∂/∂t = 0. Hence, the equation of motion becomes

0 = −∇p + j ×B + ρg, (5.1)

coupled with∇ · B = 0, j = ∇ ×B/µ, p = ρRT , (5.2)

andT satisfies an energy equation (we usually assume thatT is known).[However, before investigating any specific phenomena we consider the basic pressure bal-

ance when the magnetic field does not exert any force.]

5.1 Hydrostatic Pressure Balance

Consider a uniform vertical magnetic field. Thus,

B = B0z, g = −gz.

Hence,j = 0, (=⇒ j ×B = 0) [and so there is no Lorentz force].Pressure:p = p(z) and (5.1) becomes

dp

dz= −ρ(z)g = − g

RT (z)p(z) = − p(z)

H(z), (5.3)

where

H(z) =RT (z)

g=

p(z)

gρ(z), (5.4)

is the pressure scale height.(5.3) is a separable, 1st order ODE so that

dp

p= − 1

H(z)dz,

41

42 CHAPTER 5. MAGNETOHYDROSTATIC EQUILIBRIA

=⇒ log p = −n(z) + log p(0),

where

n(z) =

∫ z

0

1

H(u)du,

is the ‘integrated number’ of scale heights betweenz = 0 (an arbitrary level wherep = p(0))and the heightz. Therefore,

p(z) = p(0)e−n(z). (5.5)

If atmosphere is isothermal(T = const andH = const), then (5.5) gives

p(z) = p(0)e−z/H, ρ(z) = ρ(0)e−z/H , (5.6)

so that the pressure decreases exponentially on a scaleH [i.e., typical length scale given by thepressure scale heightH].

Example 5.1.1Consider typical values of the pressure scale height.N.B.: [solar gravitational constant]g = 274 m s−2 andR = 8.3 × 103 m2 s−2 K−1

1. In the photosphere,T = 6, 000 K so that

H =RT

g=

8.3 × 103 × 6 × 103

274= 182 km.

2. In the corona,T = 2 × 106 K giving

H =RT

µg=

8.3 × 103 × 2 × 106

274= 6.1 × 104 km = 61 Mm.

Extremely large - larger than the size of a typical coronal loop!

3. In the Earth’s atmosphere the gas law is slightly modified with p = ρRT/µ. HereT = 300K, g = 9.81 m s−1, µ = 29 in air and so

H =8.3 × 103 × 300

29 × 9.81= 8.8 km.

N.B. about the height of Mount Everest (8.8 km), so air pressure at the summit of Everest isabout1/e = 0.37 that of the air pressure at sea level!

5.2 The Plasma Beta -β

[If the magnetic field is now included, but gravity is neglected then we can anticipate the im-portance of the Lorentz force by comparing it with the pressure gradient.]

Consider the magnitude of the terms in (5.1) in the absence ofgravity.

∇p = j ×B → p0

l0=

B20

µl0.

5.3. THE FORCE-FREE APPROXIMATION 43

Hence, neglect [the pressure gradient term]∇p if

µp0

B20

≪ 1

and neglect [the Lorentz force]j ×B if

µp0

B20

≫ 1.

Define [the ratio of the gas pressure to the magnetic pressureas]

β ≡ gas pressuremagnetic pressure

=p0

B20/2µ

,

=⇒ β =2µp0

B20

. (5.7)

whereβ is known as theplasma beta.Therefore, ifβ ≪ 1 then neglect pressure term, ifβ ≫ 1 then neglect Lorentz force.

Example 5.2.1 1. Coronal active regions where the magnetic field is closed:

B0 = 100 G = 10−2 Tesla, µ = 4π × 10−7 Hm−1,

n0 = 1016 m−3, mp = 1.67 × 10−27kg, T0 = 2 × 106 K.

This givesρ0 = n0mp = 1.67×10−11 kg m−3 andp0 = ρ0RT0 = 0.28 Pascals. Therefore,the plasma beta is

β =2µp0

B20

= 0.007

Thus,β ≪ 1 in corona.

2. Coronal holes have a weaker magnetic field strength and a lower temperature:

B0 = 10 G = 10−3 Tesla, ρ0 = 1.67 × 10−13 kg m−3, T0 = 106 K.

Thus, in a coronal hole the plasma beta has a typical value of

β = 3.5 × 10−3,

[and is even smaller than active region value.]

5.3 The Force-Free Approximation

Consider the equation of motion (2.15)

ρ∂v

∂t+ ρ(v · ∇)v = −∇p + j× B + ρg.

Consider the magnitude of the terms in this equation


1. Neglection of gravity in favour of pressure:

p0

l0≫ ρ0g =

p0

H,

so neglect gravity [and exponential decrease in pressure with height] if

l0 ≪ H.

2. Neglection of pressure force in favour of Lorentz force:

p0

l0≪ B2

0

µl0,

so neglect [the pressure gradient term]∇p if

µp0

B20

=β

2≪ 1

3. Neglection of velocity variations in favour of Lorentz force:

ρ0v0

t0+ ρ0

v20

l0≈ 2ρ0

v0

t0≪ B2

0

µl0

so neglect velocity variations if

2v0l0t0

= 2v20 ≪ B2

0

µρ0.

Define theAlfven speedasvA = B0/√

µρ0, hence velocity variations neglected if

v0 ≪ vA.

If l0 ≪ H [we can neglect gravity],β ≪ 1[we may also neglect the gas pressure,] andv0 ≪ vA [and we may neglect velocity variations] then (2.15) reduces to theforce-free (low-βplasma) approximation, i.e.,

j× B = 0. (5.8)

and the magnetic field is called“force-free” .

Example 5.3.1 1. Coronal active regions where the magnetic field is closed:

B0 = 100G = 10−2Tesla, µ = 4π × 10−7,

ρ0 = 2 × 10−11 kg m−3, and v0 = 30 km s−1.

The Alfven speed is

vA =B0√µρ0

= 2 × 106 m s−1 = 2000 km s−1 ≫ v0 = 30 km s−1.

Thus,v0 ≪ vA in the corona.

5.4. POTENTIAL FIELDS 45

2. Coronal holes have a weaker magnetic field strength and density so

B0 = 10G = 10−3 Tesla, ρ0 = 2 × 10−13 kg m−3, v0 = 10 km s−1.

Thus, in a coronal hole the Alfven speed typical is of order

vA = 2000 km s−1 ≫ v0 = 10 km s−1,

and still we havev0 ≪ vA.

Therefore, to a good approximation the magnetic field in the solar corona is force-free since,l0 ≪ H, β ≪ 1 and v0 ≪ vA. [The next few sections will consider different force-freeequilibria.]

5.4 Potential Fields

In this section, we assume thatl0 ≪ H [so that gravity can be neglected] andβ ≪ 1 [so thatpressure gradients are unimportant] andv0 ≪ vA [so velocity variations can be ignored], thusthe magnetic field is force free and (5.8) holds

j× B = 0.

A simple solution to (5.8) is whenj = 0 [the current density is identically zero] and the mag-netic field is called‘potential’. Ampere’s law then gives,

∇ ×B = 0, (5.9)

[from vector identity no 7]. The most general solution to (5.9) is

B = ∇f, (5.10)

wheref(x, y, z) is the scalar magnetic potential. Substituting (5.10) into the solenoidal condi-tion

∇ · B = 0,

gives∇2f = 0, (5.11)

[from vector identity no 6]. So potential magnetic fields satisfy Laplace’s equation. Solutionsto (5.11) can be obtained by (i) separation of variables and (ii) complex variable theory. [Sincethis is a linear partial differential equation, solutions may be obtained by standard mathematicalmethods. For example, separation of variables may be used inboth two and three dimensionalproblems and complex variable theory may be used for two dimensional problems.]

Example 5.4.1Potential coronal arcade.[In the photosphere, magnetograms show that there are regions of opposite polarity that are

separated by a magnetic polarity inversion line. When the magnetic field joins the opposite po-larities across this inversion line the field forms a “coronal arcade”. These arcades of magneticfield lines are clearly seen in the soft x-ray images of the Suntaken by, for example, the Yohkohsatellite (Fig 5.1).] To model the field structure of a coronal arcade we assume that the length


Figure 5.1: Soft x-ray images of the solar corona showing coronal arcades.

of the arcade is much greater than the width so that the arcadefield only depends on two spatialcoordinates with variations along its length being neglected. Thus,∂/∂y = 0.

To model a potential coronal arcade we use separation of variables to find the [two-dimensional]potential magnetic field inxz-plane that satisfies the following boundary conditions

Bz → 0 as z → ∞ Bx(±l/2, z) = 0, Bz(x, 0) = G(x) (5.12)

Assuming that the functionG(x) is odd aboutx = 0, we have the additional condition

Bz(0, z) = 0.

Since field is potential use (5.11) and substitutef = X(x)Z(z) into (5.11):

X ′′Z + XZ ′′ = 0,

X ′′

X= −Z ′′

Z= constant= −k2.

Hence,Z ′′ = k2Z ⇒ Z(z) = Ae−kz + Bekz.

andX ′′ = −k2X, ⇒ X(x) = C sin kx + D cos kx,

whereA, B, C andD are arbitrary constants. Now apply the boundary conditions:

5.4. POTENTIAL FIELDS 47

Bz =∂f

∂z‖x=±l/2 = X(x)

(

−Ake−kz + Bkekz)

→ 0 as z → ∞ so B = 0.

Note that sinceBz(0, z) = 0, this implies thatX(0) = 0. Thus, the constant,D, must bezero to eliminate the cosine term inX(x). Now use the boundary conditions,Bx(±l/2, z) = 0,to determine the separation constantk.

Bx(±l/2, z) =∂f

∂x= Ae−kz(Ck cos(±lk/2) = 0

⇒ k =(2n + 1)π

l,

and the full solution tof is obtained by summing over all the possible solutions giving

f(x, z) =∑

n

An sin(2n + 1)πx

le−(2n+1)πz/l,

whereAn = AC.SinceB = ∇f we have

Bx(x, z) =∑

n

An(2n + 1)π

lcos

(2n + 1)πx

le−(2n+1)πz/l,

Bz(x, z) = −∑

n

An(2n + 1)π

lsin

(2n + 1)πx

le−(2n+1)πz/l

Finally, to calculateAn we use the boundary condition onz = 0, namely

Bz(x, 0) = G(x) = −∑

n

An(2n + 1)π

lsin

(2n + 1)πx

l.

Furthermore, if we prescribe the normal component ofB along the base to be, for example,Bz(x, 0) = G(x) = −B0 sin(πx/l) then we have only one Fourier component and all thecoefficients in the summation are zero except for the first one,which =⇒ An = 0, ∀ n ≥ 2 andA1 = B0l/π.

Thus, the scalar potential is

f(x, z) =B0l

πsin

πx

le−πz/l.

andBx(x, z) = B0 cos

πx

le−πz/l, Bz(x, z) = −B0 sin

πx

le−πz/l

FieldlinesThe fieldlines for this coronal arcade are given by

dx

Bx=

dz

Bz=⇒ −

∫

sin(πx/l)

cos(πx/l)dx =

∫

dz ,


thus, using the substitutionu = cos(πx/l), the equations of the field lines are

z =l

π

∫

1

udu =

l

πln cos

πx

l+ C .

That is they are given by plotting various different values of C where

C = cosπx

le−πz/l .

The field lines for such a potential arcade are drawn in Fig(5.2).

Figure 5.2: Magnetic field lines for a potential coronal arcade withl = 4, (k = π/4).

Clearly, there are many solutions to the equation∇2f = 0 and thus there are many possiblepotential fields. However, for a given set of boundary conditions, the solution isunique.

5.4.1 Uniqueness of Potential Fields

The potential solution inside a volumeV is unique when the normal component of the magneticfield is imposed on the boundary.

ProofConsider a closed volumeV surrounded by surfaceS, wheren is the vector normal to this

surface.Let B1 andB2 be two different potential magnetic fields that satisfy the same boundary

conditionn · B = g(x, y, z) onS. Hence,

B1 · n = B2 · n = g(x, y, z).

SinceB1 andB2 are potential we may write

B1 = ∇f1 and B2 = ∇f2,

hence,∇2f1 = 0 and ∇2f2 = 0

from the solenoidal constraint andn · ∇f1 = n · ∇f2 = g(x, y, z) on the boundary surfaceS.

5.5. FORCE-FREE FIELDS 49

Let us define a new potential magnetic field

B = B1 −B2 = ∇(f1 − f2) = ∇f .

Now f satisfies the equation

∇2f = 0,

together with the boundary condition

n · ∇f = 0

onS.Now consider the integral

∫

V

B2 dV =

∫

V

∇f · ∇f dV.

From a vector identity, we have∇f · A = ∇ · (fA) − f(∇ · A) .Hence, withA = ∇f , we have

∫

V

B2 dV =

∫

V

∇ · (f∇f) − f∇2f dV

=

∫

S

f∇f · n dS

=

∫

S

fB · n dS = 0,

since

B · n = B1 · n− B2 · n = 0.

Note that we have used∇2f = 0 and the Divergence Theorem to relate the volume integralto a surface integral. Since the integrand of the LHS can never be negative, the integral canonly ever be positive,unlessB2 = 0, ⇒ B1 = B2. Therefore, there is only one potentialmagnetic field solution to the specified boundary conditions.

5.5 Force-Free Fields

As in Section 5.4, we assume thatl0 ≪ H, β ≪ 1 and v0 ≪ vA and we again have [theforce-free field equation] (5.8)

j ×B = 0

If the magnetic field isnot potential then the general solution is where [the current density isparallel to the magnetic field]j || B. Thus,

µj = αB, ⇒ ∇ × B = αB, (5.13)

for somescalarα which may be a function of position [and time].


5.5.1 Property ofα

The scalar functionα(r) is not completely arbitrary.∇ · (∇ × B) = 0, [vector id no 9] andtherefore from (5.13)

∇ · (∇ ×B) = ∇ · (αB)

= α∇ · B + B · ∇α

= 0 .

[vector id no 5]. Hence, after applying the solenoidal constraint,

B · ∇α = 0, (5.14)

so thatα is constant along each field line (although it may vary from field line to field line). Ifα = 0, then the magnetic field is potential.

5.5.2 Constantα

If α is constant everywhere (α = constant= α0) then

∇ ×B = αB =⇒ ∇ × (∇ ×B) = ∇ × (αB) = α(∇ × B) = α2B.

However,∇ × (∇ × B) = ∇(∇ · B) −∇2B [vector id no 11] and so

∇2B = −α2B. (5.15)

Such a magnetic field is known as a linear force-free fieldor a constant-α force-free field. [Thisis a Helmholtz equation that is linear and may be solved by theusual mathematical methods.]

5.5.3 Non-constantα

If α is a function of position, i.e.α(r), then we have

∇ × (∇ ×B) = ∇ × (αB) = α∇ ×B + ∇α × B

= α2B + ∇α × B

[vector id no 8].Hence, we get two coupled equations forB andα, namely

∇2B + α2B = B × ∇α, (5.16)

and

B · ∇α = 0. (5.17)

Such a magnetic field is known as a non-linear force-free fieldor a non-constant-α force-free field.[Magnetic fields of this form will be considered in the advanced Sun course.]

5.6. THE VECTOR MAGNETIC POTENTIAL 51

5.6 The Vector Magnetic Potential

In Section 5.4, we saw that a potential magnetic field could berepresented in terms of a scalarpotential,f(x, y, z) whereB = ∇f(x, y, z).

Here, we will use the vector magnetic potential, A, which we define to be,

B = ∇ × A, (5.18)

such that (5.18) automatically satisfies [the solenoidal condition],∇ ·B = 0. [Hence, the vectormagnetic potential is often used instead ofB in numerical codes.] It is also used in a number ofsolar models.

Example 5.6.1Constant-α Coronal ArcadeHere, we work with the vector magnetic potentialA.As for the potential coronal arcade we assume that all variables only depend on the spatial

coordinatesx andz such that∂/∂y = 0 and we assume the same conditions as for the potentialarcade, namely,

Bz → 0, as z → ∞ Bx(±l/2, z) = 0, Bz(x, 0) = G(x) (5.19)

From (5.18), withA(x, z), we have

B =

(

−∂Ay

∂z,∂Ax

∂z− ∂Az

∂x,∂Ay

∂x

)

=

(

−∂Ay

∂z, By(x, z),

∂Ay

∂x

)

. (5.20)

We leave they-component ofB asBy sinceAx andAz do not appear in the other componentsof B, and one unknown is better than two!

From (5.15) we have∇2B + α2B = 0,

so

∇2

(

−∂Ay

∂z

)

− α2 ∂Ay

∂z= − ∂

∂z

(

∇2Ay + α2Ay

)

= 0, (5.21)

∇2By + α2By = 0, (5.22)

∇2

(

∂Ay

∂x

)

+ α2∂Ay

∂x=

∂

∂x

(

∇2Ay + α2Ay

)

= 0. (5.23)

(5.21) and (5.23) together imply

∇2Ay + α2Ay = constant = 0.

We can set the constant equal to zero without loss of generality.Using separation of variables we setAy(x, z) = X(x)Z(z) to give

∇2Ay + α2Ay = 0 =⇒ d2X

dx2Z + X

d2Z

dz2+ α2XZ = 0,


Therefore,1

X

d2X

dx2= − 1

Z

d2Z

dz2− α2 = −k2.

Hence,d2Z

dz2= (k2 − α2)Z =⇒ Z = Ae−mz + Bemz,

wherem =√

k2 − α2 and

d2X

dx2= −k2X =⇒ X = C cos(kx) + D sin(kx).

Applying the boundary conditions (5.19) gives:

Bz =∂Ay

∂x=

dX

dx

(

Ae−mz + Bemz)

→ 0 as z → ∞ so B = 0

Bx(±l/2, z) = −∂Ay

∂z= −dZ

dz(C cos(±lk/2) + D sin(±lk/2)) = 0

so if k =nπ

l, thenD = 0,

The final boundary condition prescribes the normal component of B along the base,Bz(x, 0) = G(x) = −B0 sin(πx/l), hence,

Bz(x, 0) =∂Ay

∂x= An

nπ

lsin

nπx

l= −B0 sin(πx/l)

=⇒ n = 1 and A1 = − lB0

π= −B0

k,

Leaving the equation in terms ofk, wherek = π/l, andm, wherem =√

k2 − α2, for easeof writing thenAy is

Ay(x, z) = −B0

kcos kxe−mz ,

and

B(x, z) =

(

B0m

kcos kxe−mz , By(x, z),−B0 sin kxe−mz

)

.

We still have to findBy(x, z). Since the magnetic field is a constant-α force-free field∇ ×B = αB, therefore

By =1

α

(

∂Bx

∂z− ∂Bz

∂x

)

=1

(k2 − m2)1/2

(

−m2B0

kcos kxe−mz + kB0 cos kxe−mz

)

=B0(k

2 − m2)

k(k2 − m2)1/2cos kxe−mz

=B0

k(k2 − m2)1/2 cos kxe−mz .

5.6. THE VECTOR MAGNETIC POTENTIAL 53

Thus, the magnetic field is

B =B0e

−mz

k

(

m cos kx,(

k2 − m2)1/2

cos kx,−k sin kx)

, (5.24)

withα =

(

k2 − m2)1/2

.

FieldlinesThe equations of the field lines are given by

dx

Bx

=dy

By

=dz

Bz

In thexz−plane,

⇒ dx

m cos kxe−mz=

dz

−k sin kxe−mz,

⇒∫

ksin kx

cos kxdx = −

∫

mdz

⇒ − log | cos kx| = −mz + constant,

⇒ cos kx = Cemz,

⇒ C = cos kxe−mz.

Compare with the expression for the field lines in Example 5.4.1. The main difference is theminstead of ak in theexp(−mz) term which stretches or compresses the loops (See Fig 5.3).In thexy−plane, [the projections of the field lines onto the photospheric surface are given by]

dx

m cos kxe−mz=

dy

(k2 − m2)1/2 cos kxe−mz

⇒∫

dy =

∫

(k2 − m2)1/2

mdx,

⇒ y =

(

k2

m2− 1

)1/2

x + C.

Thus, the projections onto the photosphere are just straight lines with gradient(k2/m2 − 1)1/2

as shown in Fig 5.3.Figure 5.4 shows two three dimensional views of the magneticfield from different angles.

The first view looks sideways on at the loops which look like squashed loop like structures. Thesecond view reveals a complex mass of loops very much like those seen in EUV images of thecorona.

Figure 5.5 shows thexz andxy projection of the field lines for a different larger, value ofalpha. Note how the field lines are now more sheared and the nested loop are made up of higherarches.


Figure 5.3: Magnetic field lines for a constantα nested arcade withk = π/4 andm = π/8 such thatα =

√3π/8. The dashed lines in thexz-plane show the coronal loops found in the potential solution for

the same boundary conditions.

-4

-2y(s)

0

2

4-2

-1

0x(s)1

0 2

1

2z(s)

3

4

x(s)

-4-2

y(s)

0240

1

2z(s)

3

4

Figure 5.4: 3D views of the magnetic field for the constantα nested arcade shown in Fig 5.3 withk = π/4 andm = π/8 andα =

√3π/8.

5.7 Grad-Shafranov Equation for 2D MHS Equilibria

Coronal arcades and prominences are examples of two dimensional equilibria with variationsin the y direction ignored,∂/∂y = 0. Expressing the magnetic field in terms of the vectormagnetic potential we have,

B = ∇ × A, (5.25)

whereA(x, z) is written as

A = (Ax(x, z), A(x, z), Az(x, z)) .

Thus, the magnetic field is expressed in general as

B =

(

−∂A

∂z, By,

∂A

∂x

)

, (5.26)

5.7. GRAD-SHAFRANOV EQUATION FOR 2D MHS EQUILIBRIA 55

Figure 5.5: The magnetic fields for a constantα arcade with anα =√

15π/16, k = π/4 andm =π/16.

with

By(x, z) =∂Ax

∂z− ∂Az

∂x. (5.27)

Notice that∇ · B = 0 is satisfied immediately. The current is

j =1

µ∇ × B =

1

µ

(

−∂By

∂z,−∇2A,

∂By

∂x

)

, (5.28)

where

∇2A =∂2A

∂x2+

∂2A

∂z2.

If the magnetic field is force-free then∇ × B = αB and so we have the three equations

−∂By

∂z= −α

∂A

∂z, (5.29)

−∇2A = αBy, (5.30)

∂By

∂x= α

∂A

∂x. (5.31)

Combining (5.29) and (5.31) to eliminateα we obtain

∂A

∂z

∂By

∂x− ∂A

∂x

∂By

∂z= J(By, A) = 0, (5.32)

whereJ is the Jacobian ofBy andA. When the Jacobian vanishes identically the most generalsolution is

By = By(A), (5.33)

so thatBy is an arbitrary function ofA.Note, (5.32) is equivalent to

∇A(x, z) × ∇By(x, z) = 0 ,

which implies that∇By = λ∇A. In other words,

∇By =dBy

dA∇A ,


Figure 5.6: Magnetic field imposed on the base (photosphere)in Example 5.7.1.

and so solving forBy we haveBy = By(A) .From (5.31) we have

∂By

∂x=

dBy

dA

∂A

∂x,

and so

α =dBy

dA. (5.34)

Thus the solutions to the Grad-Shafranov equation will, in general, be non-linear force-freefields.

Using (5.30) we can now find a nonlinear equation forA, namely

−∇2A = BydBy

dA=

d

dA

(

1

2B2

y(A)

)

(5.35)

The Grad-Shafranov equation for two dimensional magnetohydrostatic equilibria is then

∇2A +d

dA

(

1

2B2

y(A)

)

= 0, (5.36)

and

B =

(

−∂A

∂z, By(A),

∂A

∂x

)

. (5.37)

Example 5.7.1Non-linear force-free model of a coronal arcade and prominenceUsing the Grad-Shafranov equation for 2D magnetohydrostatic equilibria, find the non-

linear force-free field which satisfies the following conditions,

By = λe−A ,

with the vertical component of the magnetic field defined on the boundary as

Bz(x, 0) =2x

1 + x2.

This base boundary condition has been plotted in Figure 5.6 to help visualise it.

5.7. GRAD-SHAFRANOV EQUATION FOR 2D MHS EQUILIBRIA 57

In this case, the Grad-Shafranov equation becomes

∇2A = λ2e−2A,

which is a non-linear, elliptic, partial differential equation for A(x, z).This is not trivial to solve, however, let us assume that the field lines are circular so that

A = A(r) and r2 = x2 + (z − z0)2, x = r cos θ, z − z0 = r sin θ. Changing to the polar

coordinates we then find

∇2A =1

r

d

dr

(

rdA

dr

)

,

since we are assuming that there is noθ dependence inA. Thus, the equation we need to solveis

1

r

d

dr

(

rdA

dr

)

= λ2e−2A.

This is still not easy to solve, however, if we now consider the vertical component of themagnetic field along the base which is

Bz =∂A

∂x=

2x

1 + x2,

which translates into a boundary condition forA by integrating with respect tox to give

A(x, 0) = log(1 + x2).

Therefore, we try a solution of the form

A(x, z) = log(b2 + r2) ,

orA(x, z) = log

(

b2 + x2 + (z − z0)2)

.

Substituting into the Grad-Shafranov equation we find that we can have a solution provided

b2 =λ2

4.

ThenA(x, 0) implies that

b2 + z20 = 1, ⇒ z2

0 = 1 − λ2

4.

Hence,

z0 = ±√

1 − λ2

4.

From this result we see that there are two possible solutionsto the Grad-Shafranov equationthat have the same boundary condition, providedλ < 2. There are no possible solutions forλ > 2. This is a feature of non-linear equations.

The complete magnetic field can now be calculated fromA(x, z) and equals

B(x, z) =

( −2(z − z0)

b2 + x2 + (z − z0)2,

λ

b2 + x2 + (z − z0)2,

2x

b2 + x2 + (z − z0)2

)

,


y(s)0

-1

4

1

0

z(s)

x(s)

3

2

-212 z(s)-2

-1

2

x(s)

-1

-2

2

1

y(s)

1 0

0

-2

-1

y(s)

0

1

2 -2

-1

0x(s)

10 2

1

2z(s)

3

4

-202

1

-11

2z(s)

00y(s)x(s)

3

1-1

4

2-2

Figure 5.7: The magnetic field for the non-linear force-freefield determined in Example 5.7.1 withλ = 1 andz0 = −

√3/2. The field is viewed (a) in thexz-plane, (b) in thexy-plane, and (c) & (d) from

different angles to reveal the true 3D nature of the field.

with

α =dBy

dA= −λe−A = −By .

From Figure 5.7 in which we have taken the negative root forz0 we see that the field linesproduce a coronal arcade. However, in Figure 5.8, the positive root has been taken forz0 andwe find that nested under an arched loop system there lies a series of helical flux tubes thatcould potentially be used to model a prominence.

5.7.1 Sheared Arcades

The magnetic field lines in Example 5.6.1 and Example 5.7.1 are known asshearedsince thefootpoints of the field lines are displaced in they direction. In Example 5.6.1α is constant and

5.8. PROMINENCE EQUILIBRIA 59

y(s)

4

x(s)

0

3

-1

z(s)

1

01

2

-22 z(s)2 -102

-2

1

x(s)

-2

-1

1

0y(s)

-2-20

1

-1

2

-1

z(s)

3

4

00

x(s)y(s) 1

12

2

2 1

x(s)

0 -1 -2-2-1

y(s)

0120

1

2

3

z(s)

4

Figure 5.8: The magnetic field for the non-linear force-freefield determined in Example 5.7.1 withλ = 1andz0 =

√3/2. The field is viewed (a) in thexz-plane, (b) in thexy-plane, and (c) & (d) from different

angles to reveal the true 3D nature of the field.

the shear on each field line is the same: it has uniform shear. i.e. when looked on from abovethe field lines appear as parallel slanted lines, e.g., Figure 5.3b.

If, however, we have a non-constantα force-free field then the shear is different on eachfield line such that the magnetic field vector points in different directions at different heightssuch as in Figure 5.7 and Figure 5.9.

5.8 Prominence Equilibria

A quiescent prominence is a cool, dense sheet of plasma in thecorona that lasts for manydays (1 – 200 days). It is seen on the disk as a thin dark filamentabove a magnetic polarityinversion line. Being cool and dense it has no right to be in the hot tenuous coronal plasmaand it must be supported against the force of gravity by some mechanism. As with nearly all


x

y

z

z

x

x

y

Figure 5.9: (a) A typical non-constant-α coronal arcade made up of nested magnetic flux surfaces. (b)The arcade viewed end on. (c) The arcade viewed from above showing the projection of the magneticfield lines onto the photospheric surface.

++

+

++

++

+

−

−−−

−−

−−

prominence

50 x10 Mm

200 x10 Mm

6 x10 Mm6

6

6

prominence

Figure 5.10: (a) Sketch of how a prominence lies with respectto the line-of-sight magnetic field. (b)Sketch of the typical dimensions of a prominence.

coronal phenomena, it is the magnetic field that is responsible for the support of prominences.

Table 5.1: Prominence and coronal propertiesProminence properties Coronal properties

L ≈ 200 Mm L ≈ 50 − 100 Mmh ≈ 50 Mmw ≈ 6 Mm

T ≈ 5, 000 − 10, 000 K T ≈ 1 − 3 × 106 Kρ0 ≈ 2 × 10−10kgm−3 ρ0 ≈ 2 × 10−12kgm−3

pressure scale heightH ≈ 0.6Mm H ≈ 120Mm

Prominences have considerable “fine structure”; [which maybe a result of either the for-mation process or small-scale instabilities], “slow draining motions” of about5 km s−1, “feet”[that stretch down to the photosphere in the neighbourhood of supergranular boundaries]. They[are dynamic at the end of their lifetime and can] erupt, disappearing as coronal mass ejections


or sometimes reforming in the same place. Oscillatory motions have recently been observed [inthem and the periods of oscillation tell us something about the internal structure of the promi-nence.]

[Prominences pose us many questions that we are only now beginning to answer. Why dothey form? How are they supported? Why are they stable for long periods of time? Why dothey erupt? What are the oscillations? What are the flows? To date only partial answers areknown to some of these questions.]

Example 5.7.1, which has helical field, has a number of the characteristics required for aprominence model, such as a helical magnetic flux tube anchored under a series of coronalloops. However, since the model is force-free it is not able to support the dense plasma thatmakes up a prominence.

In this section, we consider how prominences may be supported. The equilibrium equationsare

0 = −∇p + j× B + ρg, (5.38)

j =1

µ∇ × B, (5.39)

∇ · B = 0, (5.40)

andp = ρRT , (5.41)

and the temperature is given by an energy equation.

5.8.1 Kippenhahn and Schluter Prominence Model

[The simplest model of the internal structure of a prominence is the Kippenhahn and Schlutermodel (Kippenhahn and Schluter, 1957).] The temperature is assumed to be a constant

T = T0.

Since the width≪ height≪ length, we assume that inside the prominence we may neglectvariations in they andz directions, i.e, assume∂/∂y = ∂/∂z = 0, and only consider variationsin the horizontalx direction along the length of the prominence. Furthermore,we assume onlythe height varies inx, hence,

B = (Bx0, By0, Bz(x)) , p = p(x), ρ = ρ(x), (5.42)

whereBx0 andBy0 are constants. Assume the boundary conditions are such thatthe densitytends to zero (and hence, because the temperature is constant, the pressure also tends to zero)and thatBz tends to a constant valueBz0 as we move away from the prominence. To have a dipin the magnetic field lines (so that magnetic tension can balance gravity),Bz must change signatx = 0. Thus,

p → 0 asx → ±∞ (5.43)

Bz → Bz0 asx → +∞ (5.44)

Bz → −Bz0 asx → −∞ (5.45)

[Obviously the solenoidal condition is satisfied for the magnetic field]

∇ · B =∂Bz

∂z≡ 0, j =

1

µ∇ × B =

1

µ(0,−dBz

dx, 0).


Now consider thex andz components of the force balance equation (5.38),

xdp

dx= −Bz

µ

dBz

dx, (5.46)

z 0 =Bx0

µ

dBz

dx− ρg. (5.47)

Solving (5.46) we get

p +B2

z

2µ= constant. (5.48)

From the boundary conditions the constant is found

p(|x| → ∞) +B2

z (|x| → ∞)

2µ=

B2z0

2µ.

Thus,

p =1

2µ

(

B2z0 − B2

z

)

. (5.49)

Now we consider thez-component (5.47). Since the temperature is constant we mayuse thegas law, (5.41) to eliminate the density in favour of the pressure to obtain from (5.47)

Bx0

µ

dBz

dx=

pg

RT0

=p

H, (5.50)

where the pressure scale height,H = RT0/g. Substituting (5.49) into (5.50) gives

Bx0

µ

dBz

dx=

1

2µH

(

B2z0 − B2

z

)

,∫

dBz

B2z0 − B2

z

=x

2Bx0H+ constant,

1

Bz0

tanh−1

(

Bz

Bz0

)

=x

2Bx0H+ C,

Bz = Bz0 tanh

(

Bz0

2Bx0

x

H+ C

)

.

From symmetry atx = 0, Bz(0) = 0 which givesC = 0. Therefore,

Bz = Bz0 tanh

(

Bz0

2Bx0

x

H

)

. (5.51)

and the pressure from (5.49) is

p =B2

z0

2µsech2

(

Bz0

2Bx0

x

H

)

. (5.52)

SinceT0 is constant [the temperature is constant and so the density is given by the gas law as]

ρ =1

RT0

B2z0

2µsech2

(

Bz0

2Bx0

x

H

)

. (5.53)


Figure 5.11: (a) The vertical magnetic field component,Bz, as a function of the horizontal distance,x.(b) The gas pressure,p, as a function of the horizontal distance,x.

Figure 5.12: The magnetic field lines of the Kippenhahn-Schluter prominence model.

The profile ofBz andp are shown in Figure 5.11.Fieldlines

Equation of the field lines for the Kippenhahn-Schluter prominence model.

dx

Bx=

dz

Bz,

⇒∫

Bz0

Bx0

tanh

(

Bz0

2Bx0

x

H

)

dx = z + c,

[and so integrating we obtain]

2H log

{

cosh

(

Bz0

2Bx0

x

H

)}

= z + c.

The field lines are shown in Figure 5.12. [Note, how the magnetic field lines are bent and themagnetic tension force opposes the force due to gravity. In addition, the magnetic pressureis higher away from the centre of the prominence and so there is a magnetic pressure actingtowards the centre that compresses the plasma and opposes the outward pressure gradient.]

Example 5.8.1Non-isothermal extension to K-S.


If we assume that the temperature is no longer uniform, but depends on the horizontal coor-dinate,T = T (x), then (5.49) remains the same and (5.50) becomes

Bx0

µ

dBz

dx=

1

2H(x)

(

B2z0 − B2

z

)

.

Thus, the equation is still separable and the left hand side gives the same integral. The onlydifference is that the right hand side does not simply integrate tox but instead gives

1

2Bx0

∫

dx

H(x)=

l(x)

Bx0

Thus, (5.51) becomes

Bz = Bz0 tanh

(

Bz0

2Bx0l(x)

)

.

Chapter 6

The Solar Wind

6.1 Introduction

Geomagnetic storms were first noticed in the 19th Century as periods during which the Earth’smagnetic field suddenly increased by about10−3G and then slowly decayed. Normally geomag-netic storms occur one or two days after a large solar flare. Inaddition, it was noticed that flaresand geomagnetic activity seemed to have the same 11 year periodicity. Thus, the implicationwas that there was an electrical connection between the Earth and the Sun.

6.2 Parker’s Solar Wind Model

Parker (1958) suggested that the corona could not remain in static equilibrium but must becontinually expanding, since the interstellar pressure cannot contain a static corona. Why? Thesimplest demonstration that the hot corona cannot be in hydrostatic balance follows.

Assume that the corona is maintained at a constant temperature (isothermal),T0. Thenhydrostatic balance in spherical coordinates, assuming radial symmetry and that

g = −GM◦/r2r ,

means that the radial component of the equation of motion (with v = 0 and∂/∂t = 0) gives,

dp

dr= −ρg = −ρ

GM◦

r2,

HereG is the gravitational constant andM◦ is the mass of the Sun. The inverse square lawmeans that the effect of gravity drops off asr−2. Using the gas lawp = ρRT0, we can eliminateρ and get

dp

dr= − p

RT0

GM◦

r2= − p

c2si

GM◦

r2,

where the isothermal sound speed is defined byc2si = p/ρ = RT0. Finally, we can simplify the

appearance of the equation further by defining

rc =GM◦

2RT0

=GM◦

2c2si

.

65

66 CHAPTER 6. THE SOLAR WIND

Thus, we have the separable first order differential equation for the pressure as

dp

dr= −p

2rc

r2∫

dp

p= −2rc

∫

dr

r2.

Integrating we have

log p = 2rc

r+ C

The constantC is determined by applying the condition that the pressure isp0 at the base of thecorona wherer = RO, the radius of the Sun. Therefore,

C = log p0 − 2rc

RO

Taking exponentials we havep = p0e

2rc/r−2rc/RO .

As r → ∞, the pressure tends top0e−2rc/RO rather than zero. So our isothermal static model of

the corona has a higher pressure at infinity than it should have. Therefore, there is a difference inthe pressure at the solar surface and that at infinity. This pressure difference (pressure gradient)will cause an outflow and an expansion of the corona.

This continual expansion is called thesolar wind.

• Existence of the solar wind confirmed by satellites Lunik III, Venus I in 1959 and byMariner II in early 60’s.

Assumptions are that outflow is:steady, spherically symmetricandisothermal. [It is straight-forward to relax the isothermal assumption and consider an adiabatic or polytropic atmosphere.]The basic steady equations are

∇ · (ρv) = 0, - mass continuity (6.1)

ρ (v · ∇)v = −∇p + ρg, - momentum (6.2)

p = ρRT , - gas law (6.3)

andT = T0. - energy (6.4)

The velocity is taken as purely radial,v = vr and gravitational acceleration obeys the inversesquare law,g = −(GM◦/r

2)r. In spherical coordinates, and assuming a steady flow, (6.1)gives

d

dr

(

r2ρv)

= 0 ⇒ r2ρv = D = constant, (6.5)

and the radial component of (6.2) becomes

ρvdv

dr= −dp

dr− GM◦ρ

r2. (6.6)

Defining the isothermal sound speed as(p/ρ)1/2 = csi, the gas law gives

p = c2siρ, (6.7)

6.2. PARKER’S SOLAR WIND MODEL 67

and substituting (6.7) into (6.6) we get

ρvdv

dr= −c2

si

dρ

dr− GM◦ρ

r2.

Then dividing byρ we obtain

vdv

dr= −c2

si

ρ

dρ

dr− GM◦

r2.

Now use (6.5) to expressdρ/dr andρ in terms ofr2v as

vdv

dr= −c2

sir2v

d

dr

(

1

r2v

)

− GM◦

r2

=c2si

v

dv

dr+

2c2si

r− GM◦

r2.

(ρ = D/r2v). Rearranging(

v − c2si

v

)

dv

dr= 2

c2si

r2(r − rc) , (6.8)

whererc = GM◦/2c2si.

Note,rc is important sincev = csi andr = rc is a critical pointof (6.8).At r = rc eitherdv/dr = 0 or v = csi.At v = csi eitherdv/dr = ∞ or r = rc.

(6.8) can be integrated to give a transcendental equation for the velocity in terms of the radiusas

∫

v − c2si

vdv = 2c2

si

∫

1

r− rc

r2dr,

c2si

2

(

(

v

csi

)2

− log v2

)

= 2c2si

(

log r +rc

r

)

+ C

(

v

csi

)2

− log

(

v

csi

)2

= 4 log

(

r

rc

)

+ 4rc

r+ C ′, (6.9)

whereC andC ′ are constants. The solutions for different values ofC ′ are different classesof solution (see Figure 6.1). Note, whenv/csi < 1 the flow is said to be sub-sonic, whilstv/csi > 1 is known as a super-sonic flow.

• SolutionI is double valued and so is unphysical. It is not possible for the plasma to leavethe solar surface with a velocity below the sound speed, reach a maximum radius belowrc and then turn round and return to the Sun with a super-sonic speed.

• SolutionII is also double valued and it does not even start from the solarsurface. So it isalso unphysical.

• SolutionIII starts with a velocity greater than the sound speed, but sucha fast steadyoutflow is not observed. Hence, this solution must also be neglected.


Figure 6.1: The solar wind velocity,v, as a function of the radius,r for various values of the constantC.The five different classes of solution are indicated.

• SolutionIV , the solar breeze solution, gives smallv asr −→ ∞. Using Figure 6.1, wesee thatv tends to zero asr tends to infinity. Thus, asr −→ ∞, (6.9) may be approximatedby

− log

(

v

csi

)2

≈ 4 log

(

r

rc

)

⇒ v

csi≈(rc

r

)2

⇒ r2v ≈ r2ccsi

Thus the mass continuity equation gives the density as

ρ =D

r2v=

D

r2ccsi

= const.

Since the density tends to a constant value and the plasma is isothermal,p = c2siρ, so will

the pressure. Thus, the solar breeze solution is unphysicalsince it cannot be contained bythe extremely small interstellar pressure.

• SolutionV passes through the critical point (r = rc, v = csi) also called thesonic point.For solutionV we must choose the constantC ′ so thatr = rc andv = csi and this requiresC ′ = −3.

From Fig 6.1 we see thatv is large asr −→ ∞, we may assume thatv ≫ csi so that (6.9)is approximated by

(

v

csi

)2

≈ 4 log

(

r

rc

)

⇒ v

csi≈ 2

√

log

(

r

rc

)

.

6.3. SUPER-RADIAL EXPANSION 69

Hence, from (6.5) the density is given by

ρ =D

r2v≈ D′

r2√

log(r/rc).

Thus,ρ → 0 as r → ∞.

Since the plasma is isothermal,p = c2siρ, and the pressure also tends to zero. This means

that the solution can eventually match onto the interstellar plasma at large distances fromthe Sun. Thus, solution V is a physically realistic model of the solar wind. It predicts thatthe plasma will be super-sonic beyond the critical point.

Thus, the Parker solar wind model is given by solution V. The plasma starts at the solarsurface with a small velocity that increases towards the critical point. At the critical point thespeed reaches the sound speed. Then the flow becomes (and remains) super-sonic while the gaspressure decreases.

Example 6.2.1We may calculate the critical radius,rc. Assuming a typical coronal tempera-ture of106 K the sound speed is

csi = (RT )1/2 =(

8.3 × 103 × 106)1/2

= 9.1 × 104 m s−1.

The critical radius is

rc =GM◦

2c2si

= 8 × 109 m ≈ 11R◦ ,

where the radius of the Sun,R◦ = 6.96 × 108 m.

To put this into context, the radius of the Earth’s orbit isRE ≈ 214R◦. Thus, the solar windis highly super-sonic by the time it reaches the Earth. To calculate the actual wind speed fromParker’s model we setr = RE and solve forv. Hence,

(

v

csi

)2

− log

(

v

csi

)2

= 4 log

(

214

11

)

+ 411

214− 3 = 9.078.

This may be solved using the Newton-Raphson method to give

v = 3.39csi = 309 km s−1.

Observations at1AU give the quiet solar wind as

v ≈ 320 km s−1.

Thus, Parker’s solar wind model gives quite a good estimation of the velocity.

6.3 Super-radial Expansion

The real solar wind does not come from the whole of the solar surface but only from the regionswhere there is open magnetic field such as coronal hole regions and very large strong active


regions. These regions amount to only about 20 percent of thewhole surface area. The strongmagnetic field of the other 80 percent is closed and effectively holds in the hot corona.

The field from the open regions is likely to expand super-radially above the closed regions(e.g., above helmet streamers). Such an expansion can be modelled assuming a steady isother-mal solar wind as before.

Again, letv(r) be the steady solar wind along open field lines from a coronal hole.Since here we assume that the open field expands super-radially, the continuity equation

(6.5) is replaced byd

dr(A(r)ρv) = 0 , (6.10)

whereA(r) = arn is the cross-sectional area of the open field, witha > 0 andn ≥ 2. If n = 2the magnetic field would expand radially, butn > 2 implies a super-radial expansion.

(6.10) implies

ρ =D

rnv. (6.11)

The momentum equation is purely radial as before so is the same as (6.6)

ρvdv

dr= −dp

dr− ρGM◦

r2, (6.12)

(Gravity still obeys the inverse square law as before). The plasma is isothermal as before so thegas law (6.7) is the same

p = c2siρ , (6.13)

with c2si = RT0.

These equations combine to give(

v − c2si

v

)

dv

dr=

nc2si

r− GM◦

r2. (6.14)

The sonic point in this case occurs at(rc, vc) where

vc = csi , andnc2

si

r2c

(

rc −GM◦

nc2si

)

= 0; ⇒ rc =GM◦

nc2si

.

This implies that asn increases the critical radius,rc decreases and so the sonic point movescloser into the Sun’s surface the more rapidly the open magnetic field expands.

Thus (6.14) can be written(

v − c2si

v

)

dv

dr=

nc2si

r2(r − rc) . (6.15)

Substitutingn = 2 into this equation gives ( 6.8) as one would expect.Equation (6.15) is separable inr andv and can be solved to give the following solution

which passes through the sonic point

1

2

(

v2

c2si

− 1

)

− log

(

v

csi

)

= n log

(

r

rc

)

+ n(rc

r− 1)

. (6.16)

6.4. POLYTROPIC SOLAR WIND 71

This equation is very similar to the equation for radial expansion. i.e. the only physical solarwind solution is the one that goes through the sonic point. Asr becomes very large (r → ∞)andv becomes very large (i.e.,v ≫ csi) and (6.16) maybe approximated by

v2

2c2si

≈ n logr

rc,

⇒ v ≈√

2nc2si log

r

rc

.

From (6.11) we find

ρ =D

rnv≈ D

rn√

2nc2si log(r/rc)

.

Thus, asr → ∞ , ρ → 0 , and (6.13) implies

p = c2siρ , p → 0 , as r → ∞ ,

and so again this is a physical solution for the solar wind.

6.4 Polytropic Solar Wind

Another improvement that would make the solar wind model more realistic is to drop theisothermal constraint. As an alternative a polytropic energy equation can be used instead

p = κργ .

This means that the temperature and the sound speed are no longer constant, but depend onthe radius,r. If γ = 5/3, then (6.4) would be the adiabatic energy equation. Including thisequation would produce a solar wind which remained at constant entropy rather than at constanttemperature. Instead of fixingγ as5/3 we will leave it unspecified in order that we can considerdifferent values ofγ, the only constraint is thatγ ≥ 1.

In a steady, spherically symmetric, polytropic model of thesolar wind we have

v = v(r)r , p = p(r) , ρ = ρ(r) ,

and the gravitational acceleration obeys the inverse square law,g = −(GM◦/r2)r.

Assuming that the wind expands radially, the steady continuity equation in spherical coor-dinates is

d

dr

(

r2ρv)

= 0 ⇒ ρ =D

r2v, (6.17)

whereD is a constant.The radial component of the momentum equation is

ρvdv

dr= −dp

dr− ρGM◦

r2. (6.18)

Dividing through byρ this becomes

vdv

dr= −1

ρ

dp

dr− GM◦

r2. (6.19)


The polytropic energy equation (6.4) implies that the soundspeedcs(r) =√

γp/ρ is afunction ofr.

To find a solution for the solar wind we eliminatep using (6.4) from (6.19), so

vdv

dr= −κγργ−2 dρ

dr− GM◦

r2. (6.20)

However,

κγργ−2 =γp

ρ

1

ρ=

c2s

ρ.

anddρ

dr=

D

(r2v)2

(

2rv − r2dv

dr

)

,

and therefore

vdv

dr= c2

s

(

2

r− 1

v

dv

dr

)

− GM◦

r2. (6.21)

In other words(

v − c2s(r)

v

)

dv

dr=

2c2s(r)

r− GM◦

r2. (6.22)

This equation is very similar to (6.8) and (6.14) and using this equation we see that the solarwind goes through a critical point(rc, v(rc)) where

v(rc) = cs(rc) , and 2c2s(rc)rc − GM◦ = 0 .

Now

c2s(rc) = γκ

(

D

r2ccs(rc)

)γ−1

,

which can be rearranged to give

c2s(rc) =

(

γκDγ−1

r2γ−2c

)2/(γ+1)

.

Substituting this into2c2s(rc)rc − GM◦ = 0, and rearranging, we find the radius of the critical

point is

rc =

(

GM◦

2(γκ)2/(γ+1)D(2γ−2)(γ+1)

)(γ+1)/(5−3γ)

.

Apart from finding the critical point, (6.22) does not yield an equation that can easily besolved. Instead, we reconsider the first term on the RHS of (6.19) again.

1

ρ

dp

dr=

κ

ρ

dργ

dr= κγργ−2 dρ

dr. (6.23)

However, we note thatγ

γ − 1κ

dργ−1

dr= κγργ−2 dρ

dr. (6.24)

6.4. POLYTROPIC SOLAR WIND 73

Hence, (6.19) can be written as

vdv

dr= − γ

γ − 1κ

d

dr

(

ργ−1)

− GM◦

r2. (6.25)

Since, this equation only involves derivatives (ofv2/2 andργ−1) and functions ofr it can beintegrated immediately overr to give

∫

vdv +γκ

γ − 1

∫

d(ργ−1) + GM◦

∫

1

r2dr = 0 ,

1

2v2 +

γ

γ − 1

(

p

ργ

)

ργ−1 − GM◦

r= constant . (6.26)

Substituting incs(r) = γp/ρ gives

1

2v2(r) +

c2s(r)

γ − 1− GM◦

r= constant . (6.27)

6.4.1 Nature of the Polytropic Solar Wind Solutions

There are a few interesting features of this solution. However, in order to investigate them wefirst normalise (6.27) assuming

v = cs(rc)v, cs = cs(rc)cs and r = rcr .

This gives1

2v2(r) +

cs2(r)

γ − 1− GM◦

c2s(rc)rc

1

r= C′ , (6.28)

but at the critical point2c2s(rc)rc − GM◦ = 0 and so

f(r, v) =1

2v2(r) +

cs2(r)

γ − 1− 2

r= C′ . (6.29)

Note, the value of the constant,C ′, through the critical point is given by

f(1, 1) =1

2+

1

γ − 1− 2

1= −3

2+

1

γ − 1= C′ . (6.30)

The function,f(r, v), depends on the parameterγ and so by choosing a specific value ofγwe can get a set of solutions for this equation. In Figures 6.2, we plot the set of solutions for arange of differentγ. Clearly, the nature of critical point changes asγ increases.

• If 1 < γ < 3/2 the critical point is a saddle point (Figures 6.2a-6.2b) andthere is asolution through the critical point that has positive gradient. i.e. there is a solution thatstarts off from the Sun with zero velocity and accelerates asr increases such that at thecritical point it becomes supersonic. In these cases the pressure force is always greaterthan gravity.

• If γ = 3/2 the critical point is also a saddle point (Figures 6.2c), butthe solution throughthe critical point that goes from subsonic to supersonic holds a constant speed for allr.Here, the pressure force and gravity force are in balance. This solution is unphysical asthe solar wind is known to accelerate.


Figure 6.2: Each graph contains curves that are solutions to(6.29). The difference between each graphis the value of the polytropic indexγ. The regions of subsonic (purple) and supersonic (red) flow areshaded. (a)γ = 1.2 and (b)γ = 1.3 are both possible physical solutions. (c)γ = 3/2, the solutionthrough the critical point has constant wind speed. (d)γ = 1.6, the curves through the critical point startoff from the Sun at infinite speed and continuously slow down as they leave the Sun. In (e),γ = 5/3, andthe only solution through the critical point remains at Mach1 for all r. The critical point in (f),γ = 2.0,is a minimum and therefore there are no solutions through thecritical point.

• If 3/2 < γ < 5/3 the critical point is still a saddle point (Figures 6.2d), but both solutionsthrough the critical point have negative gradient. Even though one solution goes fromsubsonic to supersonic the solution starts off from the Sun at infinite speed and constantly

6.5. THE HELIOPAUSE 75

slows down asr increases. Clearly, this is unphysical. Here, the pressureforce falls offwith r, but at such a slow rate the gravity force always dominates.

• If γ = 5/3 there is only one solution through the critical point which travels at Mach 1for all r (Figures 6.2e). It starts off from the Sun at infinite speed and continually slowsdown asr increases.

• If γ > 5/3 the critical point is a minimum and there are no solar wind solutions (Fig-ures 6.2f).

6.5 The Heliopause

Figure 6.3: Plot of the plasma temperature (top) and hydrogen density (bottom) of the solar wind in aplane though the Sun’s poles. The termination shock, heliopause and bow shock are indicated. The blueline (V1) and red line (v2) show the paths of the two voyager spacecraft, both of which have now passedthe termination shock.

The real solar wind ends at a boundary called the heliopause between the solar wind andthe local interstellar gas cloud (LIC) which is at about115 − 120 AU. Before the heliopauseis reached the supersonic solar wind must slow down to subsonic speeds. It does this at thetermination shock. The Voyager spacecraft which were launched in the early 70’s have alreadypassed the termination shock (V1 in 2004 and V2 in 2007) (Figure 6.3). They are continuingon beyond and are likely to reach the heliopause, the boundary between the solar wind and thelocal interstellar gas cloud, in about 2014 and 2017.

Beyond the heliopause there may also be another shock, a bow shock at which the localinterstellar gas cloud slows down from supersonic to subsonic due to the presence of our helio-sphere.

Chapter 7

Magnetohydrodynamic Waves

Waves are extremely important and have been observed in a wide variety of solar phenomena,such as sunspots, prominences, coronal fields and the overall oscillations of the Sun. The studyof the latter is known as helioseismology and enables the structure of the solar interior andplasma properties to be deduced from the frequency of the oscillations. The advancement ofsolar observations over the last decade has led to a new branch of solar physics called coronalseismology whereby oscillations of coronal fields are used to determine the characteristics ofthe coronal plasma.

7.1 Linearised MHD Equations

To investigate possible wave motions we assume that the amplitude of the waves is small andwe linearise the MHD equations about a particular equilibrium. We assume the basics state isstatic,

0 = −∇p0 + j0 × B0 + ρ0g, (7.1)

∇ · B0 = 0 (7.2)

p0 = ρ0RT0 (7.3)

along with an energy equation, whereB0, j0, p0, ρ0 andT0 are the equilibrium magnetic field,current density, pressure, density and temperature, respectively. Note,v0 = 0 and equilibriumquantities do not dependent on time,t. Having obtained the equilibrium we then set

B = B0 + B1(r, t), (7.4)

v = v0 + v1(r, t) = 0 + v1(r, t), (7.5)

ρ = ρ0 + ρ1(r, t), (7.6)

p = p0 + p1(r, t), (7.7)

where the perturbed quantities (subscript ‘1’) are smallerthan the equilibrium quantities. Sub-stitute these expressions into the MHD equations and neglect any products of small terms. Thus,

77

78 CHAPTER 7. MAGNETOHYDRODYNAMIC WAVES

the mass continuity equation becomes

∂ρ

∂t+ ∇ · (ρv) =

∂ρ0

∂t+

∂ρ1

∂t+ ∇ · (ρ0v1) + ∇ · (ρ1v1)

=∂ρ1

∂t+ ∇ · (ρ0v1)

= 0.

Hence, the linearised mass continuity equation is

∂ρ1

∂t+ ∇ · (ρ0v1) = 0.

In a similar manner (2.14) – (2.21) in the ideal MHD limit (i.e. Rm → ∞) reduce to (SeeTutorial sheet 8)

∂ρ1

∂t+ ∇ · (ρ0v1) = 0, (7.8)

ρ0∂v1

∂t= −∇p1 +

1

µ(∇ ×B1) ×B0 +

1

µ(∇ ×B0) × B1 + ρ1g, (7.9)

∂p1

∂t+ v1 · ∇p0 = −γp0∇ · v1, (7.10)

∂B1

∂t= ∇ × (v1 × B0) , (7.11)

p1 = ρ0RT1 + ρ1RT0 (7.12)

and∇ ·B1 = 0. (7.13)

Example 7.1.1Show that if∇ · B1 = 0 at timet = 0, (7.11) implies that (7.13) is true for alltime.

∂

∂t(∇ · B1) = ∇ · ∂B1

∂t= ∇ · (∇ × (v1 × B0)) = 0.

Therefore,∇ · B1 is constant in time. So if it is zero att = 0 then it is zero for all time.

7.2 Acoustic (Sound) Waves - Basic Waves Properties

Consider the simplest wave problem first by settingB0 ≡ 0 [the magnetic field] andg ≡ 0[gravity to zero]. This is equivalent to a high-β plasma and is valid for lengths that are muchshorter than the pressure scale height.

Thus, from (7.1) the equilibrium satisfies

∇p0 = 0 ⇒ p0 = constant.

The equilibrium isuniform. No restriction exists onρ0, [the equilibrium density] but, for sim-plicity, it is also assumed to be uniform.

p0 = constant, ρ0 = constant (7.14)

7.2. ACOUSTIC (SOUND) WAVES - BASIC WAVES PROPERTIES 79

Hence, the linearised MHD equations reduce to the followingset of equations

∂ρ1

∂t+ ρ0∇ · v1 = 0, (7.15)

ρ0∂v1

∂t= −∇p1, (7.16)

∂p1

∂t= −γp0∇ · v1, (7.17)

We look for plane waves andFourier analyseby assuming that the perturbed parameters areof the form

ρ1 = Cρei(k·r−ωt), p1 = Cpe

i(k·r−ωt), v = Cvei(k·r−ωt),

whereCρ andCp are both constants andCv is a constant vector. Also

ω – is thefrequency,

k = (kx, ky, kz) – is thewave vector,

r = (x, y, z) – is theposition vector,

such thatk·r = kxx+kyy+kzz. The frequency and the wave vector are not entirely arbitrary butare in fact related through an equation called the dispersion relation. Normally, we can choosethe wave vector and then determine what values of frequency are allowed by the dispersionrelation.

This form for the perturbed parameters is particularly useful since we note that

∂

∂t= −iω,

∂2

∂t2= −ω2,

∂

∂x= ikx,

∂2

∂x2= −k2

x,

∇· = ik·, ∇ = ik, ∇× = ik × .

Thus, defining the totalwavenumberask2 = k · k = k2x + k2

y + k2z , (7.15) (7.16) and (7.17)

become−iωρ1 + iρ0k · v1 = 0 (7.18)

−iρ0ωv1 = −ikp1 (7.19)

−iωp1 = −iγp0(k · v1) (7.20)

From (7.19) we have

v1 =p1

ωρ0k, (7.21)

thusv1||k so motions are aligned with the direction of propagation andthe wave islongitudinal.(7.18) implies that

ρ1

ρ0

=(k · v1)

ω, (7.22)


and (7.20) gives

p1 = γp0(k · v1)

ω=

γp0

ρ0

ρ1 ⇒ p1 = c2sρ1, (7.23)

wherec2s = γp0/ρ0 is the sound speed. So ifk · v1 is non zero, thenρ1 andp1 are also non zero

and we say that the motion iscompressive. Compression is a characteristic property of soundwaves.

Now take the scalar product of (7.21) withk to get

k · v1 =p1

ωρ0k2, (7.24)

and then using (7.22) and (7.23) we get

k · v1 = ωρ1

ρ0=

ωp1

c2sρ0

. (7.25)

Equating (7.24) and (7.25) implies

k2p1

ωρ0=

ωp1

c2sρ0

⇒ ω2 = k2c2s. (7.26)

This is known as the dispersion relation for sound waves and will be discussed in more detaillater.

Note, that by differentiating (7.17) in time and then substituting in (7.16) forv1 gives thetypical form for the sound wave equation

∂2p1

∂t2= −γp0∇ · ∂v1

∂t=

γp0

ρ0

∇ · ∇p1,

∂2p1

∂t2= c2

s∇2p1. (7.27)

wherecs is the characteristic speed of the propagating pressure perturbation,p1. If we Fourieranalysed this equation (7.26) immediately drops out.

Now that we know the relation betweenω andk we can determine how all the perturbedquantities vary in time for a given initial perturbation.

Example 7.2.1A medium in equilibrium with a pressure and density ofp0 andρ0 is knocked byan initial disturbance of the formp1(x, z, 0) = p0 cos(kzz). Determine what plane waves aresetup and how they effect the medium.

We look for plane wave perturbations of the form

p1(x, z, t) = Cp cos(k · r − ωt) = Re Cpei(k·r−ωt),

and since we knowp1(x, z, t) must satisfy (7.27)ω andk must satisfy (7.26), hence,

ω(k) = csk,

wherec2s = γp0/ρ0 andk2 = k · k.

7.2. ACOUSTIC (SOUND) WAVES - BASIC WAVES PROPERTIES 81

Applying the initial condition gives

p1(x, z, 0) = Cp cos(k · r) = p0 cos(kzz) =⇒ k = kzz andCp = p0.

Thus,p(x, z, t) = p0 + p1 = p0(1 + cos(kzz − cskzt)).

Obviously,v = v1, so from (7.16) we have

v =−1

ρ0

∫

∇p1dt

=1

ρ0

∫

(kzp0 sin(kzz − cskzt)z)dt

=

(

kzp0

ρ0cskz

cos(kzz − cskzt) + C

)

z

whereC is an arbitrary constant, taken here to be zero, such that

v =cs

γcos(kzz − cskzt)z

Furthermore, sincep1 = c2sρ1 we have

ρ(x, z, t) = ρ0 + ρ1 = ρ0 +p0

c2s

cos(kzz − cskzt).

So this initial disturbance gives rise to propagating soundwaves as seen in Fig 7.1.

7.2.1 Dispersion Relations

(7.26),ω2 = c2sk

2, is thedispersion relationfor sound waves. It relates the frequency withwhich the waves oscillate in time to the spatial (length) scales of the wave through the wavevector. It can be used to define two important quantities:

(i) Thephase speed,

cph =ω

k, (7.28)

which gives the speed of an individual wave component. Such waves propagate atcph in the k

direction. For sound waves,cphk = ±csk, and the phase speed is the sound speed.(ii) The group velocityis

cg =∂ω

∂k=

(

∂ω

∂kx

,∂ω

∂ky

,∂ω

∂kz

)

. (7.29)

The group velocity gives the speed and direction of transport of information and energy. Forsound waves

ω2 =(

k2x + k2

y + k2z

)

c2s.

Differentiating this gives

2ω∂ω

∂k=(

2kxc2s, 2kyc

2s, 2kzc

2s

)

,

⇒ cg =c2s

ω(kx, ky, kz) = c2

s

k

ωk = ±csk. (7.30)

on using (7.28). Hence, for sound waves the phase speed and group velocity are the same.


Figure 7.1: Plots ofp (top) andv (bottom) at timest = 0, 0.5 and1.0, where the thick dashed linerepresents a wave front which is moving at the sound speed.

7.3 Alfven Waves

What happens when there is a magnetic field present and what isthe effect of resistivity? As-sume we have an equilibrium withp0 = 0, ρ0 6= 0, andg = 0. [This allows us to see the effectof the magnetic field without having to worry about sound waves. We need to confirm that theassumed equilibrium does indeed satisfy the equilibrium equations.] From (7.1) we have

j0 × B0 = (∇ ×B0) ×B0/µ = 0.

For simplicity, we assume we have auniformequilibrium magnetic field with

B0 = B0z,

z

p = 0

B = B0 z

ρ = 0

x

0 0

^

Figure 7.2: Sketch of the assumed equilibrium magnetic field.

7.3. ALFVEN WAVES 83

Furthermore, if we assume there are no pressure or density variations then the linearisedMHD equations reduce to

∇ · v1 = 0, (7.31)

ρ0∂v1

∂t=

1

µ(∇ ×B1) ×B0, (7.32)

∂B1

∂t= ∇ × (v1 ×B0) + η∇2B1, (7.33)

and∇ ·B1 = 0. (7.34)

(7.31) implies that the motion isincompressible. Again, if we look for plane wave solutions,then we can take Fourier components and (7.31)–(7.34) reduce to

ik · v1 = 0, (7.35)

−iρ0ωv1 =1

µ(ik × B1) × B0, (7.36)

−iωB1 = ik × (v1 × B0) − ηk2B1, (7.37)

andik · B1 = 0. (7.38)

Before solving to find the dispersion relation, we look for some further characteristics of ourdisturbance (wave). From (7.36) it is clear that

v1 ·B0 = 0 (7.39)

so that the motion istransverseto the direction of the equilibrium magnetic field.By expanding the vector triple product on the right hand sideof (7.37) and collecting the

remaining terms together on the left hand side, we have

(−iω + ηk2)B1 = ik × (v1 × B0)

= i(k ·B0)v1 − i(k · v1)B0.

Sincek · v1 = 0, from (7.35), andv1 · B0 = 0, from (7.39), we have

(−iω + ηk2)(B1 · B0) = 0 ⇒ B1 · B0 = 0.

So the perturbation of the magnetic field is a right angles to the equilibrium magnetic field.Now, we can derive the disperion relation for our wave. In order to get an equation forB1

containing no other perturbed terms, we multiply (7.37) byω and then replaceωv1 using (7.36)to give

−iω2B1 = − 1

µρ0ik × {[(k ×B1) ×B0] ×B0} − ηk2ωB1.

Dividing by−i gives

ω2B1 =1

µρ0

k × {[(k × B1) × B0] × B0} − iηk2ωB1. (7.40)


Using vector identities, the right hand side of (7.40) can bebuilt up term by term.

(k × B1) × B0 = (k · B0)B1 − (B1 · B0)k = (k ·B0)B1,

sinceB1 · B0 = 0. Be careful that you recognise what is a scalar and what is a vector. (k ·B0)is a scalar that is multiplying the vectorB1. To simplify the notation we defineλ = (k ·B0) sothat

(k × B1) × B0 = λB1.

Next, we must simplify

k × {[(k × B1) × B0] × B0} = k × {λB1 ×B0} = (k ·B0)λB1 − λ(k · B1)B0.

Note, thatk · B1 = 0 from (7.38), the vanishing of the divergence of the perturbed magneticfield. Hence,

k × {[(k ×B1) ×B0] ×B0} = λ2B1 = (k · B0)2B1.

Finally, (7.40) can be expressed as

ω2B1 =(k · B0)

2

µρ0

B1 − iηk2ωB1.

Thus, ifB1 is non zero (we have a trivial solution if it is zero), then we must have

ω2 =(k · B0)

2

µρ0

− iηk2ω. (7.41)

This quadratic equation relates the frequencyω to the wave vectork and is called the dispersionrelation. We can easily solve the quadratic equation to determine the frequency, which willbe complex when resistivity is included. The imaginary partof ω, namelyωi which will benegative, corresponds to a damping of the wave in time,e−|ωi|t.

Consider the case when no resistive effects are included andthe plasma is ideal (η = 0).Then,

ω2 =(k · B0)

2

µρ0. (7.42)

We define the Alfven speed,vA, as

v2A =

B20

µρ0, (7.43)

and remembering thatB0 = B0z andk = (kx, ky, kz), (7.42) can be written as

ω = ±vA(k · z) = ±vAkz. (7.44)

(7.44) describes Alfven waves which have the following properties:

• They are anisotropic (due to thek·B0 term). That is they propagate in a preferred directionrelative to the equilibrium magnetic field.

• v1 is perpendicular to bothB0, the equilibrium magnetic field, andk, the direction ofpropagation. Thus, Alfven waves are transverse waves.


• There are no disturbances in the pressure and density and∇ · v1 = 0 so that the wavesareincompressible.

With a resistive plasma andη 6= 0, the resulting values forω are complex withω = ωr + iωi.The imaginary part means that the amplitude of the waves decay in time. This is what you wouldexpect with diffusion included. Diffusion spreads variations out and so the wave is spread outand the amplitude decays.

7.3.1 Phase and Group Speed.

From Figure 7.3a we havek · z = kz = k cos θ so that the phase speed is

cph =ω

k= vA cos θ. (7.45)

This may be represented by apolar diagramas shown in Figure 7.3b.

x

z

θ

k

B0^

x

θ

k

z

^

vA

B

v cosθ

0

A

Figure 7.3: (a) The sketch shows the angle,θ, between the direction of propagation and the direction ofthe equilibrium magnetic field. (b) The radius for the angleθ gives the magnitude of the phase speed.

(7.45) is the equation of a circle with centre atθ = 0 and radius =vA/2. Compare this withsound waves withcph = cs, which are isotropic and thus independent ofθ.

The group velocity is∂ω/∂k so from (7.44) that

cg =

(

∂ω

∂kx,

∂ω

∂ky,

∂ω

∂kz

)

= ±vAz. (7.46)

Therefore, the group velocity for the Alfven wave is alwaysin the direction of the equilibriummagnetic field and of magnitudevA.

Example 7.3.1 In the absence of gravity, an ideal (η = 0) plasma of pressure,p0 = 0 and den-sityρ0 = const has a magnetic fieldB0 = B0z. Show that this initial plasma is in equilibriumand that, if it is perturbed by an initial disturbance of the form B1(x, z, 0) = ǫB0 cos(kzz)x,Alfven waves are formed.

We look for plane wave solutions of the form

B1(x, z, t) = CB cos(k · r − ωt) = Re(

CBei(k·r−ωt))

.

Thus,B1(x, z, t) must satisfy (7.40) andω andk must satisfy (7.41). Since,B0 = B0z we have,

ω(k) = vAkz,


Figure 7.4: Plots ofB (top) andv (bottom) at timest = 0, 0.5 and1.0, where the thick dashed linerepresents a wave front which is moving at the Alfven speed.The vertical wiggly lines are magneticfield lines.

wherev2A = B2

0/µρ0 andk = (kx, ky, kz). Applying the initial condition gives

B1(x, z, 0) = CB cos(k · r) = ǫB0 cos(kzz)x =⇒ k = (0, 0, kz) andCB = (ǫB0, 0, 0).

Thus,B1(x, z, t) = ǫB0 cos(kzz − vAkzt)x.

From (7.36), we can determinev1,

−ρ0ωv1 =1

µ(k ×B1) ×B0

v1 = − ǫB20kz

µρ0kzvA

cos(kzz − vAkzt)(z × x) × z

= −ǫv2A

vA

cos(kzz − vAkzt)(y × z)

= −ǫvA cos(kzz − vAkzt)x

From (7.31) we know the disturbance is incompressible andp = 0 andρ = ρ0 for all time.So this initial disturbance gives rise to propagating Alfven waves as seen in Fig 7.4.

Example 7.3.2Given a coronal loop withB0 = 10 Gauss (10−3 Tesla),L = 50 Mm (5 × 107

m) andn = 5 × 1014 m−3 (ρ0 = 8 × 10−13 kg m−3), determine the Alfven speed and wavenumber of the fundamental mode (i.e. a wave with wavelengthL/2). The use these values tofind the frequency and period of oscillation of the fundamental mode


[Recallµ = 4π × 10−7 H m−1].The Alfven speed is given by

vA =√

B20/µρ0 =

√

10−6

4π × 10−7 × 8. × 10−13≈ 106m s−1.

The wave number isk = 2π/wavelength = 2π/(L/2) = 4π/5 × 107 = 2.5 × 10−7 m−1.Therefore, the frequency of oscillation of an Alfven wave in a coronal loop is

ω = kzvA = 2.5 × 10−7 × 106 = 0.25 s−1.

Hence, the period,τ , of oscillation is

τ =2π

ω= 25 s.

Thus, the fundamental mode of oscillation of an Alfven wave in a coronal loop with these prop-erties should have a period of approximately 25 seconds.

Example 7.3.3Consider the same coronal loop as in Example 6.3.2, withB0 = 10 Gauss(10−3 Tesla),L = 50 Mm (5 × 107 m) andn = 5 × 1014 m−3 (ρ0 = 8 × 10−13 kg m−3) andwith the same fundamental Alfven mode of oscillation, but this time assume that the plasma isresistive, withη = 1ms−2. For this situation, determine the frequency (ωr +ωi), and the periodτ of oscillation. What is the damping time of the Alfven wave?

To determine the frequency of the oscillation, we need to know the Alfven speed and wavenum-ber which are equal to the values found in example 6.3.2.

i.e., vA =√

B20/µρ0 ≈ 106 m s−1 and the wave number isk = 2π/(L/2) = 2.5 × 10−7

m−1.In resistive MHD, the fundamental frequency of oscillationof a coronal loop satisfies the

quadratic equation (7.41)

ω2 + iηk2ω − (k · B0)

µρ0

= 0

ω2 + iηk2ω − (kvA)2 = 0

ω2 + i × 1 × (2.5 × 10−7)2ω − (2.5 × 10−7 × 106)2 = 0.

ω2 + i6.25 × 10−14ω − 6.25 × 10−2 = 0.

Hence, the frequency is

ω = −6.25 × 10−14

2i ± 1

2

√

−(6.25 × 10−14)2 + 4 × 6.25 × 10−2

= ±0.5 − 3.125 × 10−14i.

Thusωr = 0.5 andωi = −3.125 × 10−14i . The period,τ , of oscillation is

τ =2π

ωr

= 12.57 s,

and the damping time (one over the damping rate = imaginary part of ω) is

τdamp = 3.2 × 1013s≈ 1 million years.

Obviously the damping of Alfven waves is negligible in this situation. Resistivity will only beimportant if the length scales are small and the wavenumberk is very large.

solar theory mt45100.4. integral theorems 9 0.4 integral theorems the two theorems that are commonly...

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