Boyd T.J.M., Sanderson J.J. The Physics of Plasmas

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5.6 MHD shocks 187

it is easily seen that the magnetic ﬁeld drops out of the jump conditions leaving

= ρ

(5.77)

+ P

= ρ

+ P

(5.78)

+ P

+ ρ

/2 = ρ

+ P

+ ρ

/2 (5.79)

where the last equation may also be written, using (5.62) and (5.77), as

γ P

(γ − 1)ρ

γ P

(γ − 1)ρ

(5.80)

This solution, therefore, corresponds to a hydrodynamic shock and it may be shown

(see Exercise 5.8) that one must have u

supersonic (relative to the sound speed in

region 1) while u

is subsonic (relative to the sound speed in region 2).

Strong shocks are deﬁned as those for which the pressure ratio R  1. In

this case, it follows (see Exercise 5.8) that the Mach number in region 1, M =

(1))  1, and the temperature ratio

≈

2γ(γ − 1)

(γ + 1)

 1 (5.81)

Since M may be as large as 100, it is clear from (5.81) that strong shocks may

be used to generate high-temperature plasmas or to obtain a plasma from a neu-

tral gas by creating a temperature T

behind the shock sufﬁciently high to cause

ionization. The conversion of ﬂow energy into thermal energy in this situation is

easily demonstrated from (5.80). In view of (5.81), the initial thermal energy may

be neglected compared with the ﬁnal thermal energy. Also (see Exercise 5.8)

≈

(γ + 1)

(γ − 1)

(5.82)

This ratio is 4 for γ = 5/3 so (5.80) may be approximated by

γ P

(γ + 1)

(5.83)

Thus, the ﬂow energy in region 1 is converted by the shock into thermal energy in

region 2.

The hydrodynamic shock is not, however, the only possible solution for prop-

agation parallel to the initial magnetic ﬁeld. It may happen that some of the ﬂow

energy is converted into magnetic energy so that |B

| > |B

|. Since the normal

component of B is conserved this means that the passage of the shock creates a

tangential component and it is said to be a switch-on shock. We shall return to this

possibility in our discussion of oblique shocks.

188 Resistive magnetohydrodynamics

5.6.3 Perpendicular shocks

Here we have

= (0, B

, 0) (5.84)

and so, using (5.54), B

may be written

= (0, B

, 0) (5.85)

Since ρu

= 0, it follows from (5.55), (5.57) and (5.59) that

= (u

/r, 0, 0)

= (0, rB

, 0)

Thus the magnetic ﬁeld is constant in direction and increased in magnitude by the

same ratio as the density. Finally, (5.58) and (5.61) are now

+ P

+ B

/2µ

= ρ

+ P

+ B

/2µ

and

γ P

(γ − 1)ρ

γ P

(γ − 1)ρ

These may be written

γ M

(1 − 1/r) = (R − 1) + (r

− 1)/β (5.86)

and

γ M



1 −



2γ

(γ − 1)



− 1



+ 4(r − 1)/β

respectively, where β = 2µ

is the plasma β ahead of the shock and the

shock Mach number M = u

, where c

= (γ P

/ρ

)

1/2

is the sound speed in

the upstream plasma. Eliminating R and excluding the solution r = R = 1, which

corresponds to no shock, we get

2(2 − γ)r

+ [2γ(β + 1) + βγ(γ − 1)M

]r − βγ(γ + 1)M

= 0

If r

and r

are the roots of this equation, then

=−βγ(γ + 1)M

/2(2 − γ)

and for γ<2 one root is negative and, therefore, non-physical. Consequently,

there is only one solution corresponding to a shock in this case. Since r > 1

βγ(γ + 1)M

> 2(2 − γ)+ 2γ(β + 1) + βγ(γ − 1)M

which reduces to

γ M

>γ+ 2/β

5.6 MHD shocks 189

and, hence,

> B

/µ

+ γ P

/ρ

= v

+ c

= (c

∗

)

where the second equality deﬁnes c

∗

. Thus, for shocks to propagate perpendicular

to a magnetic ﬁeld the shock speed must be greater than c

∗

. The speed c

∗

assumes

the role played by c

in hydrodynamic shocks (see Exercise 5.9); this is not a

surprising result since c

∗

is the speed of the fast compressional wave propagating

perpendicular to a magnetic ﬁeld (see Section 4.8). (Note that there is no shock

corresponding to the slow wave since it does not propagate perpendicular to the

magnetic ﬁeld.) We see that the effect of the magnetic ﬁeld is to increase the

effective pressure by a factor (1 + 2/γβ),

The shock strength R is reduced by the introduction of the magnetic ﬁeld (see

Exercise 5.9) since ﬂow energy is now converted into magnetic energy as well as

heat. However, since

= r <(γ + 1)/(γ − 1) (5.87)

the increase in magnetic energy is limited while from (5.86)

= R = 1 + γ M

(1 − 1/r) − (r

− 1)/β (5.88)

so that for large Mach number, relative to a ﬁxed value of β, the temperature ratio

is approximately the same as for the hydrodynamic case (see Exercise 5.9).

5.6.4 Oblique shocks

In the case of oblique propagation where in general u and B have both x and

y components it is convenient to choose a frame of reference, known as the

de Hoffmann–Teller frame, in which u

× B

= 0, that is

= u

where B

= B

by (5.54). Note that this is consistent with our choice for

parallel shocks but not for perpendicular shocks for which B

= 0. From (5.55) it

follows that

= u

and hence u

× B

= 0, i.e. u and B are parallel on both sides of the shock;

physically, this simply says E = 0 on both sides of the shock. Now

(5.89)

190 Resistive magnetohydrodynamics

Also, from (5.59), we get

− 1 =



− 1





− 1



which may be combined with (5.89) to give

− v

−rv

(5.90)

Furthermore, with this choice of reference frame the magnetic terms in (5.61)

are identically zero leaving only the hydrodynamic terms from which we get

= r +

(γ − 1)ru



1 −



= r +

(γ − 1)ru



1 −

cos

− sin



− v

−rv





(5.91)

It is clear from (5.90) that B

> B

for u

≥ rv

and, conversely,

< B

for u

≤ v

< rv

. The ﬁrst case corresponds to the fast shock and the

second to the slow shock and these are illustrated in Fig. 5.26 showing refraction

(a) away from and (b) towards the normal, respectively. It is when the equalities

hold in these relationships, i.e. u

= r v

so that B

= 0 for the fast shock, or

= v

so that B

= 0 for the slow shock, that we get (c) switch-on and (d)

switch-off shocks, respectively, corresponding to the tangential component of the

magnetic ﬁeld being switched on or off.

The switch-on shock is one of the possible solutions (the fast shock) when we

let θ → 0 (i.e. parallel propagation). The slow wave in this limit has u

= v

so that B

= 0. In this case both B

and B

are parallel to the shock normal

and B

= B

. As noted earlier when discussing (5.67) this yields the hydrody-

namic Hugoniot; in other words, the slow shock becomes a hydrodynamic shock

at parallel propagation.

5.6.5 Shock thickness

We now wish to consider the structure of a shock, i.e. the variation of pressure,

density, magnetic ﬁeld etc., within the shock itself. Even for collision-dominated

shock waves a quantitative calculation of shock structure involves considerable

effort. The procedure is to solve the appropriate transport equations inside the

shock region using either region 1 or region 2 as a set of boundary conditions. Just

what form the appropriate transport equations take is not easily decided in general.

Often the important dissipative mechanisms are viscosity, heat conductivity, and

electrical conductivity (Joule heating). Of these, only Joule heating was retained in

5.6 MHD shocks 191

(d)

switch-on

switch-off

(a) (b)

fast shock slow shock

(cc

)

shock

Fig. 5.26. Magnetic ﬁeld refraction in oblique shocks.

the resistive MHD equations. The heat conduction term must be reintroduced into

the energy equation; similarly, viscosity terms must be brought into the momentum

and energy equations. However, there is considerable doubt as to whether such

a one-ﬂuid, hydromagnetic description is appropriate for a discussion of shock

192 Resistive magnetohydrodynamics

structure. In particular, it often happens that the electrons heat up ﬁrst in a shock

and then reach an equilibrium temperature with the ions after a longer period

of time. Thus, a description involving separate ion and electron temperatures is

usually necessary. Also, depending on the conditions, other dissipative mechanisms

may be important – for instance, ionization if the unshocked gas has a zero or low

degree of ionization.

However, leaving aside a quantitative discussion of shock structure, one may ob-

tain estimates for the thickness of a shock by order of magnitude arguments. Since

the conditions on either side of the shock are given by the initial conditions together

with the solutions of the shock equations, the total rate of dissipation of energy is

known and this must occur within the shock thickness, δ. If we know (or assume)

that the dissipation is due principally to one particular mechanism, we can write an

order of magnitude relationship. For example, suppose the appropriate dissipative

process is viscosity. In the energy transport equation, the term involving viscosity

is proportional to the square of the velocity gradient. Then, if other dissipative

processes are negligible, the rate of dissipation of energy, E/t , is proportional

to ρν(u/δ)

, where ν(= µ/ρ) is the kinematic viscosity. Since t ∼ δ/u

, the

order of magnitude relationship is

E

t

∼

E

∼ ρν



− u



i.e.

δ ∼

ρν(u

− u

)

E

(5.92)

Applying this to the particular case of the strong hydrodynamic shock discussed

in Section 5.6.2, the energy dissipated E ≈

. Also, using (5.82), (5.92)

implies

δ ∼ ν/u

(5.93)

or, in other words, the Reynolds number, R = u

δ/ν, is of order unity. From kinetic

theory (see Section 8.2), one can show approximately that ν ∼ Pτ

/ρ, where τ

the ion–ion collision time. Using P ∼

+ P

) ≈

, it follows from (5.82),

(5.83), and (5.93) that

δ ∼ τ





1/2

∼ τ





1/2

Thus the shock thickness is of the order of the ion collision mean free path.

To give a further example, consider dissipation by Joule heating. Here, the dis-

sipation of energy occurs at a rate proportional to j

/σ . Since µ

j = ∇ × B, the

Exercises 193

order of magnitude relationship is

E

∼



− B



i.e.

δ ∼

− B

)

σµ

E

(5.94)

From (3.38) this may be written in terms of the magnetic Reynolds number as

= µ

σδu

∼

− B

)

E

(5.95)

Now suppose this is applied to a strong shock propagating perpendicular to a

magnetic ﬁeld (see Section 5.6.3). With E ≈

and B

≈ B

/4,

∼

/2µ

which gives the order of magnitude of the magnetic Reynolds number required if

Joule heating is to be an adequate dissipative mechanism for this shock.

As one might expect, it is found that a given dissipative mechanism can produce

the required change of state up to a certain limiting shock strength. Beyond that

other mechanisms must come into play with the result that the shock may show a

more complicated structure with more than one characteristic thickness. We return

to this question of shock structure with a speciﬁc calculation when discussing

collisionless shocks in Section 10.5.2.

Exercises

5.1 With reference to the induction equation (5.1) explain the signiﬁcance

of the magnetic Reynolds number R

. What is the relationship of the

Lundquist number S to R

(a) Why is it that even in plasmas for which S  1 resistive diffusion

cannot be completely ignored? What does a dimensional analysis of

(5.1) tell us about the relative length scales for magnetic ﬁeld changes

due to diffusion and convection?

(b) Establish from (5.4) and the conservation of mass that the Sweet–

Parker reconnection rate is S

−1/2

where S is the Lundquist number.

rizing Petschek’s model and variations, notably that due to Sonnerup

(1970).

194 Resistive magnetohydrodynamics

(d) Refer to Biskamp (1993), (Section 6.2), for a contrasting critical sum-

mary of Petschek’s model. Biskamp argues that not only is Petschek’s

concept difﬁcult to accept intuitively but it is seriously ﬂawed in the in-

appropriate treatment of the diffusion region. Numerical simulations of

driven reconnection do not produce a Petschek-like conﬁguration for

small η. In particular Biskamp (1986) found that the sheet width L in-

creased as η decreased in contrast to Petschek’s scaling of L ∼ O(η).

5.2 Explain Fig. 5.6 with reference to the heuristic arguments used in Sec-

tion 5.3.1 to obtain the parametric dependence of the growth rate (5.22) of

the tearing mode. What are the physical characteristics of the tearing mode

that make it much more of a threat to plasma stability than the gravitational

and rippling modes?

5.3 Biermann (1950) showed that stellar magnetic ﬁelds could be generated at

the expense of the thermal energy of the star. In the event the time needed

for Biermann’s mechanism to establish ﬁelds typical of those on the Sun

for example proved to be rather too long. From the generalized Ohm’s

law (3.52) if no magnetic ﬁeld is present initially show that this reduces

to E =−(m

/Zeρ)∇ p

. Then from Faraday’s law, a magnetic ﬁeld is

generated provided ∇ρ ×∇ p

= 0. For a spherically symmetric pressure

gradient, no magnetic ﬁeld is generated.

(a) Consider next the case of a rotating star for which

∇ p = ρ



g + 



(E5.1)

in which  denotes the angular velocity of the star and

r is the dis-

placement from the axis of rotation. Show that in the case of a rotating

star a toroidal magnetic ﬁeld is generated.

(b) Generally, both ohmic and Hall terms in the generalized Ohm’s law act

to limit the growth of the ﬁeld. In the case of the Sun, by balancing en-

ergy input from the battery against ohmic dissipation with appropriate

choices for parameters from Table 5.1 and using ¯r  ∼ 2 × 10

−1

show that the ﬁeld B ∼ 0.01 T and that the time needed for the ﬁeld

to evolve to this magnitude is about 10

years, the order of the age of

the Universe. However, it needs to be borne in mind that such rough

estimates are changed by allowing for convection.

netic ﬁelds generated in targets irradiated by intense laser light where

ﬁelds as large as O(100 T) have been detected over scale lengths of

O(100 µm). Check the estimate given for the strength of the magnetic

ﬁeld generated.

Exercises 195

5.4 Verify that (5.32) is a solution of (5.31) and that this corresponds to the

solution curves sketched in Fig. 5.18. Show that the constant C =−3for

the only acceptable solution for the solar wind represented by trajectory V .

What are possible explanations of why this solution does not predict the

correct density at the Earth?

5.5 Using the results in Section 5.5 for the Weber and Davis model of the

solar wind, obtain an expression for the magnitude of the interplanetary

magnetic ﬁeld B

IMF

= (B

+ B

)

1/2

and show that (i) B

IMF

∼ 1/r

near

the Alfv

en critical point r = r

and (ii) B

IMF

∼ 1/r (log r)

1/2

as r →∞.

5.6 Derive the hydromagnetic Hugoniot relation (5.67) from (5.54)–(5.61) by

eliminating the velocity variables.

[Hint: First use (5.57), in the form ρ

= ρ

= k, say, to eliminate

in favour of k and then use the resulting (5.59) and (5.60) to eliminate

and u

. Next, write (5.58) in the form k

[1/ρ]

=−[P + B

/2µ

]

and multiply by (ρ

−1

+ ρ

−1

)/2 to obtain





=−





P +

2µ



Finally, substitute this equation, together with the converted (5.55) and

(5.56) in (5.61) to eliminate k and obtain (5.67).]

5.7 In a plane shock show that the unit vector normal to the shock,

n, and

the magnetic ﬁelds on either side of the shock, B

and B

, are coplanar

provided that the mass ﬂux through the shock is non-zero (ρu ·

n = 0).

5.8 Use (5.77) to write (5.78) and (5.80) as

γ M

(1 − 1/r) = R − 1

γ M

(1 − 1/r

) =

2γ

γ − 1



− 1



By solving these equations for R, r and applying the condition R > 1

show that u

is supersonic. Rewriting the equations in terms of M

show

that u

is subsonic.

Deduce that R  1 implies M

 1.

Verify (5.82).

5.9 For the perpendicular shock discussed in Section 5.6.3 it was shown that

> c

∗

(1). By rewriting the equations in terms of the downstream vari-

ables M

and β

show that u

< c

∗

(2).

Show also that for γ<2, r < r

, where r

is the solution for zero mag-

netic ﬁeld. Hence, deduce from (5.86) that the introduction of a magnetic

ﬁeld reduces the shock strength.

196 Resistive magnetohydrodynamics

From (5.87) and (5.88) show that for ﬁxed β the temperature ratio in the

large Mach number limit is approximately the same as for the hydrody-

namic shock.