Plasma Physics - Homework Solution, 4/19/2012

21. Stability of the Z pinch: The pressure and magnetic field profile of the Z pinch is

Bθ =µ0j02r and p = −µ0j

20

4r2 + p0 for r ≤ rc =

(4p0

µ0j20

)1/2

Assume perturbations with ∂/∂θ = 0 and apply the stability criterion derived in class to the equilibriumto determine it equilibrium properties. How do these change if a constant pressure pc is superimposed.

1. Solution

Magnetic field and pressure (homework problem 18) are given by

Bθ =µ0j0r

∫ r

rdr =µ0j02r

p = −µ0j20

2

∫ r

rdr = −µ0j20

4r2 + p0 =

µ0j20

4

(r2c − r2

)with r2

c =4p0

µ0j20

Constant current j0 in the z direction in a cylindrical coordinate system.

Stability criterion (from class)

−rp

dp

dr<

4γ

2 + γβ

for stability. With

−rp

dp

dr=

r

p0 − µ0j204r2

µ0j20

2r

=2r2/r2

c

(1− r2/r2c )

and

β =2µ0p

B2=

2µ0

(p0 − µ0j20

4r2)

µ20j

20

4r2

=2 (1− r2/r2

c )

r2/r2c

Substitution in the stability condition:

2r2/r2c

(1− r2/r2c )

<4γ

2 + γβ=

2γ

1 + γ (1−r2/r2c )r2/r2c

=2γr2/r2

c

r2/r2c + γ (1− r2/r2

c )

r2/r2c + γ

(1− r2/r2

c

)< γ

(1− r2/r2

c

)

orr2/r2

c < 0

Thus this configuration is everywhere unstable. Note that this is caused by the assumed current densityprofile. Adding a constant pressure does not change this result because it means we can just increase p0

by the amount of the constant pressure

Often β is assumed to be constant or small. Assuming that β is constant the stability condition is

−rp

dp

dr=

2r2/r2c

(1− r2/r2c )<

4γ

2 + γβ= λ

or

r2 <λ

2 + λr2c

=γ

γ + 1r2c for β � 1

for stability.

Thus the small β approximation yields the correct result for of instability for sufficiently large radii butfails for smaller values of r/rc where the system is actually unstable

2nd solution:

Starting from the coefficients

a11 = γβ + 2 β =2µ0p

B2

a12 = 2

(d lnB

d ln r+β

2

d ln p

d ln r− 1

)

a22 =

(d lnB

d ln r− 1

)a12

we note that the force balance (equilibrium condition requires

d lnB

d ln r= −β

2

d ln p

d ln r− 1

which implies a12 = −4. For the special case of B = µ0j0r/2 we can see immidiately

d lnB

d ln r=

r

B

d lnB

d ln r= 1 ↪→ a22 = 0

Therefore the stability condition becomes

a11a22 − a212 = −16 > 0

Which is alway violated and therefore we find unconditional instability. Adding a constant pressure doesnot change the result because the forcebalance condition does not change and the magnetic field onlydepends on the pressure gradient such that it also does not change. The only possibilities to generatea more stable configuration are (a) to use a different current density profile or (b) to employ also a Bz

component.

21. Stability for the Harris sheet:

Determine the stability criterion for the Harris sheet equilibrium using the energy principle. Assume thatthe perturbation on the boundary is 0

Harris sheet with

B = Bey = B0 tanhx

Ley

p = p0 cosh−2 x

L

j =B0

µ0Lcosh−2 x

Lez

Solution:

The energy principle for a displacement ξ without considering surface terms is

U =1

2

∫V

[γp0 (∇ · ξ)2 +

1

µ0

(∇× (ξ ×B0))2

+ξ · ∇p0∇ · ξ −1

µ0

(ξ × (∇×B0)) · ∇ × (ξ ×B0)

]dx

(a) Show that

∇× (ξ ×B) = B

[∂ξx∂y

ex −(∂ξz∂z

+1

B

∂ξxB

∂x

)ey +

∂ξz∂y

ez

]

ξ × (∇×B) =∂B

∂x(ξyex − ξxey)

Noting that the equilibrium magnetic field has only an y component and the current density only a zcomponent and both depend on x only we get

∇× (ξ ×B) = ∇× (−ξzBex + ξxBez) = −∂zξzBey + ∂yξzBez + ∂yξxBex − ∂xξxBey

= B∂yξxex − (B∂zξz + ∂xξxB) ey +B∂yξzez

= B

[∂ξx∂y

ex −(∂ξz∂z

+1

B

∂ξxB

∂x

)ey +

∂ξz∂y

ez

]and

ξ × (∇×B) = ξ × (∂xBez) =∂B

∂x(ξyex − ξxey)

(b) With

∇ · ξ =∂ξx∂x

+∂ξy∂y

+∂ξz∂z

∇× (ξ ×B0) = B

[∂ξx∂y

ex −(∂ξz∂z

+1

B

∂ξxB

∂x

)ey +

∂ξz∂y

ez

]

ξ × (∇×B0) =∂B

∂x(ξyex − ξxey)

the potential becomes

U =1

2

∫V

[γp (∇ · ξ)2 +

∂p

∂xξx∇ · ξ

+B2

µ0

(∂ξx∂y

)2

+

(∂ξz∂z

+1

B

∂ξxB

∂x

)2

+

(∂ξz∂y

)2

− 1

2µ0

∂B2

∂x

(ξy∂ξx∂y

+ ξx

(∂ξz∂z

+1

B

∂ξxB

∂x

))]dx

For ∂/∂y = 0 and ∇ (p+B2/2µ0) = 0:

U =1

2

∫V

γp (∇ · ξ)2 +∂p

∂xξx∇ · ξ +

B2

µ0

(∂ξz∂z

+1

B

∂ξxB

∂x

)2

+∂p

∂xξx

(∂ξz∂z

+1

B

∂ξxB

∂x

)]dx

With∂ξz∂z

+1

B

∂ξxB

∂x= ∇ · ξ + ξx

1

B

dB

dx

we obtain

U =1

2

∫V

γp0 (∇ · ξ)2 +∂p

∂xξx∇ · ξ +

B2

µ0

(∇ · ξ + ξx

1

B

dB

dx

)2

+∂p

∂xξx

(∇ · ξ + ξx

1

B

dB

dx

)]dx

=1

2

∫V

[a11 (∇ · ξ)2 + 2a12ξx∇ · ξ + a22ξ

2x

]dx

with

a11 = γp+B2

µ0

a12 =dp

dx+B

µ0

dB

dx

a22 =1

µ0

(dB

dx

)2

+dp

dx

1

B

dB

dx

(c) Using the equilibrium condition for the Harris sheet:

0 =d

dx

(p+

B2

2µ0

)=dp

dx+B

µ0

dB

dx

we obtain a12 = 0 and

a22 =1

µ0

1

B

dB

dx

(BdB

dx

)+∂p

∂x

1

B

dB

dx= 0

such that the equilibrium condition reads

a11a22 − a212 = 0 ≥ 0

This is insufficient to clarify stability uniquely, but:

Compressible perturbations are always stable because a11 > 0. For incompressible perturbations theresult is inconclusive.

This result has actually not made use of the specific Harris equilibrium but applies to any one-dimenionalequilibrium where the magnetic field has only a single component.