3.2. ESQUIZOFRENIA 1 Concepto
3.2.3 Inicio y Curso
The flux rope, represented in this model by the line currentI, is in equilibrium when the force on it due to all external sources vanishes. The external magnetic fieldBextis given by subtracting the field due to the (real) line current from the full magnetic field (6.1) including the 4 sources and both line currents. On they-axis this gives a purely imaginary field,
Bxext(0, y) = 1 π 2 y2+ 1− 2λ y2+λ2 + I 2(y+h) , (6.12) Byext(0, y) = 0. (6.13)
6.2 Line Current in a Quadrupolar Field with Point Sources 128
So the only force on the flux rope aty=his vertical, given by
Fy =IBxext(0, h) =I 8h(h2−λ)(1−λ) +I(h2+ 1)(h2+λ2) 4h(h2+ 1)(h2+λ2) . (6.14)
AssumingI 6= 0, setting the numerator to zero gives the equilibrium condition
Ih4+ 8(1−λ)h3+I(1 +λ2)h2−8λ(1−λ)h+Iλ2 = 0. (6.15) This equation describes a curve in(λ, h, I)space. Treating it as a quartic equation inh, Descartes’ rule of signs shows that there can be up to 2 roots for positiveh, corresponding to 2 equilibrium branches in the(λ, h)plane. Note also that equilibrium is only possible for small enoughI. We can see this by re-writing (6.15) as a quadratic inλ,
(Ih2+ 8h+I)λ2−8h(h2+ 1)λ+Ih4+ 8h3+Ih2= 0. (6.16) The discriminant of this quadratic may be written
4h2(16−I2)(h4+ 1)−2(16 +I2)h2−16Ih(h2+ 1),
which is negative definite if I ≥ 4. Thus we see that equilibrium is not possible for I ≥ 4 (although this is not necessarily a strict upper bound).
The equilibrium curve is shown in Figure6.3(a) for the caseI = 0.5, overlayed on the bifurcation diagram from Figure6.2. The curve has been computed numerically (using Maple). We find a two-branch structure in accordance withLin and van Ballegooijen(2005, Figure 5 bottom). As the constant currentI is increased, the equilibrium curve becomes smaller and restricted to smallerλ, eventually disappearing altogether at aroundI = 3.5(in accordance with our above estimate). To determine the stability of equilibria on the two branches, we consider the vertical forceFy given
by equation (6.14). This is plotted in Figure6.3(b) as a function ofhfor three differentλ, for the caseI = 0.5. For λ= 0.25orλ = 0.5, there are two zeros of this force, corresponding to the two branches of the equilibrium curve in Figure6.3(a). The lower equilibrium is stable, because
∂Fy/∂h < 0 there, while the upper equilibrium is unstable, since ∂Fy/∂h > 0. Forλ = 0.9
(dashed curve), the force is always positive and there are no equilibria, as seen from Figure6.3(a).
This force structure always holds for this model: from equation (6.14) we see thatFy → ∞as
h → 0, Fy > 0 for large h, andFy → 0as h → ∞. In fact such a two-branch equilibrium
structure with a stable lower equilibrium and an unstable upper equilibrium is typical of the “fold catastrophe” described by the mathematical catastrophe theory of R. Thom (Poston and Stewart,
6.2 Line Current in a Quadrupolar Field with Point Sources 129
Figure 6.3: Equilibrium curve (a) and vertical force profile (b) for the quadrupolar point source model withI = 0.5. The equilibrium curve given by equation (6.15) is shown in blue on the bifurcation diagram in(λ, h)space. The vertical forceFyis shown as a function ofh, atλ= 0.25
(solid curve),λ= 0.5(dotted curve), andλ= 0.9(dashed curve).
1978;Saunders,1980). This is the elementary catastrophe appropriate for a single variable (h) and a single parameter (λ). Physically, changingλleads to a quasi-static evolution of the equilibrium configuration along the lower, stable, branch of the equilibrium curve. Asλis increased beyond a critical value (the fold point), equilibrium is no longer possible, and dynamic evolution ensues in a sudden “catastrophe”. In this model the loss of equilibrium occurs asλis increased, i.e. as the inner sources move nearer to the outer sources. This is because the effective strength of the background field at the flux rope decreases (Lin and van Ballegooijen,2005), making equilibrium more precarious.
The example in Figure6.3shows that the equilibrium curve lies only in topologies A1 and D (as defined in Figure6.2), and that equilibrium in topology A1 is always unstable (and hence unlikely to exist in practice). We now prove that these are general properties for any value ofI >0. To show that no equilibrium is possible in topologies A2 or B, note that the equilibrium condition (6.15) may be re-written in the form
h2 =λ−I
h4+ (1 +λ2)h2+λ2
8h(1−λ) . (6.17)
Since the second term is always negative forI > 0, this shows thath2 < λat any point on the equilibrium curve. Thus equilibrium is impossible in topologies A2 or B.
6.2 Line Current in a Quadrupolar Field with Point Sources 130
We may also show that equilibrium is impossible in topology C as follows. Equation (6.17) may be rearranged to give
2
I(1−λ)−h=
4h4+ (λ−h2)2+h2(λ−1)2
4h(λ−h2) . (6.18)
Sinceh2 < λ on the equilibrium curve (previous paragraph), we see that the right-hand side is positive, so that on the equilibrium curveh <(2/I)(1−λ). So the equilibrium curve always lies below the straight line in the bifurcation diagram (Figure6.2), and hence topology C cannot be in equilibrium.
We have established that equilibrium is possible only in topologies A1 or D. The example in Figure 6.3demonstrates that stable equilibrium is possible in topology D; our final step is to prove that only unstable equilibrium is possible in topology A1. To prove this, it is sufficient to show that
∂Fy/∂h > 0 everywhere in region A1 of the bifurcation diagram. Differentiating (6.14) and
substituting forI from the equilibrium condition (6.15) gives
∂Fy ∂h = 2I(1−λ) h (λ−h2)(3h4+λ2h2+h2−λ2) + 2h2(h2+ 1)(h2+λ2) (h2+ 1)2(h2+λ2)2 . (6.19)
In the case thath ≥1the result follows easily sinceλ < 1implies that h2−λ2 >0. In region A1,λ > h2, so all terms on the right-hand side are positive and ∂Fy/∂h > 0as required. The
alternative, thath < 1, is harder to deal with. We need to use the fact that we are in topology A1 and not topology D, namely thath > Iλ/(2(1−λ)). Substituting forI from the equilibrium condition gives the constraint
4λ(λ−h2)<(h2+ 1)(h2+λ2). (6.20) We use this inequality twice. Firstly it implies that
2h2(h2+ 1)(h2+λ2)>8λh2(λ−h2) (6.21) so that ∂Fy ∂h > 2I(1−λ) h (λ−h2)(3h4+λ2h2+h2+ 8λh2−λ2) (h2+ 1)2(h2+λ2)2 . (6.22)
Secondly, inequality (6.20) implies that
4λ2<(h2+ 1)(h2+λ2) + 4λh2
6.2 Line Current in a Quadrupolar Field with Point Sources 131
using our assumption thath <1. Rearranging this inequality gives
−λ2 >−h2−2λh2, (6.24)
and substituting into (6.22) gives
∂Fy ∂h > 2I(1−λ) h (λ−h2)(3h4+λ2h2+ 6λh2) (h2+ 1)2(h2+λ2)2