Transient Solution for Pre-Flare Onset Using Equivalent Circuits.

 

                         Pre-flare equivalent circuit to apply to simple loop flare 

The simplest analog to a pre-flare circuit conforms to the series circuit configuration shown above.  This presumes a simple loop with opposite magnetic polarities at each end (“foot”) bounded by a much higher density photosphere. We say the footpoints of the loop are effectively “line tied.”  We assume also, without loss of generality, that the circuit is completed below the photosphere using conductive plasma.

The resulting L-C-R circuit has a current flowing parallel to the magnetic field provided the plasma Beta b < < 1. The loop itself is a plasma tube with semi-toroidal geometry,  e.g.

Bearing plasma of resistivity: h  = R L’ / A where R is the resistance, L’ is the total circuit length (in general L’ > L) and A is the uniform cross-sectional area of the tube. The capacitance C = e L’ where L’ is as before and e is the dielectric constant of the plasma.  This is given by:

e  =   1  +  (i 4p s)/ w

where  s is the conductivity and w the plasma frequency: 

(4p n _o e /m_e)^1/2.

The self-inductance is of the order (L/c), where c is the velocity of light.

Current in the system is generated from a 'motional' emf E(t) which is time-dependent. It arises from the relative displacement of the dipole feet at points P1 and P2 on the photosphere, such that:

1)  E(t) =   ò_P1  (v X  B)  dS

where the integration is from P1 to P2 (as the unlabeled upper limit. )  Identification of circuit failure is the onset of the flare in the loop.  This requires  finding the transient solution to the differential equation:         

(2)  L (d² I/ dt² ) + R (dI/dt) + I/C =  dE/ dt

where all symbols are as previously defined. If (2) is divided through by L one obtains:

3)       (d² I/ dt² ) + R/ L (dI/dt) + I /LC =   1/L(dE/ dt)

The solution will be I(t),  a transient response function. This differential equation can be recast in terms of the associated Green’s function with I(t) º G = G(t,T) with the change in back emf  dE/ dt replaced by the Dirac Delta function  d (t – T) and the problem treated from the viewpoint of distributions. 

4)   d² G/ dt²   +  2 a (dG/dt) +   w_o² G =  d (t – T)/ L

where the response a = R/ 2L   and w_o =  1/ Ö(LC), i.e.  the natural frequency at which the dipole oscillates.  Note here that the impulse d (t – T) = 0  for t < T, hence I(t) = 0. This is the “steady state” solution: associated with the ongoing stability or “survival” of the dipole. By contrast, we are only interested in the case for t >  T, e.g. the transient or complementary solution. In particular, the solution we seek, on inverting the appropriate Laplace transforms is[1]:

5)   I(t) =  G(t, T_o) = G(t, 0) =    1/ w  [exp (- at) sin (w t)]                       t > T_o

Where w = [(w_o² -   a²)]^1/2

Not surprisingly, the current impulse becomes infinite when C = 0 and L = 0 at current interruption. In the case considered here, the failure bears a direct analogy to what happens in high power transmission lines when the current is suddenly switched off. That is, there is an explosive release of the inductive energy.

---

[1] The problem posed will resemble that for the forced oscillations of a damped harmonic oscillator such that: x’’ + 2 ax’ + w_o² x = f(t). The Laplace transform of this is analogous to the one in the text for I(t).

---