A Nonlinear Force-Free Magnetic Field Approximation Suitable for Fast Forward-Fitting to Coronal Loops. III. The Free Energy

TL;DR

This paper introduces a fast method to estimate the free magnetic energy in solar active regions using an analytical NLFFF approximation, revealing a correlation with flare activity and providing more consistent energy estimates than other codes.

Contribution

It presents a new analytical approximation and a fast forward-fitting code for NLFFF, along with two methods to estimate free energy and their validation on observed data.

Findings

Free energies range from 1% to 10% of potential energy.

The free energy correlates exponentially with GOES flare flux.

The new methods show less uncertainty compared to other NLFFF codes.

Abstract

An analytical approximation of a nonlinear force-free magnetic field (NLFFF) solution was developed in Paper I, while a numerical code that performs fast forward-fitting of this NLFFF approximation to a line-of-sight magnetogram and coronal 3D loops has been described and tested in Paper II. Here we calculate the free magnetic energy $E_{\rm free}=E_{\rm N}-E_{\rm P}$, i.e. the difference of the magnetic energies between the nonpotential field and the potential field. A second method to estimate the free energy is obtained from the mean misalignment angle change $\Delta\mu=\mu_{\rm P}-\mu_{\rm N}$ between the potential and nonpotential field, which scales as $E_{\rm free}/E_{\rm P} \approx \tan^2{(\Delta\mu)}$. For four active regions observed with STEREO in 2007 we find free energies in the range of $q_{\rm free}=(E_{\rm free}/E_{\rm P}) \approx 1%-10%$, with an uncertainty of less than $\pm 2%$ between the two methods, while the free energies obtained from 11 other NLFFF codes exhibit a larger scatter of order $\approx\pm10%$. We find also a correlation between the free magnetic energy and the GOES flux of the largest flare that occurred during the observing period, which can be quantified by an exponential relationship, $F_{\rm GOES} \propto \exp{(q_{\rm free}/0.015)}$, implying an exponentiation of the dissipated currents.