A Magnetostatic Grad–Rubin Code for Coronal Magnetic Field Extrapolations

Abstract

The coronal magnetic field cannot be directly observed, but, in principle, it can be reconstructed from the comparatively well observed photospheric magnetic field. A popular approach uses a nonlinear force-free model. Non-magnetic forces at the photosphere are significant, meaning the photospheric data are inconsistent with the force-free model, and this causes problems with the modeling (De Rosa et al., Astrophys. J. 696, 1780, 2009). In this paper we present a numerical implementation of the Grad–Rubin method for reconstructing the coronal magnetic field using a magnetostatic model. This model includes a pressure force and a non-zero magnetic Lorentz force. We demonstrate our implementation on a simple analytic test case and obtain the speed and numerical error scaling as a function of the grid size.

Appendices

Appendix A

A potential field B ₀ satisfying

$$ \nabla \times \mathbf {B}_0 = 0 $$

(47)

and the boundary conditions at z=0

$$ \mathbf {B}_0 \cdot \hat {\mathbf {z}} = B_{\mathrm{obs}} $$

(48)

is used to initiate the Grad–Rubin iteration. The appropriate field with periodic side boundary conditions is given in Equations (24)–(26) in the text. For the choice of closed side boundaries the appropriate choice is

(49)

(50)

and

$$ B_{0z}(x,y,z) = -\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} c_{mn} k \sinh \bigl(k[z-L]\bigr)\cos(k_m x)\cos(k_n y) , $$

(51)

where k _m =πm/L, k _n =πn/L, $k^{2} = k_{m}^{2}+k_{n}^{2}$, and where the coefficients c _mn are given by

$$ c_{mn} = \frac{4}{L^2} \int_0^{L} \int_0^{L} \mathrm{d}x\,\mathrm{d}y\, B_{\mathrm{obs}}(x,y) \cos(k_mx)\cos(k_ny)/\sinh(kL) . $$

(52)

Equations (49)–(52) may be evaluated on a grid with N ³ points in ∼ N ³log(N) operations using fast sine and cosine transforms (Poularikas 1996).

Appendix B

This appendix presents the solution for the current-carrying component of the test field, B _c , with closed top and side boundary conditions. The magnetic field may be calculated from a vector potential A in the Coulomb gauge (∇⋅A=0) using ∇×A=B _c . The vector potential satisfies the vector Poisson equation (Jackson 1998):

$$ \nabla^2 \mathbf {A} = -\mu_0 \mathbf {J}. $$

(53)

The boundary conditions for B _c on the six plane boundaries are

$$ \mathbf {B}_c \cdot \hat {\mathbf {n}} = 0, $$

(54)

where $\hat {\mathbf {n}}$ is the unit vector normal to the boundary. The corresponding boundary conditions for the vector potential in the Coulomb gauge are (Amari, Boulmezaoud, and Mikic 1999):

$$ \partial_n \mathbf {A} =0, $$

(55)

and

$$ \mathbf {A}_t =0, $$

(56)

where A _t denotes the component of A transverse to the boundary, and ∂ _n A denotes the normal derivative.

The vector potential A satisfying the boundary conditions Equations (55) – (56) can be written as a Fourier series:

(57)

(58)

and

$$ A_z = \sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\sum _{p=0}^{\infty }a^{(3)}_{mnp} \sin(k_mx) \sin(k_ny)\cos(k_pz), $$

(59)

where k _m =πm/L, k _n =πn/L, and k _p =πm/L. In the following, we solve the Poisson equation to find an expression for a _mnp . The process is demonstrated for the A _x component, but the approach is similar for the other components.

Substituting Equation (57) into the Poisson equation (Equation (53)) gives

$$ A_x =\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\sum _{p=0}^{\infty }a^{(1)}_{mnp} k^2 \cos(k_mx)\sin(k_ny)\sin(k_pz) = \mu_0 J_x(x,y,z), $$

(60)

where $k^{2} = k_{m}^{2}+k_{n}^{2}+k_{p}^{2}$.

Applying the standard orthogonality relations (where δ _mn is the Kronecker delta):

(61)

(62)

and

$$ \int_{0}^{L} \sin(\pi m s/L)\cos(\pi n s/L)\,\mathrm{d}s = 0 $$

(63)

yields

$$ a^{(1)}_{mnp} = \frac{8\mu_0}{L^3} \int_{0}^{L} \int_{0}^{L}\int_{0}^{L} J_x(x,y,z) \cos(k_mx)\sin(k_ny) \sin(k_pz)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z. $$

(64)

Similar expressions apply for the other two coefficients:

$$ a^{(2)}_{mnp} = \frac{8\mu_0}{L^3} \int_0^{L} \int_0^{L}\int_0^{L} J_y(x,y,z) \sin(k_m x)\cos(k_n y) \sin(k_p z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z, $$

(65)

and

$$ a^{(3)}_{mnp} = \frac{8\mu_0}{L^3} \int_0^{L} \int_0^{L}\int_0^{L} J_z(x,y,z) \sin(k_m x)\sin(k_n y) \cos(k_p z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z. $$

(66)

The magnetic field is obtained by evaluating B _c =∇×A. The components of B _c are

(67)

(68)

and

$$ B_{cz}(x,y,z) = \sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\sum _{p=0}^{\infty }\bigl[k_m a^{(2)}_{mnp}-k_n a^{(1)}_{mnp} \bigr] \cos(k_m x) \cos(k_n y)\sin(k_p z) /k^2 . $$

(69)

The solution given by Equations (67)–(69) may be computed on a grid with N ³ points in ∼ N ³log(N) operations using a combination of fast sine and cosine transforms.