On Thermal Runaway Electrons and the Polarization of X-ray Emission in Solar Flares

Abstract

A model for the thermal runaway of electrons in solar flares based on an approximate analytical solution of the kinetic equation with a linearized Landau collision integral is considered. Coulomb collisions are shown to make the flux of runaway electrons with a very high temperature nearly isotropic. The derived electron distribution function is used to estimate the polarization of the hard X-ray bremsstrahlung of solar flares. Since the anisotropy of the flux of electrons is low, the polarization of their bremsstrahlung does not exceed 3–4% at an energy of about 15 keV, which creates great difficulties for future extra-atmospheric polarization observations. Furthermore, the Compton scattering of hard X-ray emission by the photosphere and magnetic field inhomogeneity can severely distort the expected results and can make their interpretation difficult. The prospects for space experiments to measure the polarization of the hard X-ray emission from solar flares are discussed.

APPENDIX

Below we give a detailed derivation of the basic kinetic equation (5) that we solve in this paper. Let us rewrite the original equation (3) by taking into account (4) and neglecting the derivative ∂f_v/∂t (see Section 2.1):

$$\begin{gathered} {v}\cos \theta \frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial x}} - \frac{1}{{{{{v}}^{2}}}}\frac{\partial }{{\partial {v}}} \\ \, \times \left[ {{{{v}}^{2}}{{\nu }_{{{\text{coll}}}}}({v})\left( {\frac{{{{k}_{{\text{B}}}}{{T}_{2}}}}{{{{m}_{e}}}}\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial {v}}} + {v}{{f}_{{\mathbf{v}}}}} \right)} \right] \\ \, - {{\nu }_{{{\text{coll}}}}}({v})\frac{\partial }{{\partial \cos \theta }}\left( {{{{\sin }}^{2}}\theta \frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial \cos \theta }}} \right) = 0. \\ \end{gathered} $$

((A.1))

Substituting the expression for the Coulomb collision frequency of superhot electrons with electrons and protons in the hot plasma

$${{\nu }_{{{\text{coll}}}}}({v}) = \frac{{4\pi {{n}_{2}}{{e}^{4}}}}{{m_{e}^{2}{{{v}}^{3}}}}\ln \Lambda $$

into (A.1), we obtain the equation

$$\begin{gathered} {v}\cos \theta \frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial x}} - \frac{1}{{{{{v}}^{2}}}}\frac{\partial }{{\partial {v}}} \\ \, \times \left[ {{{{v}}^{2}}\frac{{4\pi {{n}_{e}}{{e}^{4}}}}{{m_{e}^{2}{{{v}}^{3}}}}\ln \Lambda \left( {\frac{{{{k}_{{\text{B}}}}{{T}_{2}}}}{{{{m}_{e}}}}\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial {v}}} + {v}{{f}_{{\mathbf{v}}}}} \right)} \right] \\ \, - \frac{{4\pi {{n}_{e}}{{e}^{4}}}}{{m_{e}^{2}{{{v}}^{3}}}}\ln \Lambda \frac{\partial }{{\partial \cos \theta }}\left( {{{{\sin }}^{2}}\theta \frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial \cos \theta }}} \right) = 0. \\ \end{gathered} $$

((A.2))

To pass from the original variables (x, ${v}$, θ) to the dimensionless variables (s, z, μ), let us make the following substitutions in (A.2):

$$\partial x = \frac{{{{{({{k}_{{\text{B}}}}{{T}_{1}})}}^{2}}}}{{\pi {{n}_{2}}{{e}^{4}}\ln \Lambda }}\partial s,\quad {v} = {{\left( {\frac{{2{{k}_{{\text{B}}}}{{T}_{1}}}}{{{{m}_{e}}}}} \right)}^{{1/2}}}{{z}^{{1/2}}},$$

$$\cos \theta = \mu .$$

We will obtain

$$\begin{gathered} {{\left( {\frac{{2{{k}_{{\text{B}}}}{{T}_{1}}}}{{{{m}_{e}}}}} \right)}^{{1/2}}}{{z}^{{1/2}}}\mu \frac{{\pi {{n}_{2}}{{e}^{4}}\ln \Lambda }}{{{{{({{k}_{{\text{B}}}}{{T}_{1}})}}^{2}}}}\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial s}} - {{\left( {\frac{{{{m}_{e}}}}{{2{{k}_{{\text{B}}}}{{T}_{1}}}}} \right)}^{{3/2}}} \\ \times 2{{z}^{{ - 1/2}}}\frac{\partial }{{\partial z}}\left[ {\left( {\frac{{2{{k}_{{\text{B}}}}{{T}_{1}}}}{{{{m}_{e}}}}} \right)\frac{{4\pi {{n}_{2}}{{e}^{4}}\ln \Lambda }}{{m_{e}^{{1/2}}{{{(2{{k}_{{\text{B}}}}{{T}_{1}})}}^{{3/2}}}}}} \right. \\ \left. { \times \left( {\left( {\frac{{2{{k}_{{\text{B}}}}{{T}_{2}}}}{{{{m}_{e}}}}} \right){{{\left( {\frac{{{{m}_{e}}}}{{2{{k}_{{\text{B}}}}{{T}_{1}}}}} \right)}}^{{1/2}}}\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial z}} + {{{\left( {\frac{{2{{k}_{{\text{B}}}}{{T}_{1}}}}{{{{m}_{e}}}}} \right)}}^{{1/2}}}{{f}_{{\mathbf{v}}}}} \right)} \right] \\ - \frac{{4\pi {{n}_{2}}{{e}^{4}}\ln \Lambda }}{{m_{e}^{{1/2}}{{{(2{{k}_{{\text{B}}}}{{T}_{1}})}}^{{3/2}}}{{z}^{{3/2}}}}}\frac{\partial }{{\partial \mu }}\left( {(1 - {{\mu }^{2}})\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial \mu }}} \right) = 0. \\ \end{gathered} $$

((A.3))

Hence we have

$$\begin{gathered} \frac{4}{{m_{e}^{{1/2}}{{{(2{{k}_{{\text{B}}}}{{T}_{1}})}}^{{3/2}}}}}{{z}^{2}}\mu \frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial s}} - 2z\left( {\frac{{{{m}_{e}}}}{{2{{k}_{{\text{B}}}}}}} \right)\frac{\partial }{{\partial z}} \\ \times \left[ {\frac{4}{{m_{e}^{{1/2}}{{{(2{{k}_{{\text{B}}}}{{T}_{1}})}}^{{3/2}}}}}\left( {\left( {\frac{{2{{k}_{{\text{B}}}}}}{{{{m}_{e}}}}} \right)\frac{{{{T}_{2}}}}{{{{T}_{1}}}}\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial z}}} \right.} \right. \\ \left. {\left. { + \left( {\frac{{2{{k}_{{\text{B}}}}}}{{{{m}_{e}}}}} \right){{f}_{{\mathbf{v}}}}} \right)} \right] - \frac{4}{{m_{e}^{{1/2}}{{{(2{{k}_{{\text{B}}}}{{T}_{1}})}}^{{3/2}}}}}\frac{\partial }{{\partial \mu }}\left( {(1\, - \,{{\mu }^{2}})\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial \mu }}} \right)\, = \,0. \\ \end{gathered} $$

((A.4))

Given τ = T₂/T₁, we transform (A.4) to

$$\begin{gathered} {{z}^{2}}\mu \frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial s}} - 2z\left( {\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial z}} + \tau \frac{{{{\partial }^{2}}{{f}_{{\mathbf{v}}}}}}{{\partial {{z}^{2}}}}} \right) \\ - \frac{\partial }{{\partial \mu }}\left( {(1 - {{\mu }^{2}})\frac{{\partial {{f}_{{\mathbf{v}}}}}}{{\partial \mu }}} \right) = 0. \\ \end{gathered} $$

((A.5))

We substitute f_v = e^–φ into Eq. (A.5) and obtain the sought-for equation in the dimensionless variables (s, z, μ) after differentiation and simple transformations:

$$\begin{gathered} {{z}^{2}}\mu \frac{{\partial \varphi }}{{\partial s}} - 2z\frac{{\partial \varphi }}{{\partial z}}\left( {1 - \tau \frac{{\partial \varphi }}{{\partial z}}} \right) - 2\tau z\frac{{{{\partial }^{2}}\varphi }}{{\partial {{z}^{2}}}} + 2\mu \frac{{\partial \varphi }}{{\partial \mu }} \\ \, + (1 - {{\mu }^{2}}){{\left( {\frac{{\partial \varphi }}{{\partial \mu }}} \right)}^{2}} - (1 - {{\mu }^{2}})\left( {\frac{{{{\partial }^{2}}\varphi }}{{\partial {{\mu }^{2}}}}} \right) = 0. \\ \end{gathered} $$

((A.6))

Q.E.D.