Some personal notes on thermal emission with thin target model

Here are notes for the thin target model of the X-ray emission in solar coronal, induced by high energetic particle.

Thermal Energy:

The Maxwell distribution in the 3d velocity space is

\[f(\vec{v})d^3\vec{v}=(\frac{m}{2\pi k_B T})^{\frac{3}{2}}\exp{(-\frac{m v^2}{2k_B T})}d^3\vec{v}\]

Thus the distribution in the velocity magnitude is,

\[f(v)dv=(\frac{m}{2\pi k_B T})^{\frac{3}{2}}\exp{(-\frac{m v^2}{2k_B T})}4\pi v^2dv\]

Therefore, the total energy in “one” averaged particle is,

\[E=\int_{0}^{\infty}f(v)\frac{1}{2}mv^2dv\\ =\int_{0}^{\infty}(\frac{m}{2\pi k_B T})^{\frac{3}{2}}\exp{(-\frac{m v^2}{2k_B T})}4\pi v^2 \frac{1}{2}mv^2 dv\\ =\frac{3}{2}k_B T\]

Then the total thermal energy for a set of \(N\) particles is \(E_{\rm thermal}=\frac{3}{2}Nk_BT\).

If we estimate the total number as \(\sqrt{EM*V}\), where emission \(EM\) is in the unit of number\(^2\) per volume. \(V\) can be estimated as \(Area^{1.5}\). The emission measure can be obtained from the fitting. \textcolor{red}{Finally, do not forget to multiply the time.}

Thin Target

model 1

The energy distribution for the non-thermal electron is,

\[f(E)dE=A(E/E_p)^{-\delta}dE, E>E_{cutoff}\]

where \(\delta=\gamma-0.5\) and \(\gamma\) is the photon spectrum slope. \(A\) is the normalization parameter for the distribution (\(\int_{E_{cutoff}}^{\infty}f(E)dE=1\)), it is obvious to find \(A=(E_{cutoff}/E_p)^{\delta-1}/{\delta-1}\). \(E_p\) is the pivot energy, you can simply fix it, e.g., 50 keV.

Here, if you use SSW package “OSPEX”, the fitting result is the index \(\delta\) for the electron spectrum. Thus has no need to care the relation of indexes between photon and electron for different model, those relations have already been considered when you select the fitting model.

If you chose the one component in the non-thermal electron with thin target model, i.e., thin1. Then, the energy of the “one” averaged particle that caused the thin target emission is,

\[E_{one}=A\int_{E_{cutoff}}^{\infty}f(E)EdE\\ =A(E_{cutoff}/E_p)^{\delta-2}/(\delta-2)\]

The final total energy is \(E=NA(E_{cutoff}/E_p)^{\delta-2}/(\delta-2)\), N is the total electron number. The estimation of the total number \(N\) is similar to that in the thermal energy section.

In practice, OSPEX gives the fitting result as \(f(E)=aE^{-\delta}\), and the integral \(\int_0^{\infty}f(E)dE=\)electron surface density, equivalently. \(a\) is the value in the unit of \(10^{55}\) cm\(^{-2}\) s\(^{-1}\). Then the final total energy is

\(E_{total}=\int_{E_{min}}^{E_{max}}aE^{-\delta}EdE\) If you want to know the energy in the unit of J or erg, you need to multiply the “AREA” and “TIME”!

model 2

If you chose the “thin2” model, the electron distribution is,

\[f(E)dE=A\left\{ \begin{array}{ll} (E/E_p)^{-\delta_1},& E_{min}\le E < E_b\\ (E/E_p)^{-\delta_2}(E_b/E_p)^{\delta_2-\delta_1},& E_{b}\le E < E_{max} \end{array} \right.\]

The normalization requires that

\[A\{\frac{1}{1-\delta_1}[(E_b/E_p)^{1-\delta_1} -(E_{min}/E_p)^{1-\delta_1}] + \frac{1}{1-\delta_2}(E_b/E_p)^{\delta_2-\delta_1}[(E_{max}/E_p)^{1-\delta_2}-(E_b/E_p)^{1-\delta_2}] \}=1\]

Then, the averaged “one” particle energy is,

\[E_{one}=\int_{E_{min}}^{E_{max}}f(E)EdE\\ =A\{\frac{1}{2-\delta_1}[(E_b/E_p)^{2-\delta_1} -(E_{min}/E_p)^{2-\delta_1}] + \frac{1}{2-\delta_2}(E_b/E_p)^{\delta_2-\delta_1}[(E_{max}/E_p)^{2-\delta_2}-(E_b/E_p)^{2-\delta_2}] \}\]

Then the total energy is the above \(E_{one}\) times the total number. The estimation of the total number \(N\) is similar to that in the thermal energy section.

Comparison

To compare the energies for the two components. You need to convert them to the same unit. For the thermal energy, if you use the SI system the unit is J m\(^{-3}\). Take care the unit of the total electron number from the OSPEX. While the non-thermal energy usually comes with the unit keV cm\(^{-3}\), as the unit of \(E_{cutoff}\) from the OSPEX fitting might be in the unit of keV. Finally, if you assume the emission comes from the same volume, you do not have to multiply the volume parameter for the ratio between the two components. Here, volume is estimated by Area\(^{1.5}\)

As the fitting from the OSPEX, if you set keyword as EFD=0, the normalized parameter a[0] is the plasma density \(\times\) non-thermal electron density \(\times\) radiation volume. Therefore, to estimate the total energy of the non-thermal electron, you need reduce the thermal plasma number. You can estimate it as \(\sqrt{EM\times volume}\), where EM comes from the thermal fitting result and volume can be estimated by Area\(^{1.5}\) Finally, \(a[0]/\{\sqrt{EM\times volume}\}\) is the total non-thermal density that be the same meaning as the a[0] from the OSPEX thick target result.

Thick target model is a little different.

\[Flux=\frac{nNA}{4\pi(AU)^2(mc^2)^2}\int f(\gamma)v_0[\int \frac{\sigma_0v}{dE/dt}dE]dE_0,\]

here, we \(n\) is the target plasma density.

In the thick target flux calculation, you need to integrate a term contains \(1/(dE/dt)\). You can find an approximation \(dE/dt=4\pi r^2 (mc^2)^2nc\ln{\Lambda}/\beta\). Here \(n\) is the target thermal plasma density. Therefore, it is obvious that \(n\) can be cancelled out through the whole integral. This is why we do not find a plasma density in the final fitting result from the thick target model.

The flux from the thin target model is,

\[Flux=\frac{nNA}{4\pi(AU)^2(mc^2)^2}\int f(\gamma)v_0dE_0.\]

Here, you cannot cancel the plasma density.

While for the thermal radiation, all the emission comes from the thermal plasma itself. Thus it is just a simple black body radiation in the X ray energy range.