Appendix D: LTE line-strength distribution for hydrogenic ions

In LTE and using Kramer's formula, the hydrogen line-strength reads

$$k_{\rm L} = \frac{\bar{\chi}_{\rm i}\lambda_{\rm i}}{\rho} \frac{1}{s_{\rm ... ...o \frac{\lambda_{\rm i}}{\rho} gf \left(\frac{n_l}{g_l}-\frac{n_u}{g_u} \right)$$ (D1)

$$\propto \frac{n_1}{\rho} \Bigl(\frac{1}{n'^2} - \frac{1}{n^2}\Bigr)^{-4} ... ...\Bigr) \Bigl({\rm e}^{-h\nu_l/kT_{\rm e}} - {\rm e}^{-h\nu_u/kT_{\rm e}}\Bigr),$$

where n₁ is the ground-state occupation number, $h\nu_l, h\nu_u$ are the excitation energies of the lower and upper level, and the dependence on the transition wavelength $\lambda_{\rm i}$ transforms into an additional power of the second factor. Realizing that the appropriate variable for a further expansion is given by x=(n/n')³ < 1, and denoting

$$\tilde k(n',T_{\rm e}) = \frac{k_{\rm L}\exp(\frac{h\nu_1}{kT_{\rm e}})}{C' n'^2}$$ (D2)

with $h\nu_1$ the ground-state ionization energy and C' a factor absorbing all "constants'' (most important: $n_1/\rho$) in the equation before, we can write

$$\tilde k(n,n',T_{\rm e}) \frac{x}{(1 - x^{2/3})^4} \bigl(K(T_{\rm e},n') - K(T_{\rm e},n')^{x^{2/3}} \bigr)$$ =

$$K(T_{\rm e},n') \exp \left(\frac{h\nu_1}{kT_{\rm e}n'^2}\right).$$ = (D3)

This equation can be expanded in powers of x, inverted and solved for n:

$$x(\tilde k,n', T_{\rm e}) \frac{\tilde k}{K -1} - \left(\frac{1}{K -1}\right)^{8/3} \times$$ =

$$\times \bigl[4(K-1) -\ln(K) \bigr] \tilde k^{5/3} +O(\tilde k)^{7/3}$$

$$n(k_{\rm L},n',T_{\rm e}) n' x^{-1/3} = n' \left(\frac{K-1}{\tilde k}\right)^{{1 \over 3}} + O(\tilde k)^{1/3}.$$ = (D4)

This expression is valid for not too large line-strengths $k_{\rm L}< C'\exp(-h\nu_1/kT_{\rm e})$. Summing up again the number of lines with line-strengths larger/equal than a given value, we finally obtain the result given in (50),

 

$$N(k_{\rm L},T_{\rm e}) = \sum_{n'=1}^{n'_{\rm max}} n(k_{\rm L},n',T_{\rm e}) - n' =$$

$$k_{\rm L}^{-{1 \over 3}} \bigl(C'{\rm e}^{-\frac{h\nu_1}{kT_{\rm e}}}\bigr)^{1 \over 3}f(T_{\rm e}) - \frac{n'_{\rm max}(n'_{\rm max}+1)}{2},$$ = (D5)

$$f(T_{\rm e}) = \sum_{n'=1}^{n'_{\rm max}} {n'}^{5 \over 3} \left(... ...r 3} \,\,\, {\rm and} \,\,\, k_{\rm L}< C'{\rm e}^{-\frac{h\nu_1}{kT_{\rm e}}}.$$