Radiation transfer

We extend here and improve the work done by LangbeinEtAl90 by resorting to a more detailed description of the radiation transfer Castor72.

Consider a radiation field; we place a surface element $d \sigma$ with a surface normal ${\bf n}$, see Fig.(2.1); the radiation energy which passes through $d \sigma$ per unit time at angle $\theta$ to ${\bf n}$ within a small range of solid angle $d \omega$ given by the directional angles $\theta$ and $\phi$ is

$$dE=I_{\nu}(\theta, \phi) \mu d\nu d \sigma d \omega$$ (53)

where $\mu = \cos \theta$.

The radiation intensity $I_{\nu}(\theta, \phi)$ is defined as the amount of energy that passes through a surface normal to the direction ($\theta$,$\phi$) per unit solid angle (1 steradian) and unit frequency range (1 Hz) in one second. The intensity of the total radiation is given by integrating over all frequencies,

$$I=\int_{0}^{\infty} I_{\nu}d \nu$$ (54)

Figure 2.1: Radiation intensity definition

The three radiation moments (the moments of order zero, one and two) are defined by:

$$J= \int_{0}^{\infty} J_{\nu} d{\nu}=\int_{0}^{\infty} d{\nu} \frac{1}{2} \int_{-1}^{+1} I_{\nu} d{\mu}$$

$$H= \int_{0}^{\infty} H_{\nu} d{\nu}=\int_{0}^{\infty} d{\nu} \frac{1}{2} \int_{-1}^{+1} I_{\nu} \mu d{\mu}$$ (55)

$$K= \int_{0}^{\infty} K_{\nu} d{\nu}=\int_{0}^{\infty} d{\nu} \frac{1}{2} \int_{-1}^{+1} I_{\nu} \mu^2 d{\mu},$$

The moment of order zero is related to the density of energy of the field of radiation $E_{\rm rad}$, the moment of order one to the flux of radiation $F_{\rm rad}$ and the moment of order one to the radiation pressure $p_{\rm rad}$,

$$E_{\rm rad}= \frac{4\pi}{c}J$$

$$F_{\rm rad}= 4\pi H$$ (56)

$$p_{\rm rad}= \frac{4 \pi}{c}K$$

The transfer of radiation in a spherically symmetric moving medium is considered taking into account the contributions which are of the order of the flow velocity divided by the velocity of light; we include also the variation from the centre up to the atmosphere of the Eddington factor $f_{\rm Edd}=K/J$, where $K$ and $J$ are the radiation moments; $f_{\rm Edd}$ is obtained from a numerical solution of the equation of radiation transfer in spherical geometry Yorke80. We get the radiation transfer equations by re-writing the frequency-integrated moment equations from Castor72:

$${\frac{1}{c} \frac{\partial J}{\partial t}+ \frac{\partial H}{\pa... ...g}- \frac {J(1+f_{\rm Edd})}{c} \frac {\partial \ln \rho_{\rm g}}{\partial t}=}$$

$$\kappa_{\rm abs}(B-J)$$ (57)

$$\frac{1}{c} \frac{\partial H}{\partial t}+\frac{\partial (J f_{\r... ...ac{2u_{\rm g}}{cr} H-\frac{2}{c}\frac{\partial \ln \rho_{\rm g}}{\partial t} H=$$

$$-\kappa_{\rm ext} \rho_{\rm g} H$$ (58)

In the equations $\kappa_{\rm abs}$ and $\kappa_{\rm ext}$ are the absorption and extinction coefficients per unit mass

$$\kappa_{\rm abs}=\frac{\rho_{\rm g} \Lambda(T)}{B},~ \kappa_{\rm ext}=\rho_{\rm g}(\kappa_{\rm abs}+\kappa_{\rm scatt}),$$ (59)

$\Lambda(T)$ is the cooling function, $B$ the Planck function and $\kappa_{\rm scatt}$ the scattering coefficient per unit mass. We have made use of $\partial M_{\rm r}/ \partial r=4 \pi^2 \rho$, $f_{\rm Edd}=K/J$, and the Kirchhoff's law, $B_{\nu}=j_{\nu}/\kappa_{\nu}$ ($j_{\nu}$ is the emission coefficient), so that the right-hand terms in Castor72 are the corresponding given here.

We now look for the logarithmic variable version of both equations; for this aim, we divide Eq.(2.63) by $J$ and multiply Eq.(2.64) by $2c/(5p_{\rm g}v_{\rm g})$,

$$\frac{1}{c}\frac{\partial \ln J}{\partial t}+\frac{5}{2J}\frac{\p... ...ac{5}{Jr}p_{\rm g}v_{\rm g}-\frac{3f_{\rm Edd}-1}{cr}u_{\rm g} -(1+f_{\rm Edd})$$

$$\frac{1}{c}\frac{\partial \ln \rho_{\rm g}} {\partial t}= \frac{\kappa_{\rm abs}}{J} (B-J)$$ (60)

$$\frac{\partial \ln v_{\rm g}}{\partial t}+\frac{\partial \ln p_{\... ...}+ \frac{2c}{5}\frac{3f_{\rm Edd}-1}{rp_{\rm g}v_{\rm g}}J-\frac{2u_{\rm g}}{r}$$

$$-2\frac{\partial \ln \rho_{\rm g}}{\partial t}= -c \kappa_{\rm ext}\rho_{\rm g}$$ (61)

Where we have substituted $H=5p_{\rm g}v_{\rm g}/2$.