Monte Carlo Simulations of Radiative Transfer: Basics of Radiative Transfer Theory (Part IIa)

SOURCES FOR CONTENT:

Chandrasekhar, S., 1960. “Radiative Transfer”. Dover. 1.
Choudhuri, A.R., 2010. “Astrophysics for Physicists”. Cambridge University Press. 2.
Boyce, W.E., and DiPrima, R.C., 2005. “Elementary Differential Equations”. John Wiley & Sons. 2.1.

Recall from last time , the radiative transfer equation

$\displaystyle \frac{1}{\epsilon \rho}\frac{dI_{\nu}}{ds}= M_{\nu}-N_{\nu}I_{\nu}, (1)$

where $M_{\nu}$ and $N_{\nu}$ are the emission and absorption coefficients, respectively. We can further define the absorption coefficient to be equivalent to $\epsilon \rho$ . Hence,

$\displaystyle N_{\nu}=\frac{d\tau_{\nu}}{ds}, (2)$

which upon rearrangement and substitution in Eq. (1) gives

$\displaystyle \frac{dI_{\nu}(\tau_{\nu})}{d\tau_{\nu}}+I_{\nu}(\tau_{\nu})= U_{\nu}(\tau_{\nu}). (3)$

We may solve this equation by using the method of integrating factors, by which we multiply Eq.(3) by some unknown function (the integrating factor) $\mu(\tau_{\nu})$ yielding

$\displaystyle \mu(\tau_{\nu})\frac{dI_{\nu}(\tau_{\nu})}{d\tau_{\nu}}+\mu(\tau_{\nu})I_{\nu}(\tau_{\nu})=\mu(\tau_{\nu})U_{\nu}(\tau_{\nu}). (4)$

Upon examining Eq.(4), we see that the left hand side is the product rule. It follows that

$\displaystyle \frac{d}{d\tau_{\nu}}\bigg\{\mu(\tau_{\nu})I_{\nu}(\tau_{\nu})\bigg\}=\mu({\tau_{\nu}})U_{\nu}(\tau_{\nu}). (5)$

This only works if $d(\mu(\tau_{\nu}))/d\tau_{\nu}=\mu(\tau_{\nu})$ . To show that this is valid, consider the equation for $\mu(\tau_{\nu})$ only:

$\displaystyle \frac{d\mu(\tau_{\nu})}{d\tau_{\nu}}=\mu(\tau_{\nu}). (6.1)$

This is a separable ordinary differential equation so we can rearrange and integrate to get

$\displaystyle \int \frac{d\mu(\tau_{\nu})}{\mu(\tau_{\nu})}=\int d\tau_{\nu}\implies \ln(\mu(\tau_{\nu}))= \tau_{\nu}+C, (6.2)$

where $C$ is some constant of integration. Let us assume that the constant of integration is $0$ , and let us also take the exponential of (6.2). This gives us

$\displaystyle \mu(\tau_{\nu})=\exp{(\tau_{\nu})}. (6.3)$

This is our integrating factor. Just as a check, let us take the derivative of our integrating factor with respect to $d\tau_{\nu}$ ,

$\displaystyle \frac{d}{d\tau_{\nu}}\exp{(\tau_{\nu})}=\exp{(\tau_{\nu})},$

Thus this requirement is satisfied. If we now return to Eq.(4) and substitute in our integrating factor we get

$\displaystyle \frac{d}{d\tau_{\nu}}\bigg\{\exp{(\tau_{\nu})}I_{\nu}(\tau_{\nu})\bigg\}=\exp{(\tau_{\nu})}U_{\nu}(\tau_{\nu}). (7)$

We can treat this as a separable differential equation so we can integrate immediately. However, we are integrating from an optical depth $0$ to some optical depth $\tau_{\nu}$ , hence we have that

$\displaystyle \int_{0}^{\tau_{\nu}}d\bigg\{\exp{(\tau_{\nu})}I_{\nu}(\tau_{\nu})\bigg\}=\int_{0}^{\tau_{\nu}}\bigg\{\exp{(\bar{\tau}_{\nu})}U_{\nu}(\bar{\tau}_{\nu})\bigg\}d\bar{\tau}_{\nu}, (8)$

We find that

$\displaystyle \exp{(\tau_{\nu})}I_{\nu}(\tau_{\nu})-I_{\nu}(0)=\int_{0}^{\tau_{\nu}}\bigg\{\exp{(\bar{\tau}_{\nu})}U_{\nu}(\bar{\tau}_{\nu})\bigg\}d\bar{\tau}_{\nu} (9),$

where if we add $I_{\nu}(0)$ and divide by $\exp{(\tau_{\nu})}$ we arrive at the general solution of the radiative transfer equation

$\displaystyle I_{\nu}(\tau_{\nu}) = I_{\nu}(0)\exp{(-\tau_{\nu})}+\int_{0}^{\tau_{\nu}}\exp{(\bar{\tau}_{\nu}-\tau_{\nu})}U_{\nu}(\bar{\tau}_{\nu})d\bar{\tau}_{\nu}. (10)$

This is the mathematically formal solution to the radiative transfer equation. While mathematically sound, much of the more interesting physical phenomena require more complicated equations and therefore more sophisticated methods of solving them (an example would be the use of quadrature formulae or $n$ -th approximation for isotropic scattering).

Recall also that in general we can write the phase function $p(\theta,\phi; \theta^{\prime},\phi^{\prime})$ via the following

$\displaystyle p(\theta,\phi;\theta^{\prime},\phi^{\prime})=\sum_{l=0}^{\infty}\gamma_{l}P_{l}(\cos{\Theta}). (11)$

Let us consider the case for which $l=0$ in the sum given by (11). This then would mean that the phase function is constant

$p(\theta,\phi;\theta^{\prime},\phi^{\prime})=\gamma_{0}=const. (12)$

Such a phase function is consistent with isotropic scattering. The term isotropic means, in this context, that radiation scattered is the same in all directions. Such a case yields a source function of the form

$\displaystyle U_{\nu}(\tau_{\nu})=\frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\gamma_{0}I_{\nu}(\tau_{\nu})\sin{\theta^{\prime}}d\theta^{\prime}d\phi^{\prime}, (13)$

where upon use in the radiative transfer equation we get the integro-differential equation

$\displaystyle \frac{dI_{\nu}(\tau_{\nu})}{d\tau_{\nu}}+I_{\nu}(\tau_{\nu})= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\gamma_{0}I_{\nu}(\tau_{\nu})\sin{\theta^{\prime}}d\theta^{\prime}d\phi^{\prime}. (14)$

Solution of this equation is beyond the scope of the project. In the next post I will discuss Rayleigh scattering and the corresponding phase function.

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