The equation for radiative transfer is $$ \omega \cdot \nabla I = \kappa(B - I) $$ where $I$ is the intensity of radiation, $\omega$ is the ray direction, $\kappa$ the absorption coefficient, $B$ the Planck function. Here, $\kappa=\kappa(\nu)$ (i.e., it depends on the radiation frequency $\nu$).

Often, though, the gray model, where $\kappa$ doesn't depend on $\nu$, is used. How can this be justified? What assumptions are used to obtain the gray model from the non-gray model?

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    $\begingroup$ I've modified your question to be more clear in hopes to reopen it. I believe that I kept the intent of your question in tact, but please check this so that we are answering the question you want answered. $\endgroup$
    – Kyle Kanos
    Jul 7, 2014 at 2:38

1 Answer 1


The gray model results from starting with the assumption that we have a plane-parallel slab:

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The light ray from the source (i.e., the star's atmosphere) travels at some angle, $\theta$, from normal, $z=0$. Since the light is coming from an angle, we need to account for that by modifying the radiative transfer equation to have a vertical optical depth, defined by $$ \tau_{\lambda,v}(z)=\int_z^0\kappa_\lambda\rho\,dz $$ which gives us $$ \omega\frac{dI_\lambda}{d\tau_{\lambda,v}}=I_\lambda-B_\lambda $$ with $\omega=\cos\theta$. Since the path length of the ray does not offer a unique geometric depth of the atmosphere, we cannot use $\nabla I$ and must use the above form for the radiative transfer equation.

In most stellar atmospheres, the primary source of opacity is the photoionization of H$^-$ ions. Fortunately, the opacity from photoionzation of hydrogen ions does not vary rapidly with wavelength, so we can approximate the opacity as $$ \kappa_\lambda\approx \bar\kappa $$ where $\bar\kappa$ is some average opacity, usually the Rosseland mean opacity.

  • $\begingroup$ How to justify the using of the gray model if, for example, some combustion system is considered but not atmosphere processes? The absorption coefficient may oscillate among spectral lines. $\endgroup$
    – jokersobak
    Jul 8, 2014 at 5:08
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    $\begingroup$ If the opacity does vary a lot with wavelength, then you cannot use the gray model as an accurate representation. I would think, though, that it still would be good for a first-order model. $\endgroup$
    – Kyle Kanos
    Jul 8, 2014 at 10:40

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