I am currently studying radiative transfer. I have seen the stationary equation of radiative transfer written in two different ways:

1.$$\dfrac{\partial{I}(\mathbf{r}, \mathbf{s})}{\partial{s}} = - \mu_t I(\mathbf{r}, \mathbf{s}) + \dfrac{\mu_s}{4 \pi} \int_{4 \pi} I(\mathbf{r}, \mathbf{s}^\prime) p(\mathbf{s}, \mathbf{s}^\prime) d \Omega^\prime,$$

2.$$\dfrac{\partial{I}(\mathbf{r}, \hat{s})}{\partial{s}} = - \mu_t I(\mathbf{r}, \hat{s}) + \mu_s \int_{4 \pi} I(\mathbf{r}, \hat{s}^\prime) p(\hat{s}, \hat{s}^\prime) d \Omega^\prime$$

$I$ is the spectral radiance.
$d\Omega^\prime$ is the unit solid angle.
$p$ is the phase function of the scattering.
$\mu_s$ is the scattering coefficient.
$\mu_t = \mu_s + \mu_a$, where $\mu_a$ is the absorption coefficient.

The only difference seems to be the presence of the factor $\dfrac{1}{4\pi}$ on the integral term. However, it isn't clear to me which one of these is the correct equation (or if both of them are correct in some way that I am not understanding, which is possible). I would greatly appreciate it if people would please take the time to clarify this.


To be clear: I have seen it most commonly (nearly always) with the $\dfrac{1}{4\pi}$ factor present.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Jul 28, 2020 at 12:09

1 Answer 1


I believe the first version of the equation is correct (although it ignores the intrinsic emission coefficient of the material, $j_\nu$, which should appear as an additive term on the RHS. However, it may boil down to how the scattering phase function has been defined.

The first term on the RHS represents light taken out of the "beam" due to scattering and absorption. The second term represents light put back into the beam due to scattering of light travelling in all other directions. Hence it is an integral over all solid angle.

Normally, the phase function would be defined, so that $$\frac{1}{4\pi}\oint p\ d\Omega = 1. $$ i.e. the scattered light must travel in some direction, and if the scattering were isotropic, $p=1$.

To work out the scattered specific intensity along a particular path, you insert the specific intensity at the point of scattering (with its own angular dependence) into the integral. That is the origin of the scattering term in your first equation.

But I suppose it is possible to define a scattering phase function where the factor $(4\pi)^{-1}$ is already part of $p$, which would lead to the second equation. Perhaps you could say where you have seen the second form of the equation?

  • $\begingroup$ Thanks for the answer, Rob. With your comment "but I suppose it is possible to define a scattering phase function where the factor $(4\pi)^{−1}$ is already part of $p$, which would lead to the second equation", I think I found the culprit. It is defined that $p(\hat{s}, \hat{s}^\prime) = p(\theta)$, and it is then said that we have the normalization $\int_0^\pi p(\theta) 2\pi \sin(\theta) \ d\theta = 1$. It is then defined that the Henyey–Greenstein function is used for the phase function: $p(\theta) = \dfrac{1}{4\pi} \cdot \dfrac{1 - g^2}{(1 + g^2 - 2g \cos(\theta))}^{3/2}$. [...] $\endgroup$ Jul 28, 2020 at 11:05
  • $\begingroup$ [...] So it indeed seems the the $\dfrac{1}{4\pi}$ is embedded in the scattering phase function $p(\hat{s}, \hat{s}^\prime) = p(\theta)$. Thanks again; I really appreciate you taking the time to clarify this. $\endgroup$ Jul 28, 2020 at 11:13
  • $\begingroup$ Typo: The Henyey–Greenstein phase function should have been $$p(\theta) = \dfrac{1}{4\pi} \cdot \dfrac{1 - g^2}{(1 + g^2 - 2g \cos(\theta))^{3/2}}$$ astro.umd.edu/~jph/HG_note.pdf $\endgroup$ Aug 4, 2020 at 0:01

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