I'm currently looking at the equation of radiation transfer given by

$$\frac{1}{c}\frac{\partial }{\partial t}I_{\gamma} + \hat{\Omega} \cdot \nabla I_{\gamma} + \left ( k_{\gamma, s} + k_{\gamma, a} \right ) = j_{\gamma} +\frac{1}{4 \pi} k_{\gamma, s}\int_{\Omega} I_{\gamma } \cdot d \Omega\tag{1.1}$$

where $j_{\gamma}$ is the emission coefficient, $k_{\gamma,s}$ is the scattering opacity and $k_{\gamma,a}$ is the absorption opacity

I'm exploring some reasonable assumption that may simplified the above expression in (1.1) for modelling radiative transfer of heat from the atmosphere to terrestrial solar cells.

For one, we may assume that the scattering opacity $k_{\gamma,s}=0$.

Further, the first and second term closely resembles that of the heat equation with the addition of $\hat{\Omega}$

Recalling the heat equation: $$\frac{\partial }{\partial t}U-\alpha\nabla U=0$$

(1.1) is reduced to

$$\nabla I_{\gamma}+\hat{\Omega} \cdot I_{\gamma} + k_{\gamma,a} I_{\gamma}=j_{\gamma}+\frac{1}{4 \pi}k_{\gamma,s}\int_{\Omega}I_{\gamma}\cdot d\Omega$$

In this reduced form of (1.1), the second term is some scalar value.

Is the reduced form of (1.1) valid given the assumptions? Can this be further reduced on physical grounds?

  • $\begingroup$ Something is definitely wrong here, since you're adding a vector to a scalar. $\endgroup$ Commented Feb 12, 2018 at 6:35
  • $\begingroup$ There are many methods to simplify and solve the exact radiative transport equation. This has applications in atmospheric physics, astrophysics, 3D graphics for computer games, nuclear weapons design... So you'll find dedicated books about that in all those fields. In Astrophysics "Mihalas & Mihalas, Foundations of Radiation hydrodynamics" is one of the standard books. I recommend having a look into those resources, a physics.SE post will necessarily be too short. $\endgroup$ Commented Jul 17, 2019 at 10:25

1 Answer 1


There seems to be a problem with the starting equation and the equation you ended up with. Your intuition is on the right track. The equation you wrote down is actually the advection equation not the diffusion equation. The radiative transfer equation is the just the continuity equation for light transport.

To recap, the continuity equation keeps track of all the gains and losses of a conserved quantity $u$. In 1D

$$ \frac{\partial u}{\partial t} = \frac{\partial \mathcal{F}}{\partial x} + s(t, x, u) $$

Where $\mathcal{F}$ is the flux of the conserved quantity and $s$ is the source function which maybe depend on the other variables.

To recover the radiative transfer equation in the limit of zero scattering, let,

$$ u = n_{\gamma} \\ \mathcal{F} = \hat{n} \cdot I_{\gamma} \\ s = - k_{\gamma, a} I_{\gamma} + j_{\gamma} $$

  • $n_{\gamma}$ $[\text{eV}^{-1} \text{cm}^{-3}]$ is the number density of photons in spectral interval $(E, E + dE)$. This is related to the photon flux by the following, where $n$ is the refractive index and $c$ is the speed of light,

$$n_{\gamma} dE = \frac{n}{c} dE \int I_{\gamma} d\Omega$$

  • $\hat{n}\cdot I_{\gamma}$ $[\text{eV}^{-1} \text{cm}^{-2} \text{Sr}^{-1} \text{s}^{-1}]$ is the photon flux in spectral interval $(E, E+dE)$, transported along direction $\hat{n}$, across an element of area, confined to element of solid angle per unit time. This is the specific intensity that Chandrasekhar refers to, but with different units (photons rather than energy).

  • $k_{\gamma, a}$ $[cm^{-1}]$ is the absorption coefficient.

  • $j_{\gamma}$ $[\text{eV}^{-1} \text{cm}^{-3} \text{Sr}^{-1} \text{s}^{-1}]$ is the emission coefficient.

It is not necessary to define the units of the quantities exactly as I have done here. You could consider using integrated values over photon energy and solid angle to remove the dependence on those dimensions.

If you apply the same logic using the nabla operator, you can recover the original equation (in the limit of zero scattering).

$$ \frac{1}{c}\frac{\partial I_{\gamma}}{\partial t} = -\hat{n}\cdot\nabla I_{\gamma} -k_{\gamma, a} I_{\gamma} + j_{\gamma} $$

For your application you can probably assume local thermodynamic equilibrium and steady-state, which will give the classic radiative transfer equation from Chandrasekhar’s book,

$$ \hat{n}\cdot\frac{1}{k_{\gamma, a}} \nabla I_{\gamma} = -I_{\gamma}+ B_{\gamma}(T) $$

The source term is derived as follows,

$$ s = -B_{\text{ein}} n_{\gamma} + j_{\gamma} $$

where $B_{\text{ein}}$ is the Einstein B-coefficient. This is related to absorption coefficient by,

$$ B_{\text{ein}} = \frac{c}{n}k_{a, \gamma} $$

and by using the definition from above,

$$ n_{\gamma} = \frac{n}{c} I_{\gamma} $$

we arrive at

$$ s = - k_{\gamma, a} I_{\gamma} + j_{\gamma} $$


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