Having taking courses in both physics and nuclear engineering, I've noticed that the two fields tend to describe photons/light in two different settings.

In nuclear engineering, the radiative transfer equation (Boltzmann transport equation for photons) is often used. For example, as noted in the Wikipedia article, simulations for radiation therapy treatments solve this equation to model the dose applied to a patient. From what I understand, this equation treats light essentially as particles.

In another more physics/E&M-oriented class that I'm taking, light is described via the Maxwell equations. I'm not a physicist so I don't know that much about those equations, but it seems like it's an entirely unrelated/different approach for describing light. Here, it seems like light is treated more as a wave as opposed to particles. In this class I'm taking, we see that, under certain assumptions, the Maxwell equations eventually lead to a Helmholtz equation, which seems very different from the Boltzmann transport equation. (For starters, there's a second order spatial derivative instead of a first-order derivative.)

Can someone help me reconcile the two approaches? i.e., how is it that we can look at light from such drastically different points of view? I know that there's the whole thing about light being both a particle and a wave, but I don't see how the two approaches are even related. Are there certain frequency ranges where the transport equation is more applicable or something?

Any relevant references or explanations would be appreciated. Thanks!

  • 1
    $\begingroup$ Doesn't the RTE describe the time-evolution of the intensity of the wave, not the photons themselves? $\endgroup$
    – Kyle Kanos
    Feb 5, 2015 at 20:16
  • $\begingroup$ Ah hm.. I see. And so, it would be physically incorrect to view the intensity of the wave as composed of individual photons? $\endgroup$
    – nukeguy
    Feb 5, 2015 at 21:31
  • $\begingroup$ Did you ever find any good references relating RTE to Maxwell? $\endgroup$
    – icurays1
    Feb 23, 2016 at 17:24
  • $\begingroup$ I did not unfortunately. I took a math class by a physicist where the RTE equations are derived from Maxwell's equations using asymptotic expansions and other techniques, and that's the basis for me asking this question here. He did it from the physics perspective, which I'm not really familiar with. A lot of the physical intuition was foreign to me (i.e., I understood the mathematical validity of all the steps in the derivation, but I had difficulty connecting the approximations used to what was going on in the physical world.) I can send you some class notes/slides if you're interested $\endgroup$
    – nukeguy
    Feb 23, 2016 at 20:42

2 Answers 2


The radiative transfer equation is a simplified model for describing light transfer. Of course it is possible to derive the radiative transfer equation by the Boltzmann equation for a photon density function $f(x,t)$: $$ \partial_t f(x,t) + v_x \partial_x f(x,t) = (\partial_t f(x,t))_{coll}. $$

Here, the term $(\partial_t f(x,t))_{coll}$ is the gain and loss of the photon density per time due to light scattering, extinction and illumination. It is not easy to derive this right hand side term (for exact calculations you need quantum mechanics). However, with some suitable assumptions (like in the kinetic theory of gases) basing on Maxwell's equations, the radiative Transfer equation can be obtained.

  • $\begingroup$ Can you be more specific about what those "suitable assumptions" are for going from Maxwell equations to the radiative transfer equation? Kyle Kanos suggests "Doesn't the RTE describe the time-evolution of the intensity of the wave, not the photons themselves?", but it sounds like you're saying the opposite -- that you CAN derive the radiative transfer equation from the Boltzmann equation for a photon density? Could the two of you clarify this? Thanks! $\endgroup$
    – nukeguy
    Feb 5, 2015 at 21:33
  • $\begingroup$ To obtin macroscopic quantities from the Boltzmann equation you multiply the Boltzmann equation with an observable that depends on the space and then you integrate over the whole space. The photon density function is normalized, i.e. \int d^3x f(x,t) = N_p = constant. When you have some electrodynamic quantities like the energy density $\frac{E^2 + B^2}{2}$ or the Poynting vector you can multiply the electrodynamic quantities with the Boltzmann equation and integrate over the whole space. When using the linearity rule of Integration, you can obtain the RTE equation. $\endgroup$
    – kryomaxim
    Feb 5, 2015 at 21:45
  • $\begingroup$ Which quantity are we multiplying the Boltzmann equation by? I understand that you can obtain macroscopic quantities by integrating them over the photon density. But I'm not sure what this has to do with deriving the radiative transfer equation. Perhaps a better question would be, how is the intensity variable in the RTE related to the photon density? Also, on a different topic, can you be more specific about what the "suitable assumptions" are for going from Maxwell equations to the RTE? $\endgroup$
    – nukeguy
    Feb 5, 2015 at 21:57

There is no settlement between the two. This is 18 century mathematics against 20 century Mathematics. Maxwell is a strictly analytical view of nature. RTE is a conservative empirical mathematics. Sometimes heuristic in nature because it relies in physical observations.This latter can effort assumptions of all king. For example, it applies Boltzmann transport theory , but with enough assumptions, it reduces to Diffusion equations.

  • $\begingroup$ Do you have any guidance on the type of situations in which RTE would not hold? From my understanding of what you are saying, Maxwell is "exact" whereas RTE relies on empirics/physical observations. $\endgroup$
    – nukeguy
    Feb 12, 2019 at 21:18
  • $\begingroup$ Yes.. it fails near the boundaries and near any source. This is particularly true when you are assuming an isotropic scattering. $\endgroup$ Feb 13, 2019 at 15:34

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