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Background information to the question

I'm a first year physics student and I have a course "Introduction to astronomy" (free translation) which until this year was thought in the third bachelor year, so a lot of the basic physics which is used has to be explained along the way because we haven't learned it yet (not a big fan). This question is about the radiative transfer equation, and specifically about how to solve it. The derivation is given in our book and I think I understand it, but the solution isn't and wherever I look for it online they always solve it in a way that isn't very clear to me. You should know that we haven't really done differential equations yet (only very basic), since that's a course for next year.

Question

So the radiative transfer equation in the general case that we derived is

$$ \dfrac{dI_\nu}{d\tau_\nu}= S_\nu - I_\nu,$$

where $S_\nu=\dfrac{j_\nu}{4\pi k_\nu}$ is the so-called source function, with $j_\nu$ an emission coefficient, and $k_\nu=\dfrac{d\tau_\nu}{ds}$. I've found the pure absorption solution where $j_\nu=0$ to be

$$I_\nu(s)=I_0e^{-\tau},$$

with $\tau_\nu=\int k_\nu ds$, the optical depth. The solution I'm looking for looks like

$$I(\tau)=I_0e^{-\tau}+\int_{0}^{\tau_\nu}S_\nu(\tau_\nu')e^{-(\tau_\nu-\tau_\nu')}d\tau_\nu'.$$

I've no idea what the physical meaning of $\tau_\nu'$ is, and I don't know how to get this solution. I've found a solution online where they define new $I$ and $S$ with "twiddle's" above them, but I've never seen that kind of way to compute solutions before and I don't understand the physical origin of $\tau_\nu'$ in their solution.

I'd really like it if someone could provide me with a general solution to this equation (I mean the steps of course) that is understandable for a first year physics student with only a basic understanding in solving differential equations, and a physical interpretation of it (can be brief since I can usually find that online).

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  • $\begingroup$ en.wikipedia.org/wiki/Optical_depth, but, do you want to solve the equation before you understand what the variables mean? $\endgroup$
    – user126422
    Jun 7, 2017 at 18:22
  • $\begingroup$ You're missing the differential in the last equation. $\endgroup$
    – Kyle Kanos
    Jun 7, 2017 at 19:22
  • $\begingroup$ $\tau_{\nu}^{\prime}$ is the variable over which you are integrating. $\endgroup$
    – ProfRob
    Jun 7, 2017 at 20:12

1 Answer 1

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In the solution you have written down, $\tau_\nu^\prime$ is a dummy variable. It is just a way of keeping track of the value of the optical depth at intermediate points along the ray you are tracing. Note that $d\tau_\nu^\prime$ should appear in your integral.

The radiative transfer equation tells us that, along a ray in a particular direction, the radiative intensity will change in response to new contributions to the radiation in that direction (emission) and depletions of the radiation as it passes through a medium (absorption).

So you can think of the solution that you've written down as follows: Along a ray with total optical depth $\tau_\nu$, the radiative intensity that was originally emitted will be depleted by a factor of $e^{-\tau_\nu}$, due to absorption (compare to your pure absorption solution). There will also be emission contributions from each point along the ray. The strength of these emission contributions as a function of optical depth is represented by the source function $S_\nu$, which can vary along the ray. We must add in these contributions, but weighted by the amount that they will be absorbed by the remaining material along the ray. Their absorption depends on the remaining optical depth from the intermediate point to the endpoint you have specified, which is $\tau_\nu - \tau_\nu^\prime$.

The solution you have written down is often referred to as the "formal" solution to the radiative transfer equation. To derive this solution, you can use an integrating factor.

The solution is called "formal" because, while generally true, it is only useful in a narrow range of situations. The difficulty lies in what knowledge you have of the source function $S_\nu$. If you are given it in advance, then all is well. But generally, for a variety of reasons that I could describe later, $S_\nu$ will depend $I_\nu$. So $I_\nu$ is actually present on both sides of the equation, and you haven't totally solved for it. To complicate matters, it is buried in an integral on the right-hand side. Such equations are difficult to solve in general, and often require numerical solutions, which are computer algorithms that chop the problem into small segments, yielding an approximate solution up to a desired level of accuracy. You can compare with other common situations where numerical methods are required, such as the equations of hydrodynamics.

On top of all that, to fully characterize the radiation field, you need to solve the radiative transfer equation for every direction the radiation can travel, at every frequency, and at every position. So this is a multi-dimensional problem: three spatial dimensions, two direction angles, and frequency, for six in total. Plus all quantities of interest could also vary in time. This is why some of the numerical techniques used to solve for the radiation field resort to Monte Carlo techniques for integration.

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