This second tutorial on radiative transfer is the follow-up to tutorial I. Let me repeat that sometimes you will find some collapsable/expandable paragraphs marked in blue (questions) or orange (quiz). These are quick tests to check your knowledge as you follow the tutorial. Click on the colored paragraph to read the hint/answer.

## Radiative Transfer Theory

In astrophysical settings, photons reach us in two steps. First, they are created by some radiative process (e.g., synchrotron, thermal emission, etc.). Then they propagate through space where they might interact with matter in several ways. They can be absorbed, scattered or more photons might be added by local sources. For example, if a star emits thermal radiation, the photons might be partially absorbed and scattered by dust and some additional infrared emission might be added from the dust itself. Absorption, scattering, and emission and therefore three fundamental steps that modify the light we receive.

In this context, transfer theory provides a mathematical framework to model these interactions, under the assumption that light moves across space in bundles of rays. The main purpose of transfer theory is to specify how specific intensity, calculated for a specific radiative process, is affected by these three aforementioned phenomena: absorption, scattering, and emission.

When coupled with the equations of hydrostatic equilibrium and those of thermal and chemical balance, transfer theory provides a way to calculate several physical properties of a body like the cooling and heating rate and the radiation field throughout the body. When dust, molecules, and/or atomic species are present in the medium or in the emitting body, radiative transfer provides also a way to calculate the emission/absorption characteristics of these bodies.

Solution to the Equation of Transport

Imagine having an object with a certain initial specific intensity, crossing a medium with both absorption and local emission. What is the *solution* to the equation of transport in this case?

Hint

Hint: Write first the equation as a function of optical depth, then use the definition of the source function, and finally integrate the equation of transport.

## Mean Free Path

The mean free path is the average distance traveled by a photon before being either scattered or absorbed. When a bundle of rays is crossing a medium with optical depth then the probability of crossing the medium up to a distance that corresponds to that optical depth

Therefore the mean optical depth can be found by integrating all the optical depths and averaging over the probabilities:

(1)

The fact that *on average*, a photon travels for a distance such that the optical depth at the time of scattering/absorption is equal to one. Since the optical depth is a dimensionless number and it is defined as the absorption coefficient times the path length, we can define an average distance traveled by the photon before the scattering/absorption process:

(2)

As an example, suppose to be in presence of a spherical source with radius

(3)

which tells us that, for a fixed optical depth, the mean free path is proportional to the size of the object being crossed. This is intuitively simple to understand since, for a fixed optical depth, a larger source will be more “diluted” than a smaller one, so that the photons will travel a longer distance before being absorbed or scattered.

Exercise – Why the Sky is Blue and the Sun is Red at Sunset

Can you provide a qualitative explanation, based on the discussion of the mean free path, of why the sky is blue during the day and why the Sun has a deep red glow at sunset/sunrise?

Hint 1

Remember how the cross-section depends on the frequency in the Rayleigh scattering law.

Hint 2

Since *n* scatterings.

Solution

Since the Rayleigh scattering law says that the cross-section is *n* is the typical number of scatterings. Since *n* is very large and

Exercise – How long before light and neutrino escape the Sun?

The Sun’s mean density is

(a) How much time does it take for a photon to escape from the core to the surface? Assume that it suffers mainly from scattering and has an effective cross-section

(b) Consider a neutrino created in the core of the Sun. Its cross-section is

Solution

(a) The number density is

with an approximate result of 3,000 yr.

(b) We can calculate the mean free path directly using the previous problem and replacing the cross-section. For its probability of interaction, we can consider that photons and neutrinos are only created in the core of the Sun and afterward they will

just interact with the solar particles (hydrogen). From previous lectures, we know that the intensity in a medium with only absorption and/or scattering is

Since

Since the probability of the neutrino interacting before escaping is extremely small we can just set

### Random Walk

Closely related to the concept of the mean free path is that of *random walk*. A random walk is a general type of random process involving a series of steps in random directions, where the choice of the random direction is determined according to some probability rule. Random walks take place in many different settings and according to many different rules. The simplest form of random walk occurs in one dimension, for example by taking the line of integer numbers and moving by one step along the positive or negative direction according to a probability rule. This can be controlled by flipping a fair coin and assigning a movement of

We can apply this intuitive explanation of a random walk to the case of photons being *isotropically* scattered many times by particles. Each interaction occurs, on average, after the photon has crossed a length equal to the mean free path. The direction the photon will have after each interaction is randomly chosen according to a probability rule associated with the type of interaction occurring during the scattering process. How far would have the photons traveled after

Let’s call

If we average

However, if we calculate the mean squared average distance (for reasons that will become clear below) we obtain a non-zero result:

The cross products are all equal to zero since these are averages over all directions (isotropic scattering) and so the end result is:

where

Let’s go back to our example of the 1-dimensional random walk with a fair coin flip. In our example, we used

This result is, however, valid only for an optically thick medium.

In the case of optically thin material, the photon will have a mean free path larger than the size occupied by the medium itself, and therefore the validity of the former equation breaks down and

Exercise – Why you should not move if you get lost

When you were a kid, your parents probably told you to not move in case you got lost. This is a piece of very wise advice, based on the notion of random-walk, as we have seen in the paragraph above. Imagine now, for simplicity, to move only along a straight line, call it *x*, and to move for a certain average distance *L* at every step. You can either move in the forward or in the backward direction with a certain random probability of 1/2. For example, you can decide to toss a coin at each step, with heads meaning move forward, and tails meaning move backward. Imagine doing *n* steps and mark the position at the *i-*th step as

- If
, how many different outcomes can exist? In other words, in how many different locations can you be? - If
, what is the probability of being at the end at exactly the same initial starting position ()?

Hint question 1

This is equivalent to asking how many different head/tail sequences can you create with three tosses of a coin.

Solution question 1

It is

Hint 1 question 2

You first need to count how many paths bring you from location 0 back to location 0 in n=10 steps. So there must be an equal number of heads and tails in this sequence. Then you need to divide this number by the total number of paths possible with 10 tosses.

Hint 2 question 2

You can use the binomial coefficient *n* choose *k*” because it represents the number of ways to choose an (unordered) subset of *k* elements from a fixed set of *n* elements. In this case

Solution question 2

The total number of paths in this problem is

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