Quick foreword: This is part one of a guide that serves as both a refresher of a class I very much enjoyed taking by Professor Massimiliano Fratoni and as a resource on something that I think is a bit harder to understand from the ground up. Before reading this I assume some knowledge of calculus, linear algebra and differential equations. However, this article could still be a fun read even if you don't know these as I will try to explain everything conceptually as well!
You probably learned in high school chemistry that the atom has three distinct constituents, the positively charged proton, the negatively charged electron and the neutral neutron.
Almost all the chemistry we learn about in high school and the first year of undergrad involves the electrons. However, the protons and neutrons inside an atom have their own interactions as well. The protons and neutrons of an atom are collectively referred to as the "nucleus" hence the name: "nuclear physics".
Neutrons, named as such because they are neutral in charge, help hold nuclei together through the strong force1×Not relevant to this article but a fun reading topic for another time!. As they are neutral they don't experience any charged particle interactions2×Neutrons do have measurable magnetic moments as a result of being uncharged particles but this will not be relevant.. This means that a neutron typically only stops or loses energy when it experiences some sort of collision, usually with nuclei.
For our purposes, you can think of a neutron as a neutrally charged ball that has the following interactions:
Click anywhere to launch neutrons. Watch how they interact with nuclei!
Imagine you have a heat camera which shows you how hot something is by looking at it. If you looked through the camera at both a car with its engine on and a house, which one would you be able to see better? Even though the house is bigger, the heat generated by the car's engine would make the car appear clearer on a thermal camera. In the same way that looking through a camera changed the apparent size of the objects, neutrons also "see" the world differently.
In a neutron's case, the question is more of how likely a neutron is to interact with an atom. If we view the neutron as a hard ball and the atoms as other hard balls, we would assume that larger atoms have higher cross sections. However, neutron energy and atomic properties can change how the neutron "sees" an atom. For example, even though Iron (Fe) is significantly larger than Boron (B), Boron's capture cross section is 200 times that of Iron.
Toggle between physical size and neutron cross-section view
The dashed circles show how large each atom appears to a neutron. Notice how Boron appears huge even though the physical atom is small!
For the purposes of this article we can interpret the cross section as how likely the neutron is to interact with something. A higher cross section for a specific interaction means that a neutron is more likely to interact with it.
Neutrons can't just appear out of nowhere, they have to appear or disappear from a system through a process. Let's now imagine a box with some nuclear material inside of it. How do we know how many free neutrons (neutrons that are just floating around) are entering or exiting the system?
Because we know that there has to be a process to add or remove a free neutron from the system, we just need to add up every process that is adding or removing neutrons and sum them up.
In order to simplify this problem, we'll just be focusing on the major interactions that a neutron will have in order to keep this manageable and realistic. In this article we're assuming three major neutron interactions:
This means that we can determine the change in neutrons in the box using the following equation, also known as the equation of continuity:
Now that we have an equation for the change of neutrons in a volume we need to figure out how to model each one mathematically.
The output of our equation, how much the neutron population changes with time, is also the easiest to model. If you remember from calculus, we already have a way to determine the change in a quantity with respect to another quantity: the derivative.
Therefore, we can write the change in neutrons with respect to time as: $\frac{\partial n}{\partial t}$.3×If you haven't taken multivariable calculus, you may not be familiar with the symbols I have used for the derivative. These are called partial derivatives and as far as this article is concerned, they only mean that you have to take the derivative with respect to t and t only. Now if we want to see that across the entire volume of the box, we have to take an integral to sum up each unit area:
For the purposes of this article, we're going to be considering all production as coming from a neutron source. This basically just means there is a beam of neutrons that is hitting the box at some constant rate.4×For those of you who are curious about how this would be done in real life, you could use a deuterium beam with a beryllium breakup target to achieve this.
This constant rate would be how many neutrons are added every second per centimeter cubed, a quantity which we will call S. Now if we want to determine how many neutrons are added every second to the entire thing, we would have to sum the neutrons in each square centimeter of the box. We can represent this as an integral:
Now it's time to introduce the concept of flux. Flux is a measure of how many neutrons are passing through a space every second. This is important because the more neutrons go through an area, the more interactions they are going to have (including absorption).
Here's a little graphic to help you visualize flux. More neutrons = higher flux = more interactions
So if we have a certain amount of neutrons going through a region, how do we know how many of those neutrons will actually absorb into the materials inside the box? If you remember, in part one we described cross sections of a material as the likelihood of interaction with that material. Therefore if we just multiply the amount of neutrons with the likelihood that a neutron will interact with a material, we should get the absorption!
If we define our cross section as $\Sigma_{abs}$ and our flux as $\phi$ then we can get our total absorption per cubic centimeter. However, if we want absorption throughout the box, we will have to take the integral across the volume meaning our absorption looks like:
Now that we know the general equations for Absorption and Production, we now need to model what happens when a neutron leaves the box. We could try and look at the flux of the neutrons through all of the sides of the box, but flux is a scalar quantity meaning it doesn't account for the direction of the neutrons. In order to account for direction, we instead use current (J) which is the flux in a specific direction. We want to see the current along all the possible directions a neutron could leave the box. Now if we consider a box, the directions which a neutron could leave are all perpendicular to the faces of the box.
These perpendicular directions can be called "normal" to the box and the current is said to be along the normal vector (n). This time because we are looking at the neutrons leaving through the faces of the box, not the volume, we instead sum across the surface area of the box. Therefore:
Now even though we have a perfectly adequate expression for leakage, it's not integrated with respect to volume like the rest of our expressions. So we'll have to use some math in order to turn it into a volume integral. Specifically, we can use divergence theorem5×See: tutorial.math.lamar.edu/classes/calciii/DivergenceTheorem.aspx to perform the conversion such that:
Previously we stated the equation as
Change in Neutrons = Production - Absorption - Leakage
Let's now put together all the terms we've calculated to get a final equation
Since we're integrating over the same volume here and none of these terms are actually volume dependent, we can get rid of the integrals to get us:
This is the Diffusion Equation and it can tell us a lot about the behavior of neutrons. From here we can develop a lot of Nuclear Reactor Theory!
In particle systems, particles in high concentration regions usually move to lower concentration regions. This is called diffusion and understanding it is key to explaining key particle behaviors. Diffusion also applies to neutrons, a region with a high flux means that there are more possible collisions there which redirect neutrons. This naturally means that neutrons will scatter towards regions where there are less regions.
Adjust the concentration gradient and diffusion coefficient to see how particles flow. Current always flows from high to low concentration!
If you remember, the current describes the amount of neutrons flowing in a specific direction. We know that the current always flows away from the regions with the highest fluxes as there is more scattering there. We can formalize this through Fick's law which is as follows:
In this equation, D is the diffusion constant which is a parameter specific to the material inside the box. We need to take the gradient6×See: betterexplained.com/articles/vector-calculus-understanding-the-gradient/ (essentially a sum of derivatives in each direction current can flow in) to account for all three dimensions.
Our current equation has two quantities that have to be calculated, $\frac{\partial n}{\partial t}$ and J. We can use Fick's law in order to put the system in terms of only $\phi$. After plugging in Fick's Law we get7×The divergence of a gradient is the Laplacian or the gradient applied to itself.
Let's say we want to solve for the values of individual terms when there is no net motion of neutrons through the box, that is when change in neutrons is zero. This is also called Steady State.
In this case we get the Time-Independent Diffusion Equation, which is when the diffusion equation reaches equilibrium and it is written as: