JCTree

Revisions for TOFE 2026 Plots

2026-05-31T00:00:00+00:00

The Nodal Expansion Method

2025-11-29T10:40:00+00:00

This page is under construction. If it seems incomplete, it is, but feel free to enjoy what is already here. This page has an accompanying toy nodal expansion method code, which lives here</a>.

Transport Equation</h1>
The neutron transport equation defines the balance of neutron creation and neutron removal processes from a volume $V$. For a system with no external source, the time rate of change of the flux is given by $$ \begin{align} \frac{1}{v}\frac{\partial\phi}{\partial t}&= \int\limits_{4\pi}\int\limits_0^\infty \Sigma_s(\vec r,E’\rightarrow E,\hat\Omega’\rightarrow\hat\Omega,t) \phi(\vec r,E’,\hat\Omega’,t)dE’d\hat\Omega’ \\ &+\frac{\chi(E)}{4\pi}\int\limits_0^\infty \nu(E’)\Sigma_f(\vec r,E’,t)\phi(\vec r,E’,\hat\Omega,t)dE’ \\ &+\frac{1}{4\pi}\sum\limits_f\chi_f(E)\lambda_fC_f(\vec r,t) \\ &-\hat\Omega\cdot\nabla\phi(\vec r,E,\hat\Omega,t) \\ &-\Sigma_t(\vec r,E,t)\phi(\vec r,E,\hat\Omega,t) \end{align} $$ where

$v$ is the neutron velocity</li>
$\phi$ is the neutron flux, and is a function of position, energy, angle, and time</li>
$\Sigma_s$ is the scattering cross section, and is a function of position, energy, angle, and time</li>
$\chi(E)$ is the probability a neutron produced from fission was produced at energy $E$</li>
$nu$ is the average number of neutrons produced in a fission event, and is a function of incident neutron energy</li>
$\Sigma_f$ is the fission cross section, and is a function of position, energy, and time</li>
$\chi_f(E)$ is the probability a decay neutron from fission product $f$ is emitted at energy $E$</li>
$\lambda_f$ is the decay constant of fission product $f$</li>
$C_f$ is the concentration of fission product $f$, and is a function of position and time</li>
$\Sigma_t$ is the total cross section, and is a function of position, neutron energy, and time</li> </ul>
On the right-hand side of the equation, there are positive terms that represent neutron gains in the system, and negative terms that represent neutron losses in the system. At this point, let us consider the flux of neutrons with a fixed energy $E$ and direction $\hat\Omega$ at the point $\vec r$ in space. The first term on the right-hand side of the transport equation is the in-scattering source, where neutrons at a different angle and energy scatter into our energy $E$ and angle $\hat\Omega$. The next term describes the neutrons produced at our position, angle, and energy from fission. The final gain term is the production of neutrons from the decay of radioactive fission products. The first loss term is the leakage of neutrons away from our position $\vec r$. The other loss term simply represents neutrons being removed from our position, angle, or energy from suffering an interaction in matter.
Discretization of the Transport Equation</h1>
The full time-dependent neutron transport equation is very difficult to solve. There are many different approximations to the transport equation that are available which allow it to be calculated. One such approximation is to assume that the angular distribution does not strongly effect the solution. To do this, we integrate the transport equation over all angles to get $$ \begin{align} \frac{1}{v}\frac{\partial\phi}{\partial t}&= \int\limits_0^\infty\Sigma_s(E’\rightarrow E)\phi(\vec r,E’,t) \\ &+\frac{\chi(E)}{4\pi}\int\limits_0^\infty \nu(E’)\Sigma_f(\vec r,E’,t)\phi(\vec r,E’,t)dE’ \\ &+\frac{1}{4\pi}\sum\limits_f\chi_f(E)\lambda_fC_f(\vec r,t) \\ &-\int\limits_{4\pi} \hat\Omega\cdot\nabla\phi(\vec r,E,\hat\Omega,t)d\hat\Omega \\ &-\Sigma_t(\vec r,E,t)\phi(\vec r,E,t) \end{align} $$ We can then use the divergence theorem on the first loss term, which yields the neutron continuity equation: $$ \begin{align} \frac{1}{v}\frac{\partial\phi}{\partial t}&= \int\limits_0^\infty\Sigma_s(E’\rightarrow E)\phi(\vec r,E’,t) \\ &+\frac{\chi(E)}{4\pi}\int\limits_0^\infty \nu(E’)\Sigma_f(\vec r,E’,t)\phi(\vec r,E’,t)dE’ \\ &+\frac{1}{4\pi}\sum\limits_f\chi_f(E)\lambda_fC_f(\vec r,t) \\ &-\nabla\cdot\vec J(\vec r,E,t) \\ &-\Sigma_t(\vec r,E,t)\phi(\vec r,E,t) \end{align} $$ where $\vec J$ is the neutron current.
Next, we discretize the equation in energy. We say that our energy spectrum is divided up into $G$ groups, where the properties within each group $g$ are approximated to be constant. This allows us to rewrite the continuity equation as $$ \frac{1}{v}\frac{\partial\phi_g}{\partial t} +\nabla\cdot\vec J_g(\vec r,t) +(\Sigma_{t,g}(\vec r,t)-\Sigma_{s,g}(\vec r,t))\phi_g(\vec r,t) =Q_g(\vec r,t) $$ where the gain terms have been combined into a single source term $Q$.
At this point, we will choose to solve a steady-state problem, meaning that the time dependence of the terms can be removed. In order to solve this, we also need to discretize our problem in space. Dividing space into cubes turns out to be convenient. Each cube in space is approximated to have constant properties. We call one of these cubes a node, and denote a specific node as $k$. This leads us to the following form of the diffusion equation: $$ \nabla\cdot\vec J^k_g+\Sigma^k_{r,g}\phi^k_g=Q^k_g $$ where we use the removal cross section, defined as $$ \Sigma^k_{r,g}=\Sigma^k_{a,g} +\sum\limits^G_{g’\neq g}\Sigma^k_{s,g\rightarrow g’} $$ and $$ Q^k_g(\vec{r})=\frac{\chi_g}{k_{eff}} \sum\limits_{g’=1}^G\nu\Sigma_{f,g’}^k\phi^k_{g’}(\vec{r})+ \sum\limits_{g’\neq g}^G\Sigma^k_{s,g’\rightarrow g}\phi^k_{g’}(\vec{r}) $$ This is the starting point of the derivation in Okumura, 1998</a>, which is roughly translated in the following section.
The Diffusion Approximation</h1>
We now introduce another approximation by assuming that the relationship between the flux and the current follow Fick’s Law</a>. This can be expressed as $$ \vec J_g(\vec r,t)=-D_g(\vec r)\nabla\phi_g(\vec r,t) $$ We will use Fick’s Law in the following sections to eliminate an unknown (the current) in a system of equations, a common result in an introductory reactor physics course.
As an aside, plugging this into the continuity equation gives the diffusion equation: $$ \frac{1}{v}\frac{\partial\phi}{\partial t}- \nabla\cdot D_g(\vec r)\nabla\phi_g(\vec r,t) +\Sigma_{a,g}(\vec r,t)\phi_g(\vec r,t)=Q_g(\vec r,t) $$
Nodal Balance</h1>
We define the center of our node $k$ to be $(0,0,0)$. Therefore, the node spans the space $-\frac{\Delta x}{2}<x<\frac{\Delta x}{2}$, $-\frac{\Delta y}{2}<y<\frac{\Delta y}{2}$, $-\frac{\Delta z}{2}<z<\frac{\Delta z}{2}$. “+” and “-” are used to denote the face of the node with a positive or negative component, respectively. Now, we expand the divergence of the current in the diffusion equation, integrate over the node, and divide by the node volume. This results in the node balance equation, given by $$ \frac{1}{\Delta_x^k}\left(J^k_{g,x+}-J^k_{g,x-}\right)+ \frac{1}{\Delta_y^k}\left(J^k_{g,y+}-J^k_{g,y-}\right)+ \frac{1}{\Delta_z^k}\left(J^k_{g,z+}-J^k_{g,z-}\right)+ \Sigma^k_{rg}\phi^k_{g,0}=Q^k_{g,0} $$ where we defined $$ \phi^k_{g,0}=\frac{1}{\Delta^k_x\Delta^k_y\Delta^k_z} \int\limits^{\Delta^k_x/2}_{-\Delta^k_x/2} \int\limits^{\Delta^k_y/2}_{-\Delta^k_y/2} \int\limits^{\Delta^k_z/2}_{-\Delta^k_z/2} \phi^k_g(x,y,z)dxdydz $$ and $$ J^k_{g,x\pm}=\frac{1}{\Delta^k_y\Delta^k_z} \int\limits^{\Delta^k_y/2}_{-\Delta^k_y/2} \int\limits^{\Delta^k_z/2}_{-\Delta^k_z/2} J^k_g\left(\pm\frac{\Delta^k_x}{2},y,z\right)dydz $$ and $$ \begin{align} Q^k_{g,0}&=\frac{1}{\Delta^k_x\Delta^k_y\Delta^k_z} \int\limits^{\Delta^k_x/2}_{-\Delta^k_x/2} \int\limits^{\Delta^k_y/2}_{-\Delta^k_y/2} \int\limits^{\Delta^k_z/2}_{-\Delta^k_z/2} Q^k_g(x,y,z)dxdydz \\ &=\frac{1}{\Delta^k_x\Delta^k_y\Delta^k_z} \int\limits^{\Delta^k_x/2}_{-\Delta^k_x/2} \int\limits^{\Delta^k_y/2}_{-\Delta^k_y/2} \int\limits^{\Delta^k_z/2}_{-\Delta^k_z/2} \left(\frac{\chi_g}{k_{eff}} \sum\limits_{g’=1}^G\nu\Sigma_{fg’}^k\phi^k_{g’}(x,y,z)+ \sum\limits_{g’\neq g}^G\Sigma^k_{sg’\rightarrow g}\phi^k_{g’}(x,y,z) \right)dxdydz \\ &=\frac{\chi_g}{k_{eff}} \sum\limits_{g’=1}^G\nu\Sigma_{fg’}^k\phi^k_{g’,0}+ \sum\limits_{g’\neq g}^G\Sigma^k_{sg’\rightarrow g}\phi^k_{g’,0} \end{align} $$ The $y$ and $z$ currents are defined in the same way as the $x$ current. In order to reduce the number of variables, we would like to reduce the number of unknowns. One approximation often made to neutron transport is the application of Fick’s Law</a>, which relates the current to the flux as $$ \vec J^k_g=-D^k_g\nabla\phi^k_g $$ Plugging this into the definition of the $x$ current yields $$ J^k_{g,x\pm}=\frac{1}{\Delta^k_y\Delta^k_z} \int\limits^{\Delta^k_y/2}_{-\Delta^k_y/2} \int\limits^{\Delta^k_z/2}_{-\Delta^k_z/2} -D^k_{g,x}\frac{\partial}{\partial x}\phi^k_g(x,y,z) \biggr|_{x=\pm\Delta^k_x/2}dydz $$
The node balance equation is expressed in terms of the volume-averaged neutron flux within a node, denoted as $\phi^k_{g,0}$, and the surface-averaged net neutron currents across the six faces of the node. To proceed, we introduce the concepts of surface-averaged neutron flux and surface-averaged partial neutron currents. Within diffusion theory, the partial neutron currents are typically defined using the P1 approximation of the angular neutron flux, as shown below. $$ \begin{align} J^{out}_g(\vec{r})&=\frac{\phi_g(\vec{r})}{4}- \frac{D_g}{2}\hat{e}_s\cdot\nabla\phi(\vec{r}) \\ J^{in}_g(\vec{r})&=\frac{\phi_g(\vec{r})}{4}+ \frac{D_g}{2}\hat{e}_s\cdot\nabla\phi(\vec{r}) \end{align} $$ Here, $\hat{e}_s$ represents the unit outward normal vector to the surface under consideration. The outgoing partial neutron current in the direction of this outward normal vector is denoted as $J^{out}$, while the incoming partial neutron current, flowing in the opposite direction, is denoted as $J^{in}$. The partial current leaving the x+ surface is therefore $$ J^{out,k}_{g,x+}=\frac{1}{4}\phi^k_{g,x+}+\frac{1}{2} \left[ \frac{1}{\Delta^k_y\Delta^k_z} \int\limits^{\Delta^k_y/2}_{-\Delta^k_y/2} \int\limits^{\Delta^k_z/2}_{-\Delta^k_z/2} -D^k_{g,x}\frac{\partial}{\partial x}\phi^k_g(x,y,z) \biggr|_{x=\Delta^k_x/2}dydz \right] $$ where $\phi^k_{g,x+}$ is the surface-averaged neutron flux on the $k+$ interface of the positive $x$ surface of the node. $$ \phi_{g,x+}^{k}=\frac{1}{\Delta_y^k\Delta_z^k} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} \phi^{k}_g\left(\frac{\Delta_x^k}{2},y,z\right)dydz $$
The term in the square brackets is the definition of $J^k_{g,x+}$ from above. Substituting this in gives that $$ J^{out,k}_{g,x+}=\frac{1}{4}\phi^k_{g,x+}+\frac{1}{2}J^k_{g,x+} \ $$ Similarly, since the only difference in the definition of the incoming current from the P1 approximation is the sign flip, we can show that $$ J^{in,k}_{g,x+}=\frac{1}{4}\phi^k_{g,x+}-\frac{1}{2}J^k_{g,x+} $$ Adding and subtracting these two equations gives expressions for the current and flux at the $x+$ surface of the node: $$ \begin{align} J^k_{g,x+}&=J^{out,k}_{g,x+}-J^{in,k}_{g,x+} \\ \phi^k_{g,x+}&=2(J^{out,k}_{g,x+}+J^{in,k}_{g,x+}) \end{align} $$ Mirroring these calculations for the $x-$ surface yields $$ \begin{align} J^k_{g,x-}&=J^{in,k}_{g,x-}-J^{out,k}_{g,x-} \\ \phi^k_{g,x-}&=2(J^{out,k}_{g,x-}+J^{in,k}_{g,x-}) \end{align} $$ This allows us to rewrite the node balance equation using the incoming and outgoing currents. $$ \begin{align} \frac{1}{\Delta^k_x} (J^{out,k}_{g,x+}-J^{in,k}_{g,x+}-J^{in,k}_{g,x-}+J^{out,k}_{g,x-}) &+\frac{1}{\Delta^k_y} (J^{out,k}_{g,y+}-J^{in,k}_{g,y+}-J^{in,k}_{g,y-}+J^{out,k}_{g,y-}) \nonumber \\ &+\frac{1}{\Delta^k_z} (J^{out,k}_{g,z+}-J^{in,k}_{g,z+}-J^{in,k}_{g,z-}+J^{out,k}_{g,z-})+ \Sigma^k_{r,g}\phi^k_{g,0}=Q^k_{g,0} \end{align} $$
We do this because it allows us to split up the calculation. I will show in a later section that we can solve for the outgoing current on each surface of a node from the values of the incoming currents. This allows a two-step iteration strategy. First, we calculate the outgoing current for each node from its incoming current. The outgoing current for each node is used to update the incoming currents of the adjacent nodes. We can iterate this until the system of nodes is balanced to converge on a solution.
Transverse Integration</h1>
In order to reduce the solution complexity, we derive a one dimensional diffusion equation by integrating over the transverse directions. This is performed for each direction, yielding three equations toward the solution. The steps to derive the diffusion equation in x are roughly outlined in this section. The same steps can be equivalently followed for integration in the y and z directions as well.
Earlier, we derived a special form of the diffusion equation. $$ \nabla\cdot\vec J^k_g+\Sigma^k_{r,g}\phi^k_g=Q^k_g $$ We can expand the divergence of the current and integrate with respect to y and z within the node. Dividing by the area $\Delta_y^k\Delta_z^k$ then yields the following: $$ \begin{align} &\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} \frac{\partial}{\partial x}J^k_{g,x}(x,y,z)dydz +\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} \frac{\partial}{\partial y}J^k_{g,y}(x,y,z)dydz \nonumber \\ +&\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} \frac{\partial}{\partial z}J^k_{g,z}(x,y,z)dydz +\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} \Sigma^k_{r,g}\phi^k_g(x,y,z)dydz \nonumber \\ =&\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} Q^k_g(x,y,z)dydz \end{align} $$ which can be simplified to: $$ \begin{equation} \frac{d}{dx}\bar J^k_{g,x}(x)+\Sigma^k_{r,g}\bar\phi^k_{g,x}(x) =\bar Q^k_{g,x}(x)-\frac{1}{\Delta^k_y}\bar L^k_{g,y}(x) -\frac{1}{\Delta^k_z}\bar L^k_{g,z}(x) \end{equation} $$ where $$ \begin{align} \bar\phi^{k}_{g,x}(x)&=\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} \phi^k_g(x,y,z)dydz \nonumber \\ \bar J^{k}_{g,x}(x)&=\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} J^k_g(x,y,z)dydz \nonumber \\ \bar Q^k_{g,x}(x)&=\frac{1}{\Delta^k_y\Delta^k_z} \int\limits_{-\Delta_y^k/2}^{\Delta_y^k/2} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2} Q^k_g(x,y,z)dydz \nonumber \\ &=\frac{\chi^k_g}{k} \sum\limits_{g’=1}^{G}\nu\Sigma^{k}_{f,g’}\bar\phi^k_{g’}(x)+ \sum\limits_{g’\neq g}^{G}\Sigma^{k}_{s,g’\rightarrow g} \bar\phi^k_{g’}(x) \\ \bar L^{k}_{g,y}(x)&=\frac{1}{\Delta^k_z} \int\limits_{-\Delta_z^k/2}^{\Delta_z^k/2}\left[ -D^k_{g,y}\frac{\partial\phi^k_g(x,y,z)}{\partial y} \right]^{\Delta^k_y/2}_{-\Delta^k_y/2}dz \nonumber \\ \bar L^{k}_{g,z}(x)&=\frac{1}{\Delta^k_y} \int\limits_{-\Delta_{y}^{k}/2}^{\Delta_{y}^{k}/2}\left[ -D^k_{g,z}\frac{\partial\phi^{k}_{g}(x,y,z)}{\partial z} \right]^{\Delta^k_z/2}_{-\Delta^k_z/2}dy \end{align} $$ are the parameters averaged over the area of the node.
Similar equations can be written for the y and z directions. This is the transverse-integrated one-dimensional neutron balance equation. The equation is a one-dimensional diffusion equation with transverse leakage terms.
We can now write Fick’s Law in the x-direction averaged over a node. $$ \begin{equation} \bar J^k_{g,x}(x)=-D^k_{g,x}\frac{d}{dx}\bar\phi^k_{g,x}(x) \end{equation} $$ and similarly for the y and z directions. Substituting these into the 1D diffusion equation with transverse leakage terms results in the following one-dimensional diffusion equation. $$ \begin{equation} -D^k_{g,x}\frac{d^2}{dx^2}\bar\phi^k_{g,x}(x)+ \Sigma^k_{r,g}\bar\phi^k_{g,x}(x) =\bar Q^k_{g,x}(x)-\frac{1}{\Delta^k_y}\bar L^k_{g,y}(x) -\frac{1}{\Delta^k_z}\bar L^k_{g,z}(x) \end{equation} $$
References</h1> Okumura, K. (1998). MOSRA-Light; high speed three-dimensional nodal diffusion code for vector computers. https://www.osti.gov/etdeweb/servlets/purl/300794 </a>

Hiking Wales

2025-08-19T00:00:00+00:00

In May of 2025, I boarded a plane bound for the UK with a friend. Our plan was to hike as much of the northern section of the Cambrian Way as we could in 10 days. The Cambrian Way traverses Wales, starting in the South in Cardiff and ending in the North in Conwy. The route makes a point of hitting as many peaks as possible. Since we only had long enough to do a single section, we chose to start in the North in Conwy and hike in reverse as far as we could. I included a small gallery of pictures along the way. The pictures are ordered to start in Conwy on day 1, and the last one is leaving Barmouth on day 7. I am making this post to share my memories from this hike.

</a> </a> </a> </a> </a> </a> </a> </a> </a> </div>

Before we started, we did not know what to expect. Neither of us had done any serious through-hiking, and my friend did not even have much experience with hiking in general. We were a little bit caught off guard, though, since our first day ended up being fairly challenging. Looking back, it was the most difficult stage we did. It was both far and steep. We were low on water, so we needed to get to a water source to pitch camp. At the end of the day, we started feeling better. All we had was about a mile to get off the ridge to finish the stage, and we had a little bit of daylight left. The descent off the ridge, however, was very challenging. There were times we had to take off our bags to climb down. About half way down, we found a stream. We were losing light, so we decided to camp now that we could.

The next day felt a lot better. It was still steep, but the distance was shorter. From the second day on, we started to develop a rhythm. On the third day, we hiked Snowdon, the tallest mountain in Wales. Neither of us realized that it was Snowdon until we reached the top and saw the visitor center. The picture in the center of the gallery shows Snowdon.

Two days later, another friend of mine met up with us for the last few days of hiking. He joined us the night before the stage that ends in Barmouth. This stage felt quite similar to our first day. The distance was very far, and the terrain was very steep. We barely made it into Barmouth, and were very happy to have a shower and a warm meal. After that stage, we started later the next day, and spent the net two days completing the next stage. The last stages to the end of the northern section were easier as well. The slower pace made the last few days of hiking much more relaxing, and also convinced my friend to stick with it until the end.

The friend who hiked the whole way with me does not really have an online presence. The other one has a website which can be found here</a>.

Radioactive Decay Chains

2025-06-16T12:00:00+00:00

Radioactive Decay</h1>
Radioactive decay is a random process. For one unstable atom, it is completely random when it decays. For a large quantity of atoms, though, radioactive decay follows certain rules. The most important rule is that a radioactive sample, a collection of atoms of the same type, decay with a constant half-life. After one half-life, half of the original amount of atoms will have decayed. We can write this mathematically as $$ N(t)=N(0)\left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} $$ where $N(t)$ is the number of atoms at time $t$, and $t_{1/2}$ is the half-life. To prove to ourselves that this works, we can plug in $t=0$, which yields $$ N(0)=N(0)\left(\frac{1}{2}\right)^0=N(0) $$ and plugging in $t=t_{1/2}$ gives $$ N(t_{1/2})=N(0)\left(\frac{1}{2}\right)^1=\frac{N(0)}{2} $$
We can rearrange this equation to make our lives easier. We start by writing that $$ N(t)=N(0)\left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}=N(0)2^{-\frac{t}{t_{1/2}}} $$ By definition, $\exp(\log(x))=x$. We use this to say that $$ \begin{align} 2^{-\frac{t}{t_{1/2}}}&= \exp\left(\log\left(2^{-\frac{t}{t_{1/2}}}\right)\right) \\ &=\exp\left(-\frac{t}{t_{1/2}}\log(2)\right) \\ &=\exp\left(-\lambda t\right) \end{align} $$ where we define $\lambda=\frac{\log(2)}{t_{1/2}}$. Plugging this into the decay equation gives an equivalent but more ubiquitous form: $$ N(t)=N(0)e^{-\lambda t} $$ $\lambda$ is called the decay constant, and represents a rate. A more natural quantity to think about, however, is the number of atoms that decay per unit of time in a sample, called the activity. The SI units for activity are the Becquerel (Bq), where 1 Bq is 1 decay per second. The number of atoms that decay per unit of time in a sample is the time rate-of-change of the number of atoms in a sample, $\frac{dN}{dt}$. We can take this derivative to get $$ \begin{align} \frac{dN(t)}{dt}&=N(0)\frac{d}{dt}\left(e^{-\lambda t}\right) \\ &=N(0)\left(-\lambda e^{-\lambda t}\right) \\ &=-\lambda\left(N(0)e^{-\lambda t}\right) \\ &=-\lambda N(t) \end{align} $$ Denoting the activity of the sample as $A$, we write $$ A(t)=-\lambda N(t) $$

Decay Chains</h1>
When atoms decay, they emit some of their mass as radiation. What remains is a different atom, either a new element or a different isotope of the old one. This new isotope may itself be unstable and decay with a different half-life into yet another atom. Depending on what isotope you start from, you can end up with long decay chains. One such decay chain is called the “thorium series”, which starts at $^{232}$Th and ends on stable $^{208}$Pb. The thorium series is shown below.
Th232 └── Ra228 └── Ac228 └── Th228 └── Ra224 └── Rn220 └── Po216 └── Pb212 └── Bi212 ├── Po212 │ └── Pb208 └── Tl208 └── Pb208 </code></pre> Each isotope in the chain has its own half-life (and therefore its own decay constant). The decay of $^{232}$Th results in the production of $^{228}$Ra, which itself decays, producing $^{228}$Ac, and so on. Looking at these chains, we can ask the question, “how much of each isotope exists at any given time?” We can formulate an answer to this question using the equations we derived above. To simplify notation, we denote $^{232}$Th as isotope 1 and $^{228}$Ra as isotope 2. Let’s say we start with $N_1(0)$ atoms of $^{232}$Th, and $N_2(0)=0$ atoms (no atoms) of $^{228}$Ra. We can immediately write the time rate-of-change of the thorium atoms: $$ \frac{dN_1(t)}{dt}=-\lambda_1N_1(t) $$ If we are losing thorium atoms at a rate of $\lambda_1N_1(t)$, then we must be gaining radium atoms at a rate of $\lambda_1N_1(t)$. For radium then, we have $$ \frac{dN_2(t)}{dt}=\lambda_1N_1(t) $$ In addition to gaining atoms from the decay of the thorium, we are also losing the radium atoms through its own decay at a rate of $\lambda_2N_2(t)$. We therefore need to subtract out this term to get $$ \frac{dN_2(t)}{dt}=\lambda_1N_1(t)-\lambda_2N_2(t) $$ We therefore get a system of two coupled differential equations to describe the number of atoms of thorium and of radium at any point in time. We can expand our analysis to the next isotope in the chain, actinium. The number of actinium atoms depends on the rate of decay of radium and the rate of decay of itself: $$ \frac{dN_3(t)}{dt}=\lambda_2N_2(t)-\lambda_3N_3(t) $$ If we keep up this numbering scheme, we can easily write an equation for each isotope in a decay chain that is $n$ isotopes long. The last element in the decay chain (lead-212 in our example) is stable, so it does not have the loss term from its own decay, only the gain term from the decay of the isotope above it in the chain. $$ \frac{dN_n(t)}{dt}=\lambda_{n-1}N_{n-1}(t) $$ Solving the Differential Equations</h1>
Numerical Integration</h2>
The goal of this method is to turn the scary derivatives into sums without doing any difficult math. We start by defining a time-step, $\delta t$, that is small compared to the half-lives of all the isotopes in the decay chain. If we really squint, the decay equations $$ \begin{align} \frac{dN_1(t)}{dt}&=-\lambda_1N_1(t) & \frac{dN_2(t)}{dt}&=\lambda_1N_1(t)-\lambda_2N_2(t) \end{align} $$ look like $$ \begin{align} \frac{\delta N_1(t)}{\delta t}&=-\lambda_1N_1(t) & \frac{\delta N_2(t)}{\delta t}&=\lambda_1N_1(t)-\lambda_2N_2(t) \end{align} $$ This does actually work. The derivative is the slope of the curve at a point. We can approximate the slope of the curve at a point by the average slope of the curve around the point. This is an approximation, but as long as the interval ($\delta t$) is short compared to how quickly the slope of the curve is changing (the half-life of the isotope), then it is not a bad one. We can multiply both sides of the equations by $\delta t$ to get $$ \begin{align} \delta N_1(t)&=-\lambda_1N_1(t)\delta t & \delta N_2(t)&=(\lambda_1N_1(t)-\lambda_2N_2(t))\delta t \end{align} $$ If we keep track of amount of atoms for each isotope at each time-step, we simply add $\delta N$ to the amount of atoms at the current time-step to get the number of atoms at the next time-step. In order to calculate this $\delta N$, we need the number of atoms of the isotope of interest and the number of atoms of the isotope one position above it in the decay chain at the current time-step. Starting at $t=0$ where we know the initial number of atoms of all isotopes, we can calculate in time-steps going forward in time the number of atoms for each isotope in the chain.
This method is very easy to implement in a computer program, and is often accurate enough. Care does have to be taken, however, that the time-step is short enough. If there are very short-lived isotopes in the decay chain, the required time-step may be so small that the method is not computationally feasible.
We can write a simple program in Python which calculates the decay of $^{232}$Th into $^{228}$Ra. First, we define the half-lives of the isotopes. The half-lives are given in years. We will also need the `numpy</code> library, which we import.`
`import numpy as np half_life_Th = 1.405e10 half_life_Ra = 5.73 </code></pre> Next, we calculate the decay constants from the half-lives.`
lambda_Th = np.log(2) / half_life_Th lambda_Ra = np.log(2) / half_life_Ra </code></pre> Since both the half-lives are multiple years, we can set our time-step as half of a year and still get reasonable results. We also need to set the initial number of atoms of thorium and radium. We put the initial numbers as the first elements in arrays, which we will append the new number of atoms to at each time-step.
dt = 0.5 N_Th = [1e15] N_Ra = [0.] </code></pre> Finally, we make a loop that calculates the number of atoms of each isotope every year for 20 years. In the loop, we first calculate the change in the number of atoms of each isotope from decay over the year. Next, we subtract the amount of atoms lost from the previous value in the array, and for radium, we add the number of atoms lost from thorium.
for i in range(20 * 2): dN_Th = lambda_Th * N_Th[i] * dt dN_Ra = lambda_Ra * N_Ra[i] * dt N_Th.append(N_Th[i] - dN_Th) N_Ra.append(N_Ra[i] + dN_Th - dN_Ra) </code></pre> This leaves us with two arrays containing the number of atoms of thorium and radium at half year increments. We can plot the data using the following code: import matplotlib.pyplot as plt times = np.linspace(0, 20, 41) plt.plot(times, N_Th, label="Thorium Atoms") plt.plot(times, N_Ra, label="Radium Atoms") plt.xlabel("Time [years]") plt.ylabel("Number of Atoms") plt.yscale("log") plt.legend() plt.show() </code></pre> The Bateman Equation</a></h2> The Bateman Equation is an exact solution to the system of coupled differential equations. It is given by $$ N_i(t)=N_1(0)\left(\prod\limits^{i-1}_{k=1}\lambda_k\right) \sum\limits^i_{k=1}\frac{e^{-\lambda_kt}} {\prod\limits^i_{j=1,j\neq k}(\lambda_j-\lambda_k)} $$ The Bateman Equation solves the decay chain exactly, and is fairly easy to implement in a computer program. It can cause issues, however, if two isotopes in the chain have decay constants $\lambda_j\approx\lambda_k$. Floating-point error can cause the difference to go to zero causing a divide by zero. The equation as written does not allow for isotopes down the chain to have non-zero starting numbers. Only the first isotope’s starting number is reflected in the solution. To treat this case, the chain can be split into components where each component is a chain with only one of the isotopes having a starting number of atoms. The component chains can be solved individually, and the results superimposed on one another. For example, given the chain A ───► B ───► C </code></pre> we can solve three separate chains A ───► 0 ───► 0 0 ───► B ───► 0 0 ───► 0 ───► C </code></pre> and sum the results to get $C(t)$. Branching</h3> If the decay chain has branching, a similar strategy can be employed. Consider the following chain: A ├───── B ───┐ │ 25% │ │ │ └───── C ───┴─ D 75% </code></pre> The goal is to solve for the number of atoms of isotope D at some time $t$. In order to do this, all possible paths to the isotope D must be found. For this simple chain, this task is fairly simple. For more complicated chains, it is beneficial to understand the structure of the chain. In a decay chain, there are no cycles. Following the path of decay, there will not be any closed loops (incidentally, this structure is known as a directed acyclic graph). This property allows us to employ a depth-first search algorithm to traverse the chain, finding all paths to the nuclide of interest. For the chain given above, the two paths are A ───► B ───► D A ───► C ───► D </code></pre> Each of these paths can be treated individually with the Bateman equation. This yields two solutions for the number of atoms of isotope D, D$_1$ and D$_2$ for the solution of the first and second paths, respectively. In this case, it is evident that 25% of the number of atoms of D come from the first path, and 75% come from the second path. The solution for the number of atoms of D is then $$ \mathrm{D=0.25D_1 + 0.75D_2} $$ For more complicated chains, with multiple non-trivial paths to the nuclide of interest, the contribution of each path can be represented as a path weight. The weight of a path is found as follows. Assume a path $p$ comprising nuclides $N_1$ to $N_n$. The branching ratio from $N_1$ to $N_2$ is represented as $b_{1\rightarrow2}$. The weight $w_p$ of the path $p$ is $$ w_p=\prod\limits_{i=2}^{n}b_{(i-1)\rightarrow i} $$ Assuming the solution to the Bateman equation for nuclide $n$ on the path $p$ is $N_{n,p}$, the total number of atoms of nuclide $n$ is $$ N_n=\sum\limits_{p}w_pN_{n,p} $$ References</h1> Bateman, H. (1908). The solution of a system of differential equations occurring in the theory of radio-active transformations. Proceedings of the Cambridge Philosophical Society, Mathematical and Physical Sciences, v.15 (1908-1910), 423–427.</a>



Monte Carlo
2025-05-02T22:52:00+00:00
This has no information at the moment. I do have a code though. It lives
here</a>.</p>


Hello World
2025-05-02T21:00:00+00:00
I am a student studying nuclear engineering. You can read about me
here</a>. I am starting this website as a way to record the projects
I am working on. I plan to describe what the projects are and how they work,
both for my own reference and in case someone finds these things interesting.</p>

JCTree

Revisions for TOFE 2026 Plots

The Nodal Expansion Method

References</h1> Okumura, K. (1998). MOSRA-Light; high speed three-dimensional nodal diffusion code for vector computers. https://www.osti.gov/etdeweb/servlets/purl/300794 </a>

Hiking Wales

Radioactive Decay Chains

Solving the Differential Equations</h1>

Branching</h3> If the decay chain has branching, a similar strategy can be employed. Consider the following chain:</p>

References</h1> Bateman, H. (1908). The solution of a system of differential equations occurring in the theory of radio-active transformations. Proceedings of the Cambridge Philosophical Society, Mathematical and Physical Sciences, v.15 (1908-1910), 423–427.</a></p>

Monte Carlo

Hello World

References</h1>
Bateman, H. (1908). The solution of a system of differential equations occurring in the theory of radio-active transformations. Proceedings of the Cambridge Philosophical Society, Mathematical and Physical Sciences, v.15 (1908-1910), 423–427.</a></p>