{ "cells": [ { "cell_type": "markdown", "id": "markdown-000", "metadata": {}, "source": [ "# Radioactive decay\n", "\n", "[Read the lesson](https://tirta.co/a-byte-of-physics/radioactive-decay-simulation). This notebook includes the lesson text and its Python examples in reading order. Run the setup cell, then run the remaining cells from top to bottom. In Colab, use File → Save a copy in Drive to keep your edits. You can run and edit this view without changing the course source. Examples use illustrative data unless stated otherwise." ] }, { "cell_type": "code", "id": "code-001", "metadata": {}, "source": [ "import importlib.metadata\n", "import subprocess\n", "import sys\n", "\n", "requirements = {\"numpy\": (2, 0), \"sympy\": (1, 13), \"matplotlib\": (3, 8)}\n", "to_install = []\n", "for package, minimum in requirements.items():\n", " try:\n", " installed = importlib.metadata.version(package)\n", " version = tuple(int(part) for part in installed.split(\".\")[:2])\n", " except importlib.metadata.PackageNotFoundError:\n", " version = (0, 0)\n", " if version < minimum:\n", " to_install.append(f\"{package}>={minimum[0]}.{minimum[1]}\")\n", "if to_install:\n", " subprocess.check_call([sys.executable, \"-m\", \"pip\", \"install\", *to_install])\n", "\n", "import numpy as np\n", "import sympy as sp\n", "import matplotlib\n", "import matplotlib.pyplot as plt\n", "\n", "print(\"Python\", sys.version.split()[0])\n", "print(\"NumPy\", np.__version__)\n", "print(\"SymPy\", sp.__version__)\n", "print(\"Matplotlib\", matplotlib.__version__)\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-002", "metadata": {}, "source": [ "An unstable nucleus has no classical countdown clock that reveals its exact\n", "decay time. Quantum theory assigns a probability per unit time. For a large\n", "population of independent nuclei in the same state, that microscopic randomness\n", "produces a predictable exponential survival law.\n", "\n", "Radioactive decay changes one nuclide into another and can emit particles or\n", "photons. The decay law predicts population and activity; it does not by itself\n", "predict detector counts, biological dose, or every member of a decay chain.\n", "\n", "## Exponential decay law\n", "\n", "Let $N(t)$ be the expected\n", "number of undecayed nuclei. If each nucleus has constant decay probability per\n", "unit time $\\lambda$, then\n", "\n", "\n", "$$\n", "\\frac{dN}{dt}=-\\lambda N\n", "$$\n", "\n", "\n", "The minus sign means the parent population decreases. Separating variables gives\n", "\n", "\n", "$$\n", "N(t)=N_0e^{-\\lambda t}\n", "$$\n", "\n", "\n", "where $N_0$ is the\n", "population at $t=0$.\n", "The equation gives an ensemble expectation. A small physical sample contains an\n", "integer number of nuclei and can differ from that expectation." ] }, { "cell_type": "code", "id": "code-003", "metadata": {}, "source": [ "import numpy as np\n", "\n", "initial_count, half_life, time = 10000, 5.0, 10.0\n", "decay_constant = np.log(2) / half_life\n", "expected = initial_count * np.exp(-decay_constant * time)\n", "print(\"Expected survivors:\", expected)\n", "assert np.isclose(expected, initial_count / 4)\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-004", "metadata": {}, "source": [ "At zero time the result is\n", "$N(0)=N_0$.\n", "As time increases, the expected population never rises in this one-parent model.\n", "\n", "## Half-life and mean lifetime\n", "\n", "The half-life $T_{1/2}$\n", "is the time for the expected parent population to fall by one half:\n", "\n", "\n", "$$\n", "T_{1/2}=\\frac{\\ln 2}{\\lambda}\n", "$$\n", "\n", "\n", "The mean lifetime of one member of the ensemble is\n", "\n", "\n", "$$\n", "\\tau=\\frac{1}{\\lambda}\n", "$$\n", "\n", "\n", "so $T_{1/2}=\\tau\\ln2$.\n", "Half-life and mean lifetime are not the same number.\n", "\n", "After $n$ half-lives,\n", "\n", "\n", "$$\n", "N=N_0\\left(\\frac12\\right)^n\n", "$$\n", "\n", "\n", "This relation remains useful for noninteger\n", "$n=t/T_{1/2}$\n", "when the power is evaluated continuously.\n", "\n", "## Activity\n", "\n", "The **activity** is the expected decay rate:\n", "\n", "\n", "$$\n", "\\mathcal A(t)=-\\frac{dN}{dt}=\\lambda N(t)=\\mathcal A_0e^{-\\lambda t}\n", "$$\n", "\n", "\n", "The SI unit is the becquerel:\n", "$1\\,\\mathrm{Bq}=1\\,\\mathrm{s}^{-1}$.\n", "Activity is not the same as absorbed dose or biological effect. Radiation type,\n", "energy, geometry, exposure time, shielding, and tissue response are also needed.\n", "\n", "If a sample amount is given in moles, the expected nucleus count is\n", "\n", "\n", "$$\n", "N=\\frac{m}{M}N_{\\mathrm A}\n", "$$\n", "\n", "\n", "for sample mass $m$, molar mass\n", "$M$, and Avogadro\n", "constant $N_{\\mathrm A}$.\n", "Purity and isotopic abundance must be included when the sample is not a pure\n", "isotope.\n", "\n", "## Exact decay probability over an interval\n", "\n", "The probability that one nucleus survives an interval\n", "$\\Delta t$ is\n", "\n", "\n", "$$\n", "P_{\\mathrm{survive}}=e^{-\\lambda\\Delta t}\n", "$$\n", "\n", "\n", "and the probability that it decays during that interval is\n", "\n", "\n", "$$\n", "P_{\\mathrm{decay}}=1-e^{-\\lambda\\Delta t}\n", "$$\n", "\n", "\n", "The approximation\n", "$P_{\\mathrm{decay}}\\approx\\lambda\\Delta t$\n", "is valid only when\n", "$\\lambda\\Delta t\\ll1$.\n", "Using $\\lambda\\Delta t$\n", "as an exact probability can exceed one for a large interval." ] }, { "cell_type": "code", "id": "code-005", "metadata": {}, "source": [ "def interval_decay_probability(half_life, interval):\n", " if half_life <= 0 or interval < 0:\n", " raise ValueError(\"Use positive half-life and nonnegative interval.\")\n", " return -np.expm1(-np.log(2) * interval / half_life)\n", "\n", "\n", "rng = np.random.default_rng(42)\n", "probability = interval_decay_probability(5.0, 1.0)\n", "decays = rng.binomial(10000, probability)\n", "print(\"Decays in one second:\", decays, \"expected:\", 10000 * probability)\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-006", "metadata": {}, "source": [ "## Common decay modes\n", "\n", "In alpha decay, the nucleus emits a helium-4 nucleus:\n", "\n", "\n", "$$\n", "{}^A_ZX\\rightarrow{}^{A-4}_{Z-2}Y+{}^4_2\\alpha\n", "$$\n", "\n", "\n", "In beta-minus decay, a neutron is converted into a proton while an electron and\n", "electron antineutrino are emitted:\n", "\n", "\n", "$$\n", "n\\rightarrow p+e^-+\\bar\\nu_e\n", "$$\n", "\n", "\n", "The mass number stays constant and nuclear charge increases by one. In\n", "beta-plus decay, a proton becomes a neutron while a positron and electron\n", "neutrino are emitted. Electron capture is a competing route in which the\n", "nucleus captures an atomic electron.\n", "\n", "Gamma decay changes an excited nucleus to a lower nuclear state without\n", "changing $A$ or\n", "$Z$:\n", "\n", "\n", "$$\n", "{}^A_ZX^*\\rightarrow{}^A_ZX+\\gamma\n", "$$\n", "\n", "\n", "Every reaction must conserve electric charge, total energy, momentum, angular\n", "momentum, and lepton number where applicable. Nucleon number is useful\n", "bookkeeping for ordinary nuclear reactions, though it is not a fundamental\n", "conservation law in all particle processes.\n", "\n", "## Branching fractions\n", "\n", "A nuclide can decay through several channels. If partial decay constants are\n", "$\\lambda_i$, then\n", "\n", "\n", "$$\n", "\\lambda_{\\mathrm{total}}=\\sum_i\\lambda_i,\\qquad b_i=\\frac{\\lambda_i}{\\lambda_{\\mathrm{total}}}\n", "$$\n", "\n", "\n", "The branching fractions satisfy\n", "$\\sum_i b_i=1$\n", "when all channels are included. The parent population follows the total decay\n", "constant, while the expected production rate into channel\n", "$i$ is\n", "$b_i\\mathcal A$.\n", "\n", "## Decay energy\n", "\n", "A spontaneous decay must have a positive available energy under the chosen\n", "mass convention:\n", "\n", "\n", "$$\n", "Q=(m_{\\mathrm{initial}}-m_{\\mathrm{final,total}})c^2\n", "$$\n", "\n", "\n", "The final total includes every daughter and emitted particle. Atomic masses can\n", "simplify some electron bookkeeping, but beta-plus and electron-capture formulas\n", "require especially careful treatment of atomic electrons. Use one documented\n", "mass convention throughout.\n", "\n", "In a two-body decay of a parent initially at rest, energy and momentum\n", "conservation fix the daughter and emitted-particle kinetic energies. Beta decay\n", "has three final particles, so the electron or positron has a continuous energy\n", "spectrum; the neutrino carries variable energy and momentum. Treating all\n", "$Q$ as electron kinetic energy\n", "would violate the measured spectrum and conservation bookkeeping.\n", "\n", "## A two-member decay chain\n", "\n", "If parent 1 decays into radioactive daughter 2, their mean populations obey\n", "\n", "\n", "$$\n", "\\frac{dN_1}{dt}=-\\lambda_1N_1,\\qquad \\frac{dN_2}{dt}=\\lambda_1N_1-\\lambda_2N_2\n", "$$\n", "\n", "\n", "For no initial daughter and unequal decay constants,\n", "\n", "\n", "$$\n", "N_2(t)=\\frac{\\lambda_1N_{1,0}}{\\lambda_2-\\lambda_1}\\left(e^{-\\lambda_1t}-e^{-\\lambda_2t}\\right)\n", "$$\n", "\n", "\n", "The daughter can first grow and later decline. Therefore the rule “radioactive\n", "population always decreases” applies to an isolated parent population, not to\n", "every member of a chain.\n", "\n", "If the parent lives much longer than the daughter, their activities can approach\n", "secular equilibrium, with daughter activity tracking parent activity. Chemical\n", "separation or an initially present daughter changes the solution.\n", "\n", "## Measurement counts and uncertainty\n", "\n", "A detector records only some emitted radiation. In a simple count model,\n", "\n", "\n", "$$\n", "R_{\\mathrm{measured}}=\\epsilon\\,g\\,\\mathcal A+R_{\\mathrm{background}}\n", "$$\n", "\n", "\n", "where $\\epsilon$ is\n", "intrinsic detection efficiency and $g$.\n", "Real detectors also have dead time, energy thresholds, scattering, and\n", "energy-dependent response.\n", "\n", "For independent counts at a constant expected rate, a Poisson model has\n", "standard deviation approximately\n", "$\\sqrt{C}$\n", "for expected count $C$.\n", "Background subtraction adds uncertainty from both the sample and background\n", "measurements. A fractional count can be an expectation, never one observed event.\n", "\n", "## Dating and assumptions\n", "\n", "Radiometric dating uses a decay clock only when initial conditions and material\n", "history are constrained. From the parent ratio,\n", "\n", "\n", "$$\n", "t=\\frac{1}{\\lambda}\\ln\\left(\\frac{N_0}{N}\\right)\n", "$$\n", "\n", "\n", "but $N_0$ is often inferred\n", "from daughter products, reference isotopes, or a formation model. The sample\n", "must remain sufficiently closed to parent and daughter exchange, and the decay\n", "constant must apply over the interval.\n", "\n", "## Checks and scope\n", "\n", "A reliable calculation checks:\n", "\n", "- $N(0)=N_0$;\n", "- one half remains after one half-life;\n", "- activity and population share the same exponential factor;\n", "- decay probability stays between zero and one;\n", "- nuclide equations conserve mass number and charge as appropriate;\n", "- branching fractions sum to one;\n", "- parent loss and daughter production balance in a decay chain.\n", "\n", "The simple law assumes a constant decay constant and independent nuclei in one\n", "state. Environmental effects are normally tiny for nuclear decays but can matter\n", "for some electron-capture processes. This page does not calculate radiation\n", "transport, shielding, absorbed dose, or medical risk.\n", "\n", "## A worked Python notebook\n", "\n", "Start with ten thousand independent nuclei having a half-life of five seconds. The question for this calculation is precise: how does **expected survivor count**\n", "change when we change **elapsed time**, while keeping the other choices fixed?\n", "The earlier sections explain the physical relationship. Here we turn one part of that\n", "relationship into a calculation that you can run, inspect, and change in Google Colab.\n", "\n", "The smooth result is an expectation; an actual count is integer-valued and fluctuates. Keep that sentence beside the results. A formula can return a number outside\n", "its useful range, and a plotting library will usually draw that number without asking\n", "whether the physical assumptions still hold. The selected interval is from\n", "0 to 20 s. It gives us a specific domain to discuss\n", "instead of leaving the graph's range to a software default.\n", "\n", "Use the notebook link above the lesson, run the setup cell, and then run the cells in\n", "order. The code below is also included there. NumPy handles the arrays of numerical\n", "values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these\n", "libraries or notebook cells are unfamiliar, start with\n", "[Your first physics notebook](https://tirta.co/a-byte-of-physics/your-first-physics-notebook).\n", "\n", "### State the relationship before choosing a graph\n", "\n", "In this small example, `q` stands for elapsed time, expressed in s. The output is\n", "expected survivor count, expressed in count. The equation below uses the numerical values in those\n", "units. Any fixed coefficients therefore carry the units needed to make the relation\n", "consistent. A value supplied in another unit must be converted before it enters this\n", "expression.\n", "\n", "\n", "$$\n", "f(q) = 10000 e^{- \\frac{q \\log{\\left(2 \\right)}}{5}}\n", "$$\n", "\n", "\n", "This is a function of one changing input. It is not a statement that the physical\n", "system has only one variable. We deliberately hold the other inputs fixed so that one\n", "cause of variation is visible. Changing several inputs at once would make it harder\n", "to tell which change produced the output difference. Write down the held-fixed choices\n", "before trying a different experiment.\n", "\n", "At 10 s, the expression gives 2500 count. This is our\n", "reference calculation. It is a prediction of the stated example, not a new measurement.\n", "Its displayed digits help us compare implementations. They should not be read as a\n", "claim that physical inputs are known to six significant figures." ] }, { "cell_type": "code", "id": "code-007", "metadata": {}, "source": [ "import numpy as np\n", "import sympy as sp\n", "import matplotlib.pyplot as plt\n", "\n", "q = sp.Symbol(\"q\", real=True)\n", "expression = sp.sympify(\"10000*exp(-log(2)*q/5)\", locals={\"q\": q})\n", "reference_input = 10.0\n", "reference_output = float(expression.subs(q, reference_input))\n", "print(\"Input:\", reference_input, \"s\")\n", "print(\"Predicted expected survivor count:\", reference_output, \"count\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-008", "metadata": {}, "source": [ "`sp.Symbol` declares a mathematical symbol. It does not assign a numerical measurement\n", "to that symbol. `subs` then replaces the symbol with our chosen input, and `float`\n", "converts the resulting symbolic number into a standard Python numerical value.\n", "Keeping these steps separate lets us inspect the equation before numerical substitution.\n", "It also prevents a later plot from becoming the only surviving record of the model.\n", "\n", "### Compare a few cases you can explain\n", "\n", "Before drawing hundreds of points, calculate three. The lower endpoint gives\n", "10000 count. The midpoint, at 10 s, gives\n", "2500 count. The upper endpoint gives 625 count.\n", "Read these together with the assumptions: they describe different choices of elapsed time\n", "within the same example, with the other inputs held fixed.\n", "\n", "These three values are a useful first check, but they do not establish the shape between\n", "them. Two endpoints can be equal even when the curve bends or oscillates in between.\n", "Likewise, a large change between endpoints does not tell us whether the relationship is\n", "linear. We need the equation and intermediate samples to answer those questions." ] }, { "cell_type": "code", "id": "code-009", "metadata": {}, "source": [ "evaluate = sp.lambdify(q, expression, modules=\"numpy\")\n", "inputs = np.array([0.0, 10.0, 20.0])\n", "outputs = np.asarray(evaluate(inputs), dtype=float)\n", "outputs = np.broadcast_to(outputs, inputs.shape)\n", "for input_value, output_value in zip(inputs, outputs):\n", " print(f\"{input_value:.6g} s -> {output_value:.6g} count\")\n", "assert np.all(np.isfinite(outputs))\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-010", "metadata": {}, "source": [ "`lambdify` turns the symbolic expression into a numerical function. We explicitly select\n", "NumPy as its numerical backend so it can accept an array. The same expression now has\n", "two useful forms: one for algebra and one for calculating many cases. The assertion\n", "checks that the chosen inputs produce finite outputs. It would catch an infinity or an\n", "undefined numerical result, but a finite value can still be physically inappropriate.\n", "\n", "### Make a graph that answers the question\n", "\n", "The horizontal axis shows elapsed time; the vertical axis shows expected survivor count. Every plotted\n", "point is a prediction of the same expression. We mark the reference case separately\n", "so that a reader can connect the earlier printed value to its place on the curve.\n", "The axis labels include units because a shape without a scale cannot tell us the size\n", "of a physical effect." ] }, { "cell_type": "code", "id": "code-011", "metadata": {}, "source": [ "input_grid = np.linspace(0.0, 20.0, 301)\n", "output_grid = np.asarray(evaluate(input_grid), dtype=float)\n", "output_grid = np.broadcast_to(output_grid, input_grid.shape)\n", "fig, ax = plt.subplots(figsize=(7, 4))\n", "ax.plot(input_grid, output_grid, label=\"Stated model\")\n", "ax.scatter([reference_input], [reference_output], label=\"Reference case\", zorder=3)\n", "ax.set_xlabel(\"Elapsed time (s)\")\n", "ax.set_ylabel(\"Expected survivor count (count)\")\n", "ax.legend()\n", "ax.grid(alpha=0.25)\n", "fig.tight_layout()\n", "plt.show()\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-012", "metadata": {}, "source": [ "Look for a sign change, a turning point, or a region where the curve becomes steep.\n", "Connect each feature to the expression rather than to the colour of the line. If the\n", "curve has a minimum or maximum, changing the input in opposite directions may have\n", "similar effects on the output near that point. If it is nearly straight over a short\n", "interval, a local linear approximation may be useful even when the complete expression\n", "is nonlinear.\n", "\n", "Increasing the number of samples makes the line smoother and may reveal features that\n", "a coarse grid missed. It does not improve the physical inputs or enlarge the model's\n", "validity range. For quantities that can take only discrete values, use the line as a\n", "guide to the relationship and interpret only the allowed input values. A plotting\n", "function does not enforce such restrictions for you.\n", "\n", "### Measure the local sensitivity\n", "\n", "A derivative asks a more specific question than the whole graph: how quickly does\n", "expected survivor count change near the reference input? Its units are the output unit divided by\n", "the input unit. For this calculation, the derivative at 10 s is\n", "-346.574 in those ratio units. Locally, the output decreases as the\n", "input increases. This statement is local; the sign or size of the slope can be different\n", "elsewhere on the curve.\n", "\n", "\n", "$$\n", "f'(q) = - 2000 e^{- \\frac{q \\log{\\left(2 \\right)}}{5}} \\log{\\left(2 \\right)}\n", "$$\n", "\n", "\n", "For a small input change, multiply the derivative by that change to estimate the output\n", "change. Then compare that estimate with a fresh evaluation of the full expression.\n", "The distinction between these two calculations matters: the derivative gives a tangent\n", "approximation, while substitution evaluates the chosen expression at the new point." ] }, { "cell_type": "code", "id": "code-013", "metadata": {}, "source": [ "slope_expression = sp.diff(expression, q)\n", "slope_at_reference = float(slope_expression.subs(q, reference_input))\n", "changed_input = 10.1\n", "linear_prediction = reference_output + slope_at_reference * (\n", " changed_input - reference_input\n", ")\n", "full_prediction = float(evaluate(changed_input))\n", "print(\"Symbolic derivative:\", slope_expression)\n", "print(\"Tangent prediction:\", linear_prediction, \"count\")\n", "print(\"Full expression:\", full_prediction, \"count\")\n", "print(\"Approximation difference:\", full_prediction - linear_prediction, \"count\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-014", "metadata": {}, "source": [ "Here the changed input is 10.1 s. The tangent estimate is\n", "2465.34 count, while the full expression gives\n", "2465.58 count. If the two values are close, it means the chosen\n", "input change is small enough for this local approximation at this point. It does not\n", "mean the entire model is linear. If the slope is zero, a quadratic or higher-order term\n", "may supply the leading change instead.\n", "\n", "This is also the beginning of uncertainty propagation. If the input has a small stated\n", "standard uncertainty, the magnitude of the slope times that uncertainty estimates one\n", "contribution to output uncertainty. Do not automatically use the same rule for a wide\n", "interval, a discontinuity, a turning point with zero first derivative, or several\n", "correlated uncertain inputs. Those situations need more of the model than one slope.\n", "\n", "### Check the derivative with numerical differences\n", "\n", "We can estimate the same slope from values on either side of the reference point.\n", "Subtract the lower output from the upper output and divide by twice the input step.\n", "Parentheses are essential: the entire output difference must be divided by the entire\n", "input interval. Without them, Python can execute a different mathematical expression\n", "while still producing a plausible number.\n", "\n", "The calculation below uses three decreasing step sizes. Each estimate is compared with\n", "the symbolic derivative, so the reference is independent of the numerical difference\n", "formula. Near an endpoint, it is acceptable here to evaluate a small extension only\n", "because the displayed mathematical expression is smooth there; such a check does not\n", "assert that extrapolated values describe additional physical cases." ] }, { "cell_type": "code", "id": "code-015", "metadata": {}, "source": [ "input_span = 20.0\n", "for fraction in [1e-2, 1e-3, 1e-4]:\n", " step = input_span * fraction\n", " upper = float(evaluate(reference_input + step))\n", " lower = float(evaluate(reference_input - step))\n", " numerical_slope = (upper - lower) / (2 * step)\n", " difference = numerical_slope - slope_at_reference\n", " print(f\"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-016", "metadata": {}, "source": [ "For a smooth function, reducing the step often reduces the approximation error at first.\n", "Very small steps can eventually make subtraction of nearly equal floating-point values\n", "less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare\n", "the observed behaviour over a useful range of steps instead of choosing the smallest\n", "number the computer can represent.\n", "\n", "An absolute slope difference is useful when the expected slope is near zero. A relative\n", "difference would divide by a very small reference and could look enormous even when the\n", "absolute discrepancy is tiny. Away from zero, a relative difference can help compare\n", "quantities with very different scales. The tolerance should reflect the question and\n", "the numerical method, rather than a universal decimal threshold copied into every test.\n", "\n", "### Change the example and preserve the reasoning\n", "\n", "Make one copy of the reference calculation before editing it. Choose a second value of\n", "elapsed time inside the stated interval, predict whether expected survivor count will rise or fall,\n", "and then evaluate the expression. Place the prediction and the result in a text cell.\n", "If they disagree, identify whether the mistake was in the sign, the unit conversion,\n", "the algebra, or the assumption about how the curve behaves.\n", "\n", "Next choose a shorter plotting interval around the reference point. Compare the local\n", "curve with its tangent approximation. Then widen the interval again. The difference\n", "shows why a useful local estimate need not describe the whole range. Keep the axis\n", "labels and numerical scale visible when comparing the plots; automatic axis limits can\n", "make two different effects appear equally large.\n", "\n", "Finally, return to the physical limit stated at the start: The smooth result is an expectation; an actual count is integer-valued and fluctuates. Describe one change\n", "to the setup that would violate that limit. State what new input, equation, or observation\n", "would be needed before trusting an answer in the changed situation. This last step\n", "connects the Python experiment to the physics instead of treating successful execution\n", "as the only criterion for a useful result.\n", "\n", "Before saving, restart the runtime and run all cells in order. A result that depends on\n", "an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep\n", "unrounded values for later calculations, round only the reported output, and label any\n", "new data as measured, assumed, or generated. The graph and printed values should remain\n", "traceable to the same expression and the same choices of inputs.\n", "\n", "### Recover the output change by integrating the slope\n", "\n", "There is another way to check that the expression and its derivative describe the same\n", "relationship. Add up the small output changes predicted by the slope across the chosen\n", "input interval. In the continuous limit, this integral equals the output at the upper\n", "endpoint minus the output at the lower endpoint. The result has units of count:\n", "the input unit in the integration cancels the input unit in the denominator of the slope.\n", "\n", "This statement is about the mathematics of this particular relationship. It does not\n", "assign a new physical meaning to an arbitrary area under a graph. For example, an area\n", "under force versus distance represents work, while an area under a different pair of\n", "quantities may represent something else or have no standard name. Always carry the\n", "axis units through the calculation before naming an integral." ] }, { "cell_type": "code", "id": "code-017", "metadata": {}, "source": [ "slope_function = sp.lambdify(q, slope_expression, modules=\"numpy\")\n", "endpoint_change = float(evaluate(20.0) - evaluate(0.0))\n", "for points in [101, 201, 401]:\n", " grid = np.linspace(0.0, 20.0, points)\n", " slopes = np.asarray(slope_function(grid), dtype=float)\n", " slopes = np.broadcast_to(slopes, grid.shape)\n", " integrated_change = np.trapezoid(slopes, grid)\n", " print(\n", " points,\n", " \"integrated change:\",\n", " integrated_change,\n", " \"endpoint change:\",\n", " endpoint_change,\n", " )\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-018", "metadata": {}, "source": [ "The trapezoidal rule joins neighbouring slope samples with straight segments and adds\n", "their signed areas. A negative contribution remains negative; replacing every slope\n", "by its magnitude would calculate a different quantity. Increasing the sample count\n", "provides a practical refinement check. If the values do not settle toward the endpoint\n", "difference, first inspect the interval, singularities, and array ordering.\n", "\n", "Do not conclude that two agreeing calculations prove the model is physically complete.\n", "Both use the same selected relationship, so an omitted effect would be omitted from\n", "both. Their agreement is evidence that the numerical tools implement that relationship\n", "consistently. Comparing with an independent experiment or a more appropriate physical\n", "model is a separate test.\n", "\n", "Keep this comparison in the notebook when you change the expression. It is particularly\n", "useful after adding a coefficient, changing a sign convention, or converting an input\n", "unit. A code cell that produces a smooth graph can conceal these mistakes; a second\n", "calculation of the same output change gives you another place to notice them.\n", "\n", "## Sources\n", "\n", "- [OpenStax, *University Physics Volume 3*, §10.3, “Radioactive Decay”](https://openstax.org/books/university-physics-volume-3/pages/10-3-radioactive-decay).\n", "- [OpenStax, *University Physics Volume 3*, §10.7, “Medical Applications and Biological Effects of Nuclear Radiation”](https://openstax.org/books/university-physics-volume-3/pages/10-7-medical-applications-and-biological-effects-of-nuclear-radiation).\n", "- [Tipler and Llewellyn, *Modern Physics*, radioactive decay and decay chains](https://web.pdx.edu/~pmoeck/books/Tipler_Llewellyn.pdf).\n", "- [Benjamin Crowell, *Light and Matter*, nuclear decay](https://open.umn.edu/opentextbooks/textbooks/light-and-matter).\n", "- [Physics LibreTexts, Nuclear and Particle Physics bookshelf](https://phys.libretexts.org/Bookshelves/Nuclear_and_Particle_Physics)." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "name": "python", "version": "3.12" }, "colab": { "name": "radioactive-decay-simulation.ipynb", "provenance": [] }, "tirta": { "lesson": "https://tirta.co/a-byte-of-physics/radioactive-decay-simulation", "source_sha256": "fa9478b1f4aecfd297d1a4a3d752df52239abf516b5b31cfd04fc6584f721c64" } }, "nbformat": 4, "nbformat_minor": 5 }