2016/2016-02-08/Solid Angle Of A Circle v4.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This notebook is an exercise in using the [**Cairo**](http://cairographics.org/) 2D graphics library via the [**Qahirah**](https://github.com/ldo/qahirah) high-level binding to construct interactive geometrical diagrams.\n",
    "\n",
    "In this case, the exercise is to derive the relationship between a solid angle on the surface of a sphere and the vertex angle of the subtending cone."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from ipywidgets import \\\n",
    "    interact\n",
    "from IPython.display import \\\n",
    "    display as ipython_display, \\\n",
    "    Latex, \\\n",
    "    display_png\n",
    "import math\n",
    "import qahirah as qah\n",
    "from qahirah import \\\n",
    "    CAIRO, \\\n",
    "    Colour, \\\n",
    "    Matrix, \\\n",
    "    Path, \\\n",
    "    Vector\n",
    "\n",
    "pix = qah.ImageSurface.create \\\n",
    "  (\n",
    "    format = CAIRO.FORMAT_RGB24,\n",
    "    dimensions = (400, 400)\n",
    "  )\n",
    "ctx = None\n",
    "\n",
    "def reset() :\n",
    "    \"(re)initializes the drawing context, wiping out any existing drawing.\"\n",
    "    global ctx\n",
    "    del ctx\n",
    "    ctx = qah.Context.create(pix)\n",
    "    (ctx\n",
    "       .save()\n",
    "       .set_source_colour(Colour.grey(.95))\n",
    "       .paint()\n",
    "       .restore()\n",
    "       .select_font_face(family = \"serif\", slant = CAIRO.FONT_SLANT_ITALIC, weight = CAIRO.FONT_WEIGHT_NORMAL)\n",
    "       .set_font_size(24)\n",
    "    )\n",
    "#end reset\n",
    "\n",
    "def display() :\n",
    "    \"(re)displays what has been drawn.\"\n",
    "    display_png(pix.to_png_bytes(), raw = True)\n",
    "#end display\n",
    "\n",
    "def display_latex(s) :\n",
    "    \"formats and displays a string as Latex markup.\"\n",
    "    ipython_display(Latex(s))\n",
    "#end display_latex\n",
    "\n",
    "reset()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "In two dimensions, the *angle subtended by an arc* can be defined as the ratio of the arc length to the radius:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "@interact(angle = (0.1, 1.5, 0.1))\n",
    "def demo_arc_angle(angle) :\n",
    "    display_latex(\"$$s = r \\\\theta$$\")\n",
    "    reset()\n",
    "    radius = abs(pix.dimensions) * 0.5\n",
    "    small_radius = radius * 0.1\n",
    "    text_mid_y = (ctx.font_extents.ascent - ctx.font_extents.descent) / 2\n",
    "    (ctx\n",
    "        .translate(Vector(0.1, 0.5) * pix.dimensions)\n",
    "        .set_line_width(2)\n",
    "        .set_source_colour(Colour.grey(0))\n",
    "        .move_to((0, 0))\n",
    "        .line_to(Vector(radius, 0).rotate(- angle / 2))\n",
    "        .arc((0, 0), radius, - angle / 2, angle / 2, False)\n",
    "        .close_path()\n",
    "        .stroke()\n",
    "        .move_to((0, 0))\n",
    "        .line_to(Vector(small_radius, 0).rotate(- angle / 2))\n",
    "        .arc((0, 0), small_radius, - angle / 2, angle / 2, False)\n",
    "        .close_path()\n",
    "        .stroke()\n",
    "        .move_to(Vector(radius, 0).rotate(- angle * 1.1 / 2) * 0.5)\n",
    "        .show_text(\"r\")\n",
    "        .move_to(Vector(small_radius * 1.3, text_mid_y))\n",
    "        .show_text(\"\\u03b8\")\n",
    "        .move_to(Vector(radius, 0) * 1.02 + Vector(0, text_mid_y))\n",
    "        .show_text(\"s\")\n",
    "    )\n",
    "    display()\n",
    "#end demo_arc_angle"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "In the above, the radius of the circle is $r$, the length of the arc is $s$, and the angle subtended at the centre of the circle is $\\theta$.\n",
    "\n",
    "In three dimensions, an analogous definition may be applied to the *solid angle subtended by a circle* on the surface of a sphere. The radius of the sphere is $r$, and consider a cone with its apex at the centre of the sphere, cutting through its surface to form a circle. Let the area of the circular portion cut out of the sphere be $a$, and let the *half-angle* subtended by the circle at the centre of the sphere (apex of the cone) be $\\theta$. Call the solid angle $\\phi$. This will be defined as\n",
    "\n",
    "$$\\phi = {a \\over r^2} $$\n",
    "\n",
    "The unit for $\\phi$ is called *steradians*, where that for flat angles like $\\theta$ is in radians. Mathematically, both kinds of angles are expressed as ratios, and hence are dimensionless numbers."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "@interact(angle = (0.1, 1.5, 0.1))\n",
    "def demo_sphere_angle(angle) :\n",
    "    display_latex(\"$$a = r^2 \\phi$$\")\n",
    "    reset()\n",
    "    radius = abs(pix.dimensions) * 0.3\n",
    "    foreshortened_radius = radius * 0.8\n",
    "    small_radius = radius * 0.2\n",
    "    text_mid_y = (ctx.font_extents.ascent - ctx.font_extents.descent) / 2\n",
    "    line_colour = Colour.grey(0)\n",
    "    (ctx\n",
    "        .translate(Vector(0.5, 0.5) * pix.dimensions)\n",
    "        .set_line_width(2)\n",
    "        .set_source_colour(line_colour)\n",
    "        .arc((0, 0), radius, 0, qah.circle, False)\n",
    "        .close_path()\n",
    "        .stroke()\n",
    "        .move_to((0, 0))\n",
    "        .line_to(Vector(foreshortened_radius, 0).rotate(- angle / 2))\n",
    "        .move_to((0, 0))\n",
    "        .line_to(Vector(foreshortened_radius, 0).rotate(angle / 2))\n",
    "        .stroke()\n",
    "        .move_to((0, 0))\n",
    "        .line_to(Vector(foreshortened_radius, 0))\n",
    "        .stroke()\n",
    "        .move_to(Vector(foreshortened_radius * math.cos(angle / 2), 0))\n",
    "        .line_to(Vector(foreshortened_radius, 0).rotate(angle / 2))\n",
    "        .move_to((0, 0))\n",
    "        .line_to(Vector(small_radius, 0))\n",
    "        .arc((0, 0), small_radius, angle / 2, angle / 2, False)\n",
    "        .close_path()\n",
    "        .stroke()\n",
    "        .append_path\n",
    "          (\n",
    "            Path.create_arc\n",
    "              (\n",
    "                centre = (0, 0),\n",
    "                radius = foreshortened_radius,\n",
    "                angle1 = 0,\n",
    "                angle2 = qah.circle,\n",
    "                negative = False\n",
    "              )\n",
    "                .transform\n",
    "                  (\n",
    "                        Matrix.translate(Vector(1, 0) * foreshortened_radius * math.cos(angle / 2))\n",
    "                    *\n",
    "                        Matrix.scale(math.sin(angle / 2))\n",
    "                    *\n",
    "                        Matrix.scale((0.5, 1))\n",
    "                  )\n",
    "          )\n",
    "        .set_source_colour(Colour.grey(0, 0.2))\n",
    "        .fill_preserve()\n",
    "        .set_source_colour(line_colour)\n",
    "        .stroke()\n",
    "        .move_to(Vector(radius, 0).rotate(- angle * 1.1 / 2) * 0.5)\n",
    "        .show_text(\"r\")\n",
    "        .move_to(Vector(foreshortened_radius, 0).rotate(angle / 4))\n",
    "        .show_text(\"r\")\n",
    "        .rel_move_to((0, ctx.font_extents.descent))\n",
    "        .show_text(\"a\")\n",
    "        .move_to(Vector(small_radius * 1.3, 0).rotate(angle / 2) + Vector(0, text_mid_y))\n",
    "        .show_text(\"\\u03b8\")\n",
    "        .move_to(Vector(foreshortened_radius, 0).rotate(- angle / 4) * 1.05 + Vector(0, text_mid_y))\n",
    "        .show_text(\"a\")\n",
    "        .move_to(Vector(0, radius)* 0.5)\n",
    "        .show_text(\"\\u03d5 = ?\")\n",
    "    )\n",
    "    display()\n",
    "#end demo_sphere_angle"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "How do we calculate $\\phi$ in terms of $\\theta$?\n",
    "\n",
    "First of all, the radius of the perimeter of the circular area is\n",
    "\n",
    "$$r_a = r \\sin \\theta$$\n",
    "\n",
    "(Note that, because of the convexity of the sphere, the centre of this perimeter circle resides *within* the sphere, not on its surface.) Now consider the incremental angle $\\mathrm{d}\\theta$ as $\\theta$ changes. The corresponding incremental change in the area of the circle is its perimeter multiplied by the incremental change in the radius, which is\n",
    "\n",
    "$$\\begin{align}\n",
    "\\mathrm{d}a & = 2 \\pi r_a r \\mathrm{d} \\theta \\\\\n",
    "& = 2 \\pi r^2 \\sin \\theta \\mathrm{d} \\theta\n",
    "\\end{align}$$\n",
    "\n",
    "Integrating this gives\n",
    "\n",
    "$$\\begin{align}\n",
    "a & = \\int 2 \\pi r^2 \\sin \\theta \\mathrm{d} \\theta \\\\\n",
    "& = C - 2 \\pi r^2 \\cos \\theta\n",
    "\\end{align}$$\n",
    "\n",
    "for some constant $C$. But $a$ has to be zero when $\\theta = 0$. Therefore,\n",
    "\n",
    "$$C = 2 \\pi r^2$$\n",
    "\n",
    "from which\n",
    "\n",
    "$$a = 2 \\pi r^2 (1 - \\cos \\theta)$$\n",
    "\n",
    "or, in other words,\n",
    "\n",
    "$$\\begin{align}\n",
    "\\phi & = {a \\over r^2} \\\\\n",
    "& = 2 \\pi (1 - \\cos \\theta)\n",
    "\\end{align}$$\n",
    "\n",
    "Let us check a couple of special cases: the surface area of a sphere of radius $r$ is $a = 4 \\pi r^2$. Therefore the complete solid angle of a sphere is ${a \\over r} = 4 \\pi$ steradians. When $\\theta = {\\pi \\over 2}$, then $cos \\theta = 0$, and the above formula should give us the solid angle of a hemisphere, which is $2 \\pi (1 - 0) = 2 \\pi$, or half that of the sphere. Which is correct.\n",
    "\n",
    "The formula also checks out if we set $\\theta = \\pi$, in which case $\\cos \\theta = -1$, and we get $\\phi = 2 \\pi (1 + 1) = 4 \\pi$, the full solid angle of the sphere."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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