{ "metadata": { "name": "schwarzschild" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Schwarzschild Solution to the Einstein's equation" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We will rederive the equations from *General Theory of Relativity*, chapter 18, by Dirac. They describe a spherically symmetric solution to the Einstein's equation in vacuum.\n", "\n", "We will not try to solve the resulting equations.\n", "\n", "This version of the notebook uses functions that explicitly calculate the Christoffel symbols, hence the work is coordinate-system-dependent." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Import the necessary modules." ] }, { "cell_type": "code", "collapsed": false, "input": [ "from sympy.diffgeom import *\n", "TP = TensorProduct" ], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 1 }, { "cell_type": "markdown", "metadata": {}, "source": [ "Define a 4D manifold with a patch and a coordinate chart on which we will work." ] }, { "cell_type": "code", "collapsed": false, "input": [ "m = Manifold('Schwarzschild', 4)\n", "p = Patch('origin', m)\n", "cs = CoordSystem('spherical', p, ['t', 'r', 'theta', 'phi'])" ], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 2 }, { "cell_type": "code", "collapsed": false, "input": [ "m, p, cs" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left ( \\mathrm{Schwarzschild}, \\quad \\mathrm{origin}_{\\mathrm{Schwarzschild}}, \\quad \\mathrm{spherical}^{\\mathrm{origin}}_{\\mathrm{Schwarzschild}}\\right )$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 3, "text": [ "(Manifold(Schwarzschild, 4), Patch(origin, Manifold(Schwarzschild, 4)), CoordS\n", "ystem(spherical, Patch(origin, Manifold(Schwarzschild, 4)), (t, r, theta, phi)\n", "))" ] } ], "prompt_number": 3 }, { "cell_type": "markdown", "metadata": {}, "source": [ "Prepare the variables containing the scalar fields and the 1-form fields." ] }, { "cell_type": "code", "collapsed": false, "input": [ "t, r, theta, phi = cs.coord_functions()\n", "t, r, theta, phi" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left ( \\boldsymbol{\\mathrm{t}}, \\quad \\boldsymbol{\\mathrm{r}}, \\quad \\boldsymbol{\\mathrm{\\theta}}, \\quad \\boldsymbol{\\mathrm{\\phi}}\\right )$$" ], "metadata": {}, "output_type": "pyout", "png": "iVBORw0KGgoAAAANSUhEUgAAAJQAAAAVBAMAAABS/tqaAAAAMFBMVEX///8AAAAAAAAAAAAAAAAA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\nHhEAAAAASUVORK5CYII=\n", "prompt_number": 4, "text": [ "(t, r, \u03b8, \u03c6)" ] } ], "prompt_number": 4 }, { "cell_type": "code", "collapsed": false, "input": [ "dt, dr, dtheta, dphi = cs.base_oneforms()\n", "dt, dr, dtheta, dphi" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left ( \\mathrm{d}t, \\quad \\mathrm{d}r, \\quad \\mathrm{d}\\theta, \\quad \\mathrm{d}\\phi\\right )$$" ], "metadata": {}, "output_type": "pyout", "png": "iVBORw0KGgoAAAANSUhEUgAAALsAAAAVBAMAAADsqILHAAAAMFBMVEX///8AAAAAAAAAAAAAAAAA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\nEvP/s0dW2SUjDGV5Ca7kdD9iAvgHv9m2uFa43G4AAAAASUVORK5CYII=\n", "prompt_number": 5, "text": [ "(\u2146 t, \u2146 r, \u2146 \u03b8, \u2146 \u03c6)" ] } ], "prompt_number": 5 }, { "cell_type": "markdown", "metadata": {}, "source": [ "The most general spherically-symmetric metric has the following form." ] }, { "cell_type": "code", "collapsed": false, "input": [ "metric = exp(2*f(r))*TP(dt, dt) - exp(2*g(r))*TP(dr, dr) - r**2*TP(dtheta, dtheta) - r**2*sin(theta)**2*TP(dphi, dphi)\n", "metric" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$e^{2 f{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} \\mathrm{d}t\\otimes\\mathrm{d}t - e^{2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} \\mathrm{d}r\\otimes\\mathrm{d}r - \\sin^{2}{\\left (\\boldsymbol{\\mathrm{\\theta}} \\right )} \\left(\\boldsymbol{\\mathrm{r}}\\right)^{2} \\mathrm{d}\\phi\\otimes\\mathrm{d}\\phi - \\left(\\boldsymbol{\\mathrm{r}}\\right)^{2} \\mathrm{d}\\theta\\otimes\\mathrm{d}\\theta$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 6, "text": [ " 2\u22c5f(r) 2\u22c5g(r) 2 2 \n", "\u212f \u22c5TensorProduct(dt, dt) - \u212f \u22c5TensorProduct(dr, dr) - sin (\u03b8)\u22c5r \u22c5Ten\n", "\n", " 2 \n", "sorProduct(dphi, dphi) - r \u22c5TensorProduct(dtheta, dtheta)" ] } ], "prompt_number": 6 }, { "cell_type": "markdown", "metadata": {}, "source": [ "The matrix $M$ representing the two-form as a bilinear map $V,U\\to V^tMU$ over the column vectors $V$ and $U$ in the canonical basis of the chosen coordinate system is:" ] }, { "cell_type": "code", "collapsed": false, "input": [ "twoform_to_matrix(metric)" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left[\\begin{matrix}e^{2 f{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} & 0 & 0 & 0\\\\0 & - e^{2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} & 0 & 0\\\\0 & 0 & - \\left(\\boldsymbol{\\mathrm{r}}\\right)^{2} & 0\\\\0 & 0 & 0 & - \\sin^{2}{\\left (\\boldsymbol{\\mathrm{\\theta}} \\right )} \\left(\\boldsymbol{\\mathrm{r}}\\right)^{2}\\end{matrix}\\right]$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 7, "text": [ "\u23a1 2\u22c5f(r) \u23a4\n", "\u23a2\u212f 0 0 0 \u23a5\n", "\u23a2 \u23a5\n", "\u23a2 2\u22c5g(r) \u23a5\n", "\u23a2 0 -\u212f 0 0 \u23a5\n", "\u23a2 \u23a5\n", "\u23a2 2 \u23a5\n", "\u23a2 0 0 -r 0 \u23a5\n", "\u23a2 \u23a5\n", "\u23a2 2 2\u23a5\n", "\u23a3 0 0 0 -sin (\u03b8)\u22c5r \u23a6" ] } ], "prompt_number": 7 }, { "cell_type": "markdown", "metadata": {}, "source": [ "Now we will calculate the components in the same basis of the Ricci tensor." ] }, { "cell_type": "code", "collapsed": false, "input": [ "ricci = metric_to_Ricci_components(metric)" ], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 8 }, { "cell_type": "code", "collapsed": false, "input": [ "ricci = [[simplify(ricci[i][j])\n", " for j in range(4)] for i in range(4)]\n" ], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 9 }, { "cell_type": "markdown", "metadata": {}, "source": [ "The diagonal components give the equations we are interested in." ] }, { "cell_type": "code", "collapsed": false, "input": [ "ricci[0][0]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\frac{1}{\\boldsymbol{\\mathrm{r}}} \\left(\\boldsymbol{\\mathrm{r}} \\left(\\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}\\right)^{2} - \\boldsymbol{\\mathrm{r}} \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} \\left. \\frac{d}{d \\xi_{1}} g{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} + \\boldsymbol{\\mathrm{r}} \\left. \\frac{d^{2}}{d \\xi_{1}^{2}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} + 2 \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}\\right) e^{2 f{\\left (\\boldsymbol{\\mathrm{r}} \\right )} - 2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}}$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 10, "text": [ "\u239b 2 \u239b 2 \n", "\u239c \u239b d \u239e\u2502 \u239b d \u239e\u2502 \u239b d \u239e\u2502 \u239c d \n", "\u239cr\u22c5\u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 - r\u22c5\u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 \u22c5\u239c\u2500\u2500\u2500(g(\u03be\u2081))\u239f\u2502 + r\u22c5\u239c\u2500\u2500\u2500\u2500(f(\u03be\u2081))\n", "\u239c \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239c 2 \n", "\u239d \u239dd\u03be\u2081 \n", "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n", " r \n", "\n", "\u239e\u2502 \u239e \n", "\u239f\u2502 \u239b d \u239e\u2502 \u239f 2\u22c5f(r) - 2\u22c5g(r)\n", "\u239f\u2502 + 2\u22c5\u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 \u239f\u22c5\u212f \n", "\u239f\u2502 \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r\u239f \n", "\u23a0\u2502\u03be\u2081=r \u23a0 \n", "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n", " " ] } ], "prompt_number": 10 }, { "cell_type": "code", "collapsed": false, "input": [ "ricci[1][1]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$- \\left(\\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}\\right)^{2} + \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} \\left. \\frac{d}{d \\xi_{1}} g{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} - \\left. \\frac{d^{2}}{d \\xi_{1}^{2}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} + \\frac{2}{\\boldsymbol{\\mathrm{r}}} \\left. \\frac{d}{d \\xi_{1}} g{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 11, "text": [ " \n", " 2 \u239b 2 \u239e\u2502 \n", " \u239b d \u239e\u2502 \u239b d \u239e\u2502 \u239b d \u239e\u2502 \u239c d \u239f\u2502 \n", "- \u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 + \u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 \u22c5\u239c\u2500\u2500\u2500(g(\u03be\u2081))\u239f\u2502 - \u239c\u2500\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 \n", " \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239c 2 \u239f\u2502 \n", " \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=\n", "\n", " \u239b d \u239e\u2502 \n", " 2\u22c5\u239c\u2500\u2500\u2500(g(\u03be\u2081))\u239f\u2502 \n", " \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r\n", " + \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n", " r \n", "r " ] } ], "prompt_number": 11 }, { "cell_type": "code", "collapsed": false, "input": [ "ricci[2][2]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left(e^{2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} - \\boldsymbol{\\mathrm{r}} \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} + \\boldsymbol{\\mathrm{r}} \\left. \\frac{d}{d \\xi_{1}} g{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} - 1\\right) e^{- 2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}}$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 12, "text": [ "\u239b 2\u22c5g(r) \u239b d \u239e\u2502 \u239b d \u239e\u2502 \u239e -2\u22c5g(r)\n", "\u239c\u212f - r\u22c5\u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 + r\u22c5\u239c\u2500\u2500\u2500(g(\u03be\u2081))\u239f\u2502 - 1\u239f\u22c5\u212f \n", "\u239d \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u23a0 " ] } ], "prompt_number": 12 }, { "cell_type": "code", "collapsed": false, "input": [ "ricci[3][3]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left(e^{2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} - \\boldsymbol{\\mathrm{r}} \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} + \\boldsymbol{\\mathrm{r}} \\left. \\frac{d}{d \\xi_{1}} g{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }} - 1\\right) e^{- 2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} \\sin^{2}{\\left (\\boldsymbol{\\mathrm{\\theta}} \\right )}$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 13, "text": [ "\u239b 2\u22c5g(r) \u239b d \u239e\u2502 \u239b d \u239e\u2502 \u239e -2\u22c5g(r) 2 \n", "\u239c\u212f - r\u22c5\u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 + r\u22c5\u239c\u2500\u2500\u2500(g(\u03be\u2081))\u239f\u2502 - 1\u239f\u22c5\u212f \u22c5sin (\u03b8)\n", "\u239d \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r \u23a0 " ] } ], "prompt_number": 13 }, { "cell_type": "markdown", "metadata": {}, "source": [ "The off-diagonal components are zero." ] }, { "cell_type": "code", "collapsed": false, "input": [ "all(ricci[i][j]==0 for i in range(4) for j in range(4) if i!=j)" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "pyout", "prompt_number": 14, "text": [ "True" ] } ], "prompt_number": 14 }, { "cell_type": "markdown", "metadata": {}, "source": [ "For completeness we can also check out the Christoffel symbol of 2nd kind. We will print only the non-zero components, and only one of the symmetric components (symmetric in the last two indices)." ] }, { "cell_type": "code", "collapsed": false, "input": [ "ch_2nd = metric_to_Christoffel_2nd(metric)\n", "filt = [((i,j,k), simplify(ch_2nd[i][j][k]))\n", " for i in range(4) for j in range(4) for k in range(j,4)\n", " if ch_2nd[i][j][k]!=0]" ], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 15 }, { "cell_type": "code", "collapsed": false, "input": [ "filt[0:3]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left [ \\left ( \\left ( 0, \\quad 0, \\quad 1\\right ), \\quad \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}\\right ), \\quad \\left ( \\left ( 1, \\quad 0, \\quad 0\\right ), \\quad e^{2 f{\\left (\\boldsymbol{\\mathrm{r}} \\right )} - 2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} \\left. \\frac{d}{d \\xi_{1}} f{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}\\right ), \\quad \\left ( \\left ( 1, \\quad 1, \\quad 1\\right ), \\quad \\left. \\frac{d}{d \\xi_{1}} g{\\left (\\xi_{1} \\right )} \\right|_{\\substack{ \\xi_{1}=\\boldsymbol{\\mathrm{r}} }}\\right )\\right ]$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 16, "text": [ "\u23a1\u239b \u239b d \u239e\u2502 \u239e \u239b 2\u22c5f(r) - 2\u22c5g(r) \u239b d \u239e\u2502 \n", "\u23a2\u239c(0, 0, 1), \u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 \u239f, \u239c(1, 0, 0), \u212f \u22c5\u239c\u2500\u2500\u2500(f(\u03be\u2081))\u239f\u2502 \n", "\u23a3\u239d \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r\u23a0 \u239d \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=\n", "\n", " \u239e \u239b \u239b d \u239e\u2502 \u239e\u23a4\n", " \u239f, \u239c(1, 1, 1), \u239c\u2500\u2500\u2500(g(\u03be\u2081))\u239f\u2502 \u239f\u23a5\n", "r\u23a0 \u239d \u239dd\u03be\u2081 \u23a0\u2502\u03be\u2081=r\u23a0\u23a6" ] } ], "prompt_number": 16 }, { "cell_type": "code", "collapsed": false, "input": [ "filt[3:6]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left [ \\left ( \\left ( 1, \\quad 2, \\quad 2\\right ), \\quad - e^{- 2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} \\boldsymbol{\\mathrm{r}}\\right ), \\quad \\left ( \\left ( 1, \\quad 3, \\quad 3\\right ), \\quad - e^{- 2 g{\\left (\\boldsymbol{\\mathrm{r}} \\right )}} \\sin^{2}{\\left (\\boldsymbol{\\mathrm{\\theta}} \\right )} \\boldsymbol{\\mathrm{r}}\\right ), \\quad \\left ( \\left ( 2, \\quad 1, \\quad 2\\right ), \\quad \\frac{1}{\\boldsymbol{\\mathrm{r}}}\\right )\\right ]$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 17, "text": [ "\u23a1\u239b -2\u22c5g(r) \u239e \u239b -2\u22c5g(r) 2 \u239e \u239b 1\u239e\u23a4\n", "\u23a2\u239d(1, 2, 2), -\u212f \u22c5r\u23a0, \u239d(1, 3, 3), -\u212f \u22c5sin (\u03b8)\u22c5r\u23a0, \u239c(2, 1, 2), \u2500\u239f\u23a5\n", "\u23a3 \u239d r\u23a0\u23a6" ] } ], "prompt_number": 17 }, { "cell_type": "code", "collapsed": false, "input": [ "filt[6:9]" ], "language": "python", "metadata": {}, "outputs": [ { "latex": [ "$$\\left [ \\left ( \\left ( 2, \\quad 3, \\quad 3\\right ), \\quad - \\frac{1}{2} \\sin{\\left (2 \\boldsymbol{\\mathrm{\\theta}} \\right )}\\right ), \\quad \\left ( \\left ( 3, \\quad 1, \\quad 3\\right ), \\quad \\frac{1}{\\boldsymbol{\\mathrm{r}}}\\right ), \\quad \\left ( \\left ( 3, \\quad 2, \\quad 3\\right ), \\quad \\frac{1}{\\tan{\\left (\\boldsymbol{\\mathrm{\\theta}} \\right )}}\\right )\\right ]$$" ], "metadata": {}, "output_type": "pyout", "png": 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"prompt_number": 18, "text": [ "\u23a1\u239b -sin(2\u22c5\u03b8) \u239e \u239b 1\u239e \u239b 1 \u239e\u23a4\n", "\u23a2\u239c(2, 3, 3), \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u239f, \u239c(3, 1, 3), \u2500\u239f, \u239c(3, 2, 3), \u2500\u2500\u2500\u2500\u2500\u2500\u239f\u23a5\n", "\u23a3\u239d 2 \u23a0 \u239d r\u23a0 \u239d tan(\u03b8)\u23a0\u23a6" ] } ], "prompt_number": 18 }, { "cell_type": "markdown", "metadata": {}, "source": [ "We can also confirm that the Christoffel symbol is symmetric." ] }, { "cell_type": "code", "collapsed": false, "input": [ "all([ch_2nd[k][i][j] == ch_2nd[k][j][i] for k in range(4) for i in range(4) for j in range(4)])" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "pyout", "prompt_number": 19, "text": [ "True" ] } ], "prompt_number": 19 } ], "metadata": {} } ] }