{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Black branes in Lifshitz-like wraped backgrounds " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "This Jupyter/SageMath \n", "worksheet implements some computations of the article\n", "- I. Ya. Aref'eva, A. A. Golubtsova & E. Gourgoulhon: *??*, in preparation\n", " \n", "These computations are based on [SageManifolds](http://sagemanifolds.obspm.fr) (v0.9)\n", "\n", "The worksheet file (ipynb format) can be downloaded from ??.\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "First we set up the notebook to display mathematical objects using LaTeX formatting:" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": true }, "outputs": [], "source": [ "%display latex" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Metric\n", "\n", "Let us declare the spacetime $M$ as a 5-dimensional manifold:" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "5-dimensional differentiable manifold M\n" ] } ], "source": [ "M = Manifold(5, 'M')\n", "print M" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We introduce a the Poincaré-type coordinate system on $M$:" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(M,(t, x, {y_1}, {y_2}, z)\\right)$ " ], "text/plain": [ "Chart (M, (t, x, y1, y2, z))" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "X. = M.chart(r't x y1:y_1 y2:y_2 z:(0,+oo)')\n", "X" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let us consider the following Lifshitz-symmetric metric, parametrized by some real numbers $\\nu$ and $c$, and the blackening function $f(z)$:" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = -\\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right)}{z^{2}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}}{z^{2}} \\mathrm{d} x\\otimes \\mathrm{d} x + \\left( z^{-\\frac{2}{{\\nu}}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\right) \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + \\left( z^{-\\frac{2}{{\\nu}}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\right) \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} + \\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}}{z^{2} f\\left(z\\right)} \\mathrm{d} z\\otimes \\mathrm{d} z$ " ], "text/plain": [ "g = -e^(1/2*c*z^2)*f(z)/z^2 dt*dt + e^(1/2*c*z^2)/z^2 dx*dx + z^(-2/nu)*e^(1/2*c*z^2) dy1*dy1 + z^(-2/nu)*e^(1/2*c*z^2) dy2*dy2 + e^(1/2*c*z^2)/(z^2*f(z)) dz*dz" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g = M.lorentzian_metric('g')\n", "var('nu', latex_name=r'\\nu', domain='real')\n", "var('c', domain='real')\n", "ff = function('f')(z)\n", "b(z) = exp(c*z^2/2)\n", "# b(z)=1 ## for checks\n", "g[0,0] = - b(z)*ff/z^2\n", "g[1,1] = b(z)/z^2\n", "g[2,2] = b(z)*z^(-2/nu)\n", "g[3,3] = b(z)*z^(-2/nu)\n", "g[4,4] = b(z)/z^2/ff\n", "g.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "A matrix view of the metric components:" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrrr}\n", "-\\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right)}{z^{2}} & 0 & 0 & 0 & 0 \\\\\n", "0 & \\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}}{z^{2}} & 0 & 0 & 0 \\\\\n", "0 & 0 & z^{-\\frac{2}{{\\nu}}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} & 0 & 0 \\\\\n", "0 & 0 & 0 & z^{-\\frac{2}{{\\nu}}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} & 0 \\\\\n", "0 & 0 & 0 & 0 & \\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}}{z^{2} f\\left(z\\right)}\n", "\\end{array}\\right)$ " ], "text/plain": [ "[ -e^(1/2*c*z^2)*f(z)/z^2 0 0 0 0]\n", "[ 0 e^(1/2*c*z^2)/z^2 0 0 0]\n", "[ 0 0 z^(-2/nu)*e^(1/2*c*z^2) 0 0]\n", "[ 0 0 0 z^(-2/nu)*e^(1/2*c*z^2) 0]\n", "[ 0 0 0 0 e^(1/2*c*z^2)/(z^2*f(z))]" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g[:]" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} g_{ \\, t \\, t }^{ \\phantom{\\, t } \\phantom{\\, t } } & = & -\\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right)}{z^{2}} \\\\ g_{ \\, x \\, x }^{ \\phantom{\\, x } \\phantom{\\, x } } & = & \\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}}{z^{2}} \\\\ g_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1} } \\phantom{\\, {y_1} } } & = & z^{-\\frac{2}{{\\nu}}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\\\ g_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2} } \\phantom{\\, {y_2} } } & = & z^{-\\frac{2}{{\\nu}}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\\\ g_{ \\, z \\, z }^{ \\phantom{\\, z } \\phantom{\\, z } } & = & \\frac{e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}}{z^{2} f\\left(z\\right)} \\end{array}$ " ], "text/plain": [ "g_t,t = -e^(1/2*c*z^2)*f(z)/z^2 \n", "g_x,x = e^(1/2*c*z^2)/z^2 \n", "g_y1,y1 = z^(-2/nu)*e^(1/2*c*z^2) \n", "g_y2,y2 = z^(-2/nu)*e^(1/2*c*z^2) \n", "g_z,z = e^(1/2*c*z^2)/(z^2*f(z)) " ] }, "execution_count": 7, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.display_comp()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Curvature\n", "\n", "The Riemann tensor is" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Tensor field Riem(g) of type (1,3) on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "Riem = g.riemann()\n", "print Riem" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{{\\left(c^{2} z^{4} - 4 \\, c z^{2} + 4\\right)} f\\left(z\\right) + {\\left(c z^{3} - 2 \\, z\\right)} \\frac{\\partial\\,f}{\\partial z}}{4 \\, z^{2}}$ " ], "text/plain": [ "-1/4*((c^2*z^4 - 4*c*z^2 + 4)*f(z) + (c*z^3 - 2*z)*d(f)/dz)/z^2" ] }, "execution_count": 10, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#Riem.display_comp(only_nonredundant=True)\n", "Riem[0,1,0,1]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The Ricci tensor:" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Field of symmetric bilinear forms Ric(g) on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "Ric = g.ricci()\n", "print Ric" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{Ric}\\left(g\\right) = \\left( \\frac{2 \\, {\\nu} z^{2} f\\left(z\\right) \\frac{\\partial^2\\,f}{\\partial z^2} + {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right)^{2} + {\\left(5 \\, c {\\nu} z^{3} - 2 \\, {\\left(3 \\, {\\nu} + 2\\right)} z\\right)} f\\left(z\\right) \\frac{\\partial\\,f}{\\partial z}}{4 \\, {\\nu} z^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{{\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right) + 2 \\, {\\left(c {\\nu} z^{3} - 2 \\, {\\nu} z\\right)} \\frac{\\partial\\,f}{\\partial z}}{4 \\, {\\nu} z^{2}} \\right) \\mathrm{d} x\\otimes \\mathrm{d} x + \\left( -\\frac{{\\left({\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right) + 2 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} z^{-\\frac{2}{{\\nu}}}}{4 \\, {\\nu}^{2}} \\right) \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + \\left( -\\frac{{\\left({\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right) + 2 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} z^{-\\frac{2}{{\\nu}}}}{4 \\, {\\nu}^{2}} \\right) \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} + \\left( -\\frac{2 \\, {\\nu}^{2} z^{2} \\frac{\\partial^2\\,f}{\\partial z^2} + 4 \\, {\\left({\\left(3 \\, c {\\nu}^{2} - c {\\nu}\\right)} z^{2} + 2 \\, {\\nu}^{2} + 2\\right)} f\\left(z\\right) + {\\left(5 \\, c {\\nu}^{2} z^{3} - 2 \\, {\\left(3 \\, {\\nu}^{2} + 2 \\, {\\nu}\\right)} z\\right)} \\frac{\\partial\\,f}{\\partial z}}{4 \\, {\\nu}^{2} z^{2} f\\left(z\\right)} \\right) \\mathrm{d} z\\otimes \\mathrm{d} z$ " ], "text/plain": [ "Ric(g) = 1/4*(2*nu*z^2*f(z)*d^2(f)/dz^2 + (3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*f(z)^2 + (5*c*nu*z^3 - 2*(3*nu + 2)*z)*f(z)*d(f)/dz)/(nu*z^2) dt*dt - 1/4*((3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*f(z) + 2*(c*nu*z^3 - 2*nu*z)*d(f)/dz)/(nu*z^2) dx*dx - 1/4*((3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*f(z) + 2*(c*nu^2*z^3 - 2*nu*z)*d(f)/dz)*z^(-2/nu)/nu^2 dy1*dy1 - 1/4*((3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*f(z) + 2*(c*nu^2*z^3 - 2*nu*z)*d(f)/dz)*z^(-2/nu)/nu^2 dy2*dy2 - 1/4*(2*nu^2*z^2*d^2(f)/dz^2 + 4*((3*c*nu^2 - c*nu)*z^2 + 2*nu^2 + 2)*f(z) + (5*c*nu^2*z^3 - 2*(3*nu^2 + 2*nu)*z)*d(f)/dz)/(nu^2*z^2*f(z)) dz*dz" ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Ric.display()" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} \\mathrm{Ric}\\left(g\\right)_{ \\, t \\, t }^{ \\phantom{\\, t } \\phantom{\\, t } } & = & \\frac{2 \\, {\\nu} z^{2} f\\left(z\\right) \\frac{\\partial^2\\,f}{\\partial z^2} + {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right)^{2} + {\\left(5 \\, c {\\nu} z^{3} - 2 \\, {\\left(3 \\, {\\nu} + 2\\right)} z\\right)} f\\left(z\\right) \\frac{\\partial\\,f}{\\partial z}}{4 \\, {\\nu} z^{2}} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, x \\, x }^{ \\phantom{\\, x } \\phantom{\\, x } } & = & -\\frac{{\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right) + 2 \\, {\\left(c {\\nu} z^{3} - 2 \\, {\\nu} z\\right)} \\frac{\\partial\\,f}{\\partial z}}{4 \\, {\\nu} z^{2}} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1} } \\phantom{\\, {y_1} } } & = & -\\frac{{\\left({\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right) + 2 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} z^{-\\frac{2}{{\\nu}}}}{4 \\, {\\nu}^{2}} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2} } \\phantom{\\, {y_2} } } & = & -\\frac{{\\left({\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} f\\left(z\\right) + 2 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} z^{-\\frac{2}{{\\nu}}}}{4 \\, {\\nu}^{2}} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, z \\, z }^{ \\phantom{\\, z } \\phantom{\\, z } } & = & -\\frac{2 \\, {\\nu}^{2} z^{2} \\frac{\\partial^2\\,f}{\\partial z^2} + 4 \\, {\\left({\\left(3 \\, c {\\nu}^{2} - c {\\nu}\\right)} z^{2} + 2 \\, {\\nu}^{2} + 2\\right)} f\\left(z\\right) + {\\left(5 \\, c {\\nu}^{2} z^{3} - 2 \\, {\\left(3 \\, {\\nu}^{2} + 2 \\, {\\nu}\\right)} z\\right)} \\frac{\\partial\\,f}{\\partial z}}{4 \\, {\\nu}^{2} z^{2} f\\left(z\\right)} \\end{array}$ " ], "text/plain": [ "Ric(g)_t,t = 1/4*(2*nu*z^2*f(z)*d^2(f)/dz^2 + (3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*f(z)^2 + (5*c*nu*z^3 - 2*(3*nu + 2)*z)*f(z)*d(f)/dz)/(nu*z^2) \n", "Ric(g)_x,x = -1/4*((3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*f(z) + 2*(c*nu*z^3 - 2*nu*z)*d(f)/dz)/(nu*z^2) \n", "Ric(g)_y1,y1 = -1/4*((3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*f(z) + 2*(c*nu^2*z^3 - 2*nu*z)*d(f)/dz)*z^(-2/nu)/nu^2 \n", "Ric(g)_y2,y2 = -1/4*((3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*f(z) + 2*(c*nu^2*z^3 - 2*nu*z)*d(f)/dz)*z^(-2/nu)/nu^2 \n", "Ric(g)_z,z = -1/4*(2*nu^2*z^2*d^2(f)/dz^2 + 4*((3*c*nu^2 - c*nu)*z^2 + 2*nu^2 + 2)*f(z) + (5*c*nu^2*z^3 - 2*(3*nu^2 + 2*nu)*z)*d(f)/dz)/(nu^2*z^2*f(z)) " ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Ric.display_comp()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The Ricci scalar:" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field r(g) on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "Rscal = g.ricci_scalar()\n", "print Rscal" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} \\mathrm{r}\\left(g\\right):& M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, z\\right) & \\longmapsto & -\\frac{{\\left({\\nu}^{2} z^{2} \\frac{\\partial^2\\,f}{\\partial z^2} + {\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 8 \\, c {\\nu} z^{2} + 6 \\, {\\nu}^{2} + 8 \\, {\\nu} + 6\\right)} f\\left(z\\right) + 4 \\, {\\left(c {\\nu}^{2} z^{3} - {\\left({\\nu}^{2} + {\\nu}\\right)} z\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{{\\nu}^{2}} \\end{array}$ " ], "text/plain": [ "r(g): M --> R\n", " (t, x, y1, y2, z) |--> -(nu^2*z^2*d^2(f)/dz^2 + (3*c^2*nu^2*z^4 - 8*c*nu*z^2 + 6*nu^2 + 8*nu + 6)*f(z) + 4*(c*nu^2*z^3 - (nu^2 + nu)*z)*d(f)/dz)*e^(-1/2*c*z^2)/nu^2" ] }, "execution_count": 15, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Rscal.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Source model\n", "Let us consider a model based on the following action, involving a dilaton scalar field $\\phi$ and a Maxwell 2-form $F$:\n", "\n", "$$ S = \\int \\left( R(g) + \\Lambda - \\frac{1}{2} \\nabla_m \\phi \\nabla^m \\phi - \\frac{1}{4} e^{\\lambda\\phi} F_{mn} F^{mn} \\right) \\sqrt{-g} \\, \\mathrm{d}^5 x \\qquad\\qquad \\mbox{(1)}$$\n", "\n", "where $R(g)$ is the Ricci scalar of metric $g$, $\\Lambda$ is the cosmological constant and $\\lambda$ is the dilatonic coupling constant." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### The dilaton scalar field\n", "\n", "We consider the following ansatz for the dilaton scalar field $\\phi$:\n", "$$ \\phi = \\frac{1}{\\lambda} \\left( \\ln\\mu - \\frac{4}{\\nu}\\ln z\\right),$$\n", "where $\\mu$ is a constant. " ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} \\phi:& M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, z\\right) & \\longmapsto & -\\frac{\\frac{4 \\, \\log\\left(z\\right)}{{\\nu}} - \\log\\left({\\mu}\\right)}{{\\lambda}} \\end{array}$ " ], "text/plain": [ "phi: M --> R\n", " (t, x, y1, y2, z) |--> -(4*log(z)/nu - log(mu))/lamb" ] }, "execution_count": 16, "metadata": {}, "output_type": "execute_result" } ], "source": [ "var('mu', latex_name=r'\\mu', domain='real')\n", "var('lamb', latex_name=r'\\lambda', domain='real')\n", "phi = M.scalar_field({X: (ln(mu) - 4/nu*ln(z))/lamb}, \n", " name='phi', latex_name=r'\\phi')\n", "phi.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The 1-form $\\mathrm{d}\\phi$ is" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "1-form dphi on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "dphi = phi.differential()\n", "print dphi" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d}\\phi = -\\frac{4}{{\\lambda} {\\nu} z} \\mathrm{d} z$ " ], "text/plain": [ "dphi = -4/(lamb*nu*z) dz" ] }, "execution_count": 18, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dphi.display()" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[0, 0, 0, 0, -\\frac{4}{{\\lambda} {\\nu} z}\\right]$ " ], "text/plain": [ "[0, 0, 0, 0, -4/(lamb*nu*z)]" ] }, "execution_count": 19, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dphi[:] # all the components in the default frame" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### The 2-form field\n", "\n", "We consider the following ansatz for $F$:\n", "$$ F = \\frac{1}{2} q \\, \\mathrm{d}y_1\\wedge \\mathrm{d}y_2, $$\n", "where $q$ is a constant. \n", "\n", "Let us first get the 1-forms $\\mathrm{d}y_1$ and $\\mathrm{d}y_2$:" ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(M ,\\left(\\mathrm{d} t,\\mathrm{d} x,\\mathrm{d} {y_1},\\mathrm{d} {y_2},\\mathrm{d} z\\right)\\right)$ " ], "text/plain": [ "Coordinate coframe (M, (dt,dx,dy1,dy2,dz))" ] }, "execution_count": 20, "metadata": {}, "output_type": "execute_result" } ], "source": [ "X.coframe()" ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "1-form dy1 on the 5-dimensional differentiable manifold M\n", "1-form dy2 on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathrm{d} {y_1}, \\mathrm{d} {y_2}\\right)$ " ], "text/plain": [ "(1-form dy1 on the 5-dimensional differentiable manifold M,\n", " 1-form dy2 on the 5-dimensional differentiable manifold M)" ] }, "execution_count": 21, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dy1 = X.coframe()[2]\n", "dy2 = X.coframe()[3]\n", "print dy1\n", "print dy2\n", "dy1, dy2" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Then we can form $F$ according to the above ansatz:" ] }, { "cell_type": "code", "execution_count": 22, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2-form F on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}F = \\frac{1}{2} \\, q \\mathrm{d} {y_1}\\wedge \\mathrm{d} {y_2}$ " ], "text/plain": [ "F = 1/2*q dy1/\\dy2" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "var('q', domain='real')\n", "F = q/2 * dy1.wedge(dy2)\n", "F.set_name('F')\n", "print F\n", "F.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "By construction, the 2-form $F$ is closed (since $q$ is constant):" ] }, { "cell_type": "code", "execution_count": 23, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "3-form dF on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "print xder(F)" ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d}F = 0$ " ], "text/plain": [ "dF = 0" ] }, "execution_count": 24, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xder(F).display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let us evaluate the square $F_{mn} F^{mn}$ of $F$:" ] }, { "cell_type": "code", "execution_count": 25, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Tensor field of type (2,0) on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2} \\, q z^{\\frac{4}{{\\nu}}} e^{\\left(-c z^{2}\\right)} \\frac{\\partial}{\\partial {y_1} }\\otimes \\frac{\\partial}{\\partial {y_2} } -\\frac{1}{2} \\, q z^{\\frac{4}{{\\nu}}} e^{\\left(-c z^{2}\\right)} \\frac{\\partial}{\\partial {y_2} }\\otimes \\frac{\\partial}{\\partial {y_1} }$ " ], "text/plain": [ "1/2*q*z^(4/nu)*e^(-c*z^2) d/dy1*d/dy2 - 1/2*q*z^(4/nu)*e^(-c*z^2) d/dy2*d/dy1" ] }, "execution_count": 25, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Fu = F.up(g)\n", "print Fu\n", "Fu.display()" ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, z\\right) & \\longmapsto & \\frac{1}{2} \\, q^{2} z^{\\frac{4}{{\\nu}}} e^{\\left(-c z^{2}\\right)} \\end{array}$ " ], "text/plain": [ "M --> R\n", "(t, x, y1, y2, z) |--> 1/2*q^2*z^(4/nu)*e^(-c*z^2)" ] }, "execution_count": 26, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F2 = F['_{mn}']*Fu['^{mn}'] # using LaTeX notations to denote contraction\n", "print F2\n", "F2.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We shall also need the tensor $\\mathcal{F}_{mn} := F_{mp} F_n^{\\ \\, p}$:" ] }, { "cell_type": "code", "execution_count": 27, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Tensor field of type (0,2) on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{4} \\, q^{2} z^{\\frac{2}{{\\nu}}} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + \\frac{1}{4} \\, q^{2} z^{\\frac{2}{{\\nu}}} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2}$ " ], "text/plain": [ "1/4*q^2*z^(2/nu)*e^(-1/2*c*z^2) dy1*dy1 + 1/4*q^2*z^(2/nu)*e^(-1/2*c*z^2) dy2*dy2" ] }, "execution_count": 27, "metadata": {}, "output_type": "execute_result" } ], "source": [ "FF = F['_mp'] * F.up(g,1)['^p_n']\n", "print FF\n", "FF.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The tensor field $\\mathcal{F}$ is symmetric:" ] }, { "cell_type": "code", "execution_count": 28, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$ " ], "text/plain": [ "True" ] }, "execution_count": 28, "metadata": {}, "output_type": "execute_result" } ], "source": [ "FF == FF.symmetrize()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Therefore, from now on, we set" ] }, { "cell_type": "code", "execution_count": 29, "metadata": { "collapsed": true }, "outputs": [], "source": [ "FF = FF.symmetrize()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Einstein equation\n", "\n", "Let us first introduce the cosmological constant:" ] }, { "cell_type": "code", "execution_count": 30, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\Lambda}$ " ], "text/plain": [ "Lamb" ] }, "execution_count": 30, "metadata": {}, "output_type": "execute_result" } ], "source": [ "var('Lamb', latex_name=r'\\Lambda', domain='real')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "From the action (1), the field equation for the metric $g$ is\n", "$$ R_{mn} + \\frac{\\Lambda}{3} \\, g - \\frac{1}{2}\\partial_m\\phi \\partial_n\\phi -\\frac{1}{2} e^{\\lambda\\phi} F_{mp} F^{\\ \\, p}_n + \\frac{1}{12} e^{\\lambda\\phi} F_{rs} F^{rs} \\, g_{mn} = 0 $$\n", "We write it as\n", "\n", " EE == 0\n", "\n", "with `EE` defined by" ] }, { "cell_type": "code", "execution_count": 31, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Field of symmetric bilinear forms E on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "EE = Ric + Lamb/3*g - 1/2* (dphi*dphi) - 1/2*exp(lamb*phi)*FF \\\n", " + 1/12*exp(lamb*phi)*F2*g\n", "EE.set_name('E')\n", "print EE" ] }, { "cell_type": "code", "execution_count": 32, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} E_{ \\, t \\, t }^{ \\phantom{\\, t } \\phantom{\\, t } } & = & \\frac{{\\left(12 \\, {\\nu} z^{2} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) \\frac{\\partial^2\\,f}{\\partial z^2} + 6 \\, {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right)^{2} + 6 \\, {\\left(5 \\, c {\\nu} z^{3} - 2 \\, {\\left(3 \\, {\\nu} + 2\\right)} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) \\frac{\\partial\\,f}{\\partial z} - {\\left({\\mu} {\\nu} q^{2} + 8 \\, {\\Lambda} {\\nu} e^{\\left(c z^{2}\\right)}\\right)} f\\left(z\\right)\\right)} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{24 \\, {\\nu} z^{2}} \\\\ E_{ \\, x \\, x }^{ \\phantom{\\, x } \\phantom{\\, x } } & = & \\frac{{\\left({\\mu} {\\nu} q^{2} + 8 \\, {\\Lambda} {\\nu} e^{\\left(c z^{2}\\right)} - 6 \\, {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) - 12 \\, {\\left(c {\\nu} z^{3} - 2 \\, {\\nu} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{24 \\, {\\nu} z^{2}} \\\\ E_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1} } \\phantom{\\, {y_1} } } & = & -\\frac{{\\left({\\mu} {\\nu}^{2} q^{2} - 4 \\, {\\Lambda} {\\nu}^{2} e^{\\left(c z^{2}\\right)} + 3 \\, {\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) + 6 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} z^{-\\frac{2}{{\\nu}}} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{12 \\, {\\nu}^{2}} \\\\ E_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2} } \\phantom{\\, {y_2} } } & = & -\\frac{{\\left({\\mu} {\\nu}^{2} q^{2} - 4 \\, {\\Lambda} {\\nu}^{2} e^{\\left(c z^{2}\\right)} + 3 \\, {\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) + 6 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} z^{-\\frac{2}{{\\nu}}} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{12 \\, {\\nu}^{2}} \\\\ E_{ \\, z \\, z }^{ \\phantom{\\, z } \\phantom{\\, z } } & = & -\\frac{{\\left(12 \\, {\\lambda}^{2} {\\nu}^{2} z^{2} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial^2\\,f}{\\partial z^2} - {\\lambda}^{2} {\\mu} {\\nu}^{2} q^{2} - 8 \\, {\\Lambda} {\\lambda}^{2} {\\nu}^{2} e^{\\left(c z^{2}\\right)} + 24 \\, {\\left(2 \\, {\\lambda}^{2} {\\nu}^{2} + {\\left(3 \\, c {\\lambda}^{2} {\\nu}^{2} - c {\\lambda}^{2} {\\nu}\\right)} z^{2} + 2 \\, {\\lambda}^{2} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) + 6 \\, {\\left(5 \\, c {\\lambda}^{2} {\\nu}^{2} z^{3} - 2 \\, {\\left(3 \\, {\\lambda}^{2} {\\nu}^{2} + 2 \\, {\\lambda}^{2} {\\nu}\\right)} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}\\right)} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{24 \\, {\\lambda}^{2} {\\nu}^{2} z^{2} f\\left(z\\right)} \\end{array}$ " ], "text/plain": [ "E_t,t = 1/24*(12*nu*z^2*e^(1/2*c*z^2)*f(z)*d^2(f)/dz^2 + 6*(3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z)^2 + 6*(5*c*nu*z^3 - 2*(3*nu + 2)*z)*e^(1/2*c*z^2)*f(z)*d(f)/dz - (mu*nu*q^2 + 8*Lamb*nu*e^(c*z^2))*f(z))*e^(-1/2*c*z^2)/(nu*z^2) \n", "E_x,x = 1/24*(mu*nu*q^2 + 8*Lamb*nu*e^(c*z^2) - 6*(3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z) - 12*(c*nu*z^3 - 2*nu*z)*e^(1/2*c*z^2)*d(f)/dz)*e^(-1/2*c*z^2)/(nu*z^2) \n", "E_y1,y1 = -1/12*(mu*nu^2*q^2 - 4*Lamb*nu^2*e^(c*z^2) + 3*(3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z) + 6*(c*nu^2*z^3 - 2*nu*z)*e^(1/2*c*z^2)*d(f)/dz)*z^(-2/nu)*e^(-1/2*c*z^2)/nu^2 \n", "E_y2,y2 = -1/12*(mu*nu^2*q^2 - 4*Lamb*nu^2*e^(c*z^2) + 3*(3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z) + 6*(c*nu^2*z^3 - 2*nu*z)*e^(1/2*c*z^2)*d(f)/dz)*z^(-2/nu)*e^(-1/2*c*z^2)/nu^2 \n", "E_z,z = -1/24*(12*lamb^2*nu^2*z^2*e^(1/2*c*z^2)*d^2(f)/dz^2 - lamb^2*mu*nu^2*q^2 - 8*Lamb*lamb^2*nu^2*e^(c*z^2) + 24*(2*lamb^2*nu^2 + (3*c*lamb^2*nu^2 - c*lamb^2*nu)*z^2 + 2*lamb^2 + 8)*e^(1/2*c*z^2)*f(z) + 6*(5*c*lamb^2*nu^2*z^3 - 2*(3*lamb^2*nu^2 + 2*lamb^2*nu)*z)*e^(1/2*c*z^2)*d(f)/dz)*e^(-1/2*c*z^2)/(lamb^2*nu^2*z^2*f(z)) " ] }, "execution_count": 32, "metadata": {}, "output_type": "execute_result" } ], "source": [ "EE.display_comp(only_nonredundant=True)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We note that `EE==0` leads to 4 independent equations:" ] }, { "cell_type": "code", "execution_count": 35, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}12 \\, {\\nu} z^{2} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial^2\\,f}{\\partial z^2} - {\\mu} {\\nu} q^{2} - 8 \\, {\\Lambda} {\\nu} e^{\\left(c z^{2}\\right)} + 6 \\, {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) + 6 \\, {\\left(5 \\, c {\\nu} z^{3} - 2 \\, {\\left(3 \\, {\\nu} + 2\\right)} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}$ " ], "text/plain": [ "12*nu*z^2*e^(1/2*c*z^2)*d^2(f)/dz^2 - mu*nu*q^2 - 8*Lamb*nu*e^(c*z^2) + 6*(3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z) + 6*(5*c*nu*z^3 - 2*(3*nu + 2)*z)*e^(1/2*c*z^2)*d(f)/dz" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq1 = EE[0,0]*24*nu*z^2/f(z)*exp(c*z^2/2)\n", "eq1" ] }, { "cell_type": "code", "execution_count": 38, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\mu} {\\nu} q^{2} + 8 \\, {\\Lambda} {\\nu} e^{\\left(c z^{2}\\right)} - 6 \\, {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) - 12 \\, {\\left(c {\\nu} z^{3} - 2 \\, {\\nu} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}$ " ], "text/plain": [ "mu*nu*q^2 + 8*Lamb*nu*e^(c*z^2) - 6*(3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z) - 12*(c*nu*z^3 - 2*nu*z)*e^(1/2*c*z^2)*d(f)/dz" ] }, "execution_count": 38, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq2 = EE[1,1]*24*nu*z^2*exp(c*z^2/2)\n", "eq2" ] }, { "cell_type": "code", "execution_count": 39, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-{\\mu} {\\nu}^{2} q^{2} + 4 \\, {\\Lambda} {\\nu}^{2} e^{\\left(c z^{2}\\right)} - 3 \\, {\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) - 6 \\, {\\left(c {\\nu}^{2} z^{3} - 2 \\, {\\nu} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}$ " ], "text/plain": [ "-mu*nu^2*q^2 + 4*Lamb*nu^2*e^(c*z^2) - 3*(3*c^2*nu^2*z^4 - 10*c*nu*z^2 + 8*nu + 8)*e^(1/2*c*z^2)*f(z) - 6*(c*nu^2*z^3 - 2*nu*z)*e^(1/2*c*z^2)*d(f)/dz" ] }, "execution_count": 39, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq3 = EE[2,2]*12*nu^2*z^(2/nu)*exp(c*z^2/2)\n", "eq3" ] }, { "cell_type": "code", "execution_count": 41, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-{\\lambda}^{2} {\\nu}^{2} z^{2} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial^2\\,f}{\\partial z^2} + \\frac{1}{12} \\, {\\lambda}^{2} {\\mu} {\\nu}^{2} q^{2} + \\frac{2}{3} \\, {\\Lambda} {\\lambda}^{2} {\\nu}^{2} e^{\\left(c z^{2}\\right)} - 2 \\, {\\left(2 \\, {\\lambda}^{2} {\\nu}^{2} + {\\left(3 \\, c {\\lambda}^{2} {\\nu}^{2} - c {\\lambda}^{2} {\\nu}\\right)} z^{2} + 2 \\, {\\lambda}^{2} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right) - \\frac{1}{2} \\, {\\left(5 \\, c {\\lambda}^{2} {\\nu}^{2} z^{3} - 2 \\, {\\left(3 \\, {\\lambda}^{2} {\\nu}^{2} + 2 \\, {\\lambda}^{2} {\\nu}\\right)} z\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z}$ " ], "text/plain": [ "-lamb^2*nu^2*z^2*e^(1/2*c*z^2)*d^2(f)/dz^2 + 1/12*lamb^2*mu*nu^2*q^2 + 2/3*Lamb*lamb^2*nu^2*e^(c*z^2) - 2*(2*lamb^2*nu^2 + (3*c*lamb^2*nu^2 - c*lamb^2*nu)*z^2 + 2*lamb^2 + 8)*e^(1/2*c*z^2)*f(z) - 1/2*(5*c*lamb^2*nu^2*z^3 - 2*(3*lamb^2*nu^2 + 2*lamb^2*nu)*z)*e^(1/2*c*z^2)*d(f)/dz" ] }, "execution_count": 41, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq4 = EE[4,4]*2*lamb^2*nu^2*z^2*f(z)*exp(c*z^2/2)\n", "eq4" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Dilaton field equation\n", "\n", "First we evaluate $\\nabla_m \\nabla^m \\phi$:" ] }, { "cell_type": "code", "execution_count": 42, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Levi-Civita connection nabla_g associated with the Lorentzian metric g on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\nabla_{g}$ " ], "text/plain": [ "Levi-Civita connection nabla_g associated with the Lorentzian metric g on the 5-dimensional differentiable manifold M" ] }, "execution_count": 42, "metadata": {}, "output_type": "execute_result" } ], "source": [ "nab = g.connection()\n", "print nab\n", "nab" ] }, { "cell_type": "code", "execution_count": 43, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, z\\right) & \\longmapsto & -\\frac{2 \\, {\\left(2 \\, {\\nu} z \\frac{\\partial\\,f}{\\partial z} + {\\left(3 \\, c {\\nu} z^{2} - 4 \\, {\\nu} - 4\\right)} f\\left(z\\right)\\right)} e^{\\left(-\\frac{1}{2} \\, c z^{2}\\right)}}{{\\lambda} {\\nu}^{2}} \\end{array}$ " ], "text/plain": [ "M --> R\n", "(t, x, y1, y2, z) |--> -2*(2*nu*z*d(f)/dz + (3*c*nu*z^2 - 4*nu - 4)*f(z))*e^(-1/2*c*z^2)/(lamb*nu^2)" ] }, "execution_count": 43, "metadata": {}, "output_type": "execute_result" } ], "source": [ "box_phi = nab(nab(phi).up(g)).trace()\n", "print box_phi\n", "box_phi.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "From the action (1), the field equation for $\\phi$ is \n", "$$ \\nabla_m \\nabla^m \\phi = \\frac{\\lambda}{4} e^{\\lambda\\phi} F_{mn} F^{mn}$$\n", "We write it as\n", "\n", " DE == 0\n", " \n", "with `DE` defined by" ] }, { "cell_type": "code", "execution_count": 44, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "DE = box_phi - lamb/4*exp(lamb*phi) * F2\n", "print DE" ] }, { "cell_type": "code", "execution_count": 45, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, z\\right) & \\longmapsto & -\\frac{{\\left({\\lambda}^{2} {\\mu} {\\nu}^{2} q^{2} + 32 \\, {\\nu} z e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z} + 16 \\, {\\left(3 \\, c {\\nu} z^{2} - 4 \\, {\\nu} - 4\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right)\\right)} e^{\\left(-c z^{2}\\right)}}{8 \\, {\\lambda} {\\nu}^{2}} \\end{array}$ " ], "text/plain": [ "M --> R\n", "(t, x, y1, y2, z) |--> -1/8*(lamb^2*mu*nu^2*q^2 + 32*nu*z*e^(1/2*c*z^2)*d(f)/dz + 16*(3*c*nu*z^2 - 4*nu - 4)*e^(1/2*c*z^2)*f(z))*e^(-c*z^2)/(lamb*nu^2)" ] }, "execution_count": 45, "metadata": {}, "output_type": "execute_result" } ], "source": [ "DE.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Hence the dilaton field equation provides a fifth equation:" ] }, { "cell_type": "code", "execution_count": 47, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{16} \\, {\\lambda}^{2} {\\mu} {\\nu}^{2} q^{2} - 2 \\, {\\nu} z e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} \\frac{\\partial\\,f}{\\partial z} - {\\left(3 \\, c {\\nu} z^{2} - 4 \\, {\\nu} - 4\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} f\\left(z\\right)$ " ], "text/plain": [ "-1/16*lamb^2*mu*nu^2*q^2 - 2*nu*z*e^(1/2*c*z^2)*d(f)/dz - (3*c*nu*z^2 - 4*nu - 4)*e^(1/2*c*z^2)*f(z)" ] }, "execution_count": 47, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq5 = DE.coord_function()*lamb*nu^2*exp(c*z^2)/2\n", "eq5" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Maxwell equation\n", "\n", "From the action (1), the field equation for $F$ is \n", "$$ \\nabla_m \\left( e^{\\lambda\\phi} F^{mn} \\right)= 0 $$\n", "We write it as\n", "\n", " ME == 0\n", " \n", "with `ME` defined by" ] }, { "cell_type": "code", "execution_count": 48, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Vector field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$ " ], "text/plain": [ "0" ] }, "execution_count": 48, "metadata": {}, "output_type": "execute_result" } ], "source": [ "ME = nab(exp(lamb*phi)*Fu).trace(0,2)\n", "print ME\n", "ME.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We get identically zero; hence the Maxwell equation do not provide any further equation." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### The solution\n", "\n", "The Einstein equation + the dilaton field equation yields a system of 5 equations (eq1, eq2, eq3, eq4, eq5). \n", "\n", "Let us show that a solution is obtained for $\\nu=2$ and $\\nu=4$ with the following specific form of the blackening function:\n", "\n", "$$ f(z) = 1 - m z^{2/\\nu+2}, $$\n", "\n", "where $m$ is a constant. \n", "\n", "To this aim, we declare" ] }, { "cell_type": "code", "execution_count": 50, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}z \\ {\\mapsto}\\ -m z^{\\frac{2}{{\\nu}} + 2} + 1$ " ], "text/plain": [ "z |--> -m*z^(2/nu + 2) + 1" ] }, "execution_count": 50, "metadata": {}, "output_type": "execute_result" } ], "source": [ "var('m', domain='real')\n", "fm(z) = 1 - m*z^(2/nu+2)\n", "fm" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "and substitute this function for $f(z)$ in all the equations:" ] }, { "cell_type": "code", "execution_count": 51, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-{\\mu} {\\nu} q^{2} - 8 \\, {\\Lambda} {\\nu} e^{\\left(c z^{2}\\right)} + 6 \\, {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} - {\\left(3 \\, c^{2} m {\\nu} z^{6} + 2 \\, {\\left(2 \\, c m {\\nu} + 3 \\, c m\\right)} z^{4}\\right)} z^{\\frac{2}{{\\nu}}} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}$ " ], "text/plain": [ "-mu*nu*q^2 - 8*Lamb*nu*e^(c*z^2) + 6*(3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 - (3*c^2*m*nu*z^6 + 2*(2*c*m*nu + 3*c*m)*z^4)*z^(2/nu) + 8*nu + 8)*e^(1/2*c*z^2)" ] }, "execution_count": 51, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq1m = eq1.expr().substitute_function(f, fm).simplify_full()\n", "eq1m" ] }, { "cell_type": "code", "execution_count": 52, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\mu} {\\nu} q^{2} + 8 \\, {\\Lambda} {\\nu} e^{\\left(c z^{2}\\right)} - 6 \\, {\\left(3 \\, c^{2} {\\nu} z^{4} - 2 \\, {\\left(3 \\, c {\\nu} + 2 \\, c\\right)} z^{2} - {\\left(3 \\, c^{2} m {\\nu} z^{6} - 2 \\, c m {\\nu} z^{4}\\right)} z^{\\frac{2}{{\\nu}}} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}$ " ], "text/plain": [ "mu*nu*q^2 + 8*Lamb*nu*e^(c*z^2) - 6*(3*c^2*nu*z^4 - 2*(3*c*nu + 2*c)*z^2 - (3*c^2*m*nu*z^6 - 2*c*m*nu*z^4)*z^(2/nu) + 8*nu + 8)*e^(1/2*c*z^2)" ] }, "execution_count": 52, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq2m = eq2.expr().substitute_function(f, fm).simplify_full()\n", "eq2m" ] }, { "cell_type": "code", "execution_count": 53, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-{\\mu} {\\nu}^{2} q^{2} + 4 \\, {\\Lambda} {\\nu}^{2} e^{\\left(c z^{2}\\right)} - 3 \\, {\\left(3 \\, c^{2} {\\nu}^{2} z^{4} - 10 \\, c {\\nu} z^{2} - {\\left(3 \\, c^{2} m {\\nu}^{2} z^{6} + 2 \\, {\\left(2 \\, c m {\\nu}^{2} - 3 \\, c m {\\nu}\\right)} z^{4}\\right)} z^{\\frac{2}{{\\nu}}} + 8 \\, {\\nu} + 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}$ " ], "text/plain": [ "-mu*nu^2*q^2 + 4*Lamb*nu^2*e^(c*z^2) - 3*(3*c^2*nu^2*z^4 - 10*c*nu*z^2 - (3*c^2*m*nu^2*z^6 + 2*(2*c*m*nu^2 - 3*c*m*nu)*z^4)*z^(2/nu) + 8*nu + 8)*e^(1/2*c*z^2)" ] }, "execution_count": 53, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq3m = eq3.expr().substitute_function(f, fm).simplify_full()\n", "eq3m" ] }, { "cell_type": "code", "execution_count": 55, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{12} \\, {\\lambda}^{2} {\\mu} {\\nu}^{2} q^{2} + \\frac{2}{3} \\, {\\Lambda} {\\lambda}^{2} {\\nu}^{2} e^{\\left(c z^{2}\\right)} - {\\left(4 \\, {\\lambda}^{2} {\\nu}^{2} + 2 \\, {\\left(3 \\, c {\\lambda}^{2} {\\nu}^{2} - c {\\lambda}^{2} {\\nu}\\right)} z^{2} + 4 \\, {\\lambda}^{2} - {\\left({\\left(11 \\, c {\\lambda}^{2} m {\\nu}^{2} + 3 \\, c {\\lambda}^{2} m {\\nu}\\right)} z^{4} - 4 \\, {\\left({\\lambda}^{2} m {\\nu} - {\\left({\\lambda}^{2} + 4\\right)} m\\right)} z^{2}\\right)} z^{\\frac{2}{{\\nu}}} + 16\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}$ " ], "text/plain": [ "1/12*lamb^2*mu*nu^2*q^2 + 2/3*Lamb*lamb^2*nu^2*e^(c*z^2) - (4*lamb^2*nu^2 + 2*(3*c*lamb^2*nu^2 - c*lamb^2*nu)*z^2 + 4*lamb^2 - ((11*c*lamb^2*m*nu^2 + 3*c*lamb^2*m*nu)*z^4 - 4*(lamb^2*m*nu - (lamb^2 + 4)*m)*z^2)*z^(2/nu) + 16)*e^(1/2*c*z^2)" ] }, "execution_count": 55, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq4m = eq4.expr().substitute_function(f, fm).simplify_full()\n", "eq4m" ] }, { "cell_type": "code", "execution_count": 56, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{16} \\, {\\lambda}^{2} {\\mu} {\\nu}^{2} q^{2} + {\\left(3 \\, c m {\\nu} z^{\\frac{2}{{\\nu}} + 4} - 3 \\, c {\\nu} z^{2} + 4 \\, {\\nu} + 4\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)}$ " ], "text/plain": [ "-1/16*lamb^2*mu*nu^2*q^2 + (3*c*m*nu*z^(2/nu + 4) - 3*c*nu*z^2 + 4*nu + 4)*e^(1/2*c*z^2)" ] }, "execution_count": 56, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq5m = eq5.expr().substitute_function(f, fm).simplify_full()\n", "eq5m" ] }, { "cell_type": "code", "execution_count": 57, "metadata": { "collapsed": false }, "outputs": [], "source": [ "eqs = [eq1m, eq2m, eq3m, eq4m, eq5m]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Solution for $\\nu = 2$" ] }, { "cell_type": "code", "execution_count": 58, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[-2 \\, {\\mu} q^{2} - 16 \\, {\\Lambda} e^{\\left(c z^{2}\\right)} - 12 \\, {\\left(3 \\, c^{2} m z^{7} + 7 \\, c m z^{5} - 3 \\, c^{2} z^{4} + 8 \\, c z^{2} - 12\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, 2 \\, {\\mu} q^{2} + 16 \\, {\\Lambda} e^{\\left(c z^{2}\\right)} + 12 \\, {\\left(3 \\, c^{2} m z^{7} - 2 \\, c m z^{5} - 3 \\, c^{2} z^{4} + 8 \\, c z^{2} - 12\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, -4 \\, {\\mu} q^{2} + 16 \\, {\\Lambda} e^{\\left(c z^{2}\\right)} + 12 \\, {\\left(3 \\, c^{2} m z^{7} + c m z^{5} - 3 \\, c^{2} z^{4} + 5 \\, c z^{2} - 6\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, \\frac{1}{3} \\, {\\lambda}^{2} {\\mu} q^{2} + \\frac{8}{3} \\, {\\Lambda} {\\lambda}^{2} e^{\\left(c z^{2}\\right)} + 2 \\, {\\left(25 \\, c {\\lambda}^{2} m z^{5} - 10 \\, c {\\lambda}^{2} z^{2} - 2 \\, {\\left({\\lambda}^{2} - 4\\right)} m z^{3} - 10 \\, {\\lambda}^{2} - 8\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, -\\frac{1}{4} \\, {\\lambda}^{2} {\\mu} q^{2} + 6 \\, {\\left(c m z^{5} - c z^{2} + 2\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0\\right]$ " ], "text/plain": [ "[-2*mu*q^2 - 16*Lamb*e^(c*z^2) - 12*(3*c^2*m*z^7 + 7*c*m*z^5 - 3*c^2*z^4 + 8*c*z^2 - 12)*e^(1/2*c*z^2) == 0,\n", " 2*mu*q^2 + 16*Lamb*e^(c*z^2) + 12*(3*c^2*m*z^7 - 2*c*m*z^5 - 3*c^2*z^4 + 8*c*z^2 - 12)*e^(1/2*c*z^2) == 0,\n", " -4*mu*q^2 + 16*Lamb*e^(c*z^2) + 12*(3*c^2*m*z^7 + c*m*z^5 - 3*c^2*z^4 + 5*c*z^2 - 6)*e^(1/2*c*z^2) == 0,\n", " 1/3*lamb^2*mu*q^2 + 8/3*Lamb*lamb^2*e^(c*z^2) + 2*(25*c*lamb^2*m*z^5 - 10*c*lamb^2*z^2 - 2*(lamb^2 - 4)*m*z^3 - 10*lamb^2 - 8)*e^(1/2*c*z^2) == 0,\n", " -1/4*lamb^2*mu*q^2 + 6*(c*m*z^5 - c*z^2 + 2)*e^(1/2*c*z^2) == 0]" ] }, "execution_count": 58, "metadata": {}, "output_type": "execute_result" } ], "source": [ "neqs = [eq.subs(nu=2).simplify_full() for eq in eqs]\n", "[eq == 0 for eq in neqs]" ] }, { "cell_type": "code", "execution_count": 59, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left[{\\lambda} = \\left(-2\\right), {\\mu} = \\frac{12}{r_{1}^{2}}, {\\Lambda} = \\left(\\frac{15}{2}\\right), q = r_{1}, m = r_{2}, c = 0\\right], \\left[{\\lambda} = 2, {\\mu} = \\frac{12}{r_{3}^{2}}, {\\Lambda} = \\left(\\frac{15}{2}\\right), q = r_{3}, m = r_{4}, c = 0\\right], \\left[{\\lambda} = 2, {\\mu} = -\\frac{24 \\, e^{3}}{r_{5}^{2}}, {\\Lambda} = 57 \\, e^{\\left(-3\\right)}, q = r_{5}, m = 0, c = \\frac{6}{z^{2}}\\right], \\left[{\\lambda} = \\left(-2\\right), {\\mu} = -\\frac{24 \\, e^{3}}{r_{6}^{2}}, {\\Lambda} = 57 \\, e^{\\left(-3\\right)}, q = r_{6}, m = 0, c = \\frac{6}{z^{2}}\\right], \\left[{\\lambda} = \\frac{2}{11} i \\, \\sqrt{11}, {\\mu} = 0, {\\Lambda} = 6 \\, e^{\\left(-1\\right)}, q = r_{7}, m = 0, c = \\frac{2}{z^{2}}\\right], \\left[{\\lambda} = -\\frac{2}{11} i \\, \\sqrt{11}, {\\mu} = 0, {\\Lambda} = 6 \\, e^{\\left(-1\\right)}, q = r_{8}, m = 0, c = \\frac{2}{z^{2}}\\right], \\left[{\\lambda} = \\frac{2}{11} i \\, \\sqrt{11}, {\\mu} = r_{9}, {\\Lambda} = 6 \\, e^{\\left(-1\\right)}, q = 0, m = 0, c = \\frac{2}{z^{2}}\\right], \\left[{\\lambda} = -\\frac{2}{11} i \\, \\sqrt{11}, {\\mu} = r_{10}, {\\Lambda} = 6 \\, e^{\\left(-1\\right)}, q = 0, m = 0, c = \\frac{2}{z^{2}}\\right]\\right]$ " ], "text/plain": [ "[[lamb == -2, mu == 12/r1^2, Lamb == (15/2), q == r1, m == r2, c == 0], [lamb == 2, mu == 12/r3^2, Lamb == (15/2), q == r3, m == r4, c == 0], [lamb == 2, mu == -24*e^3/r5^2, Lamb == 57*e^(-3), q == r5, m == 0, c == 6/z^2], [lamb == -2, mu == -24*e^3/r6^2, Lamb == 57*e^(-3), q == r6, m == 0, c == 6/z^2], [lamb == 2/11*I*sqrt(11), mu == 0, Lamb == 6*e^(-1), q == r7, m == 0, c == 2/z^2], [lamb == -2/11*I*sqrt(11), mu == 0, Lamb == 6*e^(-1), q == r8, m == 0, c == 2/z^2], [lamb == 2/11*I*sqrt(11), mu == r9, Lamb == 6*e^(-1), q == 0, m == 0, c == 2/z^2], [lamb == -2/11*I*sqrt(11), mu == r10, Lamb == 6*e^(-1), q == 0, m == 0, c == 2/z^2]]" ] }, "execution_count": 59, "metadata": {}, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, c)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "In the above solutions, $r_i$, with $i$ an integer, stands for an arbitrary parameter.\n", "\n", "The solutions for $c=0$ are those already obtained for $b(z)=1$.\n", "But there are no solution with $c\\not = 0$ and $c={\\rm const}$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Solution for $\\nu=4$" ] }, { "cell_type": "code", "execution_count": 60, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[-4 \\, {\\mu} q^{2} - 12 \\, {\\left(6 \\, c^{2} m z^{6} + 11 \\, c m z^{4}\\right)} \\sqrt{z} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} - 32 \\, {\\Lambda} e^{\\left(c z^{2}\\right)} + 24 \\, {\\left(3 \\, c^{2} z^{4} - 7 \\, c z^{2} + 10\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, 4 \\, {\\mu} q^{2} + 24 \\, {\\left(3 \\, c^{2} m z^{6} - 2 \\, c m z^{4}\\right)} \\sqrt{z} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} + 32 \\, {\\Lambda} e^{\\left(c z^{2}\\right)} - 24 \\, {\\left(3 \\, c^{2} z^{4} - 7 \\, c z^{2} + 10\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, -16 \\, {\\mu} q^{2} + 24 \\, {\\left(6 \\, c^{2} m z^{6} + 5 \\, c m z^{4}\\right)} \\sqrt{z} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} + 64 \\, {\\Lambda} e^{\\left(c z^{2}\\right)} - 24 \\, {\\left(6 \\, c^{2} z^{4} - 5 \\, c z^{2} + 5\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, \\frac{4}{3} \\, {\\lambda}^{2} {\\mu} q^{2} + \\frac{32}{3} \\, {\\Lambda} {\\lambda}^{2} e^{\\left(c z^{2}\\right)} + 4 \\, {\\left(47 \\, c {\\lambda}^{2} m z^{4} - {\\left(3 \\, {\\lambda}^{2} - 4\\right)} m z^{2}\\right)} \\sqrt{z} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} - 4 \\, {\\left(22 \\, c {\\lambda}^{2} z^{2} + 17 \\, {\\lambda}^{2} + 4\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0, 12 \\, c m z^{\\frac{9}{2}} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} - {\\lambda}^{2} {\\mu} q^{2} - 4 \\, {\\left(3 \\, c z^{2} - 5\\right)} e^{\\left(\\frac{1}{2} \\, c z^{2}\\right)} = 0\\right]$ " ], "text/plain": [ "[-4*mu*q^2 - 12*(6*c^2*m*z^6 + 11*c*m*z^4)*sqrt(z)*e^(1/2*c*z^2) - 32*Lamb*e^(c*z^2) + 24*(3*c^2*z^4 - 7*c*z^2 + 10)*e^(1/2*c*z^2) == 0,\n", " 4*mu*q^2 + 24*(3*c^2*m*z^6 - 2*c*m*z^4)*sqrt(z)*e^(1/2*c*z^2) + 32*Lamb*e^(c*z^2) - 24*(3*c^2*z^4 - 7*c*z^2 + 10)*e^(1/2*c*z^2) == 0,\n", " -16*mu*q^2 + 24*(6*c^2*m*z^6 + 5*c*m*z^4)*sqrt(z)*e^(1/2*c*z^2) + 64*Lamb*e^(c*z^2) - 24*(6*c^2*z^4 - 5*c*z^2 + 5)*e^(1/2*c*z^2) == 0,\n", " 4/3*lamb^2*mu*q^2 + 32/3*Lamb*lamb^2*e^(c*z^2) + 4*(47*c*lamb^2*m*z^4 - (3*lamb^2 - 4)*m*z^2)*sqrt(z)*e^(1/2*c*z^2) - 4*(22*c*lamb^2*z^2 + 17*lamb^2 + 4)*e^(1/2*c*z^2) == 0,\n", " 12*c*m*z^(9/2)*e^(1/2*c*z^2) - lamb^2*mu*q^2 - 4*(3*c*z^2 - 5)*e^(1/2*c*z^2) == 0]" ] }, "execution_count": 60, "metadata": {}, "output_type": "execute_result" } ], "source": [ "neqs = [eq.subs(nu=4).simplify_full() for eq in eqs]\n", "[eq == 0 for eq in neqs]" ] }, { "cell_type": "code", "execution_count": 61, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left[{\\lambda} = -\\frac{2}{3} \\, \\sqrt{3}, {\\mu} = \\frac{15}{r_{11}^{2}}, {\\Lambda} = \\left(\\frac{45}{8}\\right), q = r_{11}, m = r_{12}, c = 0\\right], \\left[{\\lambda} = \\frac{2}{3} \\, \\sqrt{3}, {\\mu} = \\frac{15}{r_{13}^{2}}, {\\Lambda} = \\left(\\frac{45}{8}\\right), q = r_{13}, m = r_{14}, c = 0\\right], \\left[{\\lambda} = \\frac{2}{3} \\, \\sqrt{3}, {\\mu} = -\\frac{39 \\, e^{3}}{r_{15}^{2}}, {\\Lambda} = \\frac{495}{8} \\, e^{\\left(-3\\right)}, q = r_{15}, m = 0, c = \\frac{6}{z^{2}}\\right], \\left[{\\lambda} = -\\frac{2}{3} \\, \\sqrt{3}, {\\mu} = -\\frac{39 \\, e^{3}}{r_{16}^{2}}, {\\Lambda} = \\frac{495}{8} \\, e^{\\left(-3\\right)}, q = r_{16}, m = 0, c = \\frac{6}{z^{2}}\\right], \\left[{\\lambda} = \\frac{2}{11} i \\, \\sqrt{3}, {\\mu} = 0, {\\Lambda} = 5 \\, e^{\\left(-\\frac{5}{6}\\right)}, q = r_{17}, m = 0, c = \\frac{5}{3 \\, z^{2}}\\right], \\left[{\\lambda} = -\\frac{2}{11} i \\, \\sqrt{3}, {\\mu} = 0, {\\Lambda} = 5 \\, e^{\\left(-\\frac{5}{6}\\right)}, q = r_{18}, m = 0, c = \\frac{5}{3 \\, z^{2}}\\right], \\left[{\\lambda} = \\frac{2}{11} i \\, \\sqrt{3}, {\\mu} = r_{19}, {\\Lambda} = 5 \\, e^{\\left(-\\frac{5}{6}\\right)}, q = 0, m = 0, c = \\frac{5}{3 \\, z^{2}}\\right], \\left[{\\lambda} = -\\frac{2}{11} i \\, \\sqrt{3}, {\\mu} = r_{20}, {\\Lambda} = 5 \\, e^{\\left(-\\frac{5}{6}\\right)}, q = 0, m = 0, c = \\frac{5}{3 \\, z^{2}}\\right]\\right]$ " ], "text/plain": [ "[[lamb == -2/3*sqrt(3), mu == 15/r11^2, Lamb == (45/8), q == r11, m == r12, c == 0], [lamb == 2/3*sqrt(3), mu == 15/r13^2, Lamb == (45/8), q == r13, m == r14, c == 0], [lamb == 2/3*sqrt(3), mu == -39*e^3/r15^2, Lamb == 495/8*e^(-3), q == r15, m == 0, c == 6/z^2], [lamb == -2/3*sqrt(3), mu == -39*e^3/r16^2, Lamb == 495/8*e^(-3), q == r16, m == 0, c == 6/z^2], [lamb == 2/11*I*sqrt(3), mu == 0, Lamb == 5*e^(-5/6), q == r17, m == 0, c == 5/3/z^2], [lamb == -2/11*I*sqrt(3), mu == 0, Lamb == 5*e^(-5/6), q == r18, m == 0, c == 5/3/z^2], [lamb == 2/11*I*sqrt(3), mu == r19, Lamb == 5*e^(-5/6), q == 0, m == 0, c == 5/3/z^2], [lamb == -2/11*I*sqrt(3), mu == r20, Lamb == 5*e^(-5/6), q == 0, m == 0, c == 5/3/z^2]]" ] }, "execution_count": 61, "metadata": {}, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, c)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "In the above solutions, $r_i$, with $i$ an integer, stands for an arbitrary parameter. \n", "\n", "The solutions for $c=0$ are those already obtained for $b(z)=1$.\n", "But there are no solution with $c\\not = 0$ and $c={\\rm const}$." ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "SageMath 6.10", "language": "", "name": "sagemath" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.10" }, "name": "Lifshitz_black_brane_warped.ipynb" }, "nbformat": 4, "nbformat_minor": 0 }