{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "# Black branes in Lifshitz-like spacetimes" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "This Jupyter/SageMath \n", "worksheet implements some computations of the article\n", "- I. Ya. Aref'eva, A. A. Golubtsova & E. Gourgoulhon: *Analytic black branes in \n", " Lifshitz-like backgrounds and thermalization*,\n", " [arXiv:1601.06046](http://arxiv.org/abs/1601.06046)\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 [here](https://cloud.sagemath.com/projects/3edbca82-97d6-41b3-9b6f-d83ea06fc1e9/files/Lifshitz_black_brane.html).\n", "\n", "\n", "#### Content\n", "1. **Five-dimensional Lifschitz-like spacetime**\n", "2. **Black brane solution**\n" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "First we set up the notebook to display mathematical objects using LaTeX formatting:" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "%display latex" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "## 1. Five-dimensional Lifshitz-like spacetime\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": { "collapsed": false }, "source": [ "We introduce a first coordinate system on $M$:" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(M,(t, x, {y_1}, {y_2}, {\\tilde{r}})\\right)$ " ] }, "execution_count": 3, "metadata": { }, "output_type": "execute_result" } ], "source": [ "X0. = M.chart(r't x y1:y_1 y2:y_2 R:\\tilde{r}:(0,+oo)')\n", "X0" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Let us consider the following Lifshitz-symmetric metric, parametrized by some real number $\\nu$:" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = -{\\tilde{r}}^{2 \\, {\\nu}} \\mathrm{d} t\\otimes \\mathrm{d} t + {\\tilde{r}}^{2 \\, {\\nu}} \\mathrm{d} x\\otimes \\mathrm{d} x + {\\tilde{r}}^{2} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + {\\tilde{r}}^{2} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} + \\frac{1}{{\\tilde{r}}^{2}} \\mathrm{d} {\\tilde{r}}\\otimes \\mathrm{d} {\\tilde{r}}$ " ] }, "execution_count": 4, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g = M.lorentzian_metric('g')\n", "var('nu', latex_name=r'\\nu', domain='real')\n", "g[0,0] = -R^(2*nu)\n", "g[1,1] = R^(2*nu)\n", "g[2,2] = R^2\n", "g[3,3] = R^2\n", "g[4,4] = 1/R^2\n", "g.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "A matrix view of the metric components:" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrrr}\n", "-{\\tilde{r}}^{2 \\, {\\nu}} & 0 & 0 & 0 & 0 \\\\\n", "0 & {\\tilde{r}}^{2 \\, {\\nu}} & 0 & 0 & 0 \\\\\n", "0 & 0 & {\\tilde{r}}^{2} & 0 & 0 \\\\\n", "0 & 0 & 0 & {\\tilde{r}}^{2} & 0 \\\\\n", "0 & 0 & 0 & 0 & \\frac{1}{{\\tilde{r}}^{2}}\n", "\\end{array}\\right)$ " ] }, "execution_count": 5, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g[:]" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "This metric is invariant under the *Lifshitz scaling*\n", "$$ (t,x,y_1,y_2,\\tilde r) \\longmapsto \\left(\\lambda^\\nu t, \\lambda^\\nu x, \\lambda y_1, \\lambda y_2, \\frac{\\tilde r}{\\lambda} \\right)$$\n", "- If $\\nu=1$ the scaling is isotropic and we recognize the metric \n", " of $\\mathrm{AdS}_5$ in Poincaré coordinates ($M$ is then the Poincaré patch of $\\mathrm{AdS}_5$)\n", "- If $\\nu\\not=1$, the scaling is anisotropic" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Let us introduce a second coordinate system on $M$:" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(M,(t, x, {y_1}, {y_2}, r)\\right)$ " ] }, "execution_count": 6, "metadata": { }, "output_type": "execute_result" } ], "source": [ "X. = M.chart('t x y1:y_1 y2:y_2 r')\n", "X" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "and relate it to the previous one by the transformation $r=\\ln\\tilde r$:" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} t & = & t \\\\ x & = & x \\\\ {y_1} & = & {y_1} \\\\ {y_2} & = & {y_2} \\\\ r & = & \\log\\left({\\tilde{r}}\\right) \\end{array}\\right.$ " ] }, "execution_count": 7, "metadata": { }, "output_type": "execute_result" } ], "source": [ "X0_to_X = X0.transition_map(X, [t, x, y1, y2, ln(R)])\n", "X0_to_X.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The inverse coordinate transition is computed by means of the method `inverse()`:" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} t & = & t \\\\ x & = & x \\\\ {y_1} & = & {y_1} \\\\ {y_2} & = & {y_2} \\\\ {\\tilde{r}} & = & e^{r} \\end{array}\\right.$ " ] }, "execution_count": 8, "metadata": { }, "output_type": "execute_result" } ], "source": [ "X_to_X0 = X0_to_X.inverse()\n", "X_to_X0.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "At this stage, the manifold's atlas defined by the user is" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left(M,(t, x, {y_1}, {y_2}, {\\tilde{r}})\\right), \\left(M,(t, x, {y_1}, {y_2}, r)\\right)\\right]$ " ] }, "execution_count": 9, "metadata": { }, "output_type": "execute_result" } ], "source": [ "M.atlas()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "and the list of defined vector frames defined is" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left(M, \\left(\\frac{\\partial}{\\partial t },\\frac{\\partial}{\\partial x },\\frac{\\partial}{\\partial {y_1} },\\frac{\\partial}{\\partial {y_2} },\\frac{\\partial}{\\partial {\\tilde{r}} }\\right)\\right), \\left(M, \\left(\\frac{\\partial}{\\partial t },\\frac{\\partial}{\\partial x },\\frac{\\partial}{\\partial {y_1} },\\frac{\\partial}{\\partial {y_2} },\\frac{\\partial}{\\partial r }\\right)\\right)\\right]$ " ] }, "execution_count": 10, "metadata": { }, "output_type": "execute_result" } ], "source": [ "M.frames()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The expression of the metric in terms of the new coordinates is" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = -e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} t\\otimes \\mathrm{d} t + e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} x\\otimes \\mathrm{d} x + e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} +\\mathrm{d} r\\otimes \\mathrm{d} r$ " ] }, "execution_count": 11, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g.display(X.frame(), X)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "or, in matrix view:" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrrr}\n", "-e^{\\left(2 \\, {\\nu} r\\right)} & 0 & 0 & 0 & 0 \\\\\n", "0 & e^{\\left(2 \\, {\\nu} r\\right)} & 0 & 0 & 0 \\\\\n", "0 & 0 & e^{\\left(2 \\, r\\right)} & 0 & 0 \\\\\n", "0 & 0 & 0 & e^{\\left(2 \\, r\\right)} & 0 \\\\\n", "0 & 0 & 0 & 0 & 1\n", "\\end{array}\\right)$ " ] }, "execution_count": 12, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g[X.frame(),:,X]" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "To access to a particular component, we have to specify (i) the frame w.r.t. which it is defined and (ii) the coordinates in which the component is expressed:" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-e^{\\left(2 \\, {\\nu} r\\right)}$ " ] }, "execution_count": 13, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g[X.frame(),0,0,X]" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-{\\tilde{r}}^{2 \\, {\\nu}}$ " ] }, "execution_count": 14, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g[X.frame(),0,0] # the default chart is used" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "From now on, let us consider the coordinates $X = (t,x,y_1,y_2,r)$ as the default ones on the manifold $M$:" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "M.set_default_chart(X)\n", "M.set_default_frame(X.frame())" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Then" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = -e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} t\\otimes \\mathrm{d} t + e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} x\\otimes \\mathrm{d} x + e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} +\\mathrm{d} r\\otimes \\mathrm{d} r$ " ] }, "execution_count": 16, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g.display()" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrrr}\n", "-e^{\\left(2 \\, {\\nu} r\\right)} & 0 & 0 & 0 & 0 \\\\\n", "0 & e^{\\left(2 \\, {\\nu} r\\right)} & 0 & 0 & 0 \\\\\n", "0 & 0 & e^{\\left(2 \\, r\\right)} & 0 & 0 \\\\\n", "0 & 0 & 0 & e^{\\left(2 \\, r\\right)} & 0 \\\\\n", "0 & 0 & 0 & 0 & 1\n", "\\end{array}\\right)$ " ] }, "execution_count": 17, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g[:]" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-e^{\\left(2 \\, {\\nu} r\\right)}$ " ] }, "execution_count": 18, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g[0,0]" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} g_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & -e^{\\left(2 \\, {\\nu} r\\right)} \\\\ g_{ \\, x \\, x }^{ \\phantom{\\, x}\\phantom{\\, x} } & = & e^{\\left(2 \\, {\\nu} r\\right)} \\\\ g_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1}}\\phantom{\\, {y_1}} } & = & e^{\\left(2 \\, r\\right)} \\\\ g_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2}}\\phantom{\\, {y_2}} } & = & e^{\\left(2 \\, r\\right)} \\\\ g_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & 1 \\end{array}$ " ] }, "execution_count": 19, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g.display_comp()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Curvature\n", "\n", "The Riemann tensor is" ] }, { "cell_type": "code", "execution_count": 20, "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": 21, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, t} \\, x \\, t \\, x }^{ \\, t \\phantom{\\, x} \\phantom{\\, t} \\phantom{\\, x} } & = & -{\\nu}^{2} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, t} \\, {y_1} \\, t \\, {y_1} }^{ \\, t \\phantom{\\, {y_1}} \\phantom{\\, t} \\phantom{\\, {y_1}} } & = & -{\\nu} e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, t} \\, {y_2} \\, t \\, {y_2} }^{ \\, t \\phantom{\\, {y_2}} \\phantom{\\, t} \\phantom{\\, {y_2}} } & = & -{\\nu} e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, t} \\, r \\, t \\, r }^{ \\, t \\phantom{\\, r} \\phantom{\\, t} \\phantom{\\, r} } & = & -{\\nu}^{2} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, x} \\, t \\, t \\, x }^{ \\, x \\phantom{\\, t} \\phantom{\\, t} \\phantom{\\, x} } & = & -{\\nu}^{2} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, x} \\, {y_1} \\, x \\, {y_1} }^{ \\, x \\phantom{\\, {y_1}} \\phantom{\\, x} \\phantom{\\, {y_1}} } & = & -{\\nu} e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, x} \\, {y_2} \\, x \\, {y_2} }^{ \\, x \\phantom{\\, {y_2}} \\phantom{\\, x} \\phantom{\\, {y_2}} } & = & -{\\nu} e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, x} \\, r \\, x \\, r }^{ \\, x \\phantom{\\, r} \\phantom{\\, x} \\phantom{\\, r} } & = & -{\\nu}^{2} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_1}} \\, t \\, t \\, {y_1} }^{ \\, {y_1} \\phantom{\\, t} \\phantom{\\, t} \\phantom{\\, {y_1}} } & = & -{\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_1}} \\, x \\, x \\, {y_1} }^{ \\, {y_1} \\phantom{\\, x} \\phantom{\\, x} \\phantom{\\, {y_1}} } & = & {\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_1}} \\, {y_2} \\, {y_1} \\, {y_2} }^{ \\, {y_1} \\phantom{\\, {y_2}} \\phantom{\\, {y_1}} \\phantom{\\, {y_2}} } & = & -e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_1}} \\, r \\, {y_1} \\, r }^{ \\, {y_1} \\phantom{\\, r} \\phantom{\\, {y_1}} \\phantom{\\, r} } & = & -1 \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_2}} \\, t \\, t \\, {y_2} }^{ \\, {y_2} \\phantom{\\, t} \\phantom{\\, t} \\phantom{\\, {y_2}} } & = & -{\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_2}} \\, x \\, x \\, {y_2} }^{ \\, {y_2} \\phantom{\\, x} \\phantom{\\, x} \\phantom{\\, {y_2}} } & = & {\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_2}} \\, {y_1} \\, {y_1} \\, {y_2} }^{ \\, {y_2} \\phantom{\\, {y_1}} \\phantom{\\, {y_1}} \\phantom{\\, {y_2}} } & = & e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, {y_2}} \\, r \\, {y_2} \\, r }^{ \\, {y_2} \\phantom{\\, r} \\phantom{\\, {y_2}} \\phantom{\\, r} } & = & -1 \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, r} \\, t \\, t \\, r }^{ \\, r \\phantom{\\, t} \\phantom{\\, t} \\phantom{\\, r} } & = & -{\\nu}^{2} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, r} \\, x \\, x \\, r }^{ \\, r \\phantom{\\, x} \\phantom{\\, x} \\phantom{\\, r} } & = & {\\nu}^{2} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, r} \\, {y_1} \\, {y_1} \\, r }^{ \\, r \\phantom{\\, {y_1}} \\phantom{\\, {y_1}} \\phantom{\\, r} } & = & e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Riem}\\left(g\\right)_{ \\phantom{\\, r} \\, {y_2} \\, {y_2} \\, r }^{ \\, r \\phantom{\\, {y_2}} \\phantom{\\, {y_2}} \\phantom{\\, r} } & = & e^{\\left(2 \\, r\\right)} \\end{array}$ " ] }, "execution_count": 21, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Riem.display_comp(only_nonredundant=True)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The Ricci tensor:" ] }, { "cell_type": "code", "execution_count": 22, "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": 23, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{Ric}\\left(g\\right) = 2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} t\\otimes \\mathrm{d} t -2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} x\\otimes \\mathrm{d} x -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} + \\left( -2 \\, {\\nu}^{2} - 2 \\right) \\mathrm{d} r\\otimes \\mathrm{d} r$ " ] }, "execution_count": 23, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Ric.display()" ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} \\mathrm{Ric}\\left(g\\right)_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & 2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, x \\, x }^{ \\phantom{\\, x}\\phantom{\\, x} } & = & -2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1}}\\phantom{\\, {y_1}} } & = & -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2}}\\phantom{\\, {y_2}} } & = & -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & -2 \\, {\\nu}^{2} - 2 \\end{array}$ " ] }, "execution_count": 24, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Ric.display_comp()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The Ricci scalar:" ] }, { "cell_type": "code", "execution_count": 25, "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": 26, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} \\mathrm{r}\\left(g\\right):& M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & -6 \\, {\\nu}^{2} - 8 \\, {\\nu} - 6 \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & -6 \\, {\\nu}^{2} - 8 \\, {\\nu} - 6 \\end{array}$ " ] }, "execution_count": 26, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Rscal.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "We note that the Ricci scalar is constant." ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "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": { "collapsed": false }, "source": [ "### The dilaton scalar field\n", "\n", "We consider the following ansatz for the dilaton scalar field $\\phi$:\n", "$$ \\phi = \\frac{1}{\\lambda} \\left( 4 r + \\ln\\mu \\right),$$\n", "where $\\mu$ is a constant. " ] }, { "cell_type": "code", "execution_count": 27, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} \\phi:& M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & \\frac{4 \\, \\log\\left({\\tilde{r}}\\right) + \\log\\left({\\mu}\\right)}{{\\lambda}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & \\frac{4 \\, r + \\log\\left({\\mu}\\right)}{{\\lambda}} \\end{array}$ " ] }, "execution_count": 27, "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: (4*r + ln(mu))/lamb}, \n", " name='phi', latex_name=r'\\phi')\n", "phi.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The 1-form $\\mathrm{d}\\phi$ is" ] }, { "cell_type": "code", "execution_count": 28, "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": 29, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d}\\phi = \\frac{4}{{\\lambda}} \\mathrm{d} r$ " ] }, "execution_count": 29, "metadata": { }, "output_type": "execute_result" } ], "source": [ "dphi.display()" ] }, { "cell_type": "code", "execution_count": 30, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[0, 0, 0, 0, \\frac{4}{{\\lambda}}\\right]$ " ] }, "execution_count": 30, "metadata": { }, "output_type": "execute_result" } ], "source": [ "dphi[:] # all the components in the default frame" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "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": 31, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(M, \\left(\\mathrm{d} t,\\mathrm{d} x,\\mathrm{d} {y_1},\\mathrm{d} {y_2},\\mathrm{d} r\\right)\\right)$ " ] }, "execution_count": 31, "metadata": { }, "output_type": "execute_result" } ], "source": [ "X.coframe()" ] }, { "cell_type": "code", "execution_count": 32, "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": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathrm{d} {y_1}, \\mathrm{d} {y_2}\\right)$ " ] }, "execution_count": 32, "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": { "collapsed": false }, "source": [ "Then we can form $F$ according to the above ansatz:" ] }, { "cell_type": "code", "execution_count": 33, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2-form F on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}F = \\frac{1}{2} \\, q \\mathrm{d} {y_1}\\wedge \\mathrm{d} {y_2}$ " ] }, "execution_count": 33, "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": { "collapsed": false }, "source": [ "By construction, the 2-form $F$ is closed (since $q$ is constant):" ] }, { "cell_type": "code", "execution_count": 34, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "3-form dF on the 5-dimensional differentiable manifold M\n" ] } ], "source": [ "print(F.exterior_derivative())" ] }, { "cell_type": "code", "execution_count": 35, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d}F = 0$ " ] }, "execution_count": 35, "metadata": { }, "output_type": "execute_result" } ], "source": [ "F.exterior_derivative().display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Let us evaluate the square $F_{mn} F^{mn}$ of $F$:" ] }, { "cell_type": "code", "execution_count": 36, "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": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2} \\, q e^{\\left(-4 \\, r\\right)} \\frac{\\partial}{\\partial {y_1} }\\otimes \\frac{\\partial}{\\partial {y_2} } -\\frac{1}{2} \\, q e^{\\left(-4 \\, r\\right)} \\frac{\\partial}{\\partial {y_2} }\\otimes \\frac{\\partial}{\\partial {y_1} }$ " ] }, "execution_count": 36, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Fu = F.up(g)\n", "print Fu\n", "Fu.display()" ] }, { "cell_type": "code", "execution_count": 37, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & \\frac{q^{2}}{2 \\, {\\tilde{r}}^{4}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & \\frac{1}{2} \\, q^{2} e^{\\left(-4 \\, r\\right)} \\end{array}$ " ] }, "execution_count": 37, "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": { "collapsed": false }, "source": [ "We shall also need the tensor $\\mathcal{F}_{mn} := F_{mp} F_n^{\\ \\, p}$:" ] }, { "cell_type": "code", "execution_count": 38, "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": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{4} \\, q^{2} e^{\\left(-2 \\, r\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + \\frac{1}{4} \\, q^{2} e^{\\left(-2 \\, r\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2}$ " ] }, "execution_count": 38, "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": { "collapsed": false }, "source": [ "The tensor field $\\mathcal{F}$ is symmetric:" ] }, { "cell_type": "code", "execution_count": 39, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$ " ] }, "execution_count": 39, "metadata": { }, "output_type": "execute_result" } ], "source": [ "FF == FF.symmetrize()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Therefore, from now on, we set" ] }, { "cell_type": "code", "execution_count": 40, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "FF = FF.symmetrize()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Einstein equation\n", "\n", "Let us first introduce the cosmological constant:" ] }, { "cell_type": "code", "execution_count": 41, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\Lambda}$ " ] }, "execution_count": 41, "metadata": { }, "output_type": "execute_result" } ], "source": [ "var('Lamb', latex_name=r'\\Lambda', domain='real')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "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": 42, "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": 43, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} E_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & -\\frac{1}{24} \\, {\\left({\\mu} q^{2} - 48 \\, {\\nu}^{2} + 8 \\, {\\Lambda} - 48 \\, {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ E_{ \\, x \\, x }^{ \\phantom{\\, x}\\phantom{\\, x} } & = & \\frac{1}{24} \\, {\\left({\\mu} q^{2} - 48 \\, {\\nu}^{2} + 8 \\, {\\Lambda} - 48 \\, {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ E_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1}}\\phantom{\\, {y_1}} } & = & -\\frac{1}{12} \\, {\\left({\\mu} q^{2} - 4 \\, {\\Lambda} + 24 \\, {\\nu} + 24\\right)} e^{\\left(2 \\, r\\right)} \\\\ E_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2}}\\phantom{\\, {y_2}} } & = & -\\frac{1}{12} \\, {\\left({\\mu} q^{2} - 4 \\, {\\Lambda} + 24 \\, {\\nu} + 24\\right)} e^{\\left(2 \\, r\\right)} \\\\ E_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & \\frac{{\\lambda}^{2} {\\mu} q^{2} - 48 \\, {\\lambda}^{2} {\\nu}^{2} + 8 \\, {\\left({\\Lambda} - 6\\right)} {\\lambda}^{2} - 192}{24 \\, {\\lambda}^{2}} \\end{array}$ " ] }, "execution_count": 43, "metadata": { }, "output_type": "execute_result" } ], "source": [ "EE.display_comp(only_nonredundant=True)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "We note that `EE==0` leads to only 3 independent equations:" ] }, { "cell_type": "code", "execution_count": 44, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{24} \\, {\\mu} q^{2} + 2 \\, {\\nu}^{2} - \\frac{1}{3} \\, {\\Lambda} + 2 \\, {\\nu}$ " ] }, "execution_count": 44, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq1 = (EE[0,0]/exp(2*nu*r)).expr()\n", "eq1" ] }, { "cell_type": "code", "execution_count": 45, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 2 \\, {\\nu} - 2$ " ] }, "execution_count": 45, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq2 = (EE[2,2]/exp(2*r)).expr()\n", "eq2" ] }, { "cell_type": "code", "execution_count": 46, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, {\\mu} q^{2} - 2 \\, {\\nu}^{2} + \\frac{1}{3} \\, {\\Lambda} - \\frac{8}{{\\lambda}^{2}} - 2$ " ] }, "execution_count": 46, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq3 = EE[4,4].expr().expand()\n", "eq3" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Dilaton field equation\n", "\n", "First we evaluate $\\nabla_m \\nabla^m \\phi$:" ] }, { "cell_type": "code", "execution_count": 47, "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": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\nabla_{g}$ " ] }, "execution_count": 47, "metadata": { }, "output_type": "execute_result" } ], "source": [ "nab = g.connection()\n", "print nab\n", "nab" ] }, { "cell_type": "code", "execution_count": 48, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & \\frac{8 \\, {\\left({\\nu} + 1\\right)}}{{\\lambda}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & \\frac{8 \\, {\\left({\\nu} + 1\\right)}}{{\\lambda}} \\end{array}$ " ] }, "execution_count": 48, "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": { "collapsed": false }, "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": 49, "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": 50, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & -\\frac{{\\lambda}^{2} {\\mu} q^{2} - 64 \\, {\\nu} - 64}{8 \\, {\\lambda}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & -\\frac{{\\lambda}^{2} {\\mu} q^{2} - 64 \\, {\\nu} - 64}{8 \\, {\\lambda}} \\end{array}$ " ] }, "execution_count": 50, "metadata": { }, "output_type": "execute_result" } ], "source": [ "DE.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Hence the dilaton field equation provides a fourth equation:" ] }, { "cell_type": "code", "execution_count": 51, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{8} \\, {\\lambda} {\\mu} q^{2} + \\frac{8 \\, {\\nu}}{{\\lambda}} + \\frac{8}{{\\lambda}}$ " ] }, "execution_count": 51, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq4 = DE.expr().expand()\n", "eq4" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "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": 52, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Vector field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$ " ] }, "execution_count": 52, "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": { "collapsed": false }, "source": [ "We get identically zero; indeed the tensor $\\nabla_p (e^{\\lambda\\phi} F^{mn})$ has a vanishing trace, as we can check:" ] }, { "cell_type": "code", "execution_count": 53, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\mu} q \\frac{\\partial}{\\partial {y_1} }\\otimes \\frac{\\partial}{\\partial {y_2} }\\otimes \\mathrm{d} r -\\frac{1}{2} \\, {\\mu} q e^{\\left(2 \\, r\\right)} \\frac{\\partial}{\\partial {y_1} }\\otimes \\frac{\\partial}{\\partial r }\\otimes \\mathrm{d} {y_2} -{\\mu} q \\frac{\\partial}{\\partial {y_2} }\\otimes \\frac{\\partial}{\\partial {y_1} }\\otimes \\mathrm{d} r + \\frac{1}{2} \\, {\\mu} q e^{\\left(2 \\, r\\right)} \\frac{\\partial}{\\partial {y_2} }\\otimes \\frac{\\partial}{\\partial r }\\otimes \\mathrm{d} {y_1} + \\frac{1}{2} \\, {\\mu} q e^{\\left(2 \\, r\\right)} \\frac{\\partial}{\\partial r }\\otimes \\frac{\\partial}{\\partial {y_1} }\\otimes \\mathrm{d} {y_2} -\\frac{1}{2} \\, {\\mu} q e^{\\left(2 \\, r\\right)} \\frac{\\partial}{\\partial r }\\otimes \\frac{\\partial}{\\partial {y_2} }\\otimes \\mathrm{d} {y_1}$ " ] }, "execution_count": 53, "metadata": { }, "output_type": "execute_result" } ], "source": [ "nab(exp(lamb*phi)*Fu).display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Summary\n", "\n", "We have 4 equations involving the constants $\\lambda$, $\\mu$, $\\nu$, $q$ and $\\Lambda$:" ] }, { "cell_type": "code", "execution_count": 54, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{24} \\, {\\mu} q^{2} + 2 \\, {\\nu}^{2} - \\frac{1}{3} \\, {\\Lambda} + 2 \\, {\\nu} = 0$ " ] }, "execution_count": 54, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq1 == 0" ] }, { "cell_type": "code", "execution_count": 55, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 2 \\, {\\nu} - 2 = 0$ " ] }, "execution_count": 55, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq2 == 0" ] }, { "cell_type": "code", "execution_count": 56, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, {\\mu} q^{2} - 2 \\, {\\nu}^{2} + \\frac{1}{3} \\, {\\Lambda} - \\frac{8}{{\\lambda}^{2}} - 2 = 0$ " ] }, "execution_count": 56, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq3 == 0" ] }, { "cell_type": "code", "execution_count": 57, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{8} \\, {\\lambda} {\\mu} q^{2} + \\frac{8 \\, {\\nu}}{{\\lambda}} + \\frac{8}{{\\lambda}} = 0$ " ] }, "execution_count": 57, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq4 == 0" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Solution for $\\nu=1$ ($\\mathrm{AdS}_5$)" ] }, { "cell_type": "code", "execution_count": 58, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "eqs = [eq1, eq2, eq3, eq4]\n", "neqs = [eq.subs(nu=1) for eq in eqs]" ] }, { "cell_type": "code", "execution_count": 59, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[-\\frac{1}{24} \\, {\\mu} q^{2} - \\frac{1}{3} \\, {\\Lambda} + 4 = 0, -\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 4 = 0, \\frac{1}{24} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - \\frac{8}{{\\lambda}^{2}} - 4 = 0, -\\frac{1}{8} \\, {\\lambda} {\\mu} q^{2} + \\frac{16}{{\\lambda}} = 0\\right]$ " ] }, "execution_count": 59, "metadata": { }, "output_type": "execute_result" } ], "source": [ "[eq == 0 for eq in neqs]" ] }, { "cell_type": "code", "execution_count": 60, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\right]$ " ] }, "execution_count": 60, "metadata": { }, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Hence there is no solution for $\\mathrm{AdS}_5$ with the above ansatz. " ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Solution for $\\nu = 2$" ] }, { "cell_type": "code", "execution_count": 61, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[-\\frac{1}{24} \\, {\\mu} q^{2} - \\frac{1}{3} \\, {\\Lambda} + 12 = 0, -\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 6 = 0, \\frac{1}{24} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - \\frac{8}{{\\lambda}^{2}} - 10 = 0, -\\frac{1}{8} \\, {\\lambda} {\\mu} q^{2} + \\frac{24}{{\\lambda}} = 0\\right]$ " ] }, "execution_count": 61, "metadata": { }, "output_type": "execute_result" } ], "source": [ "neqs = [eq.subs(nu=2) for eq in eqs]\n", "[eq == 0 for eq in neqs]" ] }, { "cell_type": "code", "execution_count": 62, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left[{\\lambda} = 2, {\\mu} = \\frac{48}{r_{1}^{2}}, {\\Lambda} = 30, q = r_{1}\\right], \\left[{\\lambda} = \\left(-2\\right), {\\mu} = \\frac{48}{r_{2}^{2}}, {\\Lambda} = 30, q = r_{2}\\right]\\right]$ " ] }, "execution_count": 62, "metadata": { }, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Hence there are two families of solutions, each famility being parametrized by e.g. $q$. \n", "Indeed, in the above writing, $r_1$ and $r_2$ stand for arbitrary parameters (nothing to do with the coordinate $r$)." ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Solution for $\\nu = 4$" ] }, { "cell_type": "code", "execution_count": 63, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[-\\frac{1}{24} \\, {\\mu} q^{2} - \\frac{1}{3} \\, {\\Lambda} + 40 = 0, -\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 10 = 0, \\frac{1}{24} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - \\frac{8}{{\\lambda}^{2}} - 34 = 0, -\\frac{1}{8} \\, {\\lambda} {\\mu} q^{2} + \\frac{40}{{\\lambda}} = 0\\right]$ " ] }, "execution_count": 63, "metadata": { }, "output_type": "execute_result" } ], "source": [ "neqs = [eq.subs(nu=4) for eq in eqs]\n", "[eq == 0 for eq in neqs]" ] }, { "cell_type": "code", "execution_count": 64, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left[{\\lambda} = \\frac{2}{3} \\, \\sqrt{3}, {\\mu} = \\frac{240}{r_{3}^{2}}, {\\Lambda} = 90, q = r_{3}\\right], \\left[{\\lambda} = -\\frac{2}{3} \\, \\sqrt{3}, {\\mu} = \\frac{240}{r_{4}^{2}}, {\\Lambda} = 90, q = r_{4}\\right]\\right]$ " ] }, "execution_count": 64, "metadata": { }, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Hence there are two families of solutions, each family being parametrized by e.g. $q$. Note that, as above, $r_i$, with $i$ an integer, stands for an arbitrary parameter (nothing to do with the coordinate $r$)." ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## 2. Black brane solution\n", "\n", "We add a blackening factor $f(r)$ to the metric; i.e. we declare a new metric $g$ according to " ] }, { "cell_type": "code", "execution_count": 65, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = -e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\mathrm{d} t\\otimes \\mathrm{d} t + e^{\\left(2 \\, {\\nu} r\\right)} \\mathrm{d} x\\otimes \\mathrm{d} x + e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + e^{\\left(2 \\, r\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2} + \\frac{1}{f\\left(r\\right)} \\mathrm{d} r\\otimes \\mathrm{d} r$ " ] }, "execution_count": 65, "metadata": { }, "output_type": "execute_result" } ], "source": [ "g = M.lorentzian_metric('g')\n", "ff = function('f')(r)\n", "g[0,0] = -ff*exp(2*nu*r)\n", "g[1,1] = exp(2*nu*r)\n", "g[2,2] = exp(2*r)\n", "g[3,3] = exp(2*r)\n", "g[4,4] = 1/ff\n", "g.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The Ricci tensor of $g$ is" ] }, { "cell_type": "code", "execution_count": 66, "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": 67, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} \\mathrm{Ric}\\left(g\\right)_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & 2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right)^{2} + {\\left(2 \\, {\\nu} + 1\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\frac{\\partial\\,f}{\\partial r} + \\frac{1}{2} \\, e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\frac{\\partial^2\\,f}{\\partial r ^ 2} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, x \\, x }^{ \\phantom{\\, x}\\phantom{\\, x} } & = & -2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) - {\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\frac{\\partial\\,f}{\\partial r} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1}}\\phantom{\\, {y_1}} } & = & -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} f\\left(r\\right) - e^{\\left(2 \\, r\\right)} \\frac{\\partial\\,f}{\\partial r} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2}}\\phantom{\\, {y_2}} } & = & -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} f\\left(r\\right) - e^{\\left(2 \\, r\\right)} \\frac{\\partial\\,f}{\\partial r} \\\\ \\mathrm{Ric}\\left(g\\right)_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & -\\frac{4 \\, {\\left({\\nu}^{2} + 1\\right)} f\\left(r\\right) + 2 \\, {\\left(2 \\, {\\nu} + 1\\right)} \\frac{\\partial\\,f}{\\partial r} + \\frac{\\partial^2\\,f}{\\partial r ^ 2}}{2 \\, f\\left(r\\right)} \\end{array}$ " ] }, "execution_count": 67, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Ric.display_comp()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "By construction, the 2-form $F$ does not depend on $g$; hence there is no need to reevaluate it:" ] }, { "cell_type": "code", "execution_count": 68, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}F = \\frac{1}{2} \\, q \\mathrm{d} {y_1}\\wedge \\mathrm{d} {y_2}$ " ] }, "execution_count": 68, "metadata": { }, "output_type": "execute_result" } ], "source": [ "F.display()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "On the contrary, we need to reevaluate its metric dual, in order to compute $F^2 := F_{mn} F^{mn}$:" ] }, { "cell_type": "code", "execution_count": 69, "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": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2} \\, q e^{\\left(-4 \\, r\\right)} \\frac{\\partial}{\\partial {y_1} }\\otimes \\frac{\\partial}{\\partial {y_2} } -\\frac{1}{2} \\, q e^{\\left(-4 \\, r\\right)} \\frac{\\partial}{\\partial {y_2} }\\otimes \\frac{\\partial}{\\partial {y_1} }$ " ] }, "execution_count": 69, "metadata": { }, "output_type": "execute_result" } ], "source": [ "Fu = F.up(g)\n", "print Fu\n", "Fu.display()" ] }, { "cell_type": "code", "execution_count": 70, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & \\frac{q^{2}}{2 \\, {\\tilde{r}}^{4}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & \\frac{1}{2} \\, q^{2} e^{\\left(-4 \\, r\\right)} \\end{array}$ " ] }, "execution_count": 70, "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": { "collapsed": false }, "source": [ "Simlarly, we need to reevaluate $\\mathcal{F}_{mn} := F_{mp} F_n^{\\ \\, p}$:" ] }, { "cell_type": "code", "execution_count": 71, "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": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{4} \\, q^{2} e^{\\left(-2 \\, r\\right)} \\mathrm{d} {y_1}\\otimes \\mathrm{d} {y_1} + \\frac{1}{4} \\, q^{2} e^{\\left(-2 \\, r\\right)} \\mathrm{d} {y_2}\\otimes \\mathrm{d} {y_2}$ " ] }, "execution_count": 71, "metadata": { }, "output_type": "execute_result" } ], "source": [ "FF = F['_mp'] * F.up(g,1)['^p_n']\n", "print FF\n", "FF.display()" ] }, { "cell_type": "code", "execution_count": 72, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$ " ] }, "execution_count": 72, "metadata": { }, "output_type": "execute_result" } ], "source": [ "FF == FF.symmetrize()" ] }, { "cell_type": "code", "execution_count": 73, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "FF = FF.symmetrize()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The new Einstein equation is" ] }, { "cell_type": "code", "execution_count": 74, "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": 75, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} E_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & 2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right)^{2} + {\\left(2 \\, {\\nu} + 1\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\frac{\\partial\\,f}{\\partial r} - \\frac{1}{24} \\, {\\left({\\mu} q^{2} + 8 \\, {\\Lambda}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) + \\frac{1}{2} \\, e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\frac{\\partial^2\\,f}{\\partial r ^ 2} \\\\ E_{ \\, x \\, x }^{ \\phantom{\\, x}\\phantom{\\, x} } & = & -2 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) - {\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\frac{\\partial\\,f}{\\partial r} + \\frac{1}{24} \\, {\\left({\\mu} q^{2} + 8 \\, {\\Lambda}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} \\\\ E_{ \\, {y_1} \\, {y_1} }^{ \\phantom{\\, {y_1}}\\phantom{\\, {y_1}} } & = & -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} f\\left(r\\right) - \\frac{1}{12} \\, {\\left({\\mu} q^{2} - 4 \\, {\\Lambda}\\right)} e^{\\left(2 \\, r\\right)} - e^{\\left(2 \\, r\\right)} \\frac{\\partial\\,f}{\\partial r} \\\\ E_{ \\, {y_2} \\, {y_2} }^{ \\phantom{\\, {y_2}}\\phantom{\\, {y_2}} } & = & -2 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} f\\left(r\\right) - \\frac{1}{12} \\, {\\left({\\mu} q^{2} - 4 \\, {\\Lambda}\\right)} e^{\\left(2 \\, r\\right)} - e^{\\left(2 \\, r\\right)} \\frac{\\partial\\,f}{\\partial r} \\\\ E_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & \\frac{{\\lambda}^{2} {\\mu} q^{2} + 8 \\, {\\Lambda} {\\lambda}^{2} - 12 \\, {\\lambda}^{2} \\frac{\\partial^2\\,f}{\\partial r ^ 2} - 48 \\, {\\left({\\lambda}^{2} {\\nu}^{2} + {\\lambda}^{2} + 4\\right)} f\\left(r\\right) - 24 \\, {\\left(2 \\, {\\lambda}^{2} {\\nu} + {\\lambda}^{2}\\right)} \\frac{\\partial\\,f}{\\partial r}}{24 \\, {\\lambda}^{2} f\\left(r\\right)} \\end{array}$ " ] }, "execution_count": 75, "metadata": { }, "output_type": "execute_result" } ], "source": [ "EE.display_comp(only_nonredundant=True)" ] }, { "cell_type": "code", "execution_count": 76, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$ " ] }, "execution_count": 76, "metadata": { }, "output_type": "execute_result" } ], "source": [ "EE[2,2] == EE[3,3]" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "There are 4 independent components:" ] }, { "cell_type": "code", "execution_count": 77, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, {\\left(48 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right)^{2} + 24 \\, {\\left(2 \\, {\\nu} + 1\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\frac{\\partial\\,f}{\\partial r} - {\\left({\\mu} q^{2} + 8 \\, {\\Lambda}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) + 12 \\, e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) \\frac{\\partial^2\\,f}{\\partial r ^ 2}\\right)} e^{\\left(-2 \\, {\\nu} r\\right)}$ " ] }, "execution_count": 77, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq0 = EE[0,0]/exp(2*nu*r)\n", "eq0" ] }, { "cell_type": "code", "execution_count": 78, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{24} \\, {\\left(48 \\, {\\left({\\nu}^{2} + {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r\\right)} f\\left(r\\right) + 24 \\, {\\nu} e^{\\left(2 \\, {\\nu} r\\right)} \\frac{\\partial\\,f}{\\partial r} - {\\left({\\mu} q^{2} + 8 \\, {\\Lambda}\\right)} e^{\\left(2 \\, {\\nu} r\\right)}\\right)} e^{\\left(-2 \\, {\\nu} r\\right)}$ " ] }, "execution_count": 78, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq1 = EE[1,1]/exp(2*nu*r)\n", "eq1" ] }, { "cell_type": "code", "execution_count": 79, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{12} \\, {\\left(24 \\, {\\left({\\nu} + 1\\right)} e^{\\left(2 \\, r\\right)} f\\left(r\\right) + {\\left({\\mu} q^{2} - 4 \\, {\\Lambda}\\right)} e^{\\left(2 \\, r\\right)} + 12 \\, e^{\\left(2 \\, r\\right)} \\frac{\\partial\\,f}{\\partial r}\\right)} e^{\\left(-2 \\, r\\right)}$ " ] }, "execution_count": 79, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq2 = EE[2,2]/exp(2*r)\n", "eq2" ] }, { "cell_type": "code", "execution_count": 80, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, {\\lambda}^{2} {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} {\\lambda}^{2} - \\frac{1}{2} \\, {\\lambda}^{2} \\frac{\\partial^2\\,f}{\\partial r ^ 2} - 2 \\, {\\left({\\lambda}^{2} {\\nu}^{2} + {\\lambda}^{2} + 4\\right)} f\\left(r\\right) - {\\left(2 \\, {\\lambda}^{2} {\\nu} + {\\lambda}^{2}\\right)} \\frac{\\partial\\,f}{\\partial r}$ " ] }, "execution_count": 80, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq3 = EE[4,4]*lamb^2*f(r)\n", "eq3" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The dilaton field equation becomes" ] }, { "cell_type": "code", "execution_count": 81, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scalar field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & \\frac{4 \\, {\\left(2 \\, {\\left({\\nu} + 1\\right)} f\\left(\\log\\left({\\tilde{r}}\\right)\\right) + \\frac{\\partial\\,f}{\\partial \\left( \\log\\left({\\tilde{r}}\\right) \\right)}\\right)}}{{\\lambda}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & \\frac{4 \\, {\\left(2 \\, {\\left({\\nu} + 1\\right)} f\\left(r\\right) + \\frac{\\partial\\,f}{\\partial r}\\right)}}{{\\lambda}} \\end{array}$ " ] }, "execution_count": 81, "metadata": { }, "output_type": "execute_result" } ], "source": [ "nab = g.connection()\n", "box_phi = nab(nab(phi).up(g)).trace()\n", "print box_phi\n", "box_phi.display()" ] }, { "cell_type": "code", "execution_count": 82, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & M & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, x, {y_1}, {y_2}, {\\tilde{r}}\\right) & \\longmapsto & -\\frac{{\\lambda}^{2} {\\mu} q^{2} - 64 \\, {\\left({\\nu} + 1\\right)} f\\left(\\log\\left({\\tilde{r}}\\right)\\right) - 32 \\, \\frac{\\partial\\,f}{\\partial \\left( \\log\\left({\\tilde{r}}\\right) \\right)}}{8 \\, {\\lambda}} \\\\ & \\left(t, x, {y_1}, {y_2}, r\\right) & \\longmapsto & -\\frac{{\\lambda}^{2} {\\mu} q^{2} - 64 \\, {\\left({\\nu} + 1\\right)} f\\left(r\\right) - 32 \\, \\frac{\\partial\\,f}{\\partial r}}{8 \\, {\\lambda}} \\end{array}$ " ] }, "execution_count": 82, "metadata": { }, "output_type": "execute_result" } ], "source": [ "DE = box_phi - lamb/4*exp(lamb*phi) * F2\n", "DE.display()" ] }, { "cell_type": "code", "execution_count": 83, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{8} \\, {\\lambda}^{2} {\\mu} q^{2} + 8 \\, {\\left({\\nu} + 1\\right)} f\\left(r\\right) + 4 \\, \\frac{\\partial\\,f}{\\partial r}$ " ] }, "execution_count": 83, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq4 = DE.coord_function()*lamb\n", "eq4" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "The Maxwell equation is still identically satisfied:" ] }, { "cell_type": "code", "execution_count": 84, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Vector field on the 5-dimensional differentiable manifold M\n" ] }, { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$ " ] }, "execution_count": 84, "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": { "collapsed": false }, "source": [ "### The solution\n", "\n", "The Einstein equation + the dilaton field equation yields a system of 5 equations (eq0, eq1, eq2, eq3, eq4). \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", "$$ f(r) = 1 - m e^{-(2\\nu +2)r}, $$\n", "\n", "where $m$ is a constant. \n", "\n", "To this aim, we declare" ] }, { "cell_type": "code", "execution_count": 85, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}r \\ {\\mapsto}\\ -m e^{\\left(-2 \\, {\\left({\\nu} + 1\\right)} r\\right)} + 1$ " ] }, "execution_count": 85, "metadata": { }, "output_type": "execute_result" } ], "source": [ "var('m', domain='real')\n", "fm(r) = 1 - m*exp(-(2*nu+2)*r)\n", "fm" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "and substitute this function for $f(r)$ in all the equations:" ] }, { "cell_type": "code", "execution_count": 86, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, {\\left(m {\\mu} q^{2} - 48 \\, m {\\nu}^{2} + 8 \\, {\\Lambda} m - 48 \\, m {\\nu} - {\\left({\\mu} q^{2} - 48 \\, {\\nu}^{2} + 8 \\, {\\Lambda} - 48 \\, {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r + 2 \\, r\\right)}\\right)} e^{\\left(-2 \\, {\\nu} r - 2 \\, r\\right)}$ " ] }, "execution_count": 86, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq0m = eq0.expr().substitute_function(f, fm).simplify_full()\n", "eq0m" ] }, { "cell_type": "code", "execution_count": 87, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, m {\\mu} q^{2} - 2 \\, m {\\nu}^{2} + \\frac{1}{3} \\, {\\Lambda} m - 2 \\, m {\\nu} - \\frac{1}{24} \\, {\\left({\\mu} q^{2} - 48 \\, {\\nu}^{2} + 8 \\, {\\Lambda} - 48 \\, {\\nu}\\right)} e^{\\left(2 \\, {\\nu} r + 2 \\, r\\right)}$ " ] }, "execution_count": 87, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq0m = (eq0m * exp(2*nu*r+2*r)).simplify_full()\n", "eq0m" ] }, { "cell_type": "code", "execution_count": 88, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{24} \\, {\\mu} q^{2} - 2 \\, {\\nu}^{2} + \\frac{1}{3} \\, {\\Lambda} - 2 \\, {\\nu}$ " ] }, "execution_count": 88, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq1m = eq1.expr().substitute_function(f, fm).simplify_full()\n", "eq1m" ] }, { "cell_type": "code", "execution_count": 89, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 2 \\, {\\nu} - 2$ " ] }, "execution_count": 89, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq2m = eq2.expr().substitute_function(f, fm).simplify_full()\n", "eq2m" ] }, { "cell_type": "code", "execution_count": 90, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{24} \\, {\\left(48 \\, {\\lambda}^{2} m {\\nu} - 48 \\, {\\left({\\lambda}^{2} + 4\\right)} m - {\\left({\\lambda}^{2} {\\mu} q^{2} - 48 \\, {\\lambda}^{2} {\\nu}^{2} + 8 \\, {\\left({\\Lambda} - 6\\right)} {\\lambda}^{2} - 192\\right)} e^{\\left(2 \\, {\\nu} r + 2 \\, r\\right)}\\right)} e^{\\left(-2 \\, {\\nu} r - 2 \\, r\\right)}$ " ] }, "execution_count": 90, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq3m = eq3.expr().substitute_function(f, fm).simplify_full()\n", "eq3m" ] }, { "cell_type": "code", "execution_count": 91, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-2 \\, {\\lambda}^{2} m {\\nu} + 2 \\, {\\left({\\lambda}^{2} + 4\\right)} m + \\frac{1}{24} \\, {\\left({\\lambda}^{2} {\\mu} q^{2} - 48 \\, {\\lambda}^{2} {\\nu}^{2} + 8 \\, {\\left({\\Lambda} - 6\\right)} {\\lambda}^{2} - 192\\right)} e^{\\left(2 \\, {\\nu} r + 2 \\, r\\right)}$ " ] }, "execution_count": 91, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq3m = (eq3m * exp(2*nu*r+2*r)).simplify_full()\n", "eq3m" ] }, { "cell_type": "code", "execution_count": 92, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{8} \\, {\\lambda}^{2} {\\mu} q^{2} + 8 \\, {\\nu} + 8$ " ] }, "execution_count": 92, "metadata": { }, "output_type": "execute_result" } ], "source": [ "eq4m = eq4.expr().substitute_function(f, fm).simplify_full()\n", "eq4m" ] }, { "cell_type": "code", "execution_count": 93, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "eqs = [eq0m, eq1m, eq2m, eq3m, eq4m]" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Solution for $\\nu = 2$" ] }, { "cell_type": "code", "execution_count": 94, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\frac{1}{24} \\, m {\\mu} q^{2} + \\frac{1}{3} \\, {\\left({\\Lambda} - 36\\right)} m - \\frac{1}{24} \\, {\\left({\\mu} q^{2} + 8 \\, {\\Lambda} - 288\\right)} e^{\\left(6 \\, r\\right)} = 0, \\frac{1}{24} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 12 = 0, -\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 6 = 0, -2 \\, {\\left({\\lambda}^{2} - 4\\right)} m + \\frac{1}{24} \\, {\\left({\\lambda}^{2} {\\mu} q^{2} + 8 \\, {\\left({\\Lambda} - 30\\right)} {\\lambda}^{2} - 192\\right)} e^{\\left(6 \\, r\\right)} = 0, -\\frac{1}{8} \\, {\\lambda}^{2} {\\mu} q^{2} + 24 = 0\\right]$ " ] }, "execution_count": 94, "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": 95, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left[{\\lambda} = 2, {\\mu} = \\frac{48}{r_{5}^{2}}, {\\Lambda} = 30, q = r_{5}, m = r_{6}, r = r_{7}\\right], \\left[{\\lambda} = \\left(-2\\right), {\\mu} = \\frac{48}{r_{8}^{2}}, {\\Lambda} = 30, q = r_{8}, m = r_{9}, r = r_{10}\\right]\\right]$ " ] }, "execution_count": 95, "metadata": { }, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, r)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "In the above solutions, $r_i$, with $i$ an integer, stands for an arbitrary parameter. \n", "We recover the same solution for $\\lambda$, $\\Lambda$ and $\\mu q^2$ as in Sec. 1 (i.e. without any black brane). The value of $m$ can be chosen arbitraly. The solution of Sec. 1 corresponds to $m=0$. " ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Solution for $\\nu=4$" ] }, { "cell_type": "code", "execution_count": 96, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\frac{1}{24} \\, m {\\mu} q^{2} + \\frac{1}{3} \\, {\\left({\\Lambda} - 120\\right)} m - \\frac{1}{24} \\, {\\left({\\mu} q^{2} + 8 \\, {\\Lambda} - 960\\right)} e^{\\left(10 \\, r\\right)} = 0, \\frac{1}{24} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 40 = 0, -\\frac{1}{12} \\, {\\mu} q^{2} + \\frac{1}{3} \\, {\\Lambda} - 10 = 0, -2 \\, {\\left(3 \\, {\\lambda}^{2} - 4\\right)} m + \\frac{1}{24} \\, {\\left({\\lambda}^{2} {\\mu} q^{2} + 8 \\, {\\left({\\Lambda} - 102\\right)} {\\lambda}^{2} - 192\\right)} e^{\\left(10 \\, r\\right)} = 0, -\\frac{1}{8} \\, {\\lambda}^{2} {\\mu} q^{2} + 40 = 0\\right]$ " ] }, "execution_count": 96, "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": 97, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left[{\\lambda} = \\frac{2}{3} \\, \\sqrt{3}, {\\mu} = \\frac{240}{r_{11}^{2}}, {\\Lambda} = 90, q = r_{11}, m = r_{12}, r = r_{13}\\right], \\left[{\\lambda} = -\\frac{2}{3} \\, \\sqrt{3}, {\\mu} = \\frac{240}{r_{14}^{2}}, {\\Lambda} = 90, q = r_{14}, m = r_{15}, r = r_{16}\\right]\\right]$ " ] }, "execution_count": 97, "metadata": { }, "output_type": "execute_result" } ], "source": [ "solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, r)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "In the above solutions, $r_i$, with $i$ an integer, stands for an arbitrary parameter. \n", "We recover the same solution for $\\lambda$, $\\Lambda$ and $\\mu q^2$ as in Sec. 1 (i.e. without any black brane). The value of $m$ can be chosen arbitraly. The solution of Sec. 1 corresponds to $m=0$. " ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ ] } ], "metadata": { "kernelspec": { "display_name": "SageMath (stable)", "language": "sagemath", "metadata": { "cocalc": { "description": "Open-source mathematical software system", "priority": 10, "url": "https://www.sagemath.org/" } }, "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": "2.7.15" }, "name": "Lifshitz_black_brane.ipynb" }, "nbformat": 4, "nbformat_minor": 0 }