{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Black branes in Lifshitz-like spacetimes"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"This Jupyter/SageMath 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://github.com/sagemanifolds/SageManifolds/raw/master/Worksheets/v0.9/SM_black_brane.ipynb)."
]
},
{
"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": [
"### Space manifold\n",
"\n",
"Let us declare $M$ 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 coordinate system on $M$:"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"Chart (M, (t, x, y1, y2, r))"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"X. = M.chart('t x y1:y_1 y2:y_2 r')\n",
"X"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Next, we define the metric tensor, which depends on some real number $\\nu$ and some arbitary function $f$:"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"g = -e^(2*nu*r)*f(r) dt*dt + e^(2*nu*r) dx*dx + e^(2*r) dy1*dy1 + e^(2*r) dy2*dy2 + 1/f(r) dr*dr"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"g = M.lorentzian_metric('g')\n",
"var('nu', latex_name=r'\\nu', domain='real')\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": {},
"source": [
"If $f(r)=1$, this is the metric of a Lifshitz spacetime; if, in addition $\\nu=1$, $(M,g)$ is a Poincaré patch of $\\mathrm{AdS}_5$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Curvature\n",
"\n",
"The Riemann tensor is"
]
},
{
"cell_type": "code",
"execution_count": 5,
"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": 6,
"metadata": {
"collapsed": false,
"scrolled": true
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"Riem(g)^t_x,t,x = -nu^2*e^(2*nu*r)*f(r) - 1/2*nu*e^(2*nu*r)*d(f)/dr \n",
"Riem(g)^t_y1,t,y1 = -nu*e^(2*r)*f(r) - 1/2*e^(2*r)*d(f)/dr \n",
"Riem(g)^t_y2,t,y2 = -nu*e^(2*r)*f(r) - 1/2*e^(2*r)*d(f)/dr \n",
"Riem(g)^t_r,t,r = -1/2*(2*nu^2*f(r) + 3*nu*d(f)/dr + d^2(f)/dr^2)/f(r) \n",
"Riem(g)^x_t,t,x = -nu^2*e^(2*nu*r)*f(r)^2 - 1/2*nu*e^(2*nu*r)*f(r)*d(f)/dr \n",
"Riem(g)^x_y1,x,y1 = -nu*e^(2*r)*f(r) \n",
"Riem(g)^x_y2,x,y2 = -nu*e^(2*r)*f(r) \n",
"Riem(g)^x_r,x,r = -1/2*(2*nu^2*f(r) + nu*d(f)/dr)/f(r) \n",
"Riem(g)^y1_t,t,y1 = -nu*e^(2*nu*r)*f(r)^2 - 1/2*e^(2*nu*r)*f(r)*d(f)/dr \n",
"Riem(g)^y1_x,x,y1 = nu*e^(2*nu*r)*f(r) \n",
"Riem(g)^y1_y2,y1,y2 = -e^(2*r)*f(r) \n",
"Riem(g)^y1_r,y1,r = -1/2*(2*f(r) + d(f)/dr)/f(r) \n",
"Riem(g)^y2_t,t,y2 = -nu*e^(2*nu*r)*f(r)^2 - 1/2*e^(2*nu*r)*f(r)*d(f)/dr \n",
"Riem(g)^y2_x,x,y2 = nu*e^(2*nu*r)*f(r) \n",
"Riem(g)^y2_y1,y1,y2 = e^(2*r)*f(r) \n",
"Riem(g)^y2_r,y2,r = -1/2*(2*f(r) + d(f)/dr)/f(r) \n",
"Riem(g)^r_t,t,r = -nu^2*e^(2*nu*r)*f(r)^2 - 3/2*nu*e^(2*nu*r)*f(r)*d(f)/dr - 1/2*e^(2*nu*r)*f(r)*d^2(f)/dr^2 \n",
"Riem(g)^r_x,x,r = nu^2*e^(2*nu*r)*f(r) + 1/2*nu*e^(2*nu*r)*d(f)/dr \n",
"Riem(g)^r_y1,y1,r = e^(2*r)*f(r) + 1/2*e^(2*r)*d(f)/dr \n",
"Riem(g)^r_y2,y2,r = e^(2*r)*f(r) + 1/2*e^(2*r)*d(f)/dr "
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Riem.display_comp(only_nonredundant=True)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The Ricci tensor:"
]
},
{
"cell_type": "code",
"execution_count": 7,
"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": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"Ric(g) = (2*(nu^2 + nu)*e^(2*nu*r)*f(r)^2 + (2*nu + 1)*e^(2*nu*r)*f(r)*d(f)/dr + 1/2*e^(2*nu*r)*f(r)*d^2(f)/dr^2) dt*dt + (-2*(nu^2 + nu)*e^(2*nu*r)*f(r) - nu*e^(2*nu*r)*d(f)/dr) dx*dx + (-2*(nu + 1)*e^(2*r)*f(r) - e^(2*r)*d(f)/dr) dy1*dy1 + (-2*(nu + 1)*e^(2*r)*f(r) - e^(2*r)*d(f)/dr) dy2*dy2 - 1/2*(4*(nu^2 + 1)*f(r) + 2*(2*nu + 1)*d(f)/dr + d^2(f)/dr^2)/f(r) dr*dr"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Ric.display()"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"Ric(g)_t,t = 2*(nu^2 + nu)*e^(2*nu*r)*f(r)^2 + (2*nu + 1)*e^(2*nu*r)*f(r)*d(f)/dr + 1/2*e^(2*nu*r)*f(r)*d^2(f)/dr^2 \n",
"Ric(g)_x,x = -2*(nu^2 + nu)*e^(2*nu*r)*f(r) - nu*e^(2*nu*r)*d(f)/dr \n",
"Ric(g)_y1,y1 = -2*(nu + 1)*e^(2*r)*f(r) - e^(2*r)*d(f)/dr \n",
"Ric(g)_y2,y2 = -2*(nu + 1)*e^(2*r)*f(r) - e^(2*r)*d(f)/dr \n",
"Ric(g)_r,r = -1/2*(4*(nu^2 + 1)*f(r) + 2*(2*nu + 1)*d(f)/dr + d^2(f)/dr^2)/f(r) "
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Ric.display_comp()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The Ricci scalar:"
]
},
{
"cell_type": "code",
"execution_count": 10,
"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": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"r(g): M --> R\n",
" (t, x, y1, y2, r) |--> -2*(3*nu^2 + 4*nu + 3)*f(r) - (5*nu + 4)*d(f)/dr - d^2(f)/dr^2"
]
},
"execution_count": 11,
"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 cosmological constant $\\bar{\\Lambda} = -\\Lambda/2$ with $\\Lambda>0$, 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 $g$ 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",
"$$ \\phi = \\frac{1}{\\lambda} \\left( 4 r + \\ln\\mu \\right) \\iff e^{\\lambda\\phi} = \\mu e^{4r},$$\n",
"where $\\mu$ is a constant. "
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"phi: M --> R\n",
" (t, x, y1, y2, r) |--> (4*r + log(mu))/lamb"
]
},
"execution_count": 12,
"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": {},
"source": [
"The 1-form $\\mathrm{d}\\phi$ is"
]
},
{
"cell_type": "code",
"execution_count": 13,
"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": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"dphi = 4/lamb dr"
]
},
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"dphi.display()"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"[0, 0, 0, 0, 4/lamb]"
]
},
"execution_count": 15,
"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",
"Let us first get the 1-forms $\\mathrm{d}y_1$ and $\\mathrm{d}y_2$:"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"Coordinate coframe (M, (dt,dx,dy1,dy2,dr))"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"X.coframe()"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"dy1 = X.coframe()[2]\n",
"dy2 = X.coframe()[3]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Then we can form $F$ according to the above ansatz:"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"2-form F on the 5-dimensional differentiable manifold M\n"
]
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"F = 1/2*q dy1/\\dy2"
]
},
"execution_count": 18,
"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": 19,
"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_der())"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"dF = 0"
]
},
"execution_count": 20,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F.exterior_der().display()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let us evaluate the square $F_{mn} F^{mn}$ of $F$:"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Tensor field of type (2,0) on the 5-dimensional differentiable manifold M\n"
]
}
],
"source": [
"Fu = F.up(g)\n",
"print(Fu)"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Scalar field on the 5-dimensional differentiable manifold M\n"
]
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"M --> R\n",
"(t, x, y1, y2, r) |--> 1/2*q^2*e^(-4*r)"
]
},
"execution_count": 22,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F2 = F['_{mn}']*Fu['^{mn}'] # using LaTeX notations for 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": 23,
"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": [
""
],
"text/plain": [
"1/4*q^2*e^(-2*r) dy1*dy1 + 1/4*q^2*e^(-2*r) dy2*dy2"
]
},
"execution_count": 23,
"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": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"True"
]
},
"execution_count": 24,
"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": 25,
"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": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"Lamb"
]
},
"execution_count": 26,
"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": 27,
"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": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"E_t,t = 2*(nu^2 + nu)*e^(2*nu*r)*f(r)^2 + (2*nu + 1)*e^(2*nu*r)*f(r)*d(f)/dr - 1/24*(mu*q^2 + 8*Lamb)*e^(2*nu*r)*f(r) + 1/2*e^(2*nu*r)*f(r)*d^2(f)/dr^2 \n",
"E_x,x = -2*(nu^2 + nu)*e^(2*nu*r)*f(r) - nu*e^(2*nu*r)*d(f)/dr + 1/24*(mu*q^2 + 8*Lamb)*e^(2*nu*r) \n",
"E_y1,y1 = -2*(nu + 1)*e^(2*r)*f(r) - 1/12*(mu*q^2 - 4*Lamb)*e^(2*r) - e^(2*r)*d(f)/dr \n",
"E_y2,y2 = -2*(nu + 1)*e^(2*r)*f(r) - 1/12*(mu*q^2 - 4*Lamb)*e^(2*r) - e^(2*r)*d(f)/dr \n",
"E_r,r = 1/24*(lamb^2*mu*q^2 + 8*Lamb*lamb^2 - 12*lamb^2*d^2(f)/dr^2 - 48*(lamb^2*nu^2 + lamb^2 + 4)*f(r) - 24*(2*lamb^2*nu + lamb^2)*d(f)/dr)/(lamb^2*f(r)) "
]
},
"execution_count": 28,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"EE.display_comp(only_nonredundant=True)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We note that `EE==0` leads to only 4 independent equations:"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"2*(nu^2 + nu)*f(r)^2 + (2*nu + 1)*f(r)*d(f)/dr - 1/24*(mu*q^2 + 8*Lamb)*f(r) + 1/2*f(r)*d^2(f)/dr^2"
]
},
"execution_count": 29,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq0 = EE[0,0]/exp(2*nu*r)\n",
"eq0"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*mu*q^2 - 2*(nu^2 + nu)*f(r) - nu*d(f)/dr + 1/3*Lamb"
]
},
"execution_count": 30,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq1 = EE[1,1]/exp(2*nu*r)\n",
"eq1"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/12*mu*q^2 - 2*(nu + 1)*f(r) + 1/3*Lamb - d(f)/dr"
]
},
"execution_count": 31,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq2 = EE[2,2]/exp(2*r)\n",
"eq2"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*lamb^2*mu*q^2 + 1/3*Lamb*lamb^2 - 1/2*lamb^2*d^2(f)/dr^2 - 2*(lamb^2*nu^2 + lamb^2 + 4)*f(r) - (2*lamb^2*nu + lamb^2)*d(f)/dr"
]
},
"execution_count": 32,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq3 = EE[4,4]*lamb^2*f(r)\n",
"eq3"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Dilaton field equation\n",
"\n",
"First we evaluate $\\nabla_m \\nabla^m \\phi$:"
]
},
{
"cell_type": "code",
"execution_count": 33,
"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"
]
}
],
"source": [
"nab = g.connection()\n",
"print(nab)"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Scalar field on the 5-dimensional differentiable manifold M\n"
]
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"M --> R\n",
"(t, x, y1, y2, r) |--> 4*(2*(nu + 1)*f(r) + d(f)/dr)/lamb"
]
},
"execution_count": 34,
"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": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Scalar field on the 5-dimensional differentiable manifold M\n"
]
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"M --> R\n",
"(t, x, y1, y2, r) |--> -1/8*(lamb^2*mu*q^2 - 64*(nu + 1)*f(r) - 32*d(f)/dr)/lamb"
]
},
"execution_count": 35,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DE = box_phi - lamb/4*exp(lamb*phi) * F2\n",
"print(DE)\n",
"DE.display()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Hence the dilaton field equation provides a fourth equation:"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/8*lamb^2*mu*q^2 + 8*(nu + 1)*f(r) + 4*d(f)/dr"
]
},
"execution_count": 36,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq4 = DE.coord_function()*lamb\n",
"eq4"
]
},
{
"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": 37,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Vector field on the 5-dimensional differentiable manifold M\n"
]
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"0"
]
},
"execution_count": 37,
"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; indeed the tensor $\\nabla_p (e^{\\lambda\\phi} F^{mn})$ has a vanishing trace, as we can check:"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"mu*q d/dy1*d/dy2*dr - 1/2*mu*q*e^(2*r)*f(r) d/dy1*d/dr*dy2 - mu*q d/dy2*d/dy1*dr + 1/2*mu*q*e^(2*r)*f(r) d/dy2*d/dr*dy1 + 1/2*mu*q*e^(2*r)*f(r) d/dr*d/dy1*dy2 - 1/2*mu*q*e^(2*r)*f(r) d/dr*d/dy2*dy1"
]
},
"execution_count": 38,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"nab(exp(lamb*phi)*Fu).display()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Solving the field equations\n",
"\n",
"The system to solve is"
]
},
{
"cell_type": "code",
"execution_count": 39,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"2*(nu^2 + nu)*f(r)^2 + (2*nu + 1)*f(r)*d(f)/dr - 1/24*(mu*q^2 + 8*Lamb)*f(r) + 1/2*f(r)*d^2(f)/dr^2 ' = 0'"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*mu*q^2 - 2*(nu^2 + nu)*f(r) - nu*d(f)/dr + 1/3*Lamb ' = 0'"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/12*mu*q^2 - 2*(nu + 1)*f(r) + 1/3*Lamb - d(f)/dr ' = 0'"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*lamb^2*mu*q^2 + 1/3*Lamb*lamb^2 - 1/2*lamb^2*d^2(f)/dr^2 - 2*(lamb^2*nu^2 + lamb^2 + 4)*f(r) - (2*lamb^2*nu + lamb^2)*d(f)/dr ' = 0'"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/8*lamb^2*mu*q^2 + 8*(nu + 1)*f(r) + 4*d(f)/dr ' = 0'"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"eqs = [eq0, eq1, eq2, eq3, eq4]\n",
"for eq in eqs:\n",
" pretty_print(eq, ' = 0')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let us solve `eq1` for $f(r)$:"
]
},
{
"cell_type": "code",
"execution_count": 40,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"_C*e^(-2*(nu + 1)*r) + 1/48*mu*q^2/((nu + 1)*nu) + 1/6*Lamb/((nu + 1)*nu)"
]
},
"execution_count": 40,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"sol_f = desolve(eq1.expr() == 0, f(r), ivar=r)\n",
"sol_f.expand()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Hence, up to some rescaling the solution is of the type\n",
"\n",
"$$ f(r) = 1 - m e^{-(2\\nu +2)r}, $$\n",
"\n",
"where $m$ is a constant. Hence we declare"
]
},
{
"cell_type": "code",
"execution_count": 41,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"r |--> -m*e^(-2*(nu + 1)*r) + 1"
]
},
"execution_count": 41,
"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": {},
"source": [
"and substitute this function for $f(r)$ in all the equations:"
]
},
{
"cell_type": "code",
"execution_count": 42,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*(m*mu*q^2 - 48*m*nu^2 + 8*Lamb*m - 48*m*nu - (mu*q^2 - 48*nu^2 + 8*Lamb - 48*nu)*e^(2*nu*r + 2*r))*e^(-2*nu*r - 2*r)"
]
},
"execution_count": 42,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq0m = eq0.expr().substitute_function(f, fm).simplify_full()\n",
"eq0m"
]
},
{
"cell_type": "code",
"execution_count": 43,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*m*mu*q^2 - 2*m*nu^2 + 1/3*Lamb*m - 2*m*nu - 1/24*(mu*q^2 - 48*nu^2 + 8*Lamb - 48*nu)*e^(2*nu*r + 2*r)"
]
},
"execution_count": 43,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq0m = (eq0m * exp(2*nu*r+2*r)).simplify_full()\n",
"eq0m"
]
},
{
"cell_type": "code",
"execution_count": 44,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"1/24*mu*q^2 - 2*nu^2 + 1/3*Lamb - 2*nu"
]
},
"execution_count": 44,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq1m = eq1.expr().substitute_function(f, fm).simplify_full()\n",
"eq1m"
]
},
{
"cell_type": "code",
"execution_count": 45,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/12*mu*q^2 + 1/3*Lamb - 2*nu - 2"
]
},
"execution_count": 45,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq2m = eq2.expr().substitute_function(f, fm).simplify_full()\n",
"eq2m"
]
},
{
"cell_type": "code",
"execution_count": 46,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/24*(48*lamb^2*m*nu - 48*(lamb^2 + 4)*m - (lamb^2*mu*q^2 - 48*lamb^2*nu^2 + 8*(Lamb - 6)*lamb^2 - 192)*e^(2*nu*r + 2*r))*e^(-2*nu*r - 2*r)"
]
},
"execution_count": 46,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq3m = eq3.expr().substitute_function(f, fm).simplify_full()\n",
"eq3m"
]
},
{
"cell_type": "code",
"execution_count": 47,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"-2*lamb^2*m*nu + 2*(lamb^2 + 4)*m + 1/24*(lamb^2*mu*q^2 - 48*lamb^2*nu^2 + 8*(Lamb - 6)*lamb^2 - 192)*e^(2*nu*r + 2*r)"
]
},
"execution_count": 47,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq3m = (eq3m * exp(2*nu*r+2*r)).simplify_full()\n",
"eq3m"
]
},
{
"cell_type": "code",
"execution_count": 48,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"-1/8*lamb^2*mu*q^2 + 8*nu + 8"
]
},
"execution_count": 48,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"eq4m = eq4.expr().substitute_function(f, fm).simplify_full()\n",
"eq4m"
]
},
{
"cell_type": "code",
"execution_count": 49,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"eqs = [eq0m, eq1m, eq2m, eq3m, eq4m]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Solution for $\\nu = 2$"
]
},
{
"cell_type": "code",
"execution_count": 50,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"[1/24*m*mu*q^2 + 1/3*(Lamb - 36)*m - 1/24*(mu*q^2 + 8*Lamb - 288)*e^(6*r) == 0,\n",
" 1/24*mu*q^2 + 1/3*Lamb - 12 == 0,\n",
" -1/12*mu*q^2 + 1/3*Lamb - 6 == 0,\n",
" -2*(lamb^2 - 4)*m + 1/24*(lamb^2*mu*q^2 + 8*(Lamb - 30)*lamb^2 - 192)*e^(6*r) == 0,\n",
" -1/8*lamb^2*mu*q^2 + 24 == 0]"
]
},
"execution_count": 50,
"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": 51,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"[[lamb == 2, mu == 48/r1^2, Lamb == 30, q == r1, m == r2, r == r3], [lamb == -2, mu == 48/r4^2, Lamb == 30, q == r4, m == r5, r == r6]]"
]
},
"execution_count": 51,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, r)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In the above solutions, $r_i$, with $i$ an integer, stands for an arbitrary parameter. In particular, we notice that $\\mu$ and $q$ are related by $\\mu q^2 = 48$ and that the value of $m$ can be chosen arbitrarily."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Solution for $\\nu=4$"
]
},
{
"cell_type": "code",
"execution_count": 52,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"[1/24*m*mu*q^2 + 1/3*(Lamb - 120)*m - 1/24*(mu*q^2 + 8*Lamb - 960)*e^(10*r) == 0,\n",
" 1/24*mu*q^2 + 1/3*Lamb - 40 == 0,\n",
" -1/12*mu*q^2 + 1/3*Lamb - 10 == 0,\n",
" -2*(3*lamb^2 - 4)*m + 1/24*(lamb^2*mu*q^2 + 8*(Lamb - 102)*lamb^2 - 192)*e^(10*r) == 0,\n",
" -1/8*lamb^2*mu*q^2 + 40 == 0]"
]
},
"execution_count": 52,
"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": 53,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/html": [
""
],
"text/plain": [
"[[lamb == 2/3*sqrt(3), mu == 240/r7^2, Lamb == 90, q == r7, m == r8, r == r9], [lamb == -2/3*sqrt(3), mu == 240/r10^2, Lamb == 90, q == r10, m == r11, r == r12]]"
]
},
"execution_count": 53,
"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": [
"As above, $r_i$, with $i$ an integer, stands for an arbitrary parameter. The constants $\\mu$ and $q$ are now related by $\\mu q^2 = 240$ and the value of $m$ is still arbitrary."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Sage 6.9",
"language": "",
"name": "sage_6_9"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
},
"name": "Lifshitz_black_brane.ipynb"
},
"nbformat": 4,
"nbformat_minor": 0
}