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Jupyter notebook Lifshitz_black_brane.ipynb

Project: Lifshitz BH

Path: Lifshitz_black_brane.ipynb

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Kernel: SageMath (stable)

Black branes in Lifshitz-like spacetimes

This Jupyter/SageMath worksheet implements some computations of the article

I. Ya. Aref'eva, A. A. Golubtsova & E. Gourgoulhon: Analytic black branes in Lifshitz-like backgrounds and thermalization, arXiv:1601.06046

These computations are based on SageManifolds (v0.9)

The worksheet file (ipynb format) can be downloaded from here.

Content

Five-dimensional Lifschitz-like spacetime
Black brane solution

First we set up the notebook to display mathematical objects using LaTeX formatting:

In [1]:

%display latex

1. Five-dimensional Lifshitz-like spacetime

Let us declare the spacetime $M$ as a 5-dimensional manifold:

In [2]:

M = Manifold(5, 'M')
print M

5-dimensional differentiable manifold M

We introduce a first coordinate system on $M$ :

In [3]:

X0.<t,x,y1,y2,R> = M.chart(r't x y1:y_1 y2:y_2 R:\tilde{r}:(0,+oo)')
X0

Let us consider the following Lifshitz-symmetric metric, parametrized by some real number $\nu$ :

In [0]:

In [0]:

In [4]:

g = M.lorentzian_metric('g')
var('nu', latex_name=r'\nu', domain='real')
g[0,0] = -R^(2*nu)
g[1,1] = R^(2*nu)
g[2,2] = R^2
g[3,3] = R^2
g[4,4] = 1/R^2
g.display()

A matrix view of the metric components:

In [5]:

g[:]

This metric is invariant under the Lifshitz scaling $(t,x,y_1,y_2,\tilde r) \longmapsto \left(\lambda^\nu t, \lambda^\nu x, \lambda y_1, \lambda y_2, \frac{\tilde r}{\lambda} \right)$

If $\nu=1$ the scaling is isotropic and we recognize the metric of $\mathrm{AdS}_5$ in Poincaré coordinates ( $M$ is then the Poincaré patch of $\mathrm{AdS}_5$ )
If $\nu\not=1$ , the scaling is anisotropic

Let us introduce a second coordinate system on $M$ :

In [6]:

X.<t,x,y1,y2,r> = M.chart('t x y1:y_1 y2:y_2 r')
X

and relate it to the previous one by the transformation $r=\ln\tilde r$ :

In [7]:

X0_to_X = X0.transition_map(X, [t, x, y1, y2, ln(R)])
X0_to_X.display()

The inverse coordinate transition is computed by means of the method inverse():

In [8]:

X_to_X0 = X0_to_X.inverse()
X_to_X0.display()

At this stage, the manifold's atlas defined by the user is

In [9]:

M.atlas()

and the list of defined vector frames defined is

In [10]:

M.frames()

The expression of the metric in terms of the new coordinates is

In [11]:

g.display(X.frame(), X)

or, in matrix view:

In [12]:

g[X.frame(),:,X]

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:

In [13]:

g[X.frame(),0,0,X]

In [14]:

g[X.frame(),0,0]  # the default chart is used

From now on, let us consider the coordinates $X = (t,x,y_1,y_2,r)$ as the default ones on the manifold $M$ :

In [15]:

M.set_default_chart(X)
M.set_default_frame(X.frame())

Then

In [16]:

g.display()

In [17]:

g[:]

In [18]:

g[0,0]

In [19]:

g.display_comp()

Curvature

The Riemann tensor is

In [20]:

Riem = g.riemann()
print Riem

Tensor field Riem(g) of type (1,3) on the 5-dimensional differentiable manifold M

In [21]:

Riem.display_comp(only_nonredundant=True)

The Ricci tensor:

In [22]:

Ric = g.ricci()
print Ric

Field of symmetric bilinear forms Ric(g) on the 5-dimensional differentiable manifold M

In [23]:

Ric.display()

In [24]:

Ric.display_comp()

The Ricci scalar:

In [25]:

Rscal = g.ricci_scalar()
print Rscal

Scalar field r(g) on the 5-dimensional differentiable manifold M

In [26]:

Rscal.display()

We note that the Ricci scalar is constant.

Source model

Let us consider a model based on the following action, involving a dilaton scalar field $\phi$ and a Maxwell 2-form $F$ :

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)}

where $R(g)$ is the Ricci scalar of metric $g$ , $\Lambda$ is the cosmological constant and $\lambda$ is the dilatonic coupling constant.

The dilaton scalar field

We consider the following ansatz for the dilaton scalar field $\phi$ : $\phi = \frac{1}{\lambda} \left( 4 r + \ln\mu \right),$ where $\mu$ is a constant.

In [27]:

var('mu', latex_name=r'\mu', domain='real')
var('lamb', latex_name=r'\lambda', domain='real')
phi = M.scalar_field({X: (4*r + ln(mu))/lamb}, 
                     name='phi', latex_name=r'\phi')
phi.display()

The 1-form $\mathrm{d}\phi$ is

In [28]:

dphi = phi.differential()
print dphi

1-form dphi on the 5-dimensional differentiable manifold M

In [29]:

dphi.display()

In [30]:

dphi[:]  # all the components in the default frame

The 2-form field

We consider the following ansatz for $F$ : $F = \frac{1}{2} q \, \mathrm{d}y_1\wedge \mathrm{d}y_2,$ where $q$ is a constant.

Let us first get the 1-forms $\mathrm{d}y_1$ and $\mathrm{d}y_2$ :

In [31]:

X.coframe()

In [32]:

dy1 = X.coframe()[2]
dy2 = X.coframe()[3]
print dy1
print dy2
dy1, dy2

1-form dy1 on the 5-dimensional differentiable manifold M
1-form dy2 on the 5-dimensional differentiable manifold M

Then we can form $F$ according to the above ansatz:

In [33]:

var('q', domain='real')
F = q/2 * dy1.wedge(dy2)
F.set_name('F')
print F
F.display()

2-form F on the 5-dimensional differentiable manifold M

By construction, the 2-form $F$ is closed (since $q$ is constant):

In [34]:

print(F.exterior_derivative())

3-form dF on the 5-dimensional differentiable manifold M

In [35]:

F.exterior_derivative().display()

Let us evaluate the square $F_{mn} F^{mn}$ of $F$ :

In [36]:

Fu = F.up(g)
print Fu
Fu.display()

Tensor field of type (2,0) on the 5-dimensional differentiable manifold M

In [37]:

F2 = F['_{mn}']*Fu['^{mn}']  # using LaTeX notations to denote contraction
print F2
F2.display()

Scalar field on the 5-dimensional differentiable manifold M

We shall also need the tensor $\mathcal{F}_{mn} := F_{mp} F_n^{\ \, p}$ :

In [38]:

FF = F['_mp'] * F.up(g,1)['^p_n']
print FF
FF.display()

Tensor field of type (0,2) on the 5-dimensional differentiable manifold M

The tensor field $\mathcal{F}$ is symmetric:

In [39]:

FF == FF.symmetrize()

Therefore, from now on, we set

In [40]:

FF = FF.symmetrize()

Einstein equation

Let us first introduce the cosmological constant:

In [41]:

var('Lamb', latex_name=r'\Lambda', domain='real')

From the action (1), the field equation for the metric $g$ is $R_{mn} + \frac{\Lambda}{3} \, g - \frac{1}{2}\partial_m\phi \partial_n\phi -\frac{1}{2} e^{\lambda\phi} F_{mp} F^{\ \, p}_n + \frac{1}{12} e^{\lambda\phi} F_{rs} F^{rs} \, g_{mn} = 0$ We write it as

EE == 0

with EE defined by

In [42]:

EE = Ric + Lamb/3*g - 1/2* (dphi*dphi) -  1/2*exp(lamb*phi)*FF \
     + 1/12*exp(lamb*phi)*F2*g
EE.set_name('E')
print EE

Field of symmetric bilinear forms E on the 5-dimensional differentiable manifold M

In [43]:

EE.display_comp(only_nonredundant=True)

We note that EE==0 leads to only 3 independent equations:

In [44]:

eq1 = (EE[0,0]/exp(2*nu*r)).expr()
eq1

In [45]:

eq2 = (EE[2,2]/exp(2*r)).expr()
eq2

In [46]:

eq3 = EE[4,4].expr().expand()
eq3

Dilaton field equation

First we evaluate $\nabla_m \nabla^m \phi$ :

In [47]:

nab = g.connection()
print nab
nab

Levi-Civita connection nabla_g associated with the Lorentzian metric g on the 5-dimensional differentiable manifold M

In [48]:

box_phi = nab(nab(phi).up(g)).trace()
print box_phi
box_phi.display()

Scalar field on the 5-dimensional differentiable manifold M

From the action (1), the field equation for $\phi$ is $\nabla_m \nabla^m \phi = \frac{\lambda}{4} e^{\lambda\phi} F_{mn} F^{mn}$ We write it as

DE == 0

with DE defined by

In [49]:

DE = box_phi - lamb/4*exp(lamb*phi) * F2
print DE

Scalar field on the 5-dimensional differentiable manifold M

In [50]:

DE.display()

Hence the dilaton field equation provides a fourth equation:

In [51]:

eq4 = DE.expr().expand()
eq4

Maxwell equation

From the action (1), the field equation for $F$ is $\nabla_m \left( e^{\lambda\phi} F^{mn} \right)= 0$ We write it as

ME == 0

with ME defined by

In [52]:

ME = nab(exp(lamb*phi)*Fu).trace(0,2)
print ME
ME.display()

Vector field on the 5-dimensional differentiable manifold M

We get identically zero; indeed the tensor $\nabla_p (e^{\lambda\phi} F^{mn})$ has a vanishing trace, as we can check:

In [53]:

nab(exp(lamb*phi)*Fu).display()

Summary

We have 4 equations involving the constants $\lambda$ , $\mu$ , $\nu$ , $q$ and $\Lambda$ :

In [54]:

eq1 == 0

In [55]:

eq2 == 0

In [56]:

eq3 == 0

In [57]:

eq4 == 0

Solution for $\nu=1$ ( $\mathrm{AdS}_5$ )

In [58]:

eqs = [eq1, eq2, eq3, eq4]
neqs = [eq.subs(nu=1) for eq in eqs]

In [59]:

[eq == 0 for eq in neqs]

In [60]:

solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q)

Hence there is no solution for $\mathrm{AdS}_5$ with the above ansatz.

Solution for $\nu = 2$

In [61]:

neqs = [eq.subs(nu=2) for eq in eqs]
[eq == 0 for eq in neqs]

In [62]:

solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q)

Hence there are two families of solutions, each famility being parametrized by e.g. $q$ . Indeed, in the above writing, $r_1$ and $r_2$ stand for arbitrary parameters (nothing to do with the coordinate $r$ ).

Solution for $\nu = 4$

In [63]:

neqs = [eq.subs(nu=4) for eq in eqs]
[eq == 0 for eq in neqs]

In [64]:

solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q)

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 $r$ ).

2. Black brane solution

We add a blackening factor $f(r)$ to the metric; i.e. we declare a new metric $g$ according to

In [65]:

g = M.lorentzian_metric('g')
ff = function('f')(r)
g[0,0] = -ff*exp(2*nu*r)
g[1,1] = exp(2*nu*r)
g[2,2] = exp(2*r)
g[3,3] = exp(2*r)
g[4,4] = 1/ff
g.display()

The Ricci tensor of $g$ is

In [66]:

Ric = g.ricci()
print Ric

Field of symmetric bilinear forms Ric(g) on the 5-dimensional differentiable manifold M

In [67]:

Ric.display_comp()

By construction, the 2-form $F$ does not depend on $g$ ; hence there is no need to reevaluate it:

In [68]:

F.display()

On the contrary, we need to reevaluate its metric dual, in order to compute $F^2 := F_{mn} F^{mn}$ :

In [69]:

Fu = F.up(g)
print Fu
Fu.display()

Tensor field of type (2,0) on the 5-dimensional differentiable manifold M

In [70]:

F2 = F['_{mn}']*Fu['^{mn}']  # using LaTeX notations to denote contraction
print F2
F2.display()

Scalar field on the 5-dimensional differentiable manifold M

Simlarly, we need to reevaluate $\mathcal{F}_{mn} := F_{mp} F_n^{\ \, p}$ :

In [71]:

FF = F['_mp'] * F.up(g,1)['^p_n']
print FF
FF.display()

Tensor field of type (0,2) on the 5-dimensional differentiable manifold M

In [72]:

FF == FF.symmetrize()

In [73]:

FF = FF.symmetrize()

The new Einstein equation is

In [74]:

EE = Ric + Lamb/3*g - 1/2* (dphi*dphi) -  1/2*exp(lamb*phi)*FF \
     + 1/12*exp(lamb*phi)*F2*g
EE.set_name('E')
print EE

Field of symmetric bilinear forms E on the 5-dimensional differentiable manifold M

In [75]:

EE.display_comp(only_nonredundant=True)

In [76]:

EE[2,2] == EE[3,3]

There are 4 independent components:

In [77]:

eq0 = EE[0,0]/exp(2*nu*r)
eq0

In [78]:

eq1 = EE[1,1]/exp(2*nu*r)
eq1

In [79]:

eq2 = EE[2,2]/exp(2*r)
eq2

In [80]:

eq3 = EE[4,4]*lamb^2*f(r)
eq3

The dilaton field equation becomes

In [81]:

nab = g.connection()
box_phi = nab(nab(phi).up(g)).trace()
print box_phi
box_phi.display()

Scalar field on the 5-dimensional differentiable manifold M

In [82]:

DE = box_phi - lamb/4*exp(lamb*phi) * F2
DE.display()

In [83]:

eq4 = DE.coord_function()*lamb
eq4

The Maxwell equation is still identically satisfied:

In [84]:

ME = nab(exp(lamb*phi)*Fu).trace(0,2)
print ME
ME.display()

Vector field on the 5-dimensional differentiable manifold M

The solution

The Einstein equation + the dilaton field equation yields a system of 5 equations (eq0, eq1, eq2, eq3, eq4).

Let us show that a solution is obtained for $\nu=2$ and $\nu=4$ with the following specific form of the blackening function:

f(r) = 1 - m e^{-(2\nu +2)r},

where $m$ is a constant.

To this aim, we declare

In [85]:

var('m', domain='real')
fm(r) = 1 - m*exp(-(2*nu+2)*r)
fm

and substitute this function for $f(r)$ in all the equations:

In [86]:

eq0m = eq0.expr().substitute_function(f, fm).simplify_full()
eq0m

In [87]:

eq0m = (eq0m * exp(2*nu*r+2*r)).simplify_full()
eq0m

In [88]:

eq1m = eq1.expr().substitute_function(f, fm).simplify_full()
eq1m

In [89]:

eq2m = eq2.expr().substitute_function(f, fm).simplify_full()
eq2m

In [90]:

eq3m = eq3.expr().substitute_function(f, fm).simplify_full()
eq3m

In [91]:

eq3m = (eq3m * exp(2*nu*r+2*r)).simplify_full()
eq3m

In [92]:

eq4m = eq4.expr().substitute_function(f, fm).simplify_full()
eq4m

In [93]:

eqs = [eq0m, eq1m, eq2m, eq3m, eq4m]

Solution for $\nu = 2$

In [94]:

neqs = [eq.subs(nu=2).simplify_full() for eq in eqs]
[eq == 0 for eq in neqs]

In [95]:

solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, r)

In the above solutions, $r_i$ , with $i$ an integer, stands for an arbitrary parameter. 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 $m=0$ .

Solution for $\nu=4$

In [96]:

neqs = [eq.subs(nu=4).simplify_full() for eq in eqs]
[eq == 0 for eq in neqs]

In [97]:

solve([eq == 0 for eq in neqs], lamb, mu, Lamb, q, m, r)

In [0]:

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Black branes in Lifshitz-like spacetimes

Content

1. Five-dimensional Lifshitz-like spacetime

Curvature

Source model

The dilaton scalar field

The 2-form field

Einstein equation

Dilaton field equation

Maxwell equation

Summary

Solution for $\nu=1$ ( $\mathrm{AdS}_5$ )

Solution for $\nu = 2$

Solution for $\nu = 4$

2. Black brane solution

The solution

Solution for $\nu = 2$

Solution for $\nu=4$

Product

Resources

Company

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Black branes in Lifshitz-like spacetimes

Content

1. Five-dimensional Lifshitz-like spacetime

Curvature

Source model

The dilaton scalar field

The 2-form field

Einstein equation

Dilaton field equation

Maxwell equation

Summary

Solution for ν=1\nu=1ν=1 (AdS5\mathrm{AdS}_5AdS5​)

Solution for ν=2\nu = 2ν=2

Solution for ν=4\nu = 4ν=4

2. Black brane solution

The solution

Solution for ν=2\nu = 2ν=2

Solution for ν=4\nu=4ν=4

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Solution for $\nu=1$ ( $\mathrm{AdS}_5$ )

Solution for $\nu = 2$

Solution for $\nu = 4$

Solution for $\nu = 2$

Solution for $\nu=4$