Rong-Gen Cai （蔡荣根）

Institute of Theoretical Physics Chinese Academy of Sciences

基金委暗能量及其基本理论高级研讨班， 2012.4.7-17，杭州

Einstein Equations & Navier-Stokes Equations

Einstein’s Field Equations:

Incompressible Navier-Stokes Equations:

18

2R g R GT

Clay Mathematics Institute (http://www.claymath.org/millennium) The seven Millennium Prize Problems (US$7 million):

1) Birch and Swinnerton-Dyer Conjecture2) Hodge Conjecture3) Navier-Stokes equations4) P vs NP5) Poincare Conjecture6) Riemann Hypothesis7) Yang-Mills theory

AdS/CFT 和七个千禧年问题 :

Waves follow our boat as we meander across the lake, and turbulent air currents follow our flight in a modern jet. Mathematicians and physicists believe that an explanation for and the prediction of both the breeze and the turbulence can be found through an understanding of solutions to the Navier-Stokes equations. Although these equations werewritten down in the 19th Century, our understanding of them remains minimal. The challenge is to make substantial progress toward a mathematical theory which will unlock the secrets hidden in the Navier-Stokes equations.

Navier-Stokes Equation:

Outline of the Talk:

1. Nonrelativistic case: vacuum Einstein gravity

2. Nonrelativistic case: AdS gravity

3. Relativistic case

1. Vacuum Einstein Gravity

Refs:1101.2451: From Navier-Stokes to Einstein1103.3022: The holographic fluid dual to vacuum Einstein gravity1105.4482: Higher curvature gravity and the holographic fluid dual to flat spacetime

a) The hydrodynamical limit and the epsilon expansion

Scaling symmetry:

Higher order

(p+2)-dimensional Einstein equations and (p+1)-dimensional incompressible Navier-Stokes Equations

Ingoing Rindler coordinates:

Consider a timelike hypersurface at r=r_c, its intrinsic metric is flat

Consider a Minkowski space in Cartesian chart

The following region is often calledRindler wedge

Defining a coordinate transformation

Then in the Rindler chart, the Minkowski space turns to be:

The Rindler coordinate chart has a coordinate singularity at x = 0. The acceleration of the Rindler observers diverges there. As we can see from the figureillustrating the Rindler wedge, the locus x = 0 in the Rindler chart corresponds tothe locus in the Cartesian chart, which consists of two null half-planes.

Unruh temperature: for a uniformly accelerating observer with acceleration a

Bulk solution:

1). Leading order

2) Nonlinear solution in the epsilon expansion

Consider the metric:

Dual fluid:

The induce metric on the cutoff surface:

The extrinsic curvature:

To solve the Einstein equations:

The first nontrivial equation appears at order e^2:

Take this to be the case, at order e^3:

Summary:

1)Consider the portion of Minkowski spacetime between a flat hypersurface Sigma_c, given by the equation, X^2-T^2=4r_c, and its future horizon H^+, the null surface X=T.

2) Keep the intrinsic metric flat, and the effects of finite perturbations of the extrinsic curvature of Sigma_c can be studied.

3) A regular Ricci flat metric exists provided that the Brown- York stress tensor on Sigma_c is that of an incompressible Navier-Stokes fluid.

4) More precisely, they work in a hydrodynamic non-relativistic limit and construct the bulk metric up to third order in the hydrodynamic expansion.

5) This provides a potential example of a holographic duality involving a flat spacetime.

In 1103.3022, Skenderis et al present an algorithm for systematically reconstructing a solution of the (d+2) dimensional vacuum Einstein equations from a (d+1) dimensional fluid, extending the non-relativistic hydro-dynamical expansion of Bredberg et al in 1101.2451.

A systematic construction of the bulk solution to all orders

Key results:

a) The metric up to order e^2 with constant velocity and pressure fields is actually flat space in disguise: it can be obtained from the Rindler metric by a linear coordinate transformation combining a boost .

b) To extend the solution to next order, promote the velocity and pressure to be spacetime dependent quantities, subject to the requirement that the Einstein equations hold up to e^3.

c) To satisfy this requirement, one needs introduce new terms of order e^3 in the metric. One finds the following holds

d) Einstein equations can in fact be satisfied to arbitrarily high order in epsilon, by adding appropriate terms to the metric and modifying the NS equations and the incompressibility condition by specific higher derivative corrections. E gets corrected at even powers of e and E_i get modified at odd powers of e.

1) Equilibrium configurations

Consider a class of Ricci flat spacetime with i) A hyprosurace with a flat induced metric

ii) The Brown-York tensor on Sigma_c:

takes the form of perfect fluid, and iii) they are stationary with respect to tau, and homogeneous in space

Ingoing Rindler metric

There are only two infinitesimal diffeomorphisms yieldingMetrics satisfying the above conditions (i),(ii), (iii) .

Consider the following two finite diffeomorphisms:(i) The first is a constant boost:

(ii) A linear shift and a rescaling

Then, one has

The Brown-York stress tensor:

And thermodynamics:

2) Seed metric for near-equilibrium configurations

View

It solves the Einstein equations up to O(e^3), provided one imposes incompressibility

associated Brown-York stress tenor

Expanding the stress tensor up to order e^2 :

3) Constructing the solution to all orders

The details are neglected! Sorry!

Take an example to order O(e^3)

The bulk solutions to order (O^5) are given. And the Correction to the BY tensor

at O(e^4)

at O(e^5)

Stress tensor:

Corrections to NS and incompressibility

The conservation of the BY tensor e^2: incom. Con.

e^3: NS eq.

Conservation of the BY stress tensor at order e^n is requiredIn order to construct the bulk metric at the same order.

at order e^4: The incompressibility gets modified:

at order e^5: The NS equations get corrected

Characterizing the dual fluid

Energy density:

Relativistic hydrodynamics for vanishing equilibrium energy density

The constraint

Up to second order in gradients

where

The energy density in local rest frame

Determining the transport coefficients

Perform the expansion of PI_{ab} in e to (e^6), and compare to the BY tensor given from the bulk solution

Note that c_5 and c_6 both vanish at O(e^6), thus fourof six coefficients can be determined.

Interesting questions:

1) the correspondence extends beyond the hydrodynamics regime?2)string embedding?3)How this correspondence changes if one adds a bulk stress tensor or consider higher derivative corrections to Einstein gravity? Will such changes modify the properties of the dual fluid?4)relativistic construction of the bulk metric?5) curved space?6) Should not be limited to flat or black hole spacetime, any spacetime should also work by the equivalent principle.

Higher derivative gravity: 1105.4482, Chirco et al.)

General setup

Induced metric:

Bulk metric:

Vacuum solution with ingoing Rindler metric:

Dual fluid:

Near-equilibrium configurations:Two diffeomorphisms, namely a boost and the translation

Expanding the metric to O(e^2), one has the seed metric

This metric is the unique singulaity-free solutions the vacuumEinstein equation up to O(e^3), provided

In GR, the momentum constrain equation on the surface can be expressed in terms of the BY tensor as

Consider the Gauss-Bonnet gravity

Many of second order transport coefficients appear At order O(e^4).

The bulk solution is found to order O(e^5).

2. Nonrelativistic case: AdS gravity (arXiv: 1104.3281, Cai et al)

The motivation is two-fold: 1) if there exists a bulk stress tensor; 2) AdS/CFT correspondence

In order to consider a (p+1)-dimensional fluid in a flat spacetime , consider (p+2)-dimensional bulk

Induced metric:

Rescaling to Minkowski spacetime

Consider two finite diffeomorphism transformations

The second one:

Here k(r) is a linear function as k(r)= br +c

This metric still solves the corresponding gravitational fieldequations, but if we promote v_i and delta k(r)=k(r)-r =(b-1)r +c to be dependent on the coordinates x^a, then the Resulted metric is no longer an exact solutions.

In order to solve the gravity equations, we take the so-called

Hydrodynamics expansion and non-relativistic limit:

We demand both (b-1) and c scale as e^2, then up to e^2,

The resulted metric: to order e^2

If we take

then …..

Now we consider

Consider the case:

Then the seed metric only solves the gravity equations at O(e^1)

at the order O(e^2)

should be added to the bulk metric. To be regular at the Horizon, one needs F(r_c)=0.

Then the final metric at O(e^2)

The Einstein gravity with a negative cosmological constant

The black brane solution

At the order O(e^2)

It solves the gravity equations at order e^2. take an example,consider p=3.

The BY stress tensor:

At order e^2,

At order e^3,

When r_c infinity, one takes

From T^(2)

The Gauss-Bonnet case

The black brane metric:

In this case

The BY stress tensor:

At order e^2

At next order

3. Relativistic case: vacuum Einstein gravity

Refs: 1201.2678 : The relativistic fluid dual to vacuum Einstein gravity 1201.2705 : The relativistic Rindler hydrodynamics

The stress tensor up to second order in gradients

The fluid dual to vacuum Einstein gravity with vanishingEquilibrium energy density :

At first order in gradients

two coefficients: shear viscosity and bulk viscosity

At second order in gradients

For vacuum Einstein gravity:

Entropy current:

A general feature of system away from equilibrium is that they posses an entropy current with non-negative divergence.

At equilibrium:

In hydrodynamics regime with vanishing energy density:

Five parameters

In the case with fluid dual to vacuum Einstein gravity

with non-negative divergence.

The entropy current:

Relativistic construction of near-equilibrium solutions

(1) Seed metric

Induced metric on the hypersurface

To obtain the zero-order seed metric, one does

Second, one performs the boost:

Associated BY tensor:

Integration scheme

We start with the zero order seed emtric

At the first order

from which one has eta=1, and xi’=0

At second order

Energy-momentum tensor

Entropy current

The constraint for necessary positivecondition.

Thank You !