de Sitter space and Holography

de Sitter space and Holography de Sitter space and Holography

hep.physics.uoc.gr
from hep.physics.uoc.gr More from this publisher
13.07.2015 Views

In d dimensions the Einstein equations with positivecosmological constant can be derived from the actionI= I bulk + I GH∫1− d d x √ −g(R + 2Λ) + 116πG8πGM∫∂Md d−1 x √ hKhere• I GH is Gibbons-Hawking surface term which is needed toget a well-defined Euler-Lagrange variation.• M is d-dimensional Manifold with Newton’s constant Gwith spatial Euclidean boundary ∂M.• g µν is the bulk metric.• h µν and K are induced metric and the trace of the extrinsiccurvature of the boundary. The extrinsic curvature is definedby K µν = −∇ (µ n ν) where n ν is outward pointing unit vector.• A useful length scale in the model is given byl =√(d − 1)(d − 2)2ΛFor example in the vacuum dS solution, l is the radius of dSspace.32

In general this action is divergent when evaluated on asolution of the equations of motion due to infinite volume ofthe spacetime.For example in the case of dS space and in the inflationarycoordinates one has (d = 3)I ∼ 18πG∫( ) −1d 2 xe 2t/l lwhich diverges as t → ∞.The divergence can be canceled by adding local boundarycounterterms that do not affect the equations of motion.In our case we haveI total = I + 18πG∫∂Md 2 x √ h 1 lFor dS space with two boundaries at t → ±∞, ∂M ± , we haveI total = I + 18πG∫∂M +d 2 x √ h 1 l + 18πG∫∂M −d 2 x √ h 1 lwhich has the same solution as the previous action but is finite.33

In general this action is divergent when evaluated on asolution of the equations of motion due to infinite volume ofthe <strong>space</strong>time.For example in the case of dS <strong>space</strong> <strong>and</strong> in the inflationarycoordinates one has (d = 3)I ∼ 18πG∫( ) −1d 2 xe 2t/l lwhich diverges as t → ∞.The divergence can be canceled by adding local boundarycounterterms that do not affect the equations of motion.In our case we haveI total = I + 18πG∫∂Md 2 x √ h 1 lFor dS <strong>space</strong> with two boundaries at t → ±∞, ∂M ± , we haveI total = I + 18πG∫∂M +d 2 x √ h 1 l + 18πG∫∂M −d 2 x √ h 1 lwhich has the same solution as the previous action but is finite.33

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!