An Introduction to General Relativity Center for Relativistic Astrophysics School of Physics Georgia Institute of Technology Notes based on textbook: Spacetime and Geometry by S.M. Carroll Spring 2013
Gravitation Einstein Equations Lagrangian Formulations Properties of the Einstein Equations Energy Conditions Equivalence Principle
Physics in Curved Spacetime We are now ready to address: how the curvature of spacetime acts on matter to manifest itself as gravity how energy and momentum influence spacetime to create curvature. Weak Principle of Equivalence (WEP) The inertial mass and gravitational mass of any object are equal. Recall Newton s Second Law. f = m i a. with m i the inertial mass. On the other hand, f g = m g Φ. with Φ the gravitational potential and m g the gravitational mass. In principle, there is no reason to believe that m g = m i. However, Galileo showed that the response of matter to gravitation was universal. That is, in Newtonian mechanics m i = m g. Therefore, a = Φ.
Minimal Coupling Principle Take a law of physics, valid in inertial coordinates in flat spacetime Write it in a coordinate-invariant (tensorial) form Assert that the resulting law remains true in curved spacetime Operationally, this principle boils down to replacing the flat metric η µν by a general metric g µν the partial derivative µ by the covariant derivative µ Example: Motion of freely-falling particles. In Flat spacetime d 2 x µ dλ 2 = 0 Rewrite Substitute Thus d 2 x µ dλ 2 dx ν µ dx dλ ν dλ d 2 x µ dλ 2 = dx ν µ dx dλ ν dλ = 0 dx ν µ dx dλ ν dλ + dx ρ dx σ Γµ ρσ dλ dλ = 0.
The Newtonian Limit Given a General Relativistic expression, one recover the Newtonian counterparts by particles move slowly with respect to the speed of light. the gravitational field is weak, namely a perturbation of spacetime. the gravitational field is static. Consider the geodesic equation. Moving slowly implies so d 2 x µ dx i dτ << dτ 2 + Γµ 00 dt dτ, ( ) dt 2 = 0. dτ Static gravitational field implies Γ µ 00 1 = 2 gµλ ( 0 g λ0 + 0 g 0λ λ g 00 ) = 1 2 gµλ λ g 00. Weakness of the gravitational field implies g µν = η µν + h µν, h µν << 1.
From g µν g νσ = δ µ σ, g µν = η µν h µν, where h µν = η µρ η νσ h ρσ. Thus Γ µ 00 = 1 2 gµλ λ g 00 = 1 2 ηµλ λ h 00. Therefore d 2 x µ dτ 2 = 1 ( ) dt 2 2 ηµλ λ h 00. dτ Using 0 h 00 = 0, the µ = 0 component of this is just d 2 t dτ 2 = 0. That is, dt is constant and dτ d can then be rewritten as 2 x i dτ 2 = 1 ( ) dt 2 i h 00. 2 dτ d 2 x i dt 2 = 1 2 i h 00 If we introduce h 00 = 2Φ or g 00 = (1 + 2Φ), we recover a = Φ
Einstein Equations Poisson equation 2 Φ = 4πGρ where Φ is the Newtonian potential, 2 = δ ij i j is the Laplacian in space and ρ is the mass density. We need a tensor equation. Recall that h 00 = 2Φ, thus 2 h 00 = 8πGT 00 where we introduce T 00 = ρ. Notice that this is only the time-time component of an equation and also h 00 is a perturbation. Let s try g µν = α αg µν = 8πGT µν but g µν = 0 because of metric compatibility. Try instead R µν = κt µν for some constant κ since R ρ σµν contains second derivatives (and first derivatives) of the metric that do not vanish. But, from conservation of energy µ T µν = 0 thus µ R µν = 0 not true in general!
Given the contracted Bianchi identities µ R µν 1 2 ν R = µ ( R µν 1 2 gµν R ) = µ G µν = 0 Einstein proposed instead Einstein Equations G µν = κt µν where T µν is the energy-momentum tensor. For a perfect fluid, this tensor is given by T µν = (ρ + p)u µu ν + p g µν with ρ and p the rest-frame energy and momentum respectively and U µ the 4-velocity of the fluid. Next, we find κ from the Newtonian limit. First rewrite the Einstein equation G µν = κt µν R µν 1 R gµν 2 = κtµν R = κt R µν = κ(t µν 1 2 T gµν ) In the Newtonian limit, the rest energy ρ = T µν U µ U ν will be much larger than the other terms in T µν, so we want to focus on the µ = 0, ν = 0
In the fluid rest frame, U µ = (U 0, 0, 0, 0) In the weak-field limit, g 00 = 1 + h 00, g 00 = 1 h 00. Thus, from g µν U µ U ν = 1, we get that U 0 = 1 and U 0 = 1 andt 00 = ρ. Therefore, T = g 00 T 00 = T 00. yields R 00 = 1 2 κt 00. We need to evaluate R 00 = R λ 0λ0 = R i 0i0, since R0 000 = 0; that is, R i 0j0 = j Γ i 00 0Γ i j0 + Γi jλ Γλ 00 Γi 0λ Γλ j0 R i 0j0 j Γ i 00 + Γi jλ Γλ 00 Γi 0λ Γλ j0 R i 0j0 j Γ i 00 Thus ( ) 1 R 00 = i 2 giλ ( 0 g λ0 + 0 g 0λ λ g 00 ) = 1 2 ηij i j h 00 = 1 2 2 h 00
Therefore but h 00 = 2Φ and ρ = T 00, yielding 2 h 00 = κt 00 2 Φ = κ 2 ρ Thus we need to set κ = 8πG to recover the Poisson equation. Einstein Equations G µν = 8π G T µν Notice: in vacuum T µν = 0, Einstein equations become R µν = 0.
Lagrangian Formulation Another approach for deriving the Einstein s equations is through the Principle of Least Action. Let s consider the action S = L d n x Since d n x is a density, L is also a density in order for S to be a scalar. Thus, define L = g L, with L a scalar. For the case of a scalar field Φ, L = 1 2 µφ µ Φ V (Φ) the variational principle yields the Euler-Lagrange equations L ( ) L Φ µ = 0 µφ yield Φ d V dφ = 0 where µ µ = g µν µ ν
Hilbert Action S H = L H d n x What scalars can we make out of the metric? Since we can locally always set g µν = η µν and Γ µ = 0, any νδ nontrivial scalar must involve αβ g µν. Therefore, the simplest choice is L H = g R thus Hilbert Action g n S H = R d x Using R = g µν R µν, δs H = = g n R d x d n [ µν x gg δrµν + gr µν δg µν + Rδ g] = (δs) 1 + (δs) 2 + (δs) 3
Let s consider first the term Recall that (δs) 1 = d n x gg µν δr µν R ρ µλν = λ Γ λ νµ + Γρ λσ Γσ νµ (λ ν). Thus, when varying the Ricci tensor, we are going to also need variations of the connection δγ ρ νµ with respect to the metric. Since the difference of two connections is a tensor, the variation δγ ρ νµ will also be a tensor. In addition, we are going to need λ (δγ ρ νµ ). To obtain this, we take its covariant derivative, λ (δγ ρ νµ ) = λ(δγ ρ νµ ) + Γρ λσ δγσ νµ Γσ λν δγρ σµ Γσ λµ δγρ νσ. It is very easy to show that δr ρ µλν = λ (δγ ρ νµ ) ν (δγρ λµ ). Therefore (δs) 1 = = d n x g g µν [ λ (δγ λ νµ ) ν (δγλ λµ ) ] d n x g σ [ g µν (δγ σ µν ) gµσ (δγ λ λµ ) ]
Next we use Stoke s theorem with Σ σ σv g d n x = n σ σv γ d n 1 x Σ V σ = g µν (δγ σ µν ) gµσ (δγ λ λµ ). Finally, δγ ρ νµ in terms of δgµν yields δγ ρ νµ = 1 2 [ g λµ ν (δg λσ ) + g λν µ(δg λσ ) g µαg νβ σ (δg αβ ] ) and thus V σ = g µν σ (δg µν ) λ (δg σλ ). If we take the boundary term to infinity and make the variation vanish at infinity, we have that (δs) 1 = 0
Next (δs) 3 = d n x R δ g. That is, we need to find δ g. We make use Tr(ln M) = ln(det M). which yields Tr(M 1 δm) = 1 δ(det M). det M Set M = g µν. Then, det M = g 1 and δ(g 1 ) = 1 gµν δgµν g thus δ g = δ[( g 1 ) 1/2 ] = 1 2 ( g 1 ) 3/2 δ( g 1 ) = 1 ggµν δg µν 2 and (δs) 3 = d n x g ( 12 ) R gµν δg µν
Finally, with the new expressions for (δs) 2 and (δs) 3, we arrive to δs = d n x g [R µν 12 ] Rgµν δg µν. Recall that the functional derivative of the action satisfies ( ) δs δs = δφ i d n x δφ i i with stationary points satisfying δs/δφ i = 0 This yields 1 g δs δg µν = Rµν 1 2 Rgµν = 0. namely the Einstein s equations in vacuum.
We need now to include matter S = 1 8πG S H + S M, where S M is the action for matter. Following through the same procedure as above leads to 1 δs g δg µν = 1 (R µν 12 ) 8πG Rgµν + 1 δs M g δg µν = 0, and we recover Einstein s equations if we set T µν = 1 g δs M δg µν. Einstein Equations G µν = 8π G T µν
Energy Conditions What metrics obey Einstein s equations? Answer: any metric is a solution if T µν is not restricted! We want solutions to Einstein s equations with realistic sources of energy and momentum. For instance, only positive energy densities are allowed. Energy Conditions: Weak Energy Condition: T µν t µ t ν 0 Null Energy Condition: T µν k µ k ν 0 Dominant Energy Condition: T µν t µ t ν 0 and (T µν t µ )(T ν αt α ) 0 Null Dominant Energy Condition: T µν k µ k ν 0 and (T µν k µ )(T ν αk α ) 0 Strong Energy Condition: T µν t µ t ν 1 2 T λ λt σ t σ where t µ and k µ are arbitrary time-like and null vectors, respectively. For T µν = (ρ + p)u µu ν + p g µν these conditions read Weak Energy Condition: ρ 0 and ρ + p 0 Null Energy Condition: ρ + p 0 Dominant Energy Condition: ρ p Null Dominant Energy Condition: ρ p and p = ρ is allowed. Strong Energy Condition: ρ + p 0 and ρ + 3 p 0
Cosmological Constant Einstein: the biggest mistake of his S = d n x g(r 2Λ), where Λ is some constant. The resulting field equations in vaccum are G µν + Λ g µν = 0 If the cosmological constant is tuned just right, it is possible to find a static but unstable solution. If instead one considers G µν = Λ g µν = 8 π T µν then T µν = Λ g µν /8 π. The cosmological constant Λ can be then interpreted as the energy density of the vacuum. Recall that T µν = (ρ + p)u µu ν + p g µν for a perfect fluid. Thus, we need ρ = p (null dominant energy condition) and ρ = Λ g µν /8 π
Alternative Theories of Gravity Generalization of the Hilbert action S = d n x g(r + α 1 R 2 + α 2 R µν R µν + α 3 g µν µr ν R + ), where the α s are coupling constants. Scalar-tensor theories: S = d n x [ g f (λ)r + 1 ] 2 gµν ( µλ)( ν λ) V (λ), where f (λ) and V (λ) are functions which define the theory.
Equivalence Principle Again The Equivalence Principle is used to justify: Principle of Covariance: Laws of physics should be expressible in a covariant form. That is, the equations are manifestly tensorial and thus coordinate invariant. Example: µf µν = J ν µf µν = J ν There exists a metric on spacetime, the curvature of which is interpreted as gravity. That is, gravitation is identified with the effects of spacetime curvature. There do not exist any other fields that resemble gravity. The interactions of matter fields to curvature are minimal. That is, there is no direct coupling of matter with the Reimann tensor. Example: F µν R ν αβγ or d 2 x µ dλ 2 + dx ρ dx σ µ dx dx σ Γµ ρσ = α σr dλ dλ dλ dλ Since dimensionally, [Γ] = L 1 and [R] = L 2, to be dimensionally consistent [α] = L 2. The only reasonable choice is α l 2 P where l P = ( G/c3 ) 1/2 = 1.626 10 33 cm (Planck s length). It is at those scales that one could in principle measure the coupling to σr