Gravitation: Gravitation - An Introduction to General

Gravitation: GravitationAn Introduction to General Relativity

Pablo Laguna

Center for Relativistic AstrophysicsSchool of Physics

Georgia Institute of Technology

Notes based on textbook: Spacetime and Geometry by S.M. CarrollSpring 2013

Pablo Laguna Gravitation: Gravitation

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 = mi a .

with mi the inertial mass.

On the other hand,fg = −mg∇Φ .

with Φ the gravitational potential and mg the gravitational mass.

In principle, there is no reason to believe that mg = mi .

However, Galileo showed that the response of matter to gravitation was universal. That is, in Newtonian mechanicsmi = mg . 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 ηµν by a general metric gµν

the partial derivative ∂µ by the covariant derivative∇µ

Example: Motion of freely-falling particles. In Flat spacetime

d2xµ

dλ2= 0

Rewrited2xµ

dλ2=

dxν

dλ∂ν

dxµ

dλ= 0

Substitutedxν

dλ∂ν

dxµ

dλ→

dxν

dλ∇ν

dxµ

dλ

Thusd2xµ

dλ2+ Γµρσ

dxρ

dλ

dxσ

dλ= 0 .


The Newtonian LimitGiven 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 impliesdx i

dτ<<

dt

dτ,

sod2xµ

dτ2+ Γ

µ00

( dt

dτ

)2= 0 .

Static gravitational field implies

Γµ00 =

1

2gµλ(∂0gλ0 + ∂0g0λ − ∂λg00)

= −1

2gµλ∂λg00 .

Weakness of the gravitational field implies

gµν = ηµν + hµν , |hµν | << 1 .


From gµνgνσ = δµσ ,gµν = η

µν − hµν ,

where hµν = ηµρηνσhρσ . Thus

Γµ00 = −

1

2gµλ∂λg00 = −

1

2ηµλ∂λh00 .

Therefored2xµ

dτ2=

1

2ηµλ∂λh00

( dt

dτ

)2.

Using ∂0h00 = 0, the µ = 0 component of this is just

d2t

dτ2= 0 .

That is, dtdτ is constant and

d2x i

dτ2=

1

2

( dt

dτ

)2∂i h00 .

can then be rewritten asd2x i

dt2=

1

2∂i h00

If we introduce h00 = −2Φ or g00 = −(1 + 2Φ), we recover

a = −∇Φ


Einstein Equations

Poisson equation∇2Φ = 4πGρ

where Φ is the Newtonian potential,∇2 = δij∂i∂j is the Laplacian in space and ρ is the mass density.

We need a tensor equation. Recall that h00 = −2Φ, thus

∇2h00 = −8πGT00

where we introduce T00 = ρ. Notice that this is only the time-time component of an equation and also h00 is aperturbation. Let’s try

�gµν = ∇α∇αgµν = −8πGTµν

but �gµν = 0 because of metric compatibility. Try instead

Rµν = κTµν

for some constant κ since Rρσµν contains second derivatives (and first derivatives) of the metric that do notvanish. But, from conservation of energy

∇µTµν = 0

thus∇µRµν = 0

not true in general!


Given the contracted Bianchi identities

∇µRµν −1

2∇νR = ∇µ

(Rµν −

1

2gµνR

)= ∇µGµν = 0

Einstein proposed instead

Einstein Equations

Gµν = κTµν

where Tµν is the energy-momentum tensor. For a perfect fluid, this tensor is given by

Tµν = (ρ + p)UµUν + p gµν

with ρ and p the rest-frame energy and momentum respectively and Uµ the 4-velocity of the fluid.

Next, we find κ from the Newtonian limit. First rewrite the Einstein equation

Gµν = κTµν

Rµν −1

2R gµν = κTµν

R = −κT

Rµν = κ(Tµν −1

2T gµν )

In the Newtonian limit, the rest energy ρ = TµνUµUν will be much larger than the other terms in Tµν , so we wantto focus on the µ = 0, ν = 0


In the fluid rest frame, Uµ = (U0, 0, 0, 0) In the weak-field limit,

g00 = −1 + h00 ,

g00 = −1− h00 .

Thus, from gµνUµUν = −1, we get that U0 = 1 and U0 = −1 andT00 = ρ. Therefore,

T = g00T00 = −T00 .

yields

R00 =1

2κT00 .

We need to evaluate R00 = Rλ0λ0 = Ri0i0, since R0

000 = 0; that is,

Ri0j0 = ∂j Γ

i00 − ∂0Γi

j0 + ΓijλΓλ00 − Γi

0λΓλj0

Ri0j0 ≈ ∂j Γ

i00 + Γi

jλΓλ00 − Γi0λΓλj0

Ri0j0 ≈ ∂j Γ

i00

Thus

R00 = ∂i

( 1

2giλ(∂0gλ0 + ∂0g0λ − ∂λg00)

)= −

1

2η

ij∂i∂j h00 = −

1

2∇2h00


Therefore∇2h00 = −κT00

but h00 = −2Φ and ρ = T00 , yielding

∇2Φ =κ

2ρ

Thus we need to set κ = 8πG to recover the Poisson equation.

Einstein Equations

Gµν = 8π G Tµν

Notice: in vacuum Tµν = 0, Einstein equations become Rµν = 0.


Lagrangian Formulation

Another approach for deriving the Einstein’s equations is through the Principle of Least Action. Let’s consider theaction

S =

∫L dnx

Since dnx is a density, L is also a density in order for S to be a scalar. Thus, define

L =√−g L̂ ,

with L̂ a scalar. For the case of a scalar field Φ,

L̂ = −1

2∇µΦ∇µΦ− V (Φ)

the variational principle yields the Euler-Lagrange equations

∂L̂∂Φ−∇µ

(∂L̂

∂∇µΦ

)= 0

yield

�Φ−d V

dφ= 0

where � ≡ ∇µ∇µ = gµν∇µ∇ν


Hilbert Action

SH =

∫LH dnx

What scalars can we make out of the metric? Since we can locally always set gµν = ηµν and Γµνδ

= 0, anynontrivial scalar must involve ∂αβgµν . Therefore, the simplest choice is

LH =√−g R

thus

Hilbert Action

SH =

∫ √−g R dnx

Using R = gµνRµν ,

δSH =

∫ √−g R dnx

=

∫dnx

[√−ggµνδRµν +

√−gRµνδgµν + Rδ

√−g]

= (δS)1 + (δS)2 + (δS)3


Let’s consider first the term(δS)1 =

∫dnx√−ggµνδRµν

Recall thatRρµλν = ∂λΓλνµ + Γ

ρλσ

Γσνµ − (λ↔ ν) .

Thus, when varying the Ricci tensor, we are going to also need variations of the connection δΓρνµ with respect tothe metric. Since the difference of two connections is a tensor, the variation δΓρνµ will also be a tensor.

In addition, we are going to need ∂λ(δΓρνµ). To obtain this, we take its covariant derivative,

∇λ(δΓρνµ) = ∂λ(δΓρνµ) + ΓρλσδΓσνµ − ΓσλνδΓρσµ − ΓσλµδΓρνσ .

It is very easy to show thatδRρµλν = ∇λ(δΓρνµ)−∇ν (δΓ

ρλµ

) .

Therefore

(δS)1 =

∫dnx√−g gµν

[∇λ(δΓλνµ)−∇ν (δΓλλµ)

]=

∫dnx√−g ∇σ

[gµν (δΓσµν )− gµσ(δΓλλµ)

]


Next we use Stoke’s theorem ∫Σ∇σVσ

√|g|dnx =

∫∂Σ

nσVσ√|γ|dn−1x

with

Vσ = gµν (δΓσµν )− gµσ(δΓλλµ) .

Finally, δΓρνµ in terms of δgµν yields

δΓρνµ = −1

2

[gλµ∇ν (δgλσ) + gλν∇µ(δgλσ)− gµαgνβ∇

σ(δgαβ )]

and thusVσ = gµν∇σ(δgµν )−∇λ(δgσλ) .

If we take the boundary term to infinity and make the variation vanish at infinity, we have that (δS)1 = 0


Next (δS)3 =∫

dnx R δ√−g. That is, we need to find δ

√−g. We make use

Tr(ln M) = ln(det M) .

which yields

Tr(M−1δM) =

1

det Mδ(det M) .

Set M = gµν . Then, det M = g−1 and

δ(g−1) =1

ggµνδgµν

thus

δ√−g = δ[(−g−1)−1/2]

= −1

2(−g−1)−3/2

δ(−g−1)

= −1

2

√−ggµνδgµν

and

(δS)3 =

∫dnx√−g(−

1

2R gµν

)δgµν


Finally, with the new expressions for (δS)2 and (δS)3, we arrive to

δS =

∫dnx√−g

[Rµν −

1

2Rgµν

]δgµν .

Recall that the functional derivative of the action satisfies

δS =

∫ ∑i

(δS

δΦiδΦi)

dnx

with stationary points satisfying δS/δΦi = 0

This yields1√−g

δS

δgµν= Rµν −

1

2Rgµν = 0 .

namely the Einstein’s equations in vacuum.


We need now to include matter

S =1

8πGSH + SM ,

where SM is the action for matter. Following through the same procedure as above leads to

1√−g

δS

δgµν=

1

8πG

(Rµν −

1

2Rgµν

)+

1√−g

δSM

δgµν= 0 ,

and we recover Einstein’s equations if we set

Tµν = −1√−g

δSM

δgµν.

Einstein Equations

Gµν = 8π G Tµν


Energy Conditions

What metrics obey Einstein’s equations?

Answer: any metric is a solution if Tµν is not restricted!

We want solutions to Einstein’s equations with realistic sources of energy and momentum. For instance, onlypositive energy densities are allowed.

Energy Conditions:

Weak Energy Condition: Tµν tµtν ≥ 0

Null Energy Condition: Tµνkµkν ≥ 0

Dominant Energy Condition: Tµν tµtν ≥ 0 and (Tµν tµ)(Tν α tα) ≤ 0

Null Dominant Energy Condition: Tµνkµkν ≥ 0 and (Tµνkµ)(Tν αkα) ≤ 0

Strong Energy Condition: Tµν tµtν ≥ 12 Tλ λtσ tσ

where tµ and kµ are arbitrary time-like and null vectors, respectively. For Tµν = (ρ + p)UµUν + p gµν theseconditions 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 =

∫dnx√−g(R − 2Λ) ,

where Λ is some constant. The resulting field equations in vaccum are

Gµν + Λ gµν = 0

If the cosmological constant is tuned just right, it is possible to find a static but unstable solution.

If instead one considersGµν = −Λ gµν = 8π Tµν

then Tµν = −Λ gµν/8π. The cosmological constant Λ can be then interpreted as the energy density of thevacuum.

Recall that Tµν = (ρ + p)UµUν + p gµν for a perfect fluid. Thus, we need ρ = −p (null dominant energycondition) and ρ = Λ gµν/8π


Alternative Theories of Gravity

Generalization of the Hilbert action

S =

∫dnx√−g(R + α1R2 + α2RµνRµν + α3gµν∇µR∇νR + · · · ) ,

where the α’s are coupling constants.

Scalar-tensor theories:

S =

∫dnx√−g[

f (λ)R +1

2gµν (∂µλ)(∂νλ)− 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 equationsare manifestly tensorial and thus coordinate invariant. Example:

∂µFµν = Jν → ∇µFµν = Jν

There exists a metric on spacetime, the curvature of which is interpreted as gravity. That is, gravitation isidentified 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 withthe Reimann tensor. Example: FµνRν αβγ or

d2xµ

dλ2+ Γµρσ

dxρ

dλ

dxσ

dλ= α∇σR

dxµ

dλ

dxσ

dλ

Since dimensionally, [Γ] = L−1 and [R] = L−2, to be dimensionally consistent [α] = L2. The onlyreasonable choice is α ∼ l2P where lP = (~G/c3)1/2 = 1.626× 10−33cm (Planck’s length). It is at thosescales that one could in principle measure the coupling to∇σR


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Gravitation: Gravitation - An Introduction to General