I. EINSTEIN FIELD EQUATION
In order to construct the equation for describing the relation between matter/energy and spacetime curvature, one has impose two requirements such that
• The equation must be reduced to Newtonian theory.
• The curvature part must contain the metric tensor gµν and its derivatives such as ρ ρ and while the matter/enegy should be proportional to the EMT . Γµν,R µσν,Rµν R Tµν
From the rst requirement, the important equation in Newtonian theory in the Poisson equation ∇2Φ = 4πGρ Which can be derived from the Gauss's law and leave for the student in Exercise. As we discussed before, the component of the metric tensor is proportional to the gravitational potential g00 ∝ Φ. Therefore, in order to reduce the master equation into the Poisson equation, the curvature part must be proportional to the second derivative of the metric. These quantities are ρ and . R µσν,Rµν R
A. Vacuum equation
For the vacuum equation, the matter/energy part vanishes and then the curvature part will be vanished
ρ (1) f(R µσν,Rµν,R) = 0.
One may rstly guess for this equation such that
ρ (2) R µσν = 0 (?).
However, one found that this may not be possible since this equation provides the at spacetime nearby the massive source. So that we can make a further guess by
Rµν = 0 (?). (3)
This is a good choice since the Ricci tensor has ten dof. like the metric tensor, hoping that ten dof. of the metric transfer to ten dof. of Ricci tensor through their second derivatives making from source nearby.
1 B. Equation with source
Now let us consider the equation with source. By adding the EMT, the equation may be written as
Rµν = kTµν (?), (4)
where k is the proportional constant. This still be the good choice since the index symmetry also satisfy. However, as we discussed before, the EMT obeys the conservation equation
µν ∇µT = 0 while the Ricci tensor does not satisfy in general. As a result, one may nd other quantities to satisfy this condition as well as maintain the mentioned requirements.
Fortunately, from the Bianchi indentity, ∇[λRρσ]µν = 0, it serve us the conservation quantity as follows
1 ∇µG = 0,G = R − Rg , (5) µν µν µν 2 µν
µ where Gµν is called Einstein tensor. Note that the derivation of ∇ Gµν = 0 from the Bianchi identity is leave in the Exercise. As a result, the equation can be constructed as
Gµν = kTµν. (6)
Next task for this construction is that we have to nd the proportional constant k as well as check whether this equation satis es the vacuum equation or not. To perform this evaluation, let us take the trace of the above equation as follows
1 R − (4)R = kT, 2 R = −kT (7)
Substituting R from this equation into Eq. (6), one obtains
1 R + kT g = kT , µν 2 µν µν 1 R = k T − T g . (8) µν µν 2 µν
From this equation, one can see that the vacuum equation still be satis ed where the source
is eliminated, Tµν = T = 0.
2 Now, we will nd the proportional constant by taking the Newtonian limit into Eq. (8). As a result, the EMT can be written as 1 v1 v2 v3 1 0 0 0 v1 v1v1 v1v2 v1v3 0 0 0 0 µν (9) T = ρ ∼ ρ . v2 v2v1 v2v2 v2v3 0 0 0 0 v3 v3v1 v3v2 v3v3 0 0 0 0 Then we have 1 0 0 0 1 1 0 1 0 0 1 µν µν µν T − T g = ρ = ρδ , 2 2 0 0 1 0 2 0 0 0 1 1 1 T − T g = ρδ . (10) µν 2 µν 2 µν Now let us consider the left hand side of Eq. (8),
ρ ρ λ ρ λ ρ (11) Rµν = ∂ρΓ µν − ∂νΓ µρ + Γ µνΓ λρ − Γ µρΓ λν.
By using the weak eld limit gµν = ηµν +hµν and then keeping only rst order perturbations, ρ does not contain the zeroth order of the perturbation metric as Γµν hµν 1 Γρ = gρσ (∂ h + ∂ h − ∂ h ) . (12) µν 2 µ νσ ν µσ σ µν Thus the last 2 terms on the left of (11) is neglected in our consideration. The the rst order of the Ricci tensor is
(1) ρ ρ Rµν = ∂ρΓµν − ∂νΓρµ, 1 1 = ∂ ηρσ (∂ h + ∂ h − ∂ h ) − ∂ ηρσ∂ h . (13) ρ 2 µ νσ ν µσ σ µν ν 2 µ ρσ
(1) Since the h00 is much more than other components, the dominant components of Rµν is 1 1 R(1) = ∂ ηρσ (∂ h + ∂ h − ∂ h ) − ∂ ηρσ∂ h , 00 ρ 2 0 0σ 0 0σ σ 00 0 2 0 ρσ 1 ' − ηρσ∂ ∂ h , 2 ρ σ 00 1 = − ∂ ∂ih , (h is time-independent.) 2 i 00 00 Φ = ∇2 . (14) c2
3 Substituting results into Eq. (8), one obtains
1 ∇2Φ = kρ. (15) 2
By comparing this equation to the Poisson equation ∇2Φ = 4πGρ, the constant k can be written as
k = 8πG. (16)
Finally, the Einstein equation is completely constructed as
1 G = R − gµνR = 8πGT . (17) µν µν 2 µν
II. LAGRANGIAN FORMULATION IN GR
Motivaton - Most of physical theories can be expressed in terms of Lagrangian formulation. - It is convenient to identify conserved quantities. - It is very useful to generalize the theory.
A. Review of Classical Field Theory
• Classical Mechanics: The action of a particle is expressed as Z S = dt L(q(t), q˙(t)). (18)
where q and q˙ are the position and velocity of the particle. • Variational Principle: in nature, the physical system will evolve along the path such that the variation of action is zero, δS = 0. Z δS = dt δL, Z ∂L ∂L = dt δq + δq˙ . (19) ∂q ∂q˙ Since the variation of path is independent of time, we can write
dq d δq˙ = δ = δq. (20) dt dt 4 Then integrating by part for the second term in parenthesis in (19), Z ∂L d ∂L d ∂L δS = dt − δq + δq . (21) ∂q dt ∂q˙ dt ∂q˙ We are interested in the last term in the square bracket. Since ∂L is continuous, we can ∂q˙ δq integrate out as Z d ∂L ∂L ∂L dt δq = δq − δq . (22) dt ∂q˙ ∂q˙ t2 ∂q˙ t1
where t1 and t2 are the initial and nal time of consideration. From the fact that we consider
the variation of the path between any two points. The position at the ends of the path (q(t1) and q(t2)) must be xed as illustrated below
Figure 1. Variation of paths
We thus have the conditions
(23) δq|t1 = δq|t2 = 0.
Eventually, the variational principle gives us the equation of motion of the particle, ∂L d ∂L δS = 0, → − = 0, (24) ∂q dt ∂q˙ which is well-known as the Euler-Lagrange equation. • Classical Field theory: The action for the eld Φa(x) can be written as Z 4 a a a S = d x L(Φ (x), ∂µΦ (x), ∂µ∂νΦ (x),...). (25)
5 Then ∂L ∂L ∂L (26) δS = 0, → a − ∂µ a + ∂µ∂ν a − ... = 0. ∂Φ ∂(∂µΦ ) ∂(∂µ∂νΦ ) It is very important to note that the condition of xing the end points in classical mechanics is equivalent to take the boundary surface to in nity and then the eld becomes zero in classical eld theory. • Classical Field theory in curved spacetime: moving to consider in curved spacetime, we have to generalize some quantities as follows
4 √ 4 ∂µ → ∇µ, ηµν → gµν, d x → −gd x. (27)
The action becomes Z 4 √ a a S = d x −gL(Φ (x), ∇µΦ (x),...), Z √ = d4x L ; L = −gL. (28)
The equation is ∂L ∂L (29) δS = 0, → a − ∇µ a + ... = 0. ∂Φ ∂(∇µΦ ) Example: A massive canonical scalar eld Z 4 √ µ 2 2 S = d x −g −∇µφ∇ φ − m φ . (30)
Then the variation with respect to φ is (gµν is xed under this consideration) Z 4 √ µν 2 δφS = d x −g −g (∇µδφ∇νφ + ∇µφ∇νδφ) − 2m φδφ , Z 4 √ µ 2 = d x −g (−2) ∇ φ∇µδφ + m φδφ , Z 4 √ µ µ 2 = d x −g (−2) ∇µ (∇ φδφ) − ∇µ∇ φ − m φ δφ . (31)
R 4 √ µ Considering the term d x −g ∇µ (∇ φδφ). The integral over proper volume, Σ of the 4-divergence can be written as the integral over its boundary, ∂Σ via the Gauss's theorem, Z Z √ 4 √ µ 3 µ d x −g ∇µ (∇ φδφ) = d x h (∇ φδφ) , (32) Σ ∂Σ
where h is the determinant of metric hµν describing the geometry on the boundary ∂Σ. Since R 4 √ µ the variation of the eld vanishes at the boundary, Then, d x −g ∇µ (∇ φδφ) = 0.
6 R 4 √ µ Moreover, it is also be considered the boundary term, d x −g ∇µ (∇ φδφ) in other form. By the relation between 4-divergence of the vector and partial derivative, 1 √ ∇ V µ = √ ∂ −g V µ , (33) µ −g µ we have Z Z 4 √ µ 4 √ µ √ µ d x −g ∇µ (∇ φδφ) = d x ∂µ −g∇ φδφ = −g∇ φδφ = 0. (34) boundary It is more familiar consideration as in classical eld theory. Finally, we obtain the equation of motion of canonical scalar eld called the Klein-Gordon equation
µ 2 δS = 0 → ∇µ∇ φ − m φ = 0. (35)
B. Einstein-Hilbert action
Requirement for constructing the action in GR - L must be covariant scalar (the action should be scalar quantity in order to con rm that the action obeys the symmetry in the theory.) - L should contain up to rst order derivative of metric tensor.
ρ µν L ∝ ∇ρgµν∇ g . (36)
Unfortunately the above Lagrangian is trivial because there exists the metric compatibility,
∇ρgµν. We should move attention to the action containing the second order derivative. But the second order derivative terms should be linear in order to obtain the boundary Z d t2 (37) t q¨ → δq˙|t1 Notice that we need a condition which is the rst derivative of the dynamical eld vanishes at boundary, . The Lagrangian in GR should be δq˙|t1 = δq˙|t2 = 0
L = L(gµν, ∂ρgµν, ∂ρ∂σgµν) (38)
As we have known that the Riemann tensor, ρ is the tensorial quantity which contains R σµν the second derivative of gµν (∼ ∂Γ). We also know the scalar quantity which is the Ricci scalar, R. So the action in GR is constructed as Z 4 √ SEH = d x −g R, (39)
7 which is called the Einstein-Hilbert action. Next, computing the variation of this action with respect to the metric tensor. Z Z 4 √ µν 4 √ µν √ µν √ µν δSEH = d xδ( −g g Rµν) = d x δ −g g Rµν + −g δg Rµν + −g g δRµν(40). | {z } | {z } | {z } δS1 δS2 δS3 From the identity
ln(det M) = Tr(ln M), det M 6= 0, (41)
ln(g) = Tr(ln[gµν]).
Then, consider the variation
δ ln(g) = δTr(ln[gµν]), 1 δg = (g−1) δg , g µν µν µν δg = g g δgµν, √ 1 1 δ −g = √ δ(−g) = − √ g gµνδg , 2 −g 2 −g µν 1√ = −g gµνδg . (42) 2 µν From
ρν ν (43) gµρg = δµ,
ρν ρν δgµρg + gµρδg = 0,
ρν ρν g δgµρ = −gµρδg ,
ρσ δgµν = −gµρgνσδg . (44)
Then, Eq. (42) can be expressed as
√ 1√ δ −g = − −g g δgµν. (45) 2 µν Substituting this result into (40), one obtains
Z √ 1 δS + δS = d4x −g R − g R δgµν, (46) 1 2 µν 2 µν Z 4 √ µν δS3 = d x −g Gµνδg .
From these two terms, one properly obtain Einstein tensor which is the left hand side of Einstein eld equation. However, we have to consider the second term which is proportional
8 to . In order to obtain , one needs to know the form of ρ . Let us start with δRµν δRµν δΓµν varying the connection as following
1 δΓρ = δgρσ (∂ g + ∂ g − ∂ g ) + gρσ (∂ δg + ∂ δg − ∂ δg ) (47) µν 2 µ σν ν σµ σ µν µ σν ν σµ σ µν 1 = gρσ (∇ δg + ∇ δg − ∇ δg ) . (48) 2 µ σν ν σµ σ µν The detail calculation from the st to the second line is left for the student in the Exercise. From the above expression, one can see that even though the connection ρ is not tensor, Γµν the variation ρ is. Then, we now move to calculate of the variation of the Ricci tensor δΓµν
δRµν as following
ρ ρ ρ σ ρ σ δRµν = δ ∂ρΓνµ − ∂µΓνρ + ΓρσΓµν − ΓµσΓρν
ρ ρ ρ σ ρ σ ρ σ ρ σ = ∂ρδΓνµ − ∂µδΓνρ + δΓρσΓµν + ΓρσδΓµν − δΓµσΓρν − ΓµσδΓρν,
ρ ρ σ ρ σ ρ ρ σ σ ρ = ∂ρδΓνµ − δΓµσΓρν + ΓρσδΓµν − (∂µδΓνρ + ΓµσδΓρν − ΓµνδΓρσ),
ρ ρ σ ρ σ σ ρ ρ ρ σ σ ρ σ ρ = ∂ρδΓνµ − δΓµσΓρν + ΓρσδΓµν − ΓµρδΓνσ − (∂µδΓνρ + ΓµσδΓρν − ΓµνδΓρσ − ΓµρδΓνσ), ρ ρ (49) = ∇ρδΓ µν − ∇νδΓ µρ.
Therefore, the second term in Eq. (40) can be expressed as Z 4 √ µν ρ ρ δS3 = d x −g g ∇ρδΓ µν − ∇νδΓ µρ , Z 4 √ να ρσ ν ρσ = d x −g ∇ν (gρσg ∇αδg − δσ∇ρδg ) , Z 4 √ ν = d x −g ∇νV . (50)
Using the same strategy, the δS3 vanishes when we require
αβ ρ ∇µδg = 0, or δΓ = 0, (51) boundary µν boundary There is no this requirement (variation of the rst derivative of metric is zero) in the previous classical eld theory. This is the important dierence among the Lagrangian formulation for GR and other theories. Therefore, the left hand side of the Einstein eld equations are obtained,
1 δ SEH = 0 → R − g R = 0. (52) g µν 2 µν Note that the requirement of vanishing of the rst derivative of metric is not the only one way to obtain the Einstein eld equations. One of other ways is to introduce the canceled
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term called the Gibbons-Hawking-York boundary term. In this approach, δgαβ = 0 boundary is enough to obtain the Einstein eld equations. Next, we will move the consider the matter sector.
C. Matter sector
The action corresponds to the Einstein eld equations with matter is written as
Z Z 4 √ 1 4 √ 1 S = SEH + S = d x −g R + L = d x −g R + L , (53) m 2κ m 2κ m
where Sm and Lm(or Lm) are the action and Lagrangian for matter. κ is a constant. Let us consider the variation with respect to gµν,
Z √ 1 δ S = d4x −g G δgµν + δ L , g 2κ µν g m Z √ 1 2κ δL = d4x −g G + √ m δgµν (54) 2κ µν −g δgµν Comparing to the Einstein eld equations (17), we have
Z √ 1 8πG δ S = d4x −g G − T δgµν (55) g 2κ µν c4 µν We also obtain that
8πG κ = (56) c4 and
−2 δL T = √ m . (57) µν −g δgµν √ Substituting Lm = −gLm to the above equation, we obtain another expression of the energy-momentum tensor as follows
−2 δ √ T = √ ( −gL ), µν −g δgµν m −2 1√ √ δL = √ − −gg L + −g m , −g 2 µν m δgµν δL = −2 m + L g . (58) δgµν m µν
10 Example: 1 µ , Lm = − 2 ∂µφ∂ φ − V (φ)
1 δ L = − δ(gµν∂ φ∂ φ), g m 2 µ ν 1 = − ∂ φ∂ φδgµν, 2 µ ν δL 1 → m = − ∂ φ∂ φ, (59) δgµν 2 µ ν 1 → T (φ) = ∂ φ∂ φ − − ∂ φ∂µφ − V g . (60) µν µ ν 2 µ µν
Example: 1 µν, Lm = − 4 FµνF
1 δ L = − δ(gµρgνσF F ), g m 4 µν ρσ 1 = − (δgµρF F ν + δgνσF F µ ), 4 µν ρ µν σ 1 = − δgµρF F ν + δgνσ(−F )(−F µ) , 4 µν ρ νµ σ 1 = − δgµρF F ν , (61) 2 µν ρ
δL 1 → m = − F F ρ, (62) δgµν 2 µρ ν 1 → T (A) = F F ρ − F F µν. (63) µν µρ ν 4 µν
Example: 1 ρ 2, Lm = 2 (∇ρA )
1 ρρ0 σσ0 L = (∇ A 0 g )(∇ A 0 g ), m 2 ρ ρ σ σ 1 γ ρρ0 γ σσ0 = (∂ A 0 − Γ 0 A )g (∂ A 0 − Γ 0 A )g . (64) 2 ρ ρ ρρ γ σ σ σσ γ
Then ρρ0 σ ρ σσ 1 ∇ρAρ0 δg ∇σA + ∇ρA ∇σAσ0 δg δgLm = , 2 γ ρρ0 σ ρ γ σσ0 +(0 − δΓρρ0 Aγ)g ∇σA + ∇ρA (0 − δΓσσ0 Aγ)g ρρ0 σ γ ρρ0 σ = ∇ρAρ0 δg ∇σA − δΓρρ0 Aγg ∇σA , ρρ0 σ 1 γγ0 ρρ0 σ = ∇ A 0 δg ∇ A − g (∇ δg 0 0 + ∇ 0 δg 0 − ∇ 0 δg 0 ) A g ∇ A , ρ ρ σ 2 ρ ρ γ ρ ργ γ ρρ γ σ σ µν 1 γγ0 µν µν µν ρρ0 σ = ∇ A ∇ A δg + g (g 0 g 0 ∇ δg + g g 0 ∇ 0 δg − g g 0 ∇ 0 δg ) A g ∇ A , µ ν σ 2 ρ µ γ ν ρ ρµ νγ ρ ρµ ρ ν γ γ σ (65)
11 Integration by part,
Z 4 √ d x −gδgLm σ µν g 0 g 0 ∇ (A ∇ A )δg ρ µ γ ν ρ γ σ Z √ 1 0 0 d4 σ µν γγ σ µν ρρ = x −g ∇µAν∇σA δg − g +gρµgνγ0 ∇ρ0 (Aγ∇σA )δg g , 2 σ µν −gρµgρ0ν∇γ0 (Aγ∇σA )δg Z √ 1 1 1 = d4x −g ∇ A ∇ Aσ − ∇ (A ∇ Aσ) − ∇ (A ∇ Aσ) + g ∇ (Aγ∇ Aσ) δgµν, µ ν σ 2 µ ν σ 2 µ ν σ 2 µν γ σ Z √ 1 = d4x −g ∇ A ∇ Aσ − ∇ (A ∇ Aσ) + g ∇ (Aγ∇ Aσ) δgµν, µ ν σ µ ν σ 2 µν γ σ Z √ 1 = d4x −g −A ∇ (∇ Aσ) + g ∇ (Aγ∇ Aσ) δgµν, ν µ σ 2 µν γ σ Z √ 1 = d4x −g −A ∇ (∇ Aσ) + g ∇ (Aγ∇ Aσ) δgµν. (66) (µ ν) σ 2 µν γ σ
We have
δL 1 m = −A ∇ (∇ Aσ) + g ∇ (Aγ∇ Aσ). (67) δgµν (µ ν) σ 2 µν γ σ
and
1 T (A) = 2A ∇ (∇ Aσ) − g ∇ (Aγ∇ Aσ) − (∇ Aρ)2 . (68) µν (µ ν) σ µν γ σ 2 ρ
It is very important to note that we can construct the energy-momentum tensor from the covariant scalar action. This energy-momentum tensor is also conserved. As we have seen that the Lagrangian for matter contains several the dynamical elds. Apart of metric, the
equation of motion for other dynamical elds, for example φ and Aµ, can be obtained from µ the conservation, ∇ Tµν = 0. The existence of energy-momentum conservation corresponds the symmetry called the in nitesimal general coordinate transformation (IGCT).
D. In nitesimal general coordinate transformation
The objective of this section is to see the relation between the conservation of matter and the variation under IGCT. This transformation for the coordinate, xµ reads
xµ → x0µ = xµ + ξµ, (69)
12 where ξµ is a in nitesimal parameter (its value is very small). The variation of action for matter is written as Z 4 √ δSm = d x −gTµνδgµν. (70)
Considering δgµν under IGCT,
0µν µν ρσ gµν → g = g + δg . (71)
Under the full GCT, the metric transforms as follows
∂x0µ ∂x0ν g0µν(x0α) = gρσ(xα). (72) ∂xρ ∂xσ
Putting (69) and keeping to the rst order of ξµ,
0µν 0α µ µ ν ν ρσ α g (x ) = (δρ + ∂ρξ )(δσ + ∂σξ )g (x ),
µν α ρν α µ ρµ α ν ' g (x ) + g (x )∂ρξ + g (x )∂ρξ ,
µν α ρ(µ ν) = g (x ) + 2g ∂ρξ . (73)
See that g0µν(x0α) and gµν(xα) are considered in dierent points. In order to obtain the quantities on the same point, doing the Taylor's expansion of g0µν(x0α) about the point xα as
0µν 0α 0µν 0µν 0ρ ρ g (x ) = g + ∂ρg (x − x ) + ..., x0=x x0=x µν α µν α ρ ' g (x ) + ∂ρg (x ) ξ . (74)
Hence, the variation of metric is
δgµν(xα) = g0µν(xα) − gµν(xα),
ρ µν ρ(µ ν) ' −ξ ∂ρg + 2g ∂ρξ . (75)
Let us compute,
µ ν µρ ν ∇ ξ = g ∇ρξ , µρ ν ν σ (76) = g (∂ρξ + Γρσξ ),
µ ν ν µ ρ(µ ν) ρ(µ ν) σ → ∇ ξ + ∇ ξ = 2g ∂ρξ + 2g Γρσξ , ρ(µ ν) µ ν ν µ ρ(µ ν) σ (77) → 2g ∂ρξ = ∇ ξ + ∇ ξ − 2g Γρσξ ,
13 and
µν ρ µν ρ µ σν ρ ν µσ ρ (78) ∇ρg ξ = 0 = ∂ρg ξ + Γρσg ξ + Γρσg ξ , ρ µν σ(µ ν) ρ (79) → −ξ ∂ρg = 2g Γρσξ .
Plugging (77) and (79) into (75),
µν σ(µ ν)ρ µ ν ν µ ρ(µ ν)σ δg = 2g Γρσξ + ∇ ξ + ∇ ξ − 2g Γρσξ , = ∇µξν + ∇νξµ, = 2∇(µξν). (80)
The variation of action for matter in (70) becomes Z 4 √ (µ ν) δSm = d x −gTµν2∇ ξ . (81)
Since Tµν is symmetric, we can write Z 4 √ µ ν δSm = d x −gTµν∇ ξ , Z 4 √ µ ν µ = d x −g [∇ (Tµνξ ) − ∇ Tµν] , Z 4 √ ν µ ν = d x −g (−ξ ∇ Tµν) + (Tµνξ ) . (82) boundary
Let the in nitesimal parameter vanishes at boundary and we consider ξµ 6= 0 in volume. So, we obtain
µ δSm = 0, → ∇ Tµν = 0. (83)
We can conclude that the conservation for matter is obtained from the variation of action for matter under IGCT.
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