Variational Theory in the Context of General Relativity Abstract

Variational theory in the context of general relativity

Hiroki Takeda1, ∗

1Department of Physics, University of Tokyo, Bunkyo, Tokyo 113-0033, Japan (Dated: October 25, 2018) Abstract This is a brief note to summarize basic concepts of variational theory in general relativity for myself. This note is mainly based on Hawking(2014)”Singularities and the geometry of spacetime” and Wald(1984)”General Relativity”.

∗ [email protected]

1 I. BRIEF REVIEW OF GENERAL RELATIVITY AND VARIATIONAL THEORY

A. Variation

defnition I.1 (Action) Let M be a differentiable manifold and Ψ be some tensor fields on M . Ψ are simply called the fields. Consider the function of Ψ, S[Ψ]. This S is a map from the set of the fields on M to the set of real numbers and called the action.

a···b Ψ actually represents some fields Ψ(i) c···d(i = 1, 2, ··· , n). I left out the indices i that indicate the number of fields and the abstract indices for the tensors. The tensor of the a···b a···b tensor fields Ψ(i) c···d(i = 1, 2, ··· , n) at x ∈ M is written by Ψ(i)|x c···d(i = 1, 2, ··· , n).

Similarly I denote the tensor of the tensor ﬁelds Ψ at x ∈ M by Ψ|x.

defnition I.2 (Variation) Let D ⊂ M be a submanifold of M and Ψ|x(u), u ∈ (−ϵ, ϵ), x ∈ M be the tensors of one-parameter family of the tensor ﬁelds Ψ(u), u ∈ (−ϵ, ϵ) at x ∈ M such that

(1) Ψ|x(0) = Ψ in M , (1)

(2) Ψ|x(u) = Ψ in M − D. (2)

Then I deﬁne the variation of the ﬁelds as follows,

dΨ(u) δΨ := . (3) du u=0

In this note, I suppose the existence of the derivatives dS/du|u=0 for all one-parameter family Ψ(u).

defnition I.3 (Functional derivative) I assume the existence of the smooth tensor ﬁelds χ which is dual to ψ such that ∫

dS = χδΨ, (4) du u=0 D for any one-parameter family. 1 Then we say that S is functionally diﬀerentiable at Ψ(0). We call χ the functional derivative of S and denote it as

δS := χ.2 (5) δΨ Ψ0

1 If the type of tensor ﬁeld of Ψ is (k, l), that of χ is (l, k). Here contraction of all indices in the integral is

understood. 2 More generally, if there exists a tensor distribution χ such that dS/du|u=0 = χ[δΨ], we say that S is

2 defnition I.4 (Lagrangian) Consider a functional form of S ∫ S[Ψ] = L[Ψ] (6) D where L is a local function of ψ and a ﬁnite number of its derivatives,

k L|x = L(Ψ(x), ∇Ψ(x), ··· , ∇ Ψ(x)). (7)

L is called by Lagrangian density.

defnition I.5 (Variational principle) Let S be functionally diﬀerentiable. We require that the ﬁelds Ψ maximize S,

dS = 0. (8) du u=0 This principle is called by variational principle. Hence, the following equation is hold by the deﬁnition of the functional derivative and the following fundamental lemma of variational calculus.

δS = 0. (9) δΨ Ψ0 This gives us the fundamental equations for the ﬁelds.

lemma I.1 (Fundamental lemma of variational calculus) Let D ⊂ M be a subman-

ifold of M and Ψ|x(u), u ∈ (−ϵ, ϵ), x ∈ M be any one-parameter family satisfying a proper boundary conditions. The following is held. ∫

dS = χδΨ = 0 ⇒ χ = 0. (10) du u=0 D This statement is called by fundamental lemma of variational calculus.

B. General Relativity

defnition I.6 (Space-time) Space-time is the set of events. In general relativity, the mathematical model of space-time is (M , g) where M is a connected four-dimensional Hausdorﬀ C∞ manifold and g is a Lorentz metric 3on M .

functionally diﬀerentiable and call χ the functional derivative of S at Ψ(0). Here the tensor distribution

is a distribution whose type is tensor, so called generalized function in Japanese. The bracket on the ∫ right hand side denotes the scalar product (distribution) which is deﬁned by χ[δΨ] =< χ, δΨ >= D χδΨ. (Schwarz, 2008) 3 Its signature is +2

3 First, I derive the fundamental equations for fields by variational method. Hereafter I a···b refer to the fields Ψ as Ψ(i) c···d, (i = 1, 2, ··· , n) without omitting the indices. I assume the form of the action is Lagrangian form and the Lagrangian depends on the fields and their first covariant derivatives. a···b Consider the variation of the fields Ψ(i) c···d, (i = 1, 2, ··· , n). Then ∫ ( ) dS ∂L ∂L a···b − a···b = Σ ··· δΨ(i) c···d ··· δ(Ψ(i) c···d;e) . (11) i a b a b du u=0 D ∂Ψ(i) c···d ∂Ψ(i) c···d;e

a···b a···b Sinceδ(Ψ(i) c···d;e) = (δΨ(i) c···d);e, the second term can be written as   ∫ ( ) ( ) ∂L ∂L  a···b − a···b  Σ a···b δΨ(i) c···d a···b δΨ(i) c···d . (12) i D ∂Ψ(i) c···d;e ∂Ψ(i) c···d;e ;e ;e

According to Stokes’ theorem, the ﬁrst term can be written as ∫ ∫ a a Q ;a = Q dσa, (13) D ∂D where e ∂L a···b Q := Σ a···b δΨ(i) c···d. (14) i ∂Ψ(i) c···d;e a···b This integral have to vanish because δΨ(i) c···d = 0 is held at the boundary ∂D from the second condition of variation Eq. (2). Finally we get the following expression,   ∫ ( ) dS ∂L ∂L  −  a···b = Σ a···b a···b δΨ(i) c···d = 0. (15) du i D ∂Ψ(i) c···d ∂Ψ(i) c···d;e u=0 ;e

Thus, we can get the following Euler-Lagrange equations from fundamental lemma of variational calculus Eq. (10), ( ) ∂L − ∂L χ = a···b a···b = 0. (16) ∂Ψ(i) c···d ∂Ψ(i) c···d;e ;e

Then we get the energy-momentum tensor from the Lagrangian by considering a variation with respect to the metric. Suppose a variation of the metric gab(u, x) does not change the a···b ﬁelds Ψ(i) c···d. Then, ∫ ( ) ∫

dS ∂L a···b ∂L ∂ϵ = Σ ··· δ(Ψ(i) c···d;e) + δgab + L δgab. (17) i a b du u=0 D ∂Ψ(i) c···d;e ∂gab D ∂gab

4 ∫ ∫ The last term comes from the fact that the integral is deﬁned by f = fϵ. Since ϵ = −1 1/2 1 2 3 4 (4!) η, ηabcd = (−g) 4!δ[a δb δc δd] and g := det(gab),

∂ηabcd 1 −1/2 ∂g 1 2 3 4 = − (−g) 4!δ[a δb δc δd] . (18) ∂gef 2 ∂gef Thus, we obtain ∂ϵ 1 = gabϵ. (19) ∂gab 2 Here I use the the fact that the cofactor of the matrix can be expressed as a product of the component of the inverse matrix and the determinant od the matrix, that is the relation ef a···b ∂g/∂gef = g g to evaluate ∂g/∂gef . The ﬁrst term arises because δ(Ψ(i) c···d;e) does a···b not necessarily become zero unlike δΨ(i) c···d due to a variation of the components of the a a a connection Γ bc induced by a variation of the metric. ∂Γ bc/∂u = δΓ bc should be a tensor because the diﬀerence between two connections should be a tensor. This can be calculated as follows, 1 δΓa = gad{(δg ) + (δg ) − (δg ) }. (20) bc 2 db ;c dc ;b bc ;d One can verify this equation by checking that this equation is held in normal coordinates at a p where the components Γ bc and the coordinate derivatives of the components gab vanish because this equation is a tensor equation. For Riemann normal coordinate, ( ) 1 ∂δg ∂δg ∂δg (l.h.s) = gad cd + bd − bc , (21) 2 ∂xb ∂xc ∂xd

a and, since the components Γ bc at p induced by the covariant derivatives vanish, 1 (r.h.s) = gad{(δg ) + (δg ) − (δg ) } = (l.h.s.). (22) 2 db ;c dc ;b bc ;d

If the tensor equation is held in a speciﬁed coordinate, it is held in any coordinates. This method is often used to derive equations in relativity. However, One should be careful to calculations including the variation and the covariant derivative. Note that δ(A + B) = δA + δB, δ(AB) = (δA)B + A(δB) and δ(∂A) = ∂(δA). Especially, the variation operator does not commute the covariant derivative for any tensors g (here its type is (0, 2) as an example), ∇ ∇ d d a (δgbc) = δ ( agbc) + gdc δΓba + gbd δΓac, (23) when one consider the variation of the metric because a change of the connection is induced. This can be easily derived by the above three equations. On the other hand, the variation

5 operator does commute the covariant derivative when you consider the variation of the ﬁelds except for the metric as I mentioned above. Eq. (19) can be also derived by this equation straightforward. One also should be careful to that the since the metricity condition holds

only for the unperturbed metric gab, ∇(g(0)) = 0 does not mean that ∇(g(∆u)) = 0. Using

to leave only terms multiplied by δgab, The first term in Eq. (17) can be expressed by the terms involving only δΓ by using the similar relation as Eq. (23). Then, using Eq. (19) and the integration by parts, it finally can be expressed by the only terms involving only δg. Thus, the energy-momentum tensor T ab can be defined as follows, ∫

dS 1 ab = (T δgab). (24) du u=0 2 D The energy-momentum tensor satisﬁes the conservation equations. Let ϕ be a diﬀeomor- phism which is is the identity except in the interior of D. Then, ∫ ∫ ∫ ∫ 1 1 1 S = L = Lη = Lη = ϕ∗(Lη). (25) D 4! D 4! ϕ(D) 4! D

Here I used the condition that ϕ is the identity except in the interior of D for the third equality, and the invariance of integrals under a differential map for the forth equality. Thus, ∫ 1 (Lη − ϕ∗(Lη)) = 0. (26) 4! D If ϕ is generated by a vector field X which is non-zero only in the interior of D, ∫ 1 LX (Lη) = 0, (27) 4! D by the definition of the Lie derivatives. On the other hand,   ∫ ( ) ∂L ∂L  −  a···b (l.h.s.) = Σ a···b a···b LX Ψ(i) c···d i D ∂Ψ(i) c···d ∂Ψ(i) c···d;e ∫ ;e 1 ab + T LX gab 2∫ D (28) 1 ab = T LX gab ∫2 D ab ab = ((T Xa);b − T ;bXa). D

I used the ﬁeld equation for the second equality and the relation LX gab = 2X(a;b) for the third equality. The ﬁrst term can be an integral over ∂D by Stokes’ theorem. It can vanish

6 because X vanishes there. Thus, ∫ ab T ;bXa = 0, (29) D is satisﬁed for all vector ﬁelds X. Hence, the conservation law of the energy-momentum tensor is derived, ab T ;b = 0. (30)

ACKNOWLEDGEMENTS

Thank you, everyone.