arXiv:physics/9802027v1 [math-ph] 14 Feb 1998 soege rmoepitt nte.I ehv etrwritt vector a have we If follows, another. as to basis point its ba one of the from of direction goes and one length as the that fact the with familiar nadffrn a hnatno tpoint at tensor a than way different a in opiainaie rmtefc htatno ttodiffer differenc two a poi at at tensor of tensor the a value is, that That limiting fact differently. a the transform space-time from as arises of complication thought A be can derivative Introduction 1 derivatives. ordinary represent and h olwn oaini sd h erctensor, metric The used: is notation following The Conventions g a ehv tensor a have we Say lob fuet rvttoa hscssa uc referen quick a as physicists gravitational to use This of ver notation. only be modern assumes also and and analysis student tensor beginning with the familiarity to Thi use of geometry. space-time be trivial should non th a account introducing of into tions taking investigated Famil are Relativity. procedures General matical in techniques mathematical tain = hspprpeet re u opeesv introduction comprehensive but brief a presents paper This | det nerto nGnrlRelativity General in Integration ( g µν ) | eicln eoecvratdrvtvswiecommas while derivatives covariant denote Semi-colons . T nrwDeBenedictis Andrew hn ieteprildrvtv,tecovariant the derivative, partial the like then, e.0,1995 03, Dec. V Abstract = 1 V α e r α + dr nplrcodntsw are we coordinates polar In . g µν a intr f+2 of signature a has , nt i etr change vectors sis transcript s complica- e ce. r a mathe- iar ae will paper iltransform will n onsin points ent basic y ocer- to ni terms in en quotient. e (1) the derivative with respect to co-ordinate xβ would be: ∂V α ∂e e + V α α . (2) ∂xβ α ∂xβ µ th We define the Christoffel symbol Γαβ as representing the coefficient of the µ ∂eα component of ∂xβ . The above derivative becomes (after relabelling dummy indices), α µ α V ,β +V Γµβ eα. (3) When it comes to integration, we are performing
the reversal of a partial differentiation and can therefore not just integrate a covariant derivative. Also, an integral over tensor components does not give a result which is a tensor whereas integration over a scalar does. ν We can convert expressions such as P;ν into an expression containing only ν partial derivatives as follows: First write out the expression P;ν in terms of the Christoffel symbol ν ν λ ν P;ν = P,ν + P Γλν. (4) Now use the fact that 1 Γν = gνα (g + g g ) (5) λν 2 να,λ αλ,ν − λν,α 1 1 = gνα (g g )+ gναg . 2 αλ,ν − λν,α 2 να,λ The first term in the last expression is equal to zero since it is gνα multiplied by a tensor which is antisymmetric in ν, α. Therefore: 1 Γν = gναg . (6) λν 2 να,λ να Using the fact that g,λ = gg gαν,λ gives,

ν 1 1 Γ = g g− (7) λν 2 ,λ √g = ,λ = (ln √g) . √g
,λ We can now write
√g P ν = P ν + P λ ,λ (8) ;ν ,ν √g
1
= (√gP ν) .(after relabelling dummy indicies) √g ,ν

2 This result is useful because it allows us to apply Gauss’ law which we know applies to partial derivatives. Gauss’ law states that the volume integral of a divergence can be re-written as an integral over the boundary surface as follows: α α P dV = P nαdS. (9) Z ,α I

Where nα is the outward unit normal to theb surface. In our case we need to integrate over proper volume and therefore must use proper surface area whose elementb is √gd3S. Therefore, in general relativity, Gauss’ law is gen- eralized to

α 4 ν 4 ν 3 P √gd x = (√gP ) d x = P nν√gd x. (10) Z ;α Z ,ν I b 1.1 Tensor Densities Ordinary tensors transform according to the following transformation law:

µ β µ ∂x′ ∂x α ′ Tν = α ν Tβ . (11) ∂x ∂x′ An object which transforms according to (11) is called a tensor density of weight zero. A tensor density of weight w transforms as follows: ℑ w µ β µ ∂x ∂x′ ∂x α ′ = , (12) ℑν ∂x ∂xα ∂x ν ℑβ ′ ′ which is similar to (11) except for the Jacobian term raised to the power of w. We can convert tensor densities of weight w to ordinary tensors by noting the transformation of the metric’s determinant.

β ′ ′ α g′ = gγ κ = gαβAγ′ Aκ′ (13) | | ∂x α ∂xβ = g | αβ| ∂x γ ∂x κ ′ ′ ∂x 2 = g. ∂x ′

Therefore we can write w w µ β w/2 µ ∂x − ∂x ∂x′ ∂x α (g′)− ′ = (14) ℑν ∂x ∂x ∂xα ∂x ν ℑβ ′ ′ ′

3 which transforms like an ordinary tensor (i.e. tensor density of weight zero). It is these types of tensor densities which we want to consider when integrat- ′ 4 ∂x 4 ing. For example, consider the volume element d x′ = ∂x d x. The corre- sponding invariant volume element (the proper element) is √ g d4x = √gd4x. ′ We see that d4x has a weight of -1 since √g has a weight of +1.

The covariant derivative of a scalar density of arbitrary weight The scalar field of weight w transforms as

∂x w Φ′ = Φ. (15) ∂x ′

Taking the derivative of this creature we get

w α w ′α 2 β ∂Φ′ ∂x ∂Φ ∂x ∂x ∂x ∂ x ′ = ′ + w ′ ′ Φ. (16) ∂x ι ∂x ∂xα ∂x ι ∂x ∂xβ ∂x ι∂x α ′ ′

Noting that the transformation property of the Christoffel symbol is

α 2 σ ′α ′α σ ∂x ∂ x ∂x Γ =Γ ′ + ′ ′ . (17) αι σα ∂x ι ∂x α∂x ι ∂xσ

This equation can be multiplied by wΦ′ and subtracted from the previous one to get w α ∂Φ′ ′α ∂x ∂Φ σ ∂x ′ wΦ′Γ = wΦΓ ′ (18) ∂x ι − αι ∂x ∂xα − σα ∂x ι ′ which displays the transformation properties of Φ and is its covariant deriv- ative. The above result can be used to find the covariant derivative of a tensor density of arbitrary weight. Let ℘µ be a contravariant tensor of weight w which we want to take the covariant derivative of. This can be written as follows,

µ w w µ − ℘;ρ = √g √g ℘ ;ρ w w
µ w w µ − − = (√g );ρ √g ℘ + √g √g ℘ ;ρ . (19)
The first term in the last expression is equal to zero from (18) and noting (7). The second term is a covariant derivative of a tensor of weight zero

4 multiplied by the factor √gw. The expression therefore equals

w w µ w λ µ − − √g √g ℘ ,ρ + √g ℘ Γλρ (20)  
√g
= ℘µ + ℘λΓµ w ,ρ ℘µ. ,ρ λρ − √g

This argument can be extended to give the covariant derivative of an arbi- trary rank tensor of arbitrary weight w,

α1α2.. α1α2.. µα2.. α1 Tβ1β2..;ρ = Tβ1β2..,ρ + Tβ1β2..Γµρ + ... (21) √g T α1α2..Γν ... w ,ρ T α1α2... − νβ2.. β1ρ − − √g β1β2..

1.2 Integrals of Second Rank Tensors Second rank tensors are most easily handled if they are antisymmet- ric. Consider an antisymmetric second rank tensor F αβ. We can take the following covariant derivative:

αβ αβ αµ β µβ α F;β = F,β + F Γµβ + F Γµβ. (22)

Since F αβ is antisymmetric and the Christoffel symbols are symmetric the µβ α F Γµβ term vanishes leaving:

αβ αβ αµ β F;β = F,β + F Γµβ. (23)

√g β ( ),µ As before, we write Γµβ = √g giving (after relabelling dummy indices)

√g F αβ = F αβ + F αβ ,β . (24) ;β ,β √g

Therefore, similar to the vector case 1 F αβ = √gF αβ . (25) ;β √g ,β

5 1.3 Killing Vectors We can exploit symmetries in the space-time to aid us in integration of second rank tensors. For example, does the metric change at all under a translation from the point x = xµ to x = xµ + ǫkµ(x)? This change is measured by the Lie derivative of the metric along k, e gµν (x) gµν(x) £kgµν = lim − . (26) ǫ 0 ǫ → e If the metric does not change under transport in the k direction then the Lie derivative vanishes. This condition implies the Killing equation

kν;µ + kµ;ν =0. (27) The solutions (if any) to this equation are called killing vectors.

Time-like Killing vector of a spherically symmetric space-time: Consider the following line element:

ds2 = α2dt2 + a2dr2 + r2dθ2 + r2 sin2(θ)dφ2 (28) − where α and a are functions of the coordinates. The metric will be stationary if the metric is time independent in some coordinate system.. That is, ∂gµν =0. (29) ∂x0 Where x0 is a time-like coordinate. We write out the full expression for the Lie derivative of the metric

γ γ γ £kgµν = k gµν,γ + gµγ k,ν + gνγ k,µ. (30) Setting this equal to zero, a time like solution satisfying this equation is the vector field α α k = δ0 . (31) Substituting this into (30) we get

γ £kgµν = δ0 gµν,γ (32)

α which equals zero from (29). Therefore δ0 is a killing vector field for the stationary spherically symmetric space-time.

6 Consider the conservation law

µν T;ν =0 (33) Where T is the stress energy tensor. We cannot integrate over this as we did in the previous section since we would not be integrating over a scalar (due to the presence of a free index). Therefore in general there is no Gauss’ law for tensor fields of rank two or higher. If we can find a killing vector field in the space we can use the Killing equation to form the following equation:

µν µν µν (kµT );ν = kµ;νT + kµT;ν =0 (34) (note that the second term equals zero from (27) and therefore the first term equals zero as well). We then proceed as follows:

µν ν (kµT );ν =0= J;ν (35) to which we can apply Gauss’ law as before.

ν 4 ν 3 (√gJ ) d x = J nν √gd x (36) Z ,ν I b The Energy of a Scalar Field: The Einstein field equations can be written in mixed form as 1 8π T µ δµT = Rµ. (37)  ν − 2 ν  ν

∂ If we choose a time-like killing vector k(t) = ∂t , we can form a mass integral of the form 1 µ µ ν M = (T δ T ) k dSµ. (38) −16π Z ν − ν (t) 3 Where dSµ = nµ√gd x has the following components:

3 1 2 3 0 2 3 0 1 3 0 1 2 nµd bx = dx dx dx , dx dx dx , dx dx dx , dx dx dx . (39)
Equation (38)b can be integrated to give the energy of the scalar field by noting that the scalar field has the same stress-energy tensor as a pressure=density perfect fluid. The stress-energy tensor of the real scalar field can also be written as 1 T αβ = (1/4π) φ;αφ;β gαβφ φ;γ . (40)  − 2 ;γ 

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