Ricci Tensor of Diagonal Metric

Efficient formulae of Ricci tensor for an arbitrary diagonal metric are presented. Introduction Calulation of the Ricci tensor is often a cumbersome task. In this note useful formulae of the Ricci tensor are presented in equations (1) and (2) for the case of the diagonal metric tensor. Application of the formulae in computing the Ricci tensor of n-sphere is also presented. Derivation The sign conventions and notation of Wald [1] will be followed. Latin indices are part of abstract index notation and Greek indices denote basis components. The Riemann curvature tensor Rabc d is defined by ∇a∇bωc − ∇b∇aωc = Rabc ωd. The Ricci tensor in a coordinate basis is Rλν = ∂ρΓ ρ λν − ∂λΓ ρ νρ − Γ σ λρΓ ρ σν + Γ ρ λνΓ σ ρσ where the Christoffel symbols are Γρλν = g ρσ (∂λgνσ + ∂νgλσ − ∂σgλν) /2. For the rest of this note we assume that gρν is diagonal unless otherwise indicated. Also to avoid having to write down “μ 6= ν” repeatedly we will always take μ 6= ν and no sum will be assumed on repeated μ and ν. We first note that Γμν = 0 if μ 6= ρ 6= ν. It means that at least two indices of Γ must be the same for it to be nonvanishing. Also ln |g| = ∑n σ=1 ln |gσσ| where g and n are determinant of the metric and dimension of manifold, respectively. Then it is easy to show that 2Γσμ = ∂σ ln |gμμ| for all σ and that 2Γ ν μμ = −g ∂νgμμ. (Again μ 6= ν and no sum on μ and ν.) These two equations give 4ΓμμΓ μ νν = g νν (∂νgμμ) g μμ (∂μgνν) = (∂μ ln |gνν |) (∂μ ln |gνν|) = 4Γ μ μνΓ ν μν Using above facts, we get, with each sum having upper limit n and lower limit 1, Rμν =∂μΓ μ μν + ∂νΓ ν μν − (1/2)∂μ∂ν ln |g| − Γ μ μρΓ ρ μν − Γ ν μρΓ ρ νν − ∑ σ 6=μ,ν ΓσμρΓ ρ σν + (1/2)Γμμν∂μ ln |g|+ (1/2)Γ ν μν∂ν ln |g| =(1/2)∂μ∂ν ln | gμμgνν g | − ΓμμμΓ μ μν − Γ μ μνΓ ν μν − Γ ν μμΓ μ νν − Γ ν νμΓ ν νν − ∑ σ 6=μ,ν ΓσσμΓ σ σν + Γμμν∂μ ln √ |gμμ|+ Γ μ μν∂μ ln √ |gνν|+ Γ ν μν∂ν ln √ |gμμ|+ Γ ν μν∂ν ln √ |gνν| + (1/2)Γμμν∂μ ln | g gμμgνν |+ (1/2)Γνμν∂ν ln | g gμμgνν | 1 =(1/2)∂μ∂ν ln | gμμgνν g | − ΓμμμΓ μ μν − Γ μ μνΓ ν μν − Γ ν μνΓ μ μν − Γ ν νμΓ ν νν − ∑ σ 6=μ,ν ΓσσμΓ σ σν + ΓμμμΓ μ μν + Γ μ μνΓ ν μν + Γ ν μνΓ μ μν + Γ ν νμΓ ν νν + (1/2)Γμμν∂μ ln | g gμμgνν |+ (1/2)Γνμν∂ν ln | g gμμgνν | =(1/2)∂μ∂ν ln | gμμgνν g | − ∑ σ 6=μ,ν ΓσσμΓ σ σν + 1 2 Γμμν∂μ ln | g gμμgνν |+ 1 2 Γνμν∂ν ln | g gμμgνν | or (remember μ 6= ν) 4Rμν = (∂μ ln |gνν| − ∂μ) ∂ν ln | g gμμgνν |+ (μ ↔ ν)− ∑ σ 6=μ,ν ∂μ ln |gσσ|∂ν ln |gσσ| (1) where (μ ↔ ν) stands for preceding terms with μ and ν interchanged.* A number of useful facts follow from the above formula. First of all Rμν = 0 if xμ is an ignorable coordinate because ∂μ is in each term of the formula. If there are m ignorable coordinates then maximum number of nonzero off-diagonal components of the Ricci tensor is (n−m)(n−m− 1)/2. Thus a sufficient condition for the Ricci tensor to be diagonal is that the (diagonal) metric be a function only of one coordinate. The Robertson-Walker metric with flat spatial sections, ds = −dt + a(t)(dx + dy + dz), satisfies this condition and its Ricci tensor is consequently diagonal. If the (diagonal) metric is a function only of two coordinates, say xμ and xν , then all off diagonal components of the Ricci tensor except Rμν are zero. For example, for Schwartzschild metric ds = −B(r)dt + A(r)dr + r ( dθ + sin θdφ ) only Rθr needs to be explicitly calculated, all other being zero. It is easily checked using equation (1) that Rθr is also zero. Symmetry arguments can also be given† as to why the Ricci tensor is diagonal in this case. Secondly second derivative terms are identically zero whenever each component of the metric is a product of functions of one single coordinate i.e. if, for all μ, gμμ = ∏n l=1 f l μ(xl) where some of f l μ may be one. A final trivial corollary * Alternative forms of equation (1) are 4Rμν = ∑ σ 6=μ,ν [(∂μ ln |gνν| − ∂μ) ∂ν ln |gσσ|+ (μ ↔ ν)− ∂μ ln |gσσ|∂ν ln |gσσ|] and 8Rμν = ∑ σ 6=μ,ν [( ∂μ ln g νν |gσσ| − 2∂μ ) ∂ν ln |gσσ|+ (μ ↔ ν) ] † Page 178 of Weinberg [2]. 2 of equation (1) is that the Ricci tensor is diagonal in 2-dimensions. This fact also follows trivially from the fact that in 2-dimensions, the Ricci tensor is the metric tensor (not necessarily diagonal) up to a factor of a scalar function. When n = 3, it is easy to find an example of a diagonal metric which results in a non-diagonal Ricci tensor. For example, ds = dx + xdy + ydz gives 4Rxy = 1/xy. We now calculate the diagonal components of Ricci tensor. We have Rμμ =∂σΓ σ μμ − ∂ 2 μ ln √ |g| − ΓσμρΓ ρ σμ + Γ ρ μμ∂ρ ln √ |g| =∂ μ ln √ |gμμ|+ ∑ σ 6=μ ∂σΓ σ μμ − ∂ 2 μ ln √ |g| − ΓμμρΓ ρ μμ − ∑ σ 6=μ ΓσμρΓ ρ μσ + Γ μ μμ∂μ ln √ |g|+ ∑ σ 6=μ Γσμμ∂σ ln √ |g| =(1/2)∂ μ ln | gμμ g |+ ∑ σ 6=μ ∂σΓ σ μμ − ( Γμμμ )2 − ∑ σ 6=μ ΓμμσΓ σ μμ − ∑ σ 6=μ ΓσμμΓ μ σμ − ∑ σ 6=μ ΓσσμΓ σ σμ + Γ μ μμ∂μ ln √ |gμμ|+ Γ μ μμ∂μ ln √