Derivation Of The Einstein Field Equations (Hilbert, 1915)

Find the original derivation of Gʲᵏ=0 in: "Die Grundlagen der Physik. (Erste Mitteilung)" in Nachrichten von der Königlichen Gesellschaft der Wissenschaften zu Göttingen. Math.-phys. Klasse. 1916. Issue 8, p. 395-407. Presented 20 November 1915.
•TYPO: Replace ∂/∂e with ∂/∂ξᵉ wherever it appears.
•NOTATION: Dᵥ ≡ ∂/∂ξᵛ v=1,2,3,4.
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•REMARKS:
1. In the derivation locally geodesic coordinates have been used. For example, gᵘᵛδRᵤᵥ=(√(-g))⁻¹Dₑ((√(-g))(gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ))=(√(-g))⁻¹(gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ)Dₑ((√(-g)) + Dₑ(gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ)=(√(-g))⁻¹(gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ)(½)((√(-g))⁻¹ggᵘᵛDₑgᵤᵥ + Dₑ(gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ)=Dₑ(gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ). Having used Dₑgᵤᵥ=0.
Since Aᵉ ≡ gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ is a tensor then we have gᵘᵛδRᵤᵥ=∇ₑAᵉ in arbitrary coordinates. Where ∇ is the covariant derivative, and ∇ₑAᵉ is the covariant derivative of Aᵉ in the direction of the e-th basis vector.
The ∂/∂ξᵉ are just ordinary partial derivatives w.r.t. the spacetime variables. For example,
∂/∂ξᵉ((ξᵘ)²)=2ξᵘδᵉᵤ, where δᵉᵤ is the Kronecker delta.
2. √(-g) dξ is the volume element in E₄, and √(-g)=1 when g=det[diag(1,-1,-1,-1)]= -1 (special relativity).
In the integrand of J= ∫ₓ R√(-g) dξ there appears the Ricci scalar (R). The Ricci scalar is the only invariant which contains the components of the line element and its derivatives only up to 2nd order, and linearly. In local coordinates, R=gᵃᵉ(Γᵘₐₑ,ᵤ - Γᵘₐᵤ,ₑ + ΓʳₐₑΓᵘᵤᵣ - ΓʳₐᵤΓᵘₑᵣ).
For the (lengthy) details about the mathematical theory of the Hilbert action see:
'Tensors, Differential Forms, and Variational Principles',
By David Lovelock, Hanno Rund (Chapter 8).
3. The presence of 'matter' produces curvature in spacetime.
The Einstein tensor (G) is defined by Gʲᵏ≐Rʲᵏ - ½gʲᵏR. The notation, Ricʲᵏ≡Rʲᵏ (`Ric` for Ricci) often appears.
Einstein assumed (which is NOT mathematically derivable) that, Gʲᵏ=(constant)Tʲᵏ. Where T (the energy-momentum/stress-energy tensor) is due to the presence of 'matter'. Constant=8πG/c⁴ (SI units), needed to fit in with the Newtonian limit.
In its most general form J= ∫ₓ [R-2∧+ℒ]√(-g) dξ, where ℒ is the 'matter' Lagrange density s.t., δ ∫ₓ ℒ√(-g) dξ=∫ₓ δ(ℒ√(-g)) dξ=∫ₓ [(constant)Tᵘᵛ]δgᵤᵥ√(-g) dξ ⇒ δ(ℒ√(-g))/δgᵤᵥ=√(-g)(constant)Tᵘᵛ (by the variational lemma). And ∧ is the 'cosmological constant', which is sometimes taken to be a scalar field.
In their most general form the Einstein field equations are (2nd order & quasilinear in the gʲᵏ),
Gʲᵏ + ∧gʲᵏ = (constant)Tʲᵏ.
a. Basically, the EFE's are a nonlinear concoction of the components of a generic metric tensor (and its first and second derivatives with respect to the spacetime variables) which if possible to determine (given additional side conditions) may, or may not, be physical. If it can be determined, and is found to be physical, then that metric tensor describes a facet of Einstein's spacetime.
b. It is important to point out that the components of the Einstein tensor in the integrand @1:18 in the clip are in fact the Euler-Lagrange equations of the variational problem. See Lovelock/Rund for the details.
4. There are many known types of energy-momentum tensors. I mention two:
i. Tʲᵏ=μuʲuᵏ. A dust cloud (diffuse, non-interacting 'matter' in vacuum): This has density (mass per volume) μ, and in an element of volume dV an observer moving with 4-velocity u=γ(c,v) (SI units) measures mass/energy μdV.
ii. Tʲᵏ=(constant)[ℱʲᵤℱᵏᵘ - ¼gʲᵏℱᵘᵛℱᵤᵥ], where ℱ is the EM field tensor. This energy-momentum tensor represents the free photon field in vacuum.
NOTE: This energy-momentum tensor shows that light can also be a source of 'gravity'.
5. One of the many interesting exact solutions of the Einstein field equations are the Weyl (Hermann Weyl (1885 -1955)) metrics,
ds²=e²ᵘdt²-e²⁽ᵛ⁻ᵘ⁾(dρ²+dz²)-ρ²e⁻²ᵘdφ² --(*).
Where c=1 (geometric units) and (ρ,φ,z) are cylindrical coordinates, as viewed in 3d Euclidean space. When viewed in 4d spacetime they`re known as Weyl`s canonical coordinates. And u,v are functions of ρ,z.
In cartesian coordinates (x=ρcosφ, y=ρsinφ) (*) takes the form,
ds²=e²ᵘdt²-ρ⁻²e²⁽ ᵘ⁻ᵛ ⁾(xdx+ydy)²-ρ⁻²e⁻²ᵛ(xdy-ydx)²-e²⁽ ᵘ⁻ᵛ ⁾dz²
6. An 'exact' solution is NOT necessarily physical ♦
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[@LeconsdAnalyse]

3. Addendum, Apply gᵤᵥ to Rᵘᵛ - ½gᵘᵛR=(constant)Tᵘᵛ -(*) Use gᵤᵥgᵘᶠ=δᵥᶠ, R=gᵤᵥRᵘᵛ, and T=gᵤᵥTᵘᵛ to get: R - 2R=(constant)T or, -R=(constant)T -(**). Substituting (**) into (*) gives: Rᵘᵛ=(constant)(Tᵘᵛ - ½gᵘᵛT). So that when Tᵘᵛ=0 (i.e., when there are no matter fields, electromagnetic fields, or any other external fields) we have Rᵘᵛ=0 (or, Ricᵘᵛ=0), which is _equivalent_ to (the Euler-Lagrange equations) Rᵘᵛ - ½gᵘᵛR=0. NOTES: 1. Since, gg⁻¹=1 or, gₐₑgᵃᶠ=δₑᶠ=diag(1,1,1,1), gₐₑgᵃᵉ= tr [diag(1,1,1,1)]=1+1+1+1=4. 2. Rᵘᵛ=(constant)(Tᵘᵛ - ½gᵘᵛT) is sometimes referred to as the *trace-reversed* form of the Einstein equations (∧=0). And, Tᵘᵛ - ½gᵘᵛT is referred to as the *trace-reversed* form of the stress-energy tensor (or, energy-momentum tensor).

[@mikevaldez7684]

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[@Dudewattimeizit]

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[@hyperbolicitylevi-civita8995]

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[@LeconsdAnalyse]

+Chen Bobby kzbin.info/www/bejne/Z3alqHqCebN9jtE @4:12

[@hyperbolicitylevi-civita8995]

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[@sylvanatup8423]

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