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The AstronomersDerivation Of The Einstein Field Equations (ℍilbert, 1915)
Derivation Of The Einstein Field Equations (ℍilbert, 1915)  • Currently 0.00/5

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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. --------------------------------------------------------------------------------­------------------------- REMARKS: 1. In the derivation use has been made of locally geodesic coordinates. 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). Since, Dₑgᵤᵥ=0. Therefore, since Aᵉ≡gᵘᵛ δΓᵉᵤᵥ - gᵉᵛ δΓᵘᵤᵥ is a tensor, gᵘᵛδRᵤᵥ=∇ₑAᵉ in arbitrary coordinates. Where ∇ is the covariant derivative, and where Dₑ≡∂﻿/∂﻿ξᵉ. The ∂﻿/∂﻿ξᵉ are just ordinary partial derivatives wrt the spacetime variables. For example, ∂﻿/∂﻿ξᵉ((ξᵘ)²)=2ξᵘδᵉᵤ, where δ 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). 3. We can take the accelerating cosmos into account, in the vacuum case given in the clip, by replacing the action "R√(-g)" with "(R-2∧)√(-g)", where ∧ is the `cosmological constant` (which is sometimes taken to be a scalar field). 4. The presence of matter/energy 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/energy. In its most general form J= ∫ₓ [R-2∧+ℒ]√(-g) dξ, where ℒ is the matter/energy 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). The result, Gʲᵏ + ∧gʲᵏ = (constant)Tʲᵏ, are the full-blown field equations. 5. There are many 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. 6. 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. 7. An `exact` solution is NOT necessarily physical ♦ ---------------------------------------------------------------------------------------------------------
Added on Nov 27, 2012 by lonewolf
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