How to derive Einstein Field Equation?
As an undergraduate student I was very attached to the 6 inch equation that recites the poetry of space time in General Relativity- the Einstein Field Equation. General Theory of Relativity, is simple if we understand how space and time behaves with matter. It tells that the space and time is not absolute quantities. It rather stretches and compresses according to the amount of matter of the body that curves. Legend has it, “Body tells space-time how to curve and space-time tells body how to move.” This line can be inscribed by a mathematical equation called Einstein Field Equation. If you want to know how to derive the equation, then it will be the note of benefit for you: Even it took 10 years for the greatest mind of all time Albert Einstein to figure it out.
To derive it you must have basic acknowledgement of General Relativity such as Cristofel symbols and Geodesics and tensors of course. If you are elligible of this lets derive it! There is 2 way of getting it-
i) Einstein’s Heuristic Method (It was done originally done by A. Einstein
ii) Hilbert Action Method (It was done by mathematician David Hilbert way before Einstein.
Einstein’s Heuristic Method
Even though it took 10 years for Einstein, his way of deriving is the only easiest way to derive the Einstein’s Field Equation. Let us start from simple equation from the topic of gravity which is taught at beginning classes of undergrad course.
∇. ∇Φ = 4πGρ
It is called Poisson equation or Gauss Law for gravity. Where ∇Φ is gravitational potential, ρ is matter field (density of object) and G is gravitational constant.
So, Divergence of gravitatoinal Potential = 4πG matter field
Geodesics equation is just acceleration of experienced by free-fall particle.
aᵘ = ∂²xᵘ/∂τ² = -Γᵨᵥᵘ x’ᵖx’ᵛ
Fᵘ/m ≡ aᵘ = -Γᵨᵥᵘ ∂²(xᵖ/∂τ)(xᵛ/∂τ) (F = ma)
Fᵘ/m ≡ aᵘ = -Γ₀₀ᵘ ∂²(x⁰/∂τ)(x⁰/∂τ) (x⁰ = t)
Fᵘ/m ≡ aᵘ = -Γ₀₀ᵘ
We can write Γ₀₀ᵘ in terms of metric tensors
-Γ₀₀ᵖ = ½gᵖᵒ(∂₀g₀ₒ + ∂₀g₀ₒ + ∂ₒg₀₀)
[∂₀g₀ₒ = 0, is Killing vector field]
Γ₀₀ᵖ = ½gᵖᵒ∂ₒg₀₀
We know from Newtonian mechanics, that gravitational force is gradient of gravitational potential.
(dΦ)ᵖ = ½gᵖᵒ∂ₒg₀₀ = Γ₀₀ᵘ
gᵨₒ(dΦ)ᵖ = ½gᵨₒgᵖᵒ∂ₒg₀₀
Using Stoke’s theorem d → ∇
∇ₒΦ = ∂ₒg₀₀
Φ = g₀₀ + constant
Comparing with Gauss’s Law.
∇. ∇g₀₀ = 4πGρ
Above equations show that metric is diverting (or converging) for ρ matter field. If ρ = 0 then ∇. ∇g₀₀ = 0 . However, LHS of above equation is similar to Ricci Tensor R₀₀, which show flow curve diverting or converging for given metric.
g₀₀ = 4πGρ
We can write matter density ρ as time component of (0, 2) tensor T.
g₀₀ = 4πGρ T₀₀
Let’s make above equation a general tensor equation for spacetime. For dimensionality we will ignore 4πG, instead we’ll use arbitrary constant ω,
Rᵤᵥ = ωTᵤᵥ
Where Tᵤᵥ, is (0, 2) tensor for matter field, called energy-momentum tensor.
The problem with this equation is when
∇. Tᵤᵥ = 0, ∇. Rᵤᵥ ≠ 0 unlike Gauss’s Law.
This creates problem of conversation of mass/energy. For any system, the amount of incoming matter should equal to amount of outgoing matter. In technical words, this implies continuity equation.
Hey Layman ! Divergence = 0 means conservation of mass.
Divergence of Ricci Tensor
Let’s compute the divergence of Ricci Tensor
∇ᵥRᵘᵛ = ∂ᵥRᵘᵛ + ΓᵥᵨᵘRᵛᵖ + ΓᵥᵨᵛRᵘᵖ
∇ᵥRᵘᵛ = ∂ᵥRᵘᵛ + ΓᵥᵨᵘRᵛᵖ + ∂ᵨ(ln√detg)Rᵘᵖ
∇ᵥRᵘᵛ = ∂ᵥRᵘᵛ + ΓᵥᵨᵘRᵛᵖ + ½∂ᵨ(ln (detg))Rᵘᵖ
Notice that the 3rd is logarithmic expansion of volume (detg) along ρ. Hence, divergence of Ricci Tensor is non-zero. Thus, we require a better LHS for field equation.
The idea is quite simple we need to kill diagonal terms of Ricci Tensor, that is, Ric — trace(Ric). {There are various ways to show this.}
Bianchi Identity
∇ₖRᵣₛᵤⁱ — ∇ₛRᵣₖᵤⁱ — ∇ᵣRₖₛᵤⁱ = 0
Contraction on Bianchi Identity by i = r
∇ₖRₛᵤ — ∇ₛRₖᵤ — ∇ᵢRₖₛᵤⁱ = 0
Multiplying by metric gˢᵘ for further contractions
gˢᵘ(∇ₖRₛᵤ — ∇ₛRₖᵤ — ∇ᵢRₖₛᵤⁱ) = 0
-∇ₖRₛˢ + ∇ₛRₖˢ + ∇ᵢRₖⁱ = 0
-∇ₖRₛˢ + 2∇ᵢRₖⁱ = 0
-∇ₖR + 2∇ᵢRₖⁱ = 0
-∇ᵢgₖⁱR + 2∇ᵢRₖⁱ = 0
∇ᵢ(Rₖⁱ — ½gₖⁱR) = 0
Re-writing in covariant form using spacetime indeces
∇ᵥ(Rᵤᵥ — ½Rgᵤᵥ) = 0
∇ᵥGᵤᵥ = 0
Where Gᵤᵥ is Einstein Tensor. And its divergence is zero, ∇.Gᵤᵥ = 0.
This implies flow of energy is same, i.e., energy is conserved.
Einstein’s Field Equation
Finally, the wait is over! Now, we can compare Gauss’s Law tensor in full generality on spacetime manifold.
∇. ∇Φ = 4πGρ
Rᵤᵥ — ½Rgᵤᵥ = 8πGTᵤᵥ
The above equation can be written in many formats. However, in cosmology this equation is written using cosmological constant Λ.
The cosmological constant was initial introduced by Einstein, but he omitted it in later stating; ‘the biggest blunder of his life’. He assumed that the universe is static. But later on Hubble showed experimentally that universe is expanding. After this discovery Friedmann solved the Einstein’s equation with cosmological constant for the expanding universe. Friedmann’s solution shows that cosmological constant is vacuum energy (dark energy) which provide the positive expansion with the current universe.
Hilbert Action
Warning!: This is not suitable for layman. Even I have written this at my own risk.
Let us build an action Sʜ, taking a scalar R such that its variation remains the invariant. From the physical point of view, we know that for the metric field g, R is Ricci Scalar. Hence, we can write action as ,
Sʜ = ∫ d⁴x√(-det g) R
where, √(-det g) is volume element. Let’s vary action Sʜ by small change δSʜ, It gives,
δSʜ = δ(d⁴x√(-det g) R)
As we know, Sʜ is invariant thus δSʜ = 0.
Re-writing above equation as trace of Ricci Curvature. Note that variation can be introduced inside integral without the loss of generality.
δSʜ = ∫ d⁴xδ(√(-det g) Rᵤᵥgᵘᵛ) = 0
Using Leibniz rule for variation we can write.
∫ dx⁴[(-½Rgᵤᵥ + Rᵤᵥ)√(-det g) δgᵘᵛ + √(-det g) δRᵤᵥgᵘᵛ] = 0
Variation in Ricci Curvature is given by,
δRᵤᵥ = ∇ₐδΓᵤᵥᵃ — ∇ᵥΓᵤₐᵃ
Thus, we can compute the later term of the action. From pervious result,
∫ d⁴x√(-det g) δRᵤᵥgᵘᵛ = ∫ d⁴x√(-det g) (∇ₐδΓᵤᵥᵃ — ∇ᵥΓᵤₐᵃ)gᵘᵛ
Using metric compatibility we can introduce metric tensor inside covariant derivative. Further relabeling the indices we get
∫ d⁴x√(-det g) δRᵤᵥgᵘᵛ = ∫ d⁴x√(-det g) ∇ₐ(gᵘᵛδΓᵤᵥᵃ — gᵘᵃΓᵤₐᵃ) (gᵘᵛδΓᵤᵥᵃ — gᵘᵃΓᵤₐᵃ)
= ∫ d⁴x√(-det g) ∇ₐZᵃ ∫ d⁴x√(-det g) δRᵤᵥgᵘᵛ
= ∫ d⁴x√(-det g) ∇ₐZᵃ
To further compute the above equations, we’ll use Stokes’ theorem on RHS to compute a boundary integral ℝ³ instead of region ℝ⁴.
Theorem: For a smooth manifold Ω with boundary ∂Ω, we can define ∫ dω over Ω = ∫ ω over ∂Ω where ω is smooth compact form.
NOTE: For ℝ³ Stokes’ theorem is usual divergence theorem.
∫ ∇. Z d³x over V = ∫ Z . ñ d²x over S
Where ñ is normal unit vector.
V is 3-region, S is boundary of V. Using Stokes’ theorem to covert the integral from 4D to 3D then,
∫ d⁴x√(-det g) ∇ₐZᵃ = ∫ d³x√(-det h) nₐZᵃ
As a consequence of variational principle boundary terms is kept constant. Hence, above equation is equivalent to zero, that is,
∫ d⁴x√(-det g) ∇ₐZᵃ = ∫ d³x√(-det h) nₐZᵃ = 0
Using above result in the action δSʜ will give,
δSʜ = ∫ dx⁴[(-½Rgᵤᵥ + Rᵤᵥ)√(-det g) δgᵘᵛ = 0
Therefore, Einstein field equation for vacuum is given as Rᵤᵥ — ½Rgᵤᵥ = 0
Let’s write using Einstein Tensor Gᵤᵥ
Gᵤᵥ = Rᵤᵥ — ½Rgᵤᵥ = 0
And Binachi Identity is given by gᵃᵛ∇ᵥGᵤᵥ = 0.
With Matter Field For matter field, let the action Sm such that total action is given by
S = Sʜ + κ Sm
Where κ is constant For any arbitrary field which depends upon metric gᵘᵛ we can write variation in action δS using field lagrangian 𝔏 as
δS = ∫ dx⁴ (δ𝔏/δgᵘᵛ) δgᵘᵛ = 0
Thus, varitaion in the total action S can be written as
(1/√(-det g))(δS/δgᵘᵛ) = (½/√(-det g))(δSʜ/δgᵘᵛ) + κ(1/√(-det g))(δSm/δgᵘᵛ) = 0
κ is appropriate constant
(½/√(-det g))(δSʜ/δgᵘᵛ) = κ(-2/√(-det g))(δSm/δgᵘᵛ)
This yields Rᵤᵥ — ½Rgᵤᵥ = κTᵤᵥ
Where Tᵤᵥ is Energy-momentum tensor. Along with Cosmological Constant For cosmological constant Λ we can write
Rᵤᵥ — ½Rgᵤᵥ + Λgᵤᵥ = κTᵤᵥ
Energy conservation is given by Binachi Identity
∇ᵥ(Gᵤᵥ — κTᵤᵥ) = 0
This may not be passing through your mind but I have told you in beginning, though Einstein took 10 years his way is far easy to understand how to derive the Einstein’s Field Equation.
