We will demonstrate that the Einstein Field Equations reduce to Newton's Law of Gravity in the case of a weak field and slow-motion of a particle (v less than the speed of light, c).

As we have already seen, Newton's Gravitation Law can be written as

(1) ∇

^{2}φ = 4πGρ, Equation (16) in Newton's Law of Gravity.

In free-fall, a particle satisfies,

(2) d

^{2}x/dt

^{2}= F/m= –∇φ, Equations (4),(15) in Newton's Law of Gravity.

In tensor notation, this is written as,

(3) φ

_{,ii}≡ 4πGρ

(4) d

^{2}x/dt

^{2}≡ –φ

_{,i}

Corresponding to (1) in General Relativity is the Einstein Field Equations, which can be written in the trace-reverse form,

(5) R

_{μν}= (8πG/c

^{4})[T

_{μν}– ½Tg

_{μν}]

And corresponding to (2), is the geodesic equation,

(6) d

^{2}x

^{α}/dτ

^{2}= –Γ

^{α}

_{βγ}(dx

^{β}/dτ)(dx

^{γ}/dτ)

Our task is to show that (5) will reduce to (1) in low gravity, low velocity.

The first approximation we make is that the particle is moving at velocity near zero,

(7)(dx

^{β}/dτ)≈ (dt/dτ,0,0,0)

The only non-zero term for the Γ's in (6) will be α =i, β = γ =0. The equation becomes,

(8) d

^{2}x

^{α}/dτ

^{2}= –Γ

^{i}

_{00}(dx

^{0}/dτ)(dx

^{0}/dτ)

Or

(9) d

^{2}x

^{α}/dτ

^{2}= –Γ

^{i}

_{00}(dt/dτ)(dt/dτ), (x

^{0}=t)

But by the chain rule of calculus,

(10) d

^{2}x

^{α}/dτ

^{2}= (dt/dτ)(dt/dτ)d

^{2}x

^{α}/dt

^{2}

Therefore,

(11) d

^{2}x

^{α}/dt

^{2}= –Γ

^{i}

_{00}

Equation (4) and (11) yields,

(12) φ

_{,i}= Γ

^{i}

_{00}

Using the Christoffel symbols of the second kind ( Torsion =0 in GR),

(13) Γ

^{α}

_{βγ}= ½ g

^{δα}(g

_{δβ,γ}+ g

_{δγ,β}– g

_{βγ,δ})

Again setting α =i, β = γ =0, equation (12) becomes,

(14) φ

_{,i}= ½ g

^{δi}(g

_{δ0,0}+ g

_{δ0,0}– g

_{00,δ})

Since the time derivative of the metric is zero (Einstein Equivalence Principle), the only surviving term is δ = i. Then,

(15) φ

_{,i}= ½ g

^{ii}(– g

_{00,i})= (–½g

_{00,i})

A solution is,

(16) g

_{00}= – c

^{2}– 2φ

Turning to the Einstein Field Equations (5), and from (16), we only need the time-time equation, μ = ν = 0

(17) R

_{00}= (8πG/c

^{4})[T

_{00}– ½Tg

_{00}]

But

(18) T

_{00}= ρc

^{4}and T = g

^{00}T

_{00}= (–1/c

^{2})ρc

^{4}= –ρc

^{2}

Substituting (18) into (17), we get,

(19) R

_{00}= 4πGρ

Now to get to Newton's Law, equation (1), we need to work on the left-hand side of equation (19).

For that we use the definition of the Ricci tensor,

(20) R

_{μν}= ∂Γ

^{γ}

_{μν}/∂x

^{γ}– ∂Γ

^{γ}

_{μγ}/∂x

^{ν}+ Γ

^{γ}

_{μν}Γ

^{δ}

_{γδ}– Γ

^{γ}

_{μδ}Γ

^{δ}

_{νγ}

From (19), μ = ν = 0,

(21) R

_{00}= ∂Γ

^{γ}

_{00}/∂x

^{γ}– ∂Γ

^{γ}

_{0γ}/∂x

^{0}+ Γ

^{γ}

_{00}Γ

^{δ}

_{γδ}– Γ

^{γ}

_{0δ}Γ

^{δ}

_{0γ}

In low gravity, the square of the Γ's is zero, and its time derivative is also zero. The only surviving term is,

(22) R

_{00}= ∂Γ

^{i}

_{00}/∂x

^{i}≡ Γ

^{i}

_{00,i}

But from (12), repeating below,

(23) Γ

^{i}

_{00}= φ

_{,i}

Take the derivative, and then use (23)

(24) Γ

^{i}

_{00,i}= φ

_{,ii}= R

_{00}

Using (19),

(25) φ

_{,ii}≡ ∇

^{2}φ = 4πGρ

And that is Newton's Law of Gravity (1).

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