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© The scientific sentence. 2010

Relativity: Equivalence Energy-Metric



1. Vectorial expression of the geodesic equation

The Geodesic equation is :

dvk/dτ + Γkijvivj = 0

In terms of coordinates, it is:

d2xα/dτ2 + Γαμν (dxμ/dτ)(dxν/dτ) = 0

Or

d2xα/dτ2 = - Γαμν (dxμ/dτ)(dxν/dτ)

We will take the following approximation and generalize after.

The motion of the particle is slow (v << c), and
The gravitational field is week and stationary (static).


The motion of the particle is slow implies neglecting the spatial part terms dx1/dτ, dx2/dτ, dx3/dτ in comparison with the term dx0/dτ = c dt/dτ. Then the geodesic equation becomes:

d2xα/dτ2 = - Γα00 c2 (dt/dτ)2

The field is week implies: ∂gμν/∂t = 0.
Therefore the affine connection, that is the Christoffel symbol:
Γkij = (1/2) gkl [dgil/∂xj + dgjl/∂xi - dgij/∂xl]
becomes:

Γα00 = (1/2) gαβ [dg/∂x0 + dg/∂x0 - dg00/∂xβ]
= (1/2) gαβ [2 dg/∂x0 - dg00/∂xβ]

We have:
2 ∂g/∂x0 = 0 (field is week). It remains:

Γα00 = (-1/2) gαβ ∂g00/∂xβ

Γα00 = (-1/2) gαβ ∂g00/∂xβ

For a week field we have:
gμν = ημν + hμν + O(h2)
and

∂gμν/∂t = c ∂gμν/∂x0 = 0

Where

hμν is the first order deviation in h from the flat Minkowski space (space time) that corresponds to the absence of gravitational field: |hμν| < 1.

Therefore:

Γα00 = (-1/2) ηαβ ∂h00/∂xβ = (-1/2) δαβ ∂h00/∂xβ

And only the spatial parts of remains. We have the Kronecker in the Minkowski space-time: δαβ = - 1 for spatial parts.

So
Γα00 = (1/2) dh00/∂xβ

Γα00 = (1/2) dh00/∂xβ

The geodesic equation, with c = 1, then becomes:
d2xα/dτ2 = (- 1/2) (dt/dτ)2 dh00/∂xβ

d2xα/dτ2 = (- 1/2) (dt/dτ)2 ∂h00/∂xβ

Vectorially, for the spacial parts:
d2xα/dτ2 = (- 1/2) (dt/dτ)2 βh00

For the approximations considered:
The motion of the particle is slow , and
the gravitational field is week and stationary, we have
dt/dτ = 1,

so
The geodesic equation, with c = 1, then becomes:

d2xα/dt2 = (- 1/2) βh00

Vectorial Geodesic equation

d2x/dt2 = (- 1/2) ∇βh00

2. Newtonian limit

Recall the Newtonian (classical) acceleration field g is:

g = g(r) = F/m = (GMm/r2)/m = GM/r2 = - ∂Φ/∂r,
where Φ = - GM/r is the Newtonian gravitational potential.

With del or gradient operator, it can be written:

g = - ∇Φ

Its gradient has the expression:

∇. g = - 4 πGρ

And the Poisson's equation for gravity is:

2Φ = 4 πGρ

Now, we want the vectorial Geodesic equation:
d2x/dt2 = (- 1/2) βh00
coincides with the newton equation:
g = - ∇Φ

That is:
(- 1/2) βh00 = - ∇Φ
or
h00 = 2Φ + constant

We have the following convention : Far from any masses, where Φ = 0, and the metric is gμν = ημν , we have h00 = 0. Then constant = 0 Hence, from the equation:
gμν = ημν + hμν + O(h2)

We have in the weak-field limit:
g00 = 1 + h00 = 1 + 2Φ

g00 = 1 + 2Φ = 1 + 2(Φ/c2)

The Poisson's equation for gravity :

2Φ = 4 πGρ
becomes:
2Φ = (1/2)∇2g00 = 4 πGρ

But ρ is the T00 component of the tensor Energy-Momentum Tμν. So
2Φ = 4 πGT00
or
(1/2)∇2g00 = 4 πGT00

That is, for a particle moving slowly in a weak gravitational field:
2g00 = 8 πGT00

Particle moving slowly in a weak gravitational field:

2g00 = 8 πGT00

Now, we make a tensor:
g002g00 = 8 πG g00T00 = 8 πG T00
And denote g002g00 by G00

Therefore:

G00 = 8 πG T00

We generalize and write:
Gμν = 8 πG Tμν

Equivalence Energy-Metric:

Gμν = gμν2gμν = 8 πGTμν = (8 πG/c4)Tμν

Recall:

Gμν = 8 πGTμν




  


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