Wolfram Alpha:

```Einstein's Field Equations
--------------------------

We will first state the Einstein field equations and then walk through
the general rationale that Einstein used to derive them (without getting
into the enormously complicated mathematics).

The equations are:

Gμν = 8πGTμν

Where,

Gμν = Rμν - (1/2)gμνR

Where,

Rμν is the RICCI tensor.
gμν is the METRIC tensor.
R is the SCALAR CURVATURE.  R = gμνRμν.  This corresponds to the trace
of the Ricci curvature tensor.
Tμν is the STRESS-ENERGY tensor.

The Einstein equation actually consists of 16 separate equations. In a
nutshell, they show how energy and momentum (aka mass) determine the
curvature of spacetime.  Energy and mass are represented on the right
hand side of the equations while curvature is represented on the left
hand side.

The Riemann tensor can be expressed entirely in terms of the Christoffel
symbols and their first partial derivatives.  In turn, the Christoffel
symbols are comprised of the partial derivatives of the metric tensor.
Therefore, the Riemann tensor is comprised of the first and second
derivatives of the metric tensor.  Thus, for the Rindler metric we
would have:

-                 -
gμν =  | (1 + 2φ)  0  0  0 |
|    0     -1  0  0 |
|    0      0 -1  0 |
|    0      0  0 -1 |
-                 -

Rμνσλ = ∂νΓλμσ - ∂μΓλνσ + ΓρμσΓλρν - ΓρνσΓλρμ

Where,

Γabc = (1/2)gad{∂gdc/∂xb + ∂gdb/∂xc - ∂gbc/∂xd}

= (1/2)gad{∂bgdc + ∂cgdb - ∂dgbc}

In Newtonian mechanics, a freely falling object accelerating towards
the earth implies a force acting between the object and the earth
corresponding to gravity.  In general relativity, however, the earth
and the object are considered to be both moving along geodesics
(along straight lines with zero proper acceleration) and it is the
acceleration produced by the convergence of these geodesics as
a result of the curvature of the spacetime that we observed as the
object falls.  For example, imagine 2 ships starting a certain distance
apart at the equator of the earth and sailing due north along different
lines of longitude (geodesics).  As they proceed, their east west
separation diminishes until they finally meet at the north pole.  We
can imagine this 'coming together' as being the result of a force
(acceleration) with both transverse and longitudinal components
acting on both objects. Thus,

Mathematially, if the separation 4-vector between the earth and
the object at the same proper time, τ, as both advance along
their respective geodesics is ξμ, the coordinate acceleration
between them is given by the GEODESIC DEVIATION EQUATION (not
proven here but fairly straightforward to work out):

D2ξμ/Dτ2 = -Rμνρσ(dxν/dτ)(dxρ/dτ)ξσ

Where D is the covariant derivative and Rμνρσ is the RIEMANN tensor.

The geodesic deviation equation assigns TIDAL forces to the Riemann
tensor.  The equation yields a component of acceleration along the
geodesic (the Γnmr(dxm/dτ)(dxr/dτ) term in the original geodesic
equation) and components transverse to the direction of motion.
Together these accelerations act together to produce tidal forces
on the objects.  The components along the geodesic result in
elongation while the tranvserse components result in compression.

Another way to see this is to imagine an object in flat spacetime
moving towards a region of curvature.  When the object encounters
the region it must 'adjust' dimensionally in the transverse and
the curved spacetime at each point.  In the process of adjusting,
the object experiences non-uniform forces and becomes stretched
in the direction of motion and squeezed in transverse direction.

Uniform gravitational fields that are 'simulated' by accelerated
reference frames do not have curvature.  Such fields do not exist
in nature (real gravitational fields involve mass and curvature).
However, the gravitational field near the surface of a large object
over a relatively small solid angle can be considered to be
approximately uniform.  Since they do not have curvature, a free
falling object in a uniform gravitational field will not experience
any tidal forces.

The Weak Field Approximation
----------------------------

The weak field approximation is analagous to perturbation theory and
allows simplifying the study of many problems in general relativity
while still producing useful approximate results.  Equations can be
obtained by assuming the spacetime metric is only slightly different
from the Minkowski metric, η, where η is constant.  Thus,

gμν = ημν + hμν and gμν = ημν - hμν

The Geodesic Equation
-----------------------

Look at acceleration of a particle along a geodesic.  In the limit
this should reduce to the familiar Newtonian force law.

d2xn/dτ2 = -Γnmr(dxm/dτ)(dxr/dτ)

x0 = t = τ, m = 1, and dτ2 = gμνdxμdxν, v << c

If the velocity is small compared to c, the spatial components can
be ignored.  Therefore,

d2x/dt2 + Γμ00(dx0/dτ)(dx0/dτ) = 0

d2x/dt2 = -Γμ00   since dx0/dτ = dτ/dτ = 1

From before we can write Γ in terms of the metric,

Γabc = (1/2)gad{∂bgdc + ∂cgdb - ∂dgbc}

Γa00 = (1/2)(ηax - hax){∂0hx0 + ∂0hx0 - ∂xh00}

Since they very small we can neglect all terms that involve products
of h and its derivatives.  Thus:

Γa00 = (1/2)ηax{∂0hx0 + ∂0h0x - ∂xh00}

= (1/2)ηax{2∂0hx0 - ∂xh00}

= -(1/2){2∂0hx0 - ∂xh00}

If the velocity is assumed to be small, the derivatives with respect
to space dominate over the derivatives with respect to time.
Therefore,

Γa00 = (1/2)∂xh00

Therefore, we can write:

d2x/dt2  = -(1/2)∂xh00

However, in Newtonian mechanics force F = -∂Φ/∂x (m = 1) where φ
is the GRAVITATIONAL POTENTIAL ENERGY.  Thus, we can write:

F = -∂Φ/∂x = d2x/dt2

Substitution yields,

2∂Φ/∂x = ∂h00/∂x

So,

h00 = 2φ + constant

In flat space (φ = 0) h00 = 1 therefore the constant equals 1.
We end up with:

h00 = g00 = 1 + 2φ

Which is the same form found in the Rindler metric associated with
accelerated reference frames.

dτ2 = (1 + 2φ)dt2 - dx2

The Field Equations
------------------

Rμν - (1/2)gμνR = 8πGTμν

gμνRμν - (1/2)gμνgμνR = gμν8πGTμν

R - (1/2)δμμR = 8πGT

-       -
| 1 0 0 0 |
Now δμμ = δνν = | 0 1 0 0 | = 4
| 0 0 1 0 |
| 0 0 0 1 |
-       -

Thus, R is just the trace of Rμν

R - 2R = 8πGT

-R = 8πGT

Substituting this back into the EFEs gives:

Rμν + (1/2)gμν8πGT = 8πGTμν

Rμν = 8πG(Tμν - (1/2)gμνT)

T = T00 + T11 + T22 + T33 ... the trace of Tμν

Consider 00 component with T00 = ρ.  Thus,

R00 = 8πG(ρ - (1/2)g00ρ)

R00 = 8πG(ρ - (1/2)ρ)

R00 = 4πGρ   ... 1.

From before, the Riemann tensor is:

Rμνσλ = ∂νΓλμσ - ∂μΓλνσ + ΓρμσΓλρν - ΓρνσΓλρμ

The ΓΓ terms can be neglected since they are the products of the
derivatives of h.  Therefore,

Rμνσλ = ∂μΓλνσ - ∂νΓλμσ

Γμνσ = (1/2)ημλ(∂νhλσ + ∂σhλν - ∂λhνσ)

Γλνσ = (1/2)ηλρ(∂νhρσ + ∂σhρν - ∂ρhνσ)

Γμμσ = (1/2)ηλρ(∂μhρσ + ∂σhρμ - ∂ρhμσ)

Rμνσλ = (1/2)ηλρ(∂μ∂νhρσ + ∂μ∂σhρν - ∂μ∂ρhνσ - ∂ν∂μhρσ - ∂ν∂σhρμ + ∂ν∂ρhμσ)

Rμνσλ = (1/2)ηλρ(∂μ∂σhρν - ∂μ∂ρhνσ - ∂ν∂σhρμ + ∂ν∂ρhμσ)

Multiply by ηνρ to get the Ricci tensor:

Rμσ = (1/2)ηνρ(∂μ∂σhρν - ∂μ∂ρhνσ - ∂ν∂σhρμ + ∂ν∂ρhμσ)

Multiply by ησρ to get the Ricci tensor:

Rμν = (1/2)ησρ(∂μ∂σhρν - ∂μ∂ρhνσ - ∂ν∂σhρμ + ∂ν∂ρhμσ)

Pick the 00 component:

R00 = (1/2)ηνρ(∂0∂0hρν - ∂0∂ρhν0 - ∂ν∂0hρ0 + ∂ν∂ρh00)

As before, if the velocity is assumed to be small, the derivatives
with respect to space dominate over the derivatives with respect
to time.  Therefore,

R00 = (1/2)ηνρ∂ν∂ρh00

R00 = -(1/2)∂i∂jh00  ... 2.

From 1. and 2. we can write:

-(1/2)∂i∂jh00 = 4πGρ

or

∇2h00 = -8πGρ

But h00 = 1 + 2φ.  Therefore,

∇2φ = 4πGρ

Which is the Newtonian equation for the gravitational field.

Einstein's Process for Developing the Field Equations
-----------------------------------------------------

From Newton's Law of Gravity:

∇2φ = 4πGρ

The solution to this equation is the familiar:

φ = -GM/r

and

F = ma = -m∇φ = GmM/r2

Einstein wanted an equation of a similar form.  Geometry and curvature
on the LHS, Energy density/momentum (the source of the gravitational
field) on the RHS.

From the weak field theory g00 = 1 + 2φ

∴ φ = (1/2)g00 - 1/2

∴ (1/2)∇2φ = 4πGρ

∴ ∇2g00 = 8πGρ

∴ ∇2g00 = 8πGT00

This was Einstein's first clue that somehow mass effects the curvature
of space.   Now, the energy-momentum tensor has 2 indeces, is symmetric
and its divergence equals 0.  Thus,

Tμν = Tνμ and DμTμν = 0

So the object representing the curvature on the LHS has to be a tensor
with the same properties.  The obvious choice was the Ricci tensor but
while it is symmetric, it does not have a divergence equal to 0.  Einstein
started with the Riemann tensor and used the second BIANCHI IDENTITIES
in conjunction of the asymmetric nature of the Riemann tensor, to find
something that did.

The second BIANCHI identities are written as:

DλRαβμν + DνRαβλμ + DμRαβνλ = 0

The Riemann tensor exhibits the following symmetries:

Rαβμν = -Rβαμν = -Rαβνμ = Rμναβ

Note:  Rαβμν = gδνRαβδμ

Also, we can make use of the fact that Dg = 0 so we can move g inside
or outside of the covariant derivatives at will.

If we multiply the above Bianchi identity by gαμ we get:

gαμDλRαβμν + gαμDνRαβλμ + gαμDμRαβνλ = 0

Now gαμRαβλμ = -gαμRαβμλ= -Rβλ

So we get:

DλRβν - DνRβλ + DμRμβνλ = 0

If we multiply the above by gβν we get:

gβνDλRβν - gβνDνRβλ + gβνDμRμβνλ = 0

∴ DλR - DβRβλ + DμβRμνβνλ = 0

∴ DλR - DβRβλ + DμβRμνβνλ = 0

Interchange μ and ν and from the asymmetry properties we get:

DλR - DβRβλ - DνβRνμβμλ = 0

∴ DλR - DβRβλ - DβRβλ = 0

∴ DλR - 2DβRβλ = 0

Alternatively,

2DβRβλ - DλR = 0

∴ DβRβλ - (1/2)DλR = 0

∴ DβRβλ - (1/2)gβλDβR = 0

∴ Dβ[Rβλ - (1/2)gβλR] = 0

Gβλ = Rβλ - (1/2)gβλR

Gβλ is both symmetric and has a divergence of 0.  Einstein could
now write:

Gβλ = 8πGρTβλ

Einstein - Hilbert Derivation
-----------------------------

The Einstein field equations can also be obtained very elegantly
through the principle of least action.  Consider an small area of
the gravitational potential energy field, φ, in curved spacetime.

√(gxx)dx
----
√(gyy)dy |  φ  |
|     |
----

Area = √(gxxgyy)dxdy

This is a special case of a metric with only diagonal components.
In reality the metric will also have gxy and gyx components so we can
generalize by writing:

dA = √(|g|)dxdy where |g| is the determinant of g (i.e. gxxgyy - gxygyx)

We can further generalize this to a small volume element in curved
spacetime by writing:

dV = √(-|gμν|)d4x

The minus sign arises because the time component is negative.

Therefore,

V = ∫√(-|gμν|)d4x

This integrand has the form of a Lagrangian density which is an invariant
quantity.  Now, it is possible to multiply the integrand by a scalar with no
loss of the invariance.  The scalar takes the form of the curvature scalar,
R, plus terms that represent the energy/matter fields, LM.  Thus we can
write:

S = ∫√(-|gμν|)(k'R + LM)d4x

where k' is defined to be c4/16πG

After a lot of work that involves starting with the Riemann tensor
(constructed out of Christoffel symbols and derivatives of Christoffel
symbols, which are turn constructed out of derivatives of the metric),
contracting indeces to get the Ricci tensor, contracting again to get the
curvature scalar, multiplying by the square root of the determinant of the
metric tensor, factoring in the mass/energy terms and then minimizing the
action using the Euler-Lagrange equations, one finally obtains the EFEs.

The Cosmological Constant
-------------------------

Einstein recognized there was an ambiguity in his equations.  In order to
achieve a stationary universe he had to include another term as:

Gμν + Λgμν = kTμν

The CC is only a factor on galactic scales since the term is otherwise
very small and can be ignored.
```