Wolfram Alpha:
Search by keyword:
Astronomy
Chemistry
Classical Mechanics
Classical Physics
Climate Change
Cosmology
Finance and Accounting
Game Theory
General Relativity
Group Theory
Lagrangian and Hamiltonian Mechanics
Macroeconomics
Mathematics
Microeconomics
Nuclear Physics
Particle Physics
Probability and Statistics
Programming and Computer Science
Quantum Computing
Quantum Field Theory
Quantum Mechanics
Semiconductor Reliability
Solid State Electronics
Special Relativity
Statistical Mechanics
String Theory
Superconductivity
Supersymmetry (SUSY) and Grand Unified Theory (GUT)
The Standard Model
Topology
Units, Constants and Useful Formulas
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:
G_{μν} = 8πGT_{μν}
Where,
G_{μν} = R_{μν} - (1/2)g_{μν}R
Where,
R_{μν} is the RICCI tensor.
g_{μν} is the METRIC tensor.
R is the SCALAR CURVATURE. R = g^{μν}R_{μν}. This corresponds to the trace
of the Ricci curvature tensor.
T_{μν} 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:
- -
g_{μν} = | (1 + 2φ) 0 0 0 |
_{ } | 0 -1 0 0 |
_{ } | 0 0 -1 0 |
_{ } | 0 0 0 -1 |
- -
R_{μνσ}^{λ} = ∂_{ν}Γ^{λ}_{μσ} - ∂_{μ}Γ^{λ}_{νσ} + Γ^{ρ}_{μσ}Γ^{λ}_{ρν} - Γ^{ρ}_{νσ}Γ^{λ}_{ρμ}
Where,
Γ^{a}_{bc} = (1/2)g^{ad}{∂g_{dc}/∂x^{b} + ∂g_{db}/∂x^{c} - ∂g_{bc}/∂x^{d}}
= (1/2)g^{ad}{∂_{b}g_{dc} + ∂_{c}g_{db} - ∂_{d}g_{bc}}
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 ξ^{μ}, the coordinate acceleration
between them is given by the GEODESIC DEVIATION EQUATION (not
proven here but fairly straightforward to work out):
D^{2}ξ^{μ}/Dτ^{2} = -R^{μ}_{νρσ}(dx^{ν}/dτ)(dx^{ρ}/dτ)ξ^{σ}
Where D is the covariant derivative and R^{μ}_{νρσ} 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 Γ^{n}_{mr}(dx^{m}/dτ)(dx^{r}/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
longitudinal (radial) directions in order to follow the metric of
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,
g_{μν} = η_{μν} + h_{μν} and g^{μν} = η^{μν} - h^{μν}
The Geodesic Equation
-----------------------
Look at acceleration of a particle along a geodesic. In the limit
this should reduce to the familiar Newtonian force law.
d^{2}x^{n}/dτ^{2} = -Γ^{n}_{mr}(dx^{m}/dτ)(dx^{r}/dτ)
x^{0} = t = τ, m = 1, and dτ^{2} = g_{μν}dx^{μ}dx^{ν}, v << c
If the velocity is small compared to c, the spatial components can
be ignored. Therefore,
d^{2}x/dt^{2} + Γ^{μ}_{00}(dx^{0}/dτ)(dx^{0}/dτ) = 0
d^{2}x/dt^{2} = -Γ^{μ}_{00} since dx^{0}/dτ = dτ/dτ = 1
From before we can write Γ in terms of the metric,
Γ^{a}_{bc} = (1/2)g^{ad}{∂_{b}g_{dc} + ∂_{c}g_{db} - ∂_{d}g_{bc}}
Γ^{a}_{00} = (1/2)(η^{ax} - h^{ax}){∂_{0}h_{x0} + ∂_{0}h_{x0} - ∂_{x}h_{00}}
Since they very small we can neglect all terms that involve products
of h and its derivatives. Thus:
Γ^{a}_{00} = (1/2)η^{ax}{∂_{0}h_{x0} + ∂_{0}h_{0x} - ∂_{x}h_{00}}
= (1/2)η^{ax}{2∂_{0}h_{x0} - ∂_{x}h_{00}}
= -(1/2){2∂_{0}h_{x0} - ∂_{x}h_{00}}
If the velocity is assumed to be small, the derivatives with respect
to space dominate over the derivatives with respect to time.
Therefore,
Γ^{a}_{00} = (1/2)∂_{x}h_{00}
Therefore, we can write:
d^{2}x/dt^{2} = -(1/2)∂_{x}h_{00}
However, in Newtonian mechanics force F = -∂Φ/∂x (m = 1) where φ
is the GRAVITATIONAL POTENTIAL ENERGY. Thus, we can write:
F = -∂Φ/∂x = d^{2}x/dt^{2}
Substitution yields,
2∂Φ/∂x = ∂h_{00}/∂x
So,
h_{00} = 2φ + constant
In flat space (φ = 0) h_{00} = 1 therefore the constant equals 1.
We end up with:
h_{00} = g_{00} = 1 + 2φ
Which is the same form found in the Rindler metric associated with
accelerated reference frames.
dτ^{2} = (1 + 2φ)dt^{2} - dx^{2}
The Field Equations
------------------
R_{μν} - (1/2)g_{μν}R = 8πGT_{μν}
g^{μν}R_{μν} - (1/2)g^{μν}g_{μν}R = g^{μν}8πGT_{μν}
R - (1/2)δ^{μ}_{μ}R = 8πGT
- -
^{ }_{ } | 1 0 0 0 |
Now δ^{μ}_{μ} = δ^{ν}_{ν} = | 0 1 0 0 | = 4
^{ }_{ } | 0 0 1 0 |
^{ }_{ } | 0 0 0 1 |
- -
Thus, R is just the trace of R_{μν}
R - 2R = 8πGT
-R = 8πGT
Substituting this back into the EFEs gives:
R_{μν} + (1/2)g_{μν}8πGT = 8πGT_{μν}
R_{μν} = 8πG(T_{μν} - (1/2)g_{μν}T)
T = T_{00} + T_{11} + T_{22} + T_{33} ... the trace of T_{μν}
Consider 00 component with T_{00} = ρ. Thus,
R_{00} = 8πG(ρ - (1/2)g_{00}ρ)
R_{00} = 8πG(ρ - (1/2)ρ)
R_{00} = 4πGρ ... 1.
From before, the Riemann tensor is:
R_{μνσ}^{λ} = ∂_{ν}Γ^{λ}_{μσ} - ∂_{μ}Γ^{λ}_{νσ} + Γ^{ρ}_{μσ}Γ^{λ}_{ρν} - Γ^{ρ}_{νσ}Γ^{λ}_{ρμ}
The ΓΓ terms can be neglected since they are the products of the
derivatives of h. Therefore,
R_{μνσ}^{λ} = ∂_{μ}Γ^{λ}_{νσ} - ∂_{ν}Γ^{λ}_{μσ}
Γ^{μ}_{νσ} = (1/2)η^{μλ}(∂_{ν}h_{λσ} + ∂_{σ}h_{λν} - ∂_{λ}h_{νσ})
Γ^{λ}_{νσ} = (1/2)η^{λρ}(∂_{ν}h_{ρσ} + ∂_{σ}h_{ρν} - ∂_{ρ}h_{νσ})
Γ^{μ}_{μσ} = (1/2)η^{λρ}(∂_{μ}h_{ρσ} + ∂_{σ}h_{ρμ} - ∂_{ρ}h_{μσ})
R_{μνσ}^{λ} = (1/2)η^{λρ}(∂_{μ}∂_{ν}h_{ρσ} + ∂_{μ}∂_{σ}h_{ρν} - ∂_{μ}∂_{ρ}h_{νσ} - ∂_{ν}∂_{μ}h_{ρσ} - ∂_{ν}∂_{σ}h_{ρμ} + ∂_{ν}∂_{ρ}h_{μσ})
R_{μνσ}^{λ} = (1/2)η^{λρ}(∂_{μ}∂_{σ}h_{ρν} - ∂_{μ}∂_{ρ}h_{νσ} - ∂_{ν}∂_{σ}h_{ρμ} + ∂_{ν}∂_{ρ}h_{μσ})
Multiply by η^{νρ} to get the Ricci tensor:
R_{μσ} = (1/2)η^{νρ}(∂_{μ}∂_{σ}h_{ρν} - ∂_{μ}∂_{ρ}h_{νσ} - ∂_{ν}∂_{σ}h_{ρμ} + ∂_{ν}∂_{ρ}h_{μσ})
Multiply by η^{σρ} to get the Ricci tensor:
R_{μν} = (1/2)η^{σρ}(∂_{μ}∂_{σ}h_{ρν} - ∂_{μ}∂_{ρ}h_{νσ} - ∂_{ν}∂_{σ}h_{ρμ} + ∂_{ν}∂_{ρ}h_{μσ})
Pick the 00 component:
R_{00} = (1/2)η^{νρ}(∂_{0}∂_{0}h_{ρν} - ∂_{0}∂_{ρ}h_{ν0} - ∂_{ν}∂_{0}h_{ρ0} + ∂_{ν}∂_{ρ}h_{00})
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,
R_{00} = (1/2)η^{νρ}∂_{ν}∂_{ρ}h_{00}
R_{00} = -(1/2)∂_{i}∂_{j}h_{00} ... 2.
From 1. and 2. we can write:
-(1/2)∂_{i}∂_{j}h_{00} = 4πGρ
or
∇^{2}h_{00} = -8πGρ
But h_{00} = 1 + 2φ. Therefore,
∇^{2}φ = 4πGρ
Which is the Newtonian equation for the gravitational field.
Einstein's Process for Developing the Field Equations
-----------------------------------------------------
From Newton's Law of Gravity:
∇^{2}φ = 4πGρ
The solution to this equation is the familiar:
φ = -GM/r
and
F = ma = -m∇φ = GmM/r^{2}
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 g_{00} = 1 + 2φ
∴ φ = (1/2)g_{00} - 1/2
∴ (1/2)∇^{2}φ = 4πGρ
∴ ∇^{2}g_{00} = 8πGρ
∴ ∇^{2}g_{00} = 8πGT_{00}
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,
T_{μν} = T_{νμ} and D_{μ}T_{μν} = 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:
D_{λ}R_{αβμν} + D_{ν}R_{αβλμ} + D_{μ}R_{αβνλ} = 0
The Riemann tensor exhibits the following symmetries:
R_{αβμν} = -R_{βαμν} = -R_{αβνμ} = R_{μναβ}
Note: R_{αβμν} = g_{δν}R_{αβ}^{δ}_{μ}
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 g^{αμ} we get:
g^{αμ}D_{λ}R_{αβμν} + g^{αμ}D_{ν}R_{αβλμ} + g^{αμ}D_{μ}R_{αβνλ} = 0
Now g^{αμ}R_{αβλμ} = -g^{αμ}R_{αβμλ}= -R_{βλ}
So we get:
D_{λ}R_{βν} - D_{ν}R_{βλ} + D_{μ}R^{μ}_{βνλ} = 0
If we multiply the above by g^{βν} we get:
g^{βν}D_{λ}R_{βν} - g^{βν}D_{ν}R_{βλ} + g^{βν}D_{μ}R^{μ}_{βνλ} = 0
∴ D_{λ}R - D^{β}R_{βλ} + D_{μ}^{β}R^{μν}_{βνλ} = 0
∴ D_{λ}R - D^{β}R_{βλ} + D_{μ}^{β}R^{μν}_{βνλ} = 0
Interchange μ and ν and from the asymmetry properties we get:
D_{λ}R - D^{β}R_{βλ} - D_{ν}^{β}R^{νμ}_{βμλ} = 0
∴ D_{λ}R - D^{β}R_{βλ} - D^{β}R_{βλ} = 0
∴ D_{λ}R - 2D^{β}R_{βλ} = 0
Alternatively,
2D^{β}R_{βλ} - D_{λ}R = 0
∴ D^{β}R_{βλ} - (1/2)D_{λ}R = 0
∴ D^{β}R_{βλ} - (1/2)g_{βλ}D^{β}R = 0
∴ D^{β}[R_{βλ} - (1/2)g_{βλ}R] = 0
G_{βλ} = R_{βλ} - (1/2)g_{βλ}R
G_{βλ} is both symmetric and has a divergence of 0. Einstein could
now write:
G_{βλ} = 8πGρT_{βλ}
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.
√(g_{xx})dx
----
√(g_{yy})dy | φ |
_{ } | |
----
Area = √(g_{xx}g_{yy})dxdy
This is a special case of a metric with only diagonal components.
In reality the metric will also have g_{xy} and g_{yx} components so we can
generalize by writing:
dA = √(|g|)dxdy where |g| is the determinant of g (i.e. g_{xx}g_{yy} - g_{xy}g_{yx})
We can further generalize this to a small volume element in curved
spacetime by writing:
dV = √(-|g_{μν}|)d^{4}x
The minus sign arises because the time component is negative.
Therefore,
V = ∫√(-|g_{μν}|)d^{4}x
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, L_{M}. Thus we can
write:
S = ∫√(-|g_{μν}|)(k'R + L_{M})d^{4}x
where k' is defined to be c^{4}/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:
G_{μν} + Λg_{μν} = kT_{μν}
The CC is only a factor on galactic scales since the term is otherwise
very small and can be ignored.