In finite dimensions, a space of symmetric bilinear forms with a non-degenerate signature is a submanifold of a space of symmetric bilinear forms, and its tangent space can be embedded into , with as the notation for the tangent vector at
In manifold , a space of symmetric bilinear forms with a non-degenerate signature metric is a submanifold of a space of field of symmetric bilinear forms, and its tangent space can be embedded into , with as the notation for the tangent vector at
[Einstein-Lagrangian] := . In coordinates
Question Is there a good explanation for using scalar curvature in the action?
The action contains second derivatives of , so it cannot be treated with general first-order differential action theory.
Scalar curvature is not homology-scalar-curvature; the integral of the latter (proportional to ?) is homology invariant, so it always varies to zero, and has trivial eq.
Prop Variation with respect to :
So the product rule gives
Prop In Einstein-Lagrangian,
Proof
Prop The derivative of is
Proof
and . So
So the variation of the volume form is
Treating as a matrix, the adjoint can be written as
Prop The differential of is . Proof Using
So the variation of scalar-curvature is
Perform tedious calculations on the following
This might be useful for calculations and
is a divergence (cf. Laplacian-of-tensor-field.typ for )
==>
Let the variation of the action be zero
forall , therefore
[Einstein-equation] [Einstein-metric]
is equivalent to (by taking )
with
i.e. is constant proportionally to and scalar-curvature is constant
equivalently
i.e. trace-free Ricci-curvature is zero, and scalar-curvature is constant
Einstein-equation is a second-order nonlinear PDE for
when , with
When there is interaction, although , the divergence is still zero
Proof
does not need to be Einstein-metric
ฮด diffeomorphism generates a first-order infinitesimal of the metric
Because Einstein action is diffeomorphism invariant, the result of the ฮด diffeomorphism variation is zero
forall , therefore
This will give
Prop For Einstein action, the energy-momentum tensor of ฮด-isometry will be zero
moduli-space-of-Einstein-metric := the orbit space of the metric space acted upon by diffeomorphisms, restricted to the Einstein-metric space. isotropy-group is isometry
Question Even if we know all Einstein-metrics for every manifold, we still don't know which manifold to choose
Question The classification of manifolds with constant-sectional-curvature or simple-symmetric-space seems satisfactory
When dimension there exist manifolds that do not allow constant-sectional-curvature metric but allow Einstein-metric
[Schwarzschild-metric] in := asymptotically flat static spherically symmetric, as the simplest generalization of non-relativity gravity in . Use spherical coordinates in space
with and . Therefore only conformal curvature
Generalize to ?
[Schwarzschild-metric-derivation] (ref-9, ch.4)
Assume metric is spherically symmetric
Point particle gravity source i.e. outside the point particle Einstein equations with give
Asymptotically flat i.e. approaches metric when , gives , then Einstein equation gives
[Schwarzschild-metric-approximate-to-Newton-gravity]
To approximate non-relativity, restore some constants . At this point Schwarzschild-metric
In time coordinates, for this metric, approximate from relativistic point particle action to non-relativistic
- is rest energy, which will vary to
- is the kinetic energy of a non-relativistic point particle
- is the non-relativistic gravitational potential energy
- disappears in the limit
Question If the gravitational source is or , what is the metric?
Some Einstein-metric examples
Einstein ==> harmonics. Einstein-equation satisfy eq defined by Lagrangian