note-math

cf. #link(<curvature-of-metric.typ>)[]

Einstein-Lagrangian_(tag) := $(scal - 2 Λ) 𝑑 Vol$ . In coordinates $𝑑 Vol = | 𝑔 | 𝑑 𝑥$

Question Is there a good explanation for using scalar curvature in the action?

The action contains second derivatives of $𝑔$ , 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 $𝜒 (𝑀)$ ?) is homology invariant, so it always varies to zero, and has trivial eq.

Prop Variation with respect to $𝑔$ : $Δ scal \sim - Ric + divergence term, Δ | 𝑔 | \sim \frac{1}{2} | 𝑔 | 𝑔$

So the product rule gives

Prop In Einstein-Lagrangian, $Δ (scal | 𝑔 |) \sim | 𝑔 | (Ric - \frac{1}{2} scal 𝑔) + divergence term$

Proof

Prop The derivative of $det$ is $\partial det (𝐴) = det 𝐴 tr (𝐴^{- 1} \partial 𝐴) = det 𝐴 tr (\partial log 𝐴)$

Proof

$det (𝑋) = det (𝐴) det (𝐴^{- 1} 𝑋)$ and $\partial det (𝟙) = tr$ . So

\begin{matrix} \partial det (𝐴^{- 1} 𝑋) (𝐴 : base, Δ 𝑋 : vector) \\ = & \partial det (𝟙 : base, \partial (𝐴^{- 1} 𝑋) (𝐴 : base, Δ 𝑋 : vector) : vector) \\ = & tr (𝐴^{- 1} Δ 𝑋) \end{matrix}

So the variation of the volume form is

\begin{matrix} Δ | 𝑔 | & = & Δ | det 𝑔 |^{\frac{1}{2}} \\ = & \frac{1}{2} | det 𝑔 |^{\frac{1}{2}} tr (𝑔^{- 1} Δ 𝑔) \end{matrix}

Treating $Ric$ as a matrix, the adjoint ${(𝑔 \cdot)}^{†}$ can be written as

\begin{matrix} scal & = & {(𝑔 \cdot)}^{†} Ric \\ = & tr (𝑔^{- 1} Ric) \end{matrix}

Prop The differential of $𝐴 ⇝ 𝐴^{- 1}$ is $- 𝐴^{- 1} (\partial 𝐴) 𝐴^{- 1}$ . Proof Using $0 = \partial 𝟙 = \partial (𝐴 𝐴^{- 1}) = \partial 𝐴 \cdot 𝐴^{- 1} + 𝐴 \cdot \partial (𝐴^{- 1})$

So the variation of scalar-curvature is

Δ scal = tr (- 𝑔^{- 1} (Δ 𝑔) 𝑔^{- 1} Ric) + tr (𝑔^{- 1} Δ Ric)

Perform tedious calculations on the following

$Δ Ric = Δ ({(𝑔 ⧀)}^{†} 𝑅)$
$Δ 𝑅$
$Δ Γ$

This might be useful for calculations $\nabla {(𝑔 ⧀)}^{†} = {(𝑔 ⧀)}^{†} \nabla$ and $\nabla {(𝑔 \cdot)}^{†} = {(𝑔 \cdot)}^{†} \nabla$

tr (𝑔^{- 1} Δ Ric) = \nabla^{†} \nabla tr (𝑔^{- 1} Δ 𝑔) + \nabla_{⊙}^{†} \nabla_{⊙}^{†} Δ 𝑔

is a divergence (cf. #link(<Laplacian-of-tensor-field.typ>)[] for $\nabla^{†}, \nabla_{⊙}, \nabla_{⊙}^{†}$ )

$tr (𝑔^{- 1} Δ 𝑔) = 𝑔 (Δ 𝑔, 𝑔)$
$tr (- 𝑔^{- 1} (Δ 𝑔) 𝑔^{- 1} Ric) = 𝑔 (Δ 𝑔, - Ric)$

==>

$Δ | 𝑔 | = \frac{1}{2} | 𝑔 | 𝑔 (Δ 𝑔, 𝑔)$
$Δ scal = 𝑔 (Δ 𝑔, - Ric) + divergence term$

Let the variation of the action be zero

0 = - \int 𝑑 𝑥 | 𝑔 | 𝑔 (Δ 𝑔, Ric - (\frac{1}{2} \cdot scal - Λ) \cdot 𝑔)

forall $Δ 𝑔$ , therefore

Einstein-equation_(tag) Einstein-metric_(tag)

Ric - (\frac{1}{2} \cdot scal - Λ) \cdot 𝑔 = 0

is equivalent to (by taking ${(𝑔 \cdot)}^{†}$ )

\begin{matrix} Ric & = & \frac{2 Λ}{𝑛 - 2} \cdot 𝑔 \\ = & \frac{1}{𝑛} \cdot scal \cdot 𝑔 \end{matrix}

with $Λ = (\frac{1}{2} - \frac{1}{𝑛}) scal$

i.e. $Ric$ is constant proportionally to $𝑔$ and scalar-curvature is constant

equivalently

\begin{matrix} tr-free-Ric & = & 0 \\ scal & = & \frac{2 Λ}{𝑛 - 2} \end{matrix}

i.e. trace-free Ricci-curvature is zero, and scalar-curvature is constant

Einstein-equation is a second-order nonlinear PDE for $𝑔$

when $𝑛 = 4$ , $Ric = Λ \cdot 𝑔$ with $Λ = \frac{1}{4} scal$

When there is interaction, although $𝑇 = Ric - (\frac{1}{2} \cdot scal - Λ) \cdot 𝑔 \neq 0$ , the divergence is still zero $\nabla_{⊙}^{†} 𝑇 = 0$

Proof

$𝑔$ does not need to be Einstein-metric

δ diffeomorphism $𝑋$ generates a first-order infinitesimal of the metric $Δ 𝑔 = \nabla_{⊙} 𝑋$

Because Einstein action is diffeomorphism invariant, the result of the δ diffeomorphism $𝑋$ variation is zero

\begin{matrix} 0 & = & \int 𝑔 (Δ 𝑔, 𝑇) 𝑑 Vol (𝑔) \\ = & \int 𝑔 (\nabla_{⊙} 𝑋, 𝑇) 𝑑 Vol (𝑔) \\ = & \int 𝑔 (𝑋, \nabla_{⊙}^{†} 𝑇) 𝑑 Vol (𝑔) \end{matrix}

forall $𝑋$ , therefore $\nabla_{⊙}^{†} 𝑇 = 0$

\nabla_{⊙}^{†} (Ric - (\frac{1}{2} \cdot scal - Λ) \cdot 𝑔) = 0

This will give

\nabla_{⊙}^{†} Ric = \nabla_{⊙} scal

Prop For Einstein action, the energy-momentum tensor of δ-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 $\geq 4$ there exist manifolds that do not allow constant-sectional-curvature metric but allow Einstein-metric

Schwarzschild-metric_(tag) in $ℝ^{1, 3}$ := asymptotically flat static spherically symmetric, as the simplest generalization of non-relativity gravity in $ℝ^{3}$ . Use spherical coordinates in space $ℝ^{3}$

𝑔 = (1 - \frac{2 𝑚}{𝑟}) 𝑑 𝑡^{2} - ({(1 - \frac{2 𝑚}{𝑟})}^{- 1} 𝑑 𝑟^{2} + 𝑟^{2} 𝑔_{𝕊^{2}})

with $scal = 0$ and $Ric = 0$ . Therefore only conformal curvature

Generalize to $ℝ^{1, 𝑛 - 1}$ ?

𝑔 = (1 - \frac{2 𝑚}{𝑟^{𝑛 - 3}}) 𝑑 𝑡^{2} - ({(1 - \frac{2 𝑚}{𝑟^{𝑛 - 3}})}^{- 1} 𝑑 𝑟^{2} + 𝑟^{2} 𝑔_{𝕊^{𝑛 - 2}})

Schwarzschild-metric-derivation_(tag) (ref-9, ch.4)

Assume metric is spherically symmetric

𝑔 = 𝑓_{𝑡} {(𝑟)}^{2} 𝑑 𝑡^{2} - (𝑓_{𝑟} {(𝑟)}^{2} 𝑑 𝑟^{2} + 𝑓_{𝕊^{2}} {(𝑟)}^{2} 𝑔_{𝕊^{2}})

Point particle gravity source i.e. outside the point particle Einstein equations with $Λ = 0$ give $𝑓_{𝑟} = {(𝑎 𝑓_{𝑡})}^{- 1}, 𝑓_{𝕊^{2}} = 𝑟$

Asymptotically flat i.e. approaches $ℝ^{1, 3}$ metric $𝑑 𝑡^{2} - (𝑑 𝑟^{2} + 𝑟^{2} 𝑔_{𝕊^{2}})$ when $𝑟 \to \infty$ , gives $𝑎 = 1$ , then Einstein equation gives $𝑓_{𝑡} {(𝑟)}^{2} = 1 - \frac{2 𝑚}{𝑟}$

Schwarzschild-metric-approximate-to-Newton-gravity_(tag)

To approximate non-relativity, restore some constants $𝐺, 𝑐, 𝑥_{0} = 𝑐 𝑡$ . At this point Schwarzschild-metric

𝑔 = (1 - \frac{2 𝐺 𝑀}{𝑐^{2} 𝑟}) 𝑐^{2} 𝑑 𝑡^{2} - ({(1 - \frac{2 𝐺 𝑀}{𝑐^{2} 𝑟})}^{- 1} 𝑑 𝑟^{2} + 𝑟^{2} 𝑔_{𝕊^{2}})

In time coordinates, for this metric, approximate from relativistic point particle action to non-relativistic

\begin{matrix} 𝑚 𝑐 | \dot{𝑥} | & = & 𝑚 𝑐^{2} {(1 - \frac{2 𝐺 𝑀}{𝑐^{2} 𝑟} - \frac{1}{𝑐^{2}} ({(1 - \frac{2 𝐺 𝑀}{𝑐^{2} 𝑟})}^{- 1} 𝑣_{𝑟}^{2} + 𝑟^{2} 𝑣_{𝕊^{2}}^{2}))}^{\frac{1}{2}} \\ = & 𝑚 𝑐^{2} (1 - \frac{1}{2} (\frac{2 𝐺 𝑀}{𝑐^{2} 𝑟} + \frac{1}{𝑐^{2}} (𝑣_{𝑟}^{2} + 𝑟^{2} 𝑣_{𝕊^{2}}^{2})) + 𝑜 (\frac{1}{𝑐^{2}})) \\ = & 𝑚 𝑐^{2} - (\frac{1}{2} 𝑚 𝑣^{2} - (- \frac{𝐺 𝑀 𝑚}{𝑟})) + 𝑜 (1) \end{matrix}

$𝑚 𝑐^{2}$ is rest energy, which will vary to $0$
$\frac{1}{2} 𝑚 𝑣^{2}$ is the kinetic energy of a non-relativistic point particle
$- \frac{𝐺 𝑀 𝑚}{𝑟}$ is the non-relativistic gravitational potential energy
$𝑜 (1)$ disappears in the limit $lim_{𝑐 \to \infty}$

Question If the gravitational source is $𝑇 = constant$ or $𝑇_{00} = constant$ , what is the metric?

Some Einstein-metric examples

#link(<constant-sectional-curvature-imply-Einstein-metric>)[constant sectional curvature]
#link(<simple-symmetric-space>)[]

Einstein ==> harmonics. Einstein-equation satisfy eq defined by Lagrangian $| 𝑅 |^{2} 𝑑 Vol$