note-math

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

geodesic_(tag)

Geodesics as possible "shortest length paths". Action

\begin{matrix} \int 𝑑 𝑙 & = & \int 𝑑 𝑡 | \dot{𝑥} | = \int 𝑑 𝑡 | {𝑔 (\dot{𝑥})}^{2} |^{\frac{1}{2}} \\ or & = & \int 𝑑 𝑡 | 𝑔_{𝑖 𝑖^{'}} \cdot {\dot{𝑥}}^{𝑖} \cdot {\dot{𝑥}}^{𝑖^{'}} |^{\frac{1}{2}} \\ or & = & \int 𝑑 𝑡 | {\dot{𝑥}}^{⊺} 𝑔 \dot{𝑥} |^{\frac{1}{2}} \end{matrix}

ODE initial value $𝑥 (0), \dot{𝑥} (0)$ . The definition of geodesics does not depend on coordinate selection

$𝑑 𝑙$ is the metric-volume-form $𝑑 Vol = | 𝑔 | 𝑑 𝑥 = | det 𝑔 |^{\frac{1}{2}} 𝑑 𝑥$ restricted to a 1 dimension path. $det 𝑔 = det (𝑔_{𝑖 𝑗})$ is the induced quadratic-form ${𝑔 (𝑑 𝑥^{1} \land \dots \land 𝑑 𝑥^{𝑛})}^{2} = det (\begin{matrix} 𝑔 (𝑑 𝑥^{1}, 𝑑 𝑥^{1}) & \dots & 𝑔 (𝑑 𝑥^{1}, 𝑑 𝑥^{𝑛}) \\ ⋮ & ⋱ & ⋮ \\ 𝑔 (𝑑 𝑥^{𝑛}, 𝑑 𝑥^{1}) & \dots & 𝑔 (𝑑 𝑥^{𝑛}, 𝑑 𝑥^{𝑛}) \end{matrix})$

The Lagrange equation is

\begin{matrix} \frac{𝑑}{𝑑 𝑡} (\frac{𝑔 \dot{𝑥}}{| {𝑔 (\dot{𝑥})}^{2} |^{\frac{1}{2}}}) & = & \frac{1}{2} \frac{{\dot{𝑥}}^{⊺} (\partial 𝑔) \dot{𝑥}}{| {𝑔 (\dot{𝑥})}^{2} |^{\frac{1}{2}}} & where \partial 𝑔 ≃ (\begin{matrix} \partial_{1} 𝑔 \\ ⋮ \\ \partial_{𝑛} 𝑔 \end{matrix}) \\ or \frac{𝑑}{𝑑 𝑡} (\frac{𝑔_{𝑗 𝑖^{'}} 𝑥^{𝑖^{'}}}{| 𝑔_{𝑖 𝑖^{'}} {\dot{𝑥}}^{𝑖} {\dot{𝑥}}^{𝑖^{'}} |^{\frac{1}{2}}}) & = & \frac{1}{2} \frac{(\partial_{𝑗} 𝑔_{𝑖 𝑖^{'}}) {\dot{𝑥}}^{𝑖} {\dot{𝑥}}^{𝑖^{'}}}{| 𝑔_{𝑖 𝑖^{'}} {\dot{𝑥}}^{𝑖} {\dot{𝑥}}^{𝑖^{'}} |^{\frac{1}{2}}} \end{matrix}

For unit length parameter of the path, $| \dot{𝑥} | = | {𝑔 (\dot{𝑥})}^{2} |^{\frac{1}{2}} = 1$ , the equation becomes

\begin{matrix} \frac{𝑑}{𝑑 𝑡} (𝑔 \dot{𝑥}) & = & \frac{1}{2} {\dot{𝑥}}^{⊺} (\partial 𝑔) \dot{𝑥} \\ or \frac{𝑑}{𝑑 𝑡} (𝑔_{𝑗 𝑖^{'}} 𝑥^{𝑖^{'}}) & = & \frac{1}{2} (\partial_{𝑗} 𝑔_{𝑖 𝑖^{'}}) {\dot{𝑥}}^{𝑖} {\dot{𝑥}}^{𝑖^{'}} \end{matrix}

product-rule expansion $\frac{𝑑}{𝑑 𝑡} (𝑔 \dot{𝑥}) = (\partial (\dot{𝑥}) 𝑔) \dot{𝑥} + 𝑔 \ddot{𝑥}$ , where $\partial (\dot{𝑥}) 𝑔 = \frac{\partial}{\partial \dot{𝑥}} 𝑔 = \frac{𝑑}{𝑑 𝑡} 𝑔 (𝑥 (𝑡))$ . Transposing terms and using $𝑔^{- 1}$ , the equation becomes

\ddot{𝑥} = 𝑔^{- 1} (\frac{1}{2} {\dot{𝑥}}^{⊺} (\partial 𝑔) \dot{𝑥} - (\partial (\dot{𝑥}) 𝑔) \dot{𝑥})

Or written as

\begin{matrix} \ddot{𝑥} + {\dot{𝑥}}^{⊺} Γ \dot{𝑥} & = & 0 \\ or {\ddot{𝑥}}^{𝑗} + Γ_{𝑖 𝑖^{'}}^{𝑗} \cdot {\dot{𝑥}}^{𝑖} \cdot {\dot{𝑥}}^{𝑖^{'}} & = & 0 \end{matrix}

where $Γ$ is metric-connection_(tag) alias Levi-Civita-connection_(tag) alias Christoffel-symbols_(tag)

\begin{matrix} 𝑣^{⊺} Γ 𝑣 & = & 𝑔^{- 1} ((\partial (𝑣) 𝑔) 𝑣 - \frac{1}{2} 𝑣^{⊺} (\partial 𝑔) 𝑣) \in ℝ^{𝑝, 𝑞} \\ or & 𝑣^{' ⊺} Γ 𝑣 & = & \frac{1}{2} 𝑔^{- 1} ((\partial (𝑣^{'}) 𝑔) 𝑣 + (\partial (𝑣) 𝑔) 𝑣^{'} - 𝑣^{' ⊺} (\partial 𝑔) 𝑣) & (cf. #link(<difference-symmetric-tensor>)[difference]) \\ or & Γ_{𝑖 𝑖^{'}}^{𝑗} & = & \frac{1}{2} \sum_{𝑖^{″}} 𝑔^{𝑗 𝑖^{″}} (\partial_{𝑖} 𝑔_{𝑖^{'} 𝑖^{″}} + \partial_{𝑖^{'}} 𝑔_{𝑖 𝑖^{″}} - \partial_{𝑖^{″}} 𝑔_{𝑖 𝑖^{'}}) \end{matrix}

metric-connection is not a tensor. The transformation of metric-connection connection-transformations_(tag)

Γ (𝑦) = \frac{\partial 𝑦}{\partial 𝑥} \cdot Γ (𝑥) \cdot \frac{\partial 𝑥}{\partial 𝑦} + \frac{\partial 𝑦}{\partial 𝑥} \cdot \frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦})

by the transformation of metric $𝑔$ in the definition of $Γ$

$𝑔 (𝑦) = \frac{\partial 𝑥}{\partial 𝑦}^{⊺} \cdot 𝑔 (𝑥) \cdot \frac{\partial 𝑥}{\partial 𝑦}$
$𝑔^{- 1} (𝑦) = \frac{\partial 𝑦}{\partial 𝑥} \cdot 𝑔^{- 1} (𝑥) \cdot \frac{\partial 𝑦}{\partial 𝑥}^{⊺}$

geodesic-exponential_(tag) …

geodesic-coordinate_(tag)

Geodesic $𝑡 ⇝ 𝑣 𝑡$ or $exp (𝑣 𝑡)$ with $𝑡 = 1$ gives coordinate $𝑣 \to exp (𝑣)$

It is a local diffeomorphism by $\partial exp = 𝟙$ at $𝑝$ , by

The solution of the ODE analytically depends on the initial values $𝑝, 𝑣$
$\frac{\partial}{\partial 𝑣} exp = \frac{\partial}{\partial 𝑡} (𝑡 = 0) 𝑥 (𝑡, 𝑝, 𝑣) = 𝑣$

In geodesic coordinates, the geodesic equation is $\ddot{𝑥} = 0$ . Proof The geodesic is $𝑡 ⇝ 𝑣 𝑡$

In geodesic coordinates at point $𝑝$ , the connection at point $𝑝$ is zero, $Γ (𝑝) = 0$

Proof

ODE $\ddot{𝑥} + {\dot{𝑥}}^{⊺} Γ \dot{𝑥} = 0$

Initial value $\ddot{𝑥} = 0$ and $\dot{𝑥} = 𝑣$

Substitute the solution $𝑡 𝑣$ of the ODE into the ODE to get $𝑣^{⊺} Γ 𝑣 = 0$

Thus, at point $𝑝$ , for all directions $𝑣$ , $𝑣^{⊺} Γ 𝑣 = 0$ ==> $Γ = 0$ at $𝑝$

Taylor-expansion-of-metric-in-geodesic-coordinate_(tag)

In geodesic coordinates, the Taylor expansion of the metric is $𝑔 (𝑝 + 𝑣) = \sum \frac{1}{𝑛!} (\partial^{𝑛} 𝑔) (𝑝) (𝑣^{𝑛})$

The zeroth-order term is the standard metric $𝜂$
The first-order term is zero i.e. the first derivative is zero

Combining the two, $𝑔 (𝑝 + 𝑣) = 𝜂 + 𝑜 (𝑣)$

Proof

In geodesic coordinates at point $𝑝$

0th. For geodesic coordinates, at point $𝑝$ , $\partial exp = 𝟙$ , which copies the orthonormal basis $𝑒_{1}, \dots, 𝑒_{𝑛}$ of the origin in geodesic coordinates to the coordinate-frame $\partial_{1}, \dots, \partial_{𝑛}$ of the tangent space at point $𝑝$
1st.

differenial-of-metric-vs-connection_(tag) Prop $𝑣^{⊺} (\partial (𝑣^{″}) 𝑔) 𝑣^{'} = 𝑣^{' ⊺} 𝑔 (𝑣^{″ ⊺} Γ 𝑣) + 𝑣^{' ⊺} 𝑔 (𝑣^{″ ⊺} Γ 𝑣)$ Proof Directly substitute the #link(<metric-connection>)[definition] of $Γ$ into the formula

Then use $Γ (𝑝) = 0$ to get $\partial 𝑔 (𝑝) = 0$ at point $𝑝$

In coordinates $\partial_{𝑖^{″}} 𝑔_{𝑖 𝑖^{'}} = Γ_{𝑖^{″} 𝑖 𝑖^{'}} + Γ_{𝑖^{″} 𝑖^{'} 𝑖} = 𝑔_{𝑖^{'} 𝑗} Γ_{𝑖^{″} 𝑖}^{𝑗} + 𝑔_{𝑖 𝑗} Γ_{𝑖^{″} 𝑖^{'}}^{𝑗}$

Can also be written as $(\partial (𝑣^{″}) 𝑔) (𝑣, 𝑣^{'}) = 𝑔 (Γ (𝑣^{″}, 𝑣), 𝑣^{'}) + 𝑔 (𝑣, Γ (𝑣^{″}, 𝑣^{'}))$

For the inverse matrix, there is a similar one

differenial-of-metric-inverse-vs-connection_(tag) Prop $𝛼^{⊺} (\partial (𝛼^{″}) 𝑔^{- 1}) 𝛼^{'} = - (𝛼^{' ⊺} 𝑔^{- 1} (𝛼^{″ ⊺} Γ 𝛼) + 𝛼^{' ⊺} 𝑔^{- 1} (𝛼^{″ ⊺} Γ 𝛼))$

Proof Using $𝑔 𝑔^{- 1} = 𝟙 ⟹ \partial (𝑔 𝑔^{- 1}) = 0 ⟹ \partial (𝑔^{- 1}) = - 𝑔^{- 1} (\partial 𝑔) 𝑔^{- 1}$ and $𝑔 = 𝑔 𝑔^{- 1} 𝑔$

In coordinates $\partial_{𝑖^{″}} 𝑔^{𝑖 𝑖^{'}} = - Γ_{𝑖^{″}}^{𝑖 𝑖^{'}} - Γ_{𝑖^{″}}^{𝑖^{'} 𝑖} = - 𝑔^{𝑖 𝑗} Γ_{𝑖^{″} 𝑗}^{𝑖^{'}} - 𝑔^{𝑖^{'} 𝑗} Γ_{𝑖^{″} 𝑗}^{𝑖}$

Can also be written as $(\partial (𝛼^{″}) 𝑔^{- 1}) (𝛼, 𝛼^{'}) = 𝑔^{- 1} (- Γ (𝛼^{″}, 𝛼), 𝛼^{'}) + 𝑔^{- 1} (𝛼, - Γ (𝛼^{″}, 𝛼^{'}))$