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Geodesic Coordinate Patch Mapping and Christoffel Symbols 📂Geometry

Geodesic Coordinate Patch Mapping and Christoffel Symbols

Theorem1

[gij]=[100h2](h>0) \left[ g_{ij} \right] = \begin{bmatrix} 1 & 0 \\ 0 & h^{2} \end{bmatrix} \quad (h \gt 0)

Then, the Christoffel symbols of x\mathbf{x} are as follows, and except for the below, all are 00.

Γ221=−hh1,Γ122=Γ212=h1h,Γ222=h2h \Gamma_{22}^{1} = -hh_{1},\quad \Gamma_{12}^{2} = \Gamma_{21}^{2} = \dfrac{h_{1}}{h},\quad \Gamma_{22}^{2} = \dfrac{h_{2}}{h}

At this time, (u1,u2)(u^{1}, u^{2}) is the coordinate of UU, and hi=∂h∂uih_{i} = \dfrac{\partial h}{\partial u^{i}} is valid.

Proof

Before proving, let’s do the necessary calculations. Since g11=⟨x1,x1⟩=1g_{11} = \left\langle\mathbf{x}_{1}, \mathbf{x}_{1}\right\rangle = 1,

∂g11∂ui=∂⟨x1,x1⟩∂ui=2⟨x1i,x1⟩=0 \dfrac{\partial g_{11}}{\partial u_{i}} = \dfrac{\partial \left\langle\mathbf{x}_{1}, \mathbf{x}_{1}\right\rangle}{\partial u_{i}} = 2\left\langle\mathbf{x}_{1i}, \mathbf{x}_{1}\right\rangle = 0

  ⟹  ⟨x11,x1⟩=0 and ⟨x12,x1⟩=0(1) \implies \left\langle\mathbf{x}_{11}, \mathbf{x}_{1}\right\rangle = 0 \quad \text{ and } \quad \left\langle\mathbf{x}_{12}, \mathbf{x}_{1}\right\rangle = 0 \tag{1}

Similarly, since g12=⟨x1,x2⟩=0g_{12} = \left\langle\mathbf{x}_{1}, \mathbf{x}_{2}\right\rangle = 0,

∂g12∂ui=∂⟨x1,x2⟩∂ui=⟨x1i,x2⟩+⟨x1,x2i⟩=0 \dfrac{\partial g_{12}}{\partial u_{i}} = \dfrac{\partial \left\langle\mathbf{x}_{1}, \mathbf{x}_{2}\right\rangle}{\partial u_{i}} = \left\langle\mathbf{x}_{1i}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{2i}\right\rangle = 0

When looking at the case of i=1i=1, ⟨x11,x2⟩+⟨x1,x21⟩=0\left\langle \mathbf{x}_{11}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{21}\right\rangle = 0, and by (1)(1), it becomes ⟨x11,x2⟩=0\left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle = 0. When i=2i=2,

⟨x12,x2⟩+⟨x1,x22⟩=0(2) \left\langle\mathbf{x}_{12}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{22}\right\rangle = 0 \tag{2}

Also, since h=g22=⟨x2,x2⟩h = \sqrt{g_{22}} = \sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle},

hi=∂h∂ui=∂⟨x2,x2⟩∂ui=12⟨x2,x2⟩2⟨x2i,x2⟩=1h⟨x2i,x2⟩ h_{i} = \dfrac{\partial h}{\partial u^{i}} = \dfrac{\partial \sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}}{\partial u^{i}} = \dfrac{1}{2\sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}} 2\left\langle \mathbf{x}_{2i}, \mathbf{x}_{2} \right\rangle = \dfrac{1}{h}\left\langle \mathbf{x}_{2i}, \mathbf{x}_{2} \right\rangle

  ⟹  ⟨x21,x2⟩=hh1 and ⟨x22,x2⟩=hh2(3) \implies \left\langle \mathbf{x}_{21}, \mathbf{x}_{2} \right\rangle = hh_{1} \quad \text{ and } \quad \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle = hh_{2}\tag{3}

Christoffel Symbol

Γijk:=∑l=12⟨xij,xl⟩glk \Gamma_{ij}^{k} := \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk}

Now, just substituting what we calculated above into the definition of the Christoffel symbol gives the result.

Γ111=⟨x11,x1⟩g11+⟨x11,x2⟩g21=0⋅1+0⋅0=0 \begin{align*} \Gamma_{11}^{1} &= \left\langle \mathbf{x}_{11}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= 0 \cdot 1 + 0 \cdot 0 = 0 \end{align*}

Γ121=Γ211=⟨x12,x1⟩g11+⟨x12,x2⟩g21=0⋅1+⟨x12,x2⟩⋅0=0 \begin{align*} \Gamma_{12}^{1} = \Gamma_{21}^{1} &= \left\langle \mathbf{x}_{12}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= 0 \cdot 1 + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle \cdot 0 = 0 \end{align*}

Γ221=⟨x22,x1⟩g11+⟨x22,x2⟩g21=⟨x22,x1⟩⋅1+⟨x22,x2⟩⋅0=⟨x22,x1⟩=−⟨x12,x2⟩ by (2)=−hh1 by (3) \begin{align*} \Gamma_{22}^{1} &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle \cdot 1 + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle \cdot 0 \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle = - \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle & \text{ by } (2)\\ &= - hh_{1} & \text{ by } (3)\\ \end{align*}

Γ112=⟨x11,x1⟩g12+⟨x11,x2⟩g22=0⋅0+0⋅1h2=0 \begin{align*} \Gamma_{11}^{2} &= \left\langle \mathbf{x}_{11}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= 0 \cdot 0 + 0 \cdot \dfrac{1}{h^{2}} = 0 \\ \end{align*}

Γ122=Γ212=⟨x12,x1⟩g12+⟨x12,x2⟩g22=0⋅0+hh1⋅1h2=h1h \begin{align*} \Gamma_{12}^{2} = \Gamma_{21}^{2} &= \left\langle \mathbf{x}_{12}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= 0 \cdot 0 + hh_{1} \cdot \dfrac{1}{h^{2}} \\ &= \dfrac{h_{1}}{h} \\ \end{align*}

Γ222=⟨x22,x1⟩g12+⟨x22,x2⟩g22=⟨x22,x1⟩⋅0+hh2⋅1h2=h2h \begin{align*} \Gamma_{22}^{2} &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle \cdot 0 + hh_{2} \cdot \dfrac{1}{h^{2}} \\ &= \dfrac{h_{2}}{h} \\ \end{align*}

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  1. Richard S. Millman and George D. Parker, Elements of Differential Geometry (1977), p179 problem 2.3 ↩︎