Geodesic on a Surface of Revolution
📂Geometry Geodesic on a Surface of Revolution Theorem
Let us assume that M M M surface of revolution generated by the unit-speed curve α ( t ) = ( r ( t ) , z ( t ) ) \alpha (t) = (r(t), z(t)) α ( t ) = ( r ( t ) , z ( t ))
(a) All meridians are geodesics.
(b) The condition for parallels to be geodesics is that x t \mathbf{x}_{t} x t
The circle of latitude is a geodesic. ⟺ x t is parallel to the axis of revolution at all point on the circle of latitude.
\text{The circle of latitude is a geodesic.}
\\ \iff \mathbf{x}_{t} \text{ is parallel to the axis of revolution at all point on the circle of latitude.}
The circle of latitude is a geodesic. ⟺ x t is parallel to the axis of revolution at all point on the circle of latitude.
Explanation Although geodesics are not always the globally shortest paths, they can locally be the shortest paths.
(b) implies that parallels are geodesics only at the most convex or concave parts of the surface of revolution.
Proof First, calculate the part that is common across all scenarios, then apply it specifically to each case.
Let’s denote the surface of revolution as x ( t , θ ) = ( r ( t ) cos θ , r ( t ) sin θ , z ( t ) ) \mathbf{x}(t, \theta) = \left( r(t)\cos\theta, r(t)\sin\theta, z(t) \right) x ( t , θ ) = ( r ( t ) cos θ , r ( t ) sin θ , z ( t ) )
x t ( t , θ ) = x 1 = ( r ˙ ( t ) cos θ , r ˙ ( t ) sin θ , z ˙ ( t ) )
\mathbf{x}_{t}(t,\theta) = \mathbf{x}_{1} = \left( \dot{r}(t) \cos \theta, \dot{r}(t) \sin \theta, \dot{z}(t) \right)
x t ( t , θ ) = x 1 = ( r ˙ ( t ) cos θ , r ˙ ( t ) sin θ , z ˙ ( t ) )
x θ ( t , θ ) = x 2 = ( − r ( t ) sin θ , r ( t ) cos θ , 0 )
\mathbf{x}_{\theta}(t, \theta) = \mathbf{x}_{2} = \left( -r(t) \sin\theta, r(t)\cos\theta , 0\right)
x θ ( t , θ ) = x 2 = ( − r ( t ) sin θ , r ( t ) cos θ , 0 )
Then, the Riemann metric matrix and its inverse are as follows. Since α \alpha α ∣ α ˙ ∣ 2 = r ˙ 2 + z ˙ 2 = 1 \left| \dot{\alpha} \right|^{2} = \dot{r}^{2} + \dot{z}^{2} = 1 ∣ α ˙ ∣ 2 = r ˙ 2 + z ˙ 2 = 1
[ g i j ] = [ ⟨ x 1 , x 1 ⟩ ⟨ x 1 , x 2 ⟩ ⟨ x 2 , x 1 ⟩ ⟨ x 2 , x 2 ⟩ ] = [ r ˙ 2 + z ˙ 2 0 0 r 2 ] = [ 1 0 0 r 2 ]
\left[ g_{ij} \right] = \begin{bmatrix} \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle & \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle \\ \left\langle \mathbf{x}_{2}, \mathbf{x}_{1} \right\rangle & \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle \end{bmatrix}
= \begin{bmatrix} \dot{r}^{2} + \dot{z}^{2} & 0 \\ 0 & r^{2} \end{bmatrix}
= \begin{bmatrix} 1 & 0 \\ 0 & r^{2} \end{bmatrix}
[ g ij ] = [ ⟨ x 1 , x 1 ⟩ ⟨ x 2 , x 1 ⟩ ⟨ x 1 , x 2 ⟩ ⟨ x 2 , x 2 ⟩ ] = [ r ˙ 2 + z ˙ 2 0 0 r 2 ] = [ 1 0 0 r 2 ]
[ g i j ] − 1 = [ g l k ] = 1 r 2 [ r 2 0 0 1 ] = [ 1 0 0 1 r 2 ]
\left[ g_{ij} \right]^{-1} = [g^{lk}] = \dfrac{1}{r^{2}}\begin{bmatrix} r^{2} & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \dfrac{1}{r^{2}} \end{bmatrix}
[ g ij ] − 1 = [ g l k ] = r 2 1 [ r 2 0 0 1 ] = [ 1 0 0 r 2 1 ]
Christoffel symbols
Γ i j k = 1 2 ∑ l = 1 2 g l k ( ∂ g l j ∂ u i − ∂ g i j ∂ u l + ∂ g i l ∂ u j )
\Gamma_{ij}^{k} = \dfrac{1}{2} \sum \limits_{l=1}^{2} g^{lk} \left( \dfrac{\partial g_{lj}}{\partial u_{i}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{il}}{\partial u_{j}} \right)
Γ ij k = 2 1 l = 1 ∑ 2 g l k ( ∂ u i ∂ g l j − ∂ u l ∂ g ij + ∂ u j ∂ g i l )
[ g i j ] \left[ g_{ij} \right] [ g ij ] g l k g^{lk} g l k l = k l=k l = k Γ i j k \Gamma_{ij}^{k} Γ ij k
Γ 11 1 = 1 2 g 11 ∂ g 11 ∂ u 1 = 0 Γ 12 1 = Γ 21 1 = 1 2 g 11 ( ∂ g 21 ∂ u 1 − ∂ g 12 ∂ u 1 + ∂ g 12 ∂ u 2 ) = 0 Γ 22 1 = 1 2 g 11 ( ∂ g 12 ∂ u 2 − ∂ g 22 ∂ u 1 + ∂ g 21 ∂ u 2 ) = − 1 2 g 11 ∂ g 22 ∂ u 1 = − 1 2 1 ⋅ ∂ g 22 ∂ t = − 1 2 ∂ ∂ t ( r 2 ) = − r r ˙ Γ 11 2 = 1 2 g 22 ( ∂ g 12 ∂ u 1 − ∂ g 11 ∂ u 2 + ∂ g 12 ∂ u 1 ) = 0 Γ 12 2 = Γ 21 2 = 1 2 g 22 ( ∂ g 22 ∂ u 1 − ∂ g 12 ∂ u 2 + ∂ g 12 ∂ u 2 ) = 1 2 1 r 2 ∂ ∂ t r 2 = 1 2 1 r 2 2 r r ˙ = r ˙ r Γ 22 2 = 1 2 g 22 ( ∂ g 22 ∂ u 2 ) = 1 2 1 r 2 ∂ ∂ θ r 2 = 0
\begin{align*}
\Gamma_{11}^{1} =&\ \dfrac{1}{2}g^{11}\dfrac{\partial g_{11}}{\partial u_{1}} = 0
\\ \Gamma_{12}^{1} =&\ \Gamma_{21}^{1} = \dfrac{1}{2}g^{11} \left( \dfrac{\partial g_{21}}{\partial u_{1}} - \dfrac{\partial g_{12}}{\partial u_{1}} + \dfrac{\partial g_{12}}{\partial u_{2}}\right) = 0
\\ \Gamma_{22}^{1} =&\ \dfrac{1}{2} g^{11} \left( \dfrac{\partial g_{12}}{\partial u_{2}} - \dfrac{\partial g_{22}}{\partial u_{1}} + \dfrac{\partial g_{21}}{\partial u_{2}}\right) = -\dfrac{1}{2} g^{11} \dfrac{\partial g_{22}}{\partial u_{1}} = -\dfrac{1}{2} 1 \cdot \dfrac{\partial g_{22}}{\partial t} = -\dfrac{1}{2}\dfrac{\partial}{\partial t}(r^{2}) = -r\dot{r}
\\ \Gamma_{11}^{2} =&\ \dfrac{1}{2} g^{22} \left( \dfrac{\partial g_{12}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{2}} + \dfrac{\partial g_{12}}{\partial u_{1}} \right)=0
\\ \Gamma_{12}^{2} =&\ \Gamma_{21}^{2} = \dfrac{1}{2} g^{22} \left( \dfrac{\partial g_{22}}{\partial u_{1}} - \dfrac{\partial g_{12}}{\partial u_{2}} + \dfrac{\partial g_{12}}{\partial u_{2}} \right) = \dfrac{1}{2} \dfrac{1}{r^{2}} \dfrac{\partial}{\partial t}r^{2} = \dfrac{1}{2} \dfrac{1}{r^{2}} 2 r \dot{r} = \dfrac{\dot{r}}{r}
\\ \Gamma_{22}^{2} =&\ \dfrac{1}{2} g^{22} \left( \dfrac{\partial g_{22}}{\partial u_{2}} \right) = \dfrac{1}{2}\dfrac{1}{r^{2}} \dfrac{\partial}{\partial \theta}r^{2} = 0
\end{align*}
Γ 11 1 = Γ 12 1 = Γ 22 1 = Γ 11 2 = Γ 12 2 = Γ 22 2 = 2 1 g 11 ∂ u 1 ∂ g 11 = 0 Γ 21 1 = 2 1 g 11 ( ∂ u 1 ∂ g 21 − ∂ u 1 ∂ g 12 + ∂ u 2 ∂ g 12 ) = 0 2 1 g 11 ( ∂ u 2 ∂ g 12 − ∂ u 1 ∂ g 22 + ∂ u 2 ∂ g 21 ) = − 2 1 g 11 ∂ u 1 ∂ g 22 = − 2 1 1 ⋅ ∂ t ∂ g 22 = − 2 1 ∂ t ∂ ( r 2 ) = − r r ˙ 2 1 g 22 ( ∂ u 1 ∂ g 12 − ∂ u 2 ∂ g 11 + ∂ u 1 ∂ g 12 ) = 0 Γ 21 2 = 2 1 g 22 ( ∂ u 1 ∂ g 22 − ∂ u 2 ∂ g 12 + ∂ u 2 ∂ g 12 ) = 2 1 r 2 1 ∂ t ∂ r 2 = 2 1 r 2 1 2 r r ˙ = r r ˙ 2 1 g 22 ( ∂ u 2 ∂ g 22 ) = 2 1 r 2 1 ∂ θ ∂ r 2 = 0
Necessary and sufficient condition for a geodesic
The necessary and sufficient condition for the unit-speed curve α ( s ) = x ( α 1 ( s ) , α 2 ( s ) ) \alpha (s) = \mathbf{x}(\alpha^{1}(s), \alpha^{2}(s)) α ( s ) = x ( α 1 ( s ) , α 2 ( s ))
α is geodesic ⟺ ( α k ) ′ ′ + ∑ i = 1 2 ∑ j = 1 2 Γ i j k ( α i ) ′ ( α j ) ′ = 0 , ∀ k = 1 , 2
\alpha \text{ is geodesic} \iff (\alpha^{k})^{\prime \prime} + \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} {\Gamma _{ij}}^{k} (\alpha^{i})^{\prime} (\alpha^{j})^{\prime} = 0, \quad \forall k=1,2
α is geodesic ⟺ ( α k ) ′′ + i = 1 ∑ 2 j = 1 ∑ 2 Γ ij k ( α i ) ′ ( α j ) ′ = 0 , ∀ k = 1 , 2
As α 1 ( s ) = t ( s ) , α 2 ( s ) = θ ( s ) \alpha^{1}(s)=t(s), \alpha^{2}(s)=\theta (s) α 1 ( s ) = t ( s ) , α 2 ( s ) = θ ( s )
When k = 1 k=1 k = 1
( α 1 ) ′ ′ + Γ 22 1 ( α 2 ) ′ ( α 2 ) ′ = t ′ ′ − r r ˙ θ ′ θ ′ = 0
\begin{equation}
(\alpha^{1})^{\prime \prime} + \Gamma_{22}^{1}(\alpha^{2})^{\prime}(\alpha^{2})^{\prime} = t^{\prime \prime} -r\dot{r}\theta^{\prime}\theta^{\prime}= 0
\end{equation}
( α 1 ) ′′ + Γ 22 1 ( α 2 ) ′ ( α 2 ) ′ = t ′′ − r r ˙ θ ′ θ ′ = 0
When k = 2 k=2 k = 2
( α 2 ) ′ ′ + 2 Γ 12 2 ( α 1 ) ′ ( α 2 ) ′ = θ ′ ′ + 2 r ˙ r t ′ θ ′ = 0
\begin{equation}
(\alpha^{2})^{\prime \prime} + 2\Gamma_{12}^{2}(\alpha^{1})^{\prime} (\alpha^{2})^{\prime} = \theta^{\prime \prime} + 2\dfrac{\dot{r}}{r} t^{\prime} \theta^{\prime} = 0
\end{equation}
( α 2 ) ′′ + 2 Γ 12 2 ( α 1 ) ′ ( α 2 ) ′ = θ ′′ + 2 r r ˙ t ′ θ ′ = 0
(a) A meridian α \alpha α parameter curve on the surface, hence, for a fixed θ 0 \theta_{0} θ 0
α ( s ) = x ( t ( s ) , θ 0 ) = ( r ( t ) cos θ 0 , r ( t ) sin θ 0 , z ( t ) ) ,
\alpha (s) = \mathbf{x}(t(s), \theta_{0}) = \left( r(t) \cos \theta_{0}, r(t)\sin\theta_{0}, z(t) \right),
α ( s ) = x ( t ( s ) , θ 0 ) = ( r ( t ) cos θ 0 , r ( t ) sin θ 0 , z ( t ) ) ,
Since θ 0 \theta_{0} θ 0 θ ′ = θ ′ ′ = 0 \theta^{\prime}=\theta^{\prime \prime}=0 θ ′ = θ ′′ = 0 ( 2 ) (2) ( 2 ) ( 1 ) (1) ( 1 )
t ′ ′ = 0
t^{\prime \prime} = 0
t ′′ = 0
As the meridian is the case of s = t s=t s = t t ′ = d t d t = 1 t^{\prime} = \dfrac{dt}{dt} =1 t ′ = d t d t = 1 t ′ ′ = 0 t^{\prime \prime}=0 t ′′ = 0 ( 1 ) (1) ( 1 )
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(b) A parallel is when t t t t ′ = t ′ ′ = 0 t^{\prime} = t^{\prime \prime} = 0 t ′ = t ′′ = 0 α ( θ ) \alpha (\theta) α ( θ ) x ( t , θ ) \mathbf{x}(t, \theta) x ( t , θ ) t t t
α ( θ ) = ( r ( t 0 ) cos θ , r ( t 0 ) sin θ , z ( t 0 ) )
\alpha (\theta) = \left( r(t_{0}) \cos \theta, r(t_{0})\sin \theta, z(t_{0}) \right)
α ( θ ) = ( r ( t 0 ) cos θ , r ( t 0 ) sin θ , z ( t 0 ) )
∣ d α d θ ∣ = ∣ ( − r ( t 0 ) sin θ , r ( t 0 ) cos θ , 0 ) ∣ = r ( t 0 )
\left| \dfrac{d\alpha}{d\theta} \right| = \left| (-r(t_{0})\sin\theta, r(t_{0})\cos\theta, 0) \right| = r(t_{0})
d θ d α = ∣ ( − r ( t 0 ) sin θ , r ( t 0 ) cos θ , 0 ) ∣ = r ( t 0 )
Therefore, consider the reparametrized parallel as described in α ( s ) = x ( t ( s ) , θ ( s ) ) \alpha (s) = \mathbf{x}\left( t(s), \theta (s) \right) α ( s ) = x ( t ( s ) , θ ( s ) )
1 = ∣ α ′ ( s ) ∣ 2 = ∣ ∂ x ∂ t t ′ + ∂ x ∂ θ θ ′ ∣ 2 = ∣ ∂ x ∂ θ θ ′ ∣ 2 = ⟨ x θ , x θ ⟩ ( θ ′ ) 2 = g 22 ( θ ′ ) 2
1 = \left| \alpha^{\prime}(s) \right|^{2} = \left| \dfrac{\partial \mathbf{x}}{\partial t}t^{\prime} + \dfrac{\partial \mathbf{x}}{\partial \theta}\theta^{\prime} \right|^{2} = \left| \dfrac{\partial \mathbf{x}}{\partial \theta}\theta^{\prime} \right|^{2} = \left\langle \mathbf{x}_{\theta} , \mathbf{x}_{\theta} \right\rangle (\theta^{\prime})^{2} = g_{22} (\theta^{\prime})^{2}
1 = ∣ α ′ ( s ) ∣ 2 = ∂ t ∂ x t ′ + ∂ θ ∂ x θ ′ 2 = ∂ θ ∂ x θ ′ 2 = ⟨ x θ , x θ ⟩ ( θ ′ ) 2 = g 22 ( θ ′ ) 2
As g 22 = r 2 g_{22} = r^{2} g 22 = r 2
1 = r 2 ( θ ′ ) 2
1 = r^{2} (\theta^{\prime})^{2}
1 = r 2 ( θ ′ ) 2
Thus, θ ′ ≠ 0 \theta^{\prime} \ne 0 θ ′ = 0
0 ≠ θ ′ = ± 1 r
0 \ne \theta^{\prime} = \pm \dfrac{1}{r}
0 = θ ′ = ± r 1
At this point, on the parallel, r r r θ ′ ′ = 0 \theta^{\prime \prime} = 0 θ ′′ = 0 ( 2 ) (2) ( 2 ) ( 1 ) (1) ( 1 )
( 1 ) is hold. ⟺ r ˙ = 0 ⟺ x t = ( r ˙ cos θ , r ˙ sin θ , z ˙ ) = ( 0 , 0 , z ˙ ) ⟺ x t is parallel to z − axis.
\begin{align*}
(1) \text{ is hold.} \iff& \dot{r}=0
\\ \iff& \mathbf{x}_{t} = (\dot{r} \cos \theta, \dot{r}\sin\theta, \dot{z}) = (0,0,\dot{z})
\\ \iff& \mathbf{x}_{t} \text{ is parallel to } z- \text{axis.}
\end{align*}
( 1 ) is hold. ⟺ ⟺ ⟺ r ˙ = 0 x t = ( r ˙ cos θ , r ˙ sin θ , z ˙ ) = ( 0 , 0 , z ˙ ) x t is parallel to z − axis.
Since ( 1 ) (1) ( 1 )
The circle of latitude is a geodesic. ⟺ x t is parallel to the axis of revolution at all point on the circle of latitude.
\text{The circle of latitude is a geodesic.}
\\ \iff \mathbf{x}_{t} \text{ is parallel to the axis of revolution at all point on the circle of latitude.}
The circle of latitude is a geodesic. ⟺ x t is parallel to the axis of revolution at all point on the circle of latitude.
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