Specific Examples of Calculations Using the Riemann Metric
📂Geometry Specific Examples of Calculations Using the Riemann Metric Notation Let’s say we have coordinates for ( u , v ) (u,v) ( u , v ) U U U simple surface x : U → R 3 \mathbf{x} : U \to \mathbb{R}^{3} x : U → R 3
x 1 : = ∂ x ∂ u and x 2 : = ∂ x ∂ v
\mathbf{x}_{1} := \dfrac{\partial \mathbf{x}}{\partial u}\quad \text{and} \quad \mathbf{x}_{2} := \dfrac{\partial \mathbf{x}}{\partial v}
x 1 := ∂ u ∂ x and x 2 := ∂ v ∂ x
Let’s represent the coefficients of the Riemannian metric as follows.
g i j = ⟨ x i , x j ⟩ g 11 = ⟨ x 1 , x 1 ⟩ = E g 12 = g 21 = ⟨ x 1 , x 2 ⟩ = F g 22 = ⟨ x 2 , x 2 ⟩ = G
\begin{align*}
g_{ij} =&\ \left\langle \mathbf{x}_{i}, \mathbf{x}_{j} \right\rangle
\\ g_{11} =&\ \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = E
\\ g_{12} =&\ g_{21} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = F
\\ g_{22} =&\ \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle = G
\end{align*}
g ij = g 11 = g 12 = g 22 = ⟨ x i , x j ⟩ ⟨ x 1 , x 1 ⟩ = E g 21 = ⟨ x 1 , x 2 ⟩ = F ⟨ x 2 , x 2 ⟩ = G
Example Length of a Curve Let’s say U = { ( u , v ) : u 2 + v 2 < 1 } U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} U = { ( u , v ) : u 2 + v 2 < 1 } simple surface x : U → R 3 \mathbf{x} : U \to \mathbb{R}^{3} x : U → R 3
x ( u , v ) = ( u , v , 1 − u 2 − v 2 )
\mathbf{x}(u,v) = (u, v, \sqrt{1-u^{2}-v^{2}})
x ( u , v ) = ( u , v , 1 − u 2 − v 2 )
Then
x u = x 1 = ( 1 , 0 , − u 1 − u 2 − v 2 ) and x v = x 2 = ( 0 , 1 , − v 1 − u 2 − v 2 )
\mathbf{x}_{u} = \mathbf{x}_{1} = \left( 1, 0, \dfrac{-u}{\sqrt{1-u^{2}-v^{2}}} \right) \quad \text{and} \quad \mathbf{x}_{v} = \mathbf{x}_{2} = \left( 0, 1, \dfrac{-v}{\sqrt{1-u^{2}-v^{2}}} \right)
x u = x 1 = ( 1 , 0 , 1 − u 2 − v 2 − u ) and x v = x 2 = ( 0 , 1 , 1 − u 2 − v 2 − v )
And let’s say u , v u, v u , v
α 1 ( t ) = u ( t ) = t and α 2 ( t ) = v ( t ) = t 2
\alpha_{1}(t) = u(t) = t \quad \text{and} \quad \alpha_{2}(t) = v(t) = t^{2}
α 1 ( t ) = u ( t ) = t and α 2 ( t ) = v ( t ) = t 2
Then, the path in each region according to 0 < t < 5 − 1 2 0 \lt t \lt \sqrt{\dfrac{\sqrt{5}-1}{2}} 0 < t < 2 5 − 1
The formula for calculating the length of the blue curve on the hemisphere is as follows.
∫ a b g i j α i ′ α j ′ d t = ∫ a b E ( u ′ ) 2 + 2 F u ′ v ′ + G ( v ′ ) 2 d t
\int_{a}^{b} \sqrt{ g_{ij} \alpha_{i}^{\prime} \alpha_{j}^{\prime} } dt = \int_{a}^{b} \sqrt{ E(u^{\prime})^{2} + 2F u^{\prime} v^{\prime} + G(v^{\prime})^{2}}dt
∫ a b g ij α i ′ α j ′ d t = ∫ a b E ( u ′ ) 2 + 2 F u ′ v ′ + G ( v ′ ) 2 d t
Calculating each gives the following.
u ′ = 1 and v ′ = 2 t
u^{\prime} = 1 \quad \text{and} \quad v^{\prime} = 2t
u ′ = 1 and v ′ = 2 t
E = g 11 = ⟨ x 1 , x 1 ⟩ = 1 + u 2 1 − u 2 − v 2 = 1 − v 2 1 − u 2 − v 2 = 1 − t 4 1 − t 2 − t 4 F = g 12 = ⟨ x 1 , x 2 ⟩ = u v 1 − u 2 − v 2 = t 3 1 − t 2 − t 4 G = g 22 = ⟨ x 2 , x 2 ⟩ = 1 + v 2 1 − u 2 − v 2 = 1 − u 2 1 − u 2 − v 2 = 1 − t 2 1 − t 2 − t 4
\begin{align*}
E =&\ g_{11} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = 1 + \dfrac{u^{2}}{1-u^{2}-v^{2}} = \dfrac{1-v^{2}}{1-u^{2}-v^{2}} = \dfrac{1-t^{4}}{1-t^{2}-t^{4}}
\\ F =&\ g_{12} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = \dfrac{uv}{1-u^{2}-v^{2}} = \dfrac{t^{3}}{1-t^{2}-t^{4}}
\\ G =&\ g_{22} = \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle = 1 + \dfrac{v^{2}}{1-u^{2}-v^{2}} = \dfrac{1-u^{2}}{1-u^{2}-v^{2}} = \dfrac{1-t^{2}}{1-t^{2}-t^{4}}
\end{align*}
E = F = G = g 11 = ⟨ x 1 , x 1 ⟩ = 1 + 1 − u 2 − v 2 u 2 = 1 − u 2 − v 2 1 − v 2 = 1 − t 2 − t 4 1 − t 4 g 12 = ⟨ x 1 , x 2 ⟩ = 1 − u 2 − v 2 uv = 1 − t 2 − t 4 t 3 g 22 = ⟨ x 2 , x 2 ⟩ = 1 + 1 − u 2 − v 2 v 2 = 1 − u 2 − v 2 1 − u 2 = 1 − t 2 − t 4 1 − t 2
Therefore, the length of the curve is as follows.
∫ 0 ( 5 − 1 ) / 2 E ( u ′ ) 2 + 2 F u ′ v ′ + G ( v ′ ) 2 d t = ∫ 0 ( 5 − 1 ) / 2 1 − t 4 1 − t 2 − t 4 ( 1 ) 2 + 2 t 3 1 − t 2 − t 4 ⋅ 1 ⋅ 2 t + 1 − t 2 1 − t 2 − t 4 ( 2 t ) 2 d t = ∫ 0 ( 5 − 1 ) / 2 − t 4 + 4 t 2 + 1 1 − t 2 − t 4 d t
\begin{align*}
& \int_{0}^{\sqrt{(\sqrt{5}-1) / 2}} \sqrt{ E(u^{\prime})^{2} + 2F u^{\prime} v^{\prime} + G(v^{\prime})^{2}}dt
\\ =&\ \int_{0}^{\sqrt{(\sqrt{5}-1) / 2}} \sqrt{ \dfrac{1-t^{4}}{1-t^{2}-t^{4}}(1)^{2} + 2\dfrac{t^{3}}{1-t^{2}-t^{4}} \cdot 1 \cdot 2t + \dfrac{1-t^{2}}{1-t^{2}-t^{4}}(2t)^{2}}dt
\\ =&\ \int_{0}^{\sqrt{(\sqrt{5}-1) / 2}} \sqrt{ \dfrac{-t^{4} + 4t^{2} +1}{1-t^{2}-t^{4}}}dt
\end{align*}
= = ∫ 0 ( 5 − 1 ) /2 E ( u ′ ) 2 + 2 F u ′ v ′ + G ( v ′ ) 2 d t ∫ 0 ( 5 − 1 ) /2 1 − t 2 − t 4 1 − t 4 ( 1 ) 2 + 2 1 − t 2 − t 4 t 3 ⋅ 1 ⋅ 2 t + 1 − t 2 − t 4 1 − t 2 ( 2 t ) 2 d t ∫ 0 ( 5 − 1 ) /2 1 − t 2 − t 4 − t 4 + 4 t 2 + 1 d t
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Area of a Surface In the same situation, let’s say Q = { ( u , v ) ∈ U : 0 ≤ u , v < 1 } Q= \left\{ (u,v) \in U : 0\le u,v \lt 1 \right\} Q = { ( u , v ) ∈ U : 0 ≤ u , v < 1 } u , v u, v u , v u = r cos θ , v = r sin θ u = r\cos \theta, v = r\sin \theta u = r cos θ , v = r sin θ
x u = x 1 = ( 1 , 0 , − u 1 − u 2 − v 2 ) = ( 1 , 0 , − r cos θ 1 − r 2 ) x v = x 2 = ( 0 , 1 , − v 1 − u 2 − v 2 ) = ( 1 , 0 , − r sin θ 1 − r 2 )
\mathbf{x}_{u} = \mathbf{x}_{1} = \left( 1, 0, \dfrac{-u}{\sqrt{1-u^{2}-v^{2}}} \right) = \left( 1, 0, \dfrac{-r\cos\theta}{\sqrt{1-r^{2}}} \right)
\\ \mathbf{x}_{v} = \mathbf{x}_{2} = \left( 0, 1, \dfrac{-v}{\sqrt{1-u^{2}-v^{2}}} \right) = \left( 1, 0, \dfrac{-r\sin\theta}{\sqrt{1-r^{2}}} \right)
x u = x 1 = ( 1 , 0 , 1 − u 2 − v 2 − u ) = ( 1 , 0 , 1 − r 2 − r cos θ ) x v = x 2 = ( 0 , 1 , 1 − u 2 − v 2 − v ) = ( 1 , 0 , 1 − r 2 − r sin θ )
E = g 11 = ⟨ x 1 , x 1 ⟩ = 1 − v 2 1 − u 2 − v 2 = 1 − r 2 sin 2 θ 1 − r 2 F = g 12 = ⟨ x 1 , x 2 ⟩ = r 2 cos θ sin θ 1 − r 2 G = g 22 = ⟨ x 2 , x 2 ⟩ = 1 − u 2 1 − u 2 − v 2 = 1 − r 2 cos 2 θ 1 − r 2
\begin{align*}
E =&\ g_{11} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = \dfrac{1-v^{2}}{1-u^{2}-v^{2}} = \dfrac{1-r^{2}\sin^{2}\theta}{1-r^{2}}
\\ F =&\ g_{12} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = \dfrac{r^{2}\cos\theta \sin\theta}{1-r^{2}}
\\ G =&\ g_{22} = \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle = \dfrac{1-u^{2}}{1-u^{2}-v^{2}} = \dfrac{1-r^{2}\cos^{2}\theta}{1-r^{2}}
\end{align*}
E = F = G = g 11 = ⟨ x 1 , x 1 ⟩ = 1 − u 2 − v 2 1 − v 2 = 1 − r 2 1 − r 2 sin 2 θ g 12 = ⟨ x 1 , x 2 ⟩ = 1 − r 2 r 2 cos θ sin θ g 22 = ⟨ x 2 , x 2 ⟩ = 1 − u 2 − v 2 1 − u 2 = 1 − r 2 1 − r 2 cos 2 θ
The formula for calculating the area of region R = x ( Q ) R = \mathbf{x}(Q) R = x ( Q )
∬ Q g d u d v = ∬ Q ∣ x 1 × x 2 ∣ d u d v = ∬ Q E G − F 2 d u d v
\iint _{Q} \sqrt{g} du dv = \iint _{Q} \left| \mathbf{x}_{1} \times \mathbf{x}_{2} \right| du dv = \iint _{Q} \sqrt{EG-F^{2}} du dv
∬ Q g d u d v = ∬ Q ∣ x 1 × x 2 ∣ d u d v = ∬ Q EG − F 2 d u d v
Since R R R 1 8 \dfrac{1}{8} 8 1 4 π 8 = π 2 \dfrac{4\pi}{8}=\dfrac{\pi}{2} 8 4 π = 2 π Jacobian is r r r d u d v = r d r d θ dudv=rdrd\theta d u d v = r d r d θ
∬ Q E G − F 2 d u d v = ∫ θ = 0 π / 2 ∫ r = 0 1 E G − F 2 r d r d θ = ∫ θ = 0 π / 2 ∫ r = 0 1 ( 1 − r 2 cos 2 θ ) ( 1 − r 2 sin 2 θ ) − r 4 cos 2 θ sin 2 θ 1 − r 2 r d r d θ = ∫ θ = 0 π / 2 ∫ r = 0 1 − r 2 cos 2 θ − r 2 sin 2 θ + 1 1 − r 2 r d r d θ = ∫ θ = 0 π / 2 d θ ∫ r = 0 1 1 − r 2 1 − r 2 r d r = π 2 ∫ r = 0 1 r 1 − r 2 d r = π 2 [ − 1 − r 2 ] 0 1 = π 2
\begin{align*}
& \iint _{Q} \sqrt{EG-F^{2}} du dv
\\ =&\ \int \limits_{\theta = 0}^{\pi / 2} \int \limits_{r=0}^{1} \sqrt{EG-F^{2}} rdr d\theta
\\ =&\ \int \limits_{\theta = 0}^{\pi / 2} \int \limits_{r=0}^{1} \dfrac{\sqrt{(1-r^{2}\cos^{2}\theta)(1-r^{2}\sin^{2}\theta)-r^{4}\cos^{2}\theta \sin^{2}\theta}}{1-r^{2}} rdr d\theta
\\ =&\ \int \limits_{\theta = 0}^{\pi / 2} \int \limits_{r=0}^{1} \dfrac{\sqrt{-r^{2}\cos^{2}\theta - r^{2}\sin^{2}\theta + 1}}{1-r^{2}} rdr d\theta
\\ =&\ \int _{\theta = 0}^{\pi / 2}d\theta \int _{r=0}^{1} \dfrac{\sqrt{1- r^{2}}}{1-r^{2}} rdr
\\ =&\ \dfrac{\pi}{2} \int _{r=0}^{1} \dfrac{r}{\sqrt{1-r^{2}}}dr
\\ =&\ \dfrac{\pi}{2} \left[ -\sqrt{1-r^{2}} \right]_{0}^{1}
\\ =&\ \dfrac{\pi}{2}
\end{align*}
= = = = = = = ∬ Q EG − F 2 d u d v θ = 0 ∫ π /2 r = 0 ∫ 1 EG − F 2 r d r d θ θ = 0 ∫ π /2 r = 0 ∫ 1 1 − r 2 ( 1 − r 2 cos 2 θ ) ( 1 − r 2 sin 2 θ ) − r 4 cos 2 θ sin 2 θ r d r d θ θ = 0 ∫ π /2 r = 0 ∫ 1 1 − r 2 − r 2 cos 2 θ − r 2 sin 2 θ + 1 r d r d θ ∫ θ = 0 π /2 d θ ∫ r = 0 1 1 − r 2 1 − r 2 r d r 2 π ∫ r = 0 1 1 − r 2 r d r 2 π [ − 1 − r 2 ] 0 1 2 π
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