Projection onto the tangent plane

https://math.stackexchange.com/questions/2275019/geodesic-curvature-and-projection-onto-the-tangent-plane

https://math.stackexchange.com/questions/578680/projection-of-a-curve-on-a-plane

Let’s have a unit speed parameters sphere surface \[\bar r = \{r \cos(\theta/r),r\sin(\theta/r), \sqrt{a^2 - r^2}\}\]

Sphere surface normal \[N = \{\frac{r}{a} \cos(\frac{\theta}{r}),\frac{r}{a}\sin(\frac{\theta}{r}), \frac{\sqrt{a^2 - r^2}}{a}\}\]

A small circle is curve \(\gamma\) on sphere \[\gamma(\theta) = \left\{r \cos\frac{\theta}{r}, \;r \sin\frac{\theta}{r}, \; \sqrt{a^2 - r^2}\right\}\]

We can have first derivative and second derivative of \(\gamma\) w.r.t \(\theta\) \[\gamma'(\theta) = \left\{- \sin\frac{\theta}{r}, \; \cos\frac{\theta}{r}, \; 0\right\}\] \[\gamma''(\theta) = \left\{-\frac{1}{r} \cos\frac{\theta}{r}, \;-\frac{1}{r} \sin\frac{\theta}{r}, \; 0\right\}\]

We need to choose a point \(p\) on sphere surface to have a tangent plane. And projection curve \(\gamma(\theta)\) onto tangent plane is \[\beta(s) = \gamma(s)+\langle \gamma(p)-\gamma(s),N(p)\rangle N(p)\]

Let \(p=\{r=r_0,\theta=\theta_0\}\)

gamma=c("r*cos(theta/r)","r*sin(theta/r)","sqrt(a^2-r^2)")
gammap=c("r0*cos(theta0/r0)","r0*sin(theta0/r0)","sqrt(a^2-r0^2)")
Normalp=c("r0 * cos(theta0/r0)/a", "r0 * sin(theta0/r0)/a", "sqrt(a^2 - r0^2)/a")

beta<-function(r,t,a,r0,t0,a0){
  theta0=t0
  theta=t
 gt=sapply(gamma,function(m) eval(parse(text=m)))
 gp=sapply(gammap,function(m) eval(parse(text=m)))
 Np=sapply(Normalp,function(m) eval(parse(text=m)))
 
 beta=gt+(gp-gt)%*%Np%*%Np
 return(beta)
}

For the equator curve at \(r=1,a=1\), we project it to the ceiling x-y plane, we see a circle of radius 1.

t=runif(100,0,100)
projection=sapply(t,function(m) beta(r=1,t=m,a=1,r0=.01,t0=pi/2,a0=1))
plot(x=projection[1,],y=projection[2,],xlim=c(-2,5),ylim=c(-2,2))

For the equator curve at \(r=1,a=1\), we project it to the side a plane perpendicular to the equator, the x-z plane, we see a line segment.

projection=sapply(t,function(m) beta(r=1,t=m,a=1,r0=1,t0=pi/2,a0=1))
plot(x=projection[1,],y=projection[3,],xlim=c(-2,5),ylim=c(-2,2))

Geodesic curvature from projection

Say we have a sphere of radius 3, and there is a smaller circle with radius 2 on the sphere. The sphere has curvature \(1/3\) and the small circle has curvature \(1/2\). The geodesic curvature is just \(\sqrt{(\frac{1}{2})^2-(\frac{1}{3})^2}=0.372678\) by relation \[k^2=k_n^2+k_g^2\]

In this example we went over unit speed parameters to get geodesic curvature https://rpubs.com/lizhi1800/Diff_Geom11 \[k_g = \gamma'' \cdot (\mathbf{N}\times \gamma ') =\frac{h}{ra} = \frac{\sqrt{a^2 - r^2}}{ra}\]

r=2;a=3;t=1
sqrt(1/4-1/9)
## [1] 0.372678
sqrt(a^2-r^2)/r/a
## [1] 0.372678

From the projected curve \(\beta(s)\) we take first and second derivatives. The second derivative is the so called covariant derivative \(D\alpha '/ds\). \[\beta'(s) = \alpha'(s) - \langle \alpha'(s),N(p)\rangle N(p)\], \[\beta''(s) = \alpha''(s) - k_n(\alpha'(s)) N(p)=\alpha''(s) - \langle \alpha ''(s),N(p)\rangle N(p)\] Notation \(k_n(\alpha'(s))\) is surface normal curvature in \(\alpha'\) direction.

Geodesic curvature is computed as \[k_g(p) = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\cos \theta_2\] where \(\theta_1 = \angle(\alpha'(0), (D\alpha'/{\rm d}s)(0))\) and \(\theta_1+\theta_2 = \pi/2\)

gamma1=c("-sin(theta/r)","cos(theta/r)","0")
gamma2=c("-1/r*cos(theta/r)","-1/r*sin(theta/r)","0")
beta2<-function(r,t,a,r0,t0,a0){
  theta0=t0
  theta=t
 g1=sapply(gamma1,function(m) eval(parse(text=m)))
 g2=sapply(gamma2,function(m) eval(parse(text=m)))
 Np=sapply(Normalp,function(m) eval(parse(text=m)))
 
 b2=g2-g2%*%Np%*%Np
 ang=b2%*%g1/sqrt(sum(g1^2))/sqrt(sum(b2^2))
 ang=acos(ang)
 return(c(b2,ang))
}

out=beta2(r=2,t=pi/2,a=3,r0=2,t0=pi/2,a0=3)
coD=out[1:3]
theta1=out[4]
theta2=pi/2-theta1

sqrt(sum(coD^2))*cos(theta2)#kg
## [1] 0.372678

Curvature of the projected curve is \[k_{\beta}(p) = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\sin \theta_1\]

sqrt(sum(coD^2))*sin(theta1)#k beta
## [1] 0.372678

And we can compute the \(k_g,k_{\beta}\) close to the equator. Equator is a geodesic, we get a \(k_g\) close to 0. Projection equator to x-z plane is a line segment, so \(k_{\beta}\) is close to 0.

out=beta2(r=2.999,t=pi/2,a=3,r0=2.999,t0=pi/2,a0=3)
coD=out[1:3]
theta1=out[4]
theta2=pi/2-theta1

sqrt(sum(coD^2))*cos(theta2)#kg
## [1] 0.008608782
sqrt(sum(coD^2))*sin(theta1)#k beta
## [1] 0.008608782

Covariant derivative and geodesic equations

Covariant derivative and geodesic equations are like that we want to keep things in 2D without considering 3D embedding space.

Sphere is a 2 dimensional surface, and tangent plane is also a 2 dimensional surface. We want to project a vector from the 2D sphere surface onto the 2D tangent space \(T_pS\), without considering the embedding 3D Euclidean space. The vector is usually a tangent vector of a surface curve. Here is the formula, let \(X^i\) be a vector field along a curve \(x^i(t)\). Covariant derivative is computed as \[\frac{DX^i}{dt}=\frac{dX^i}{dt}+\Gamma^i_{pq}X^p\frac{dx^q}{dt}\] I really would convert the 2D curve into 3D, and project onto tangent plane in 3D, and then convert it back to 2D. But the text book has a sophisticated proof for the formula.

The sphere surface is \[\bar r=\{x=a\cos\theta \sin\phi,y=a\sin\theta \sin\phi,z=a\cos\phi\}\]

Unit speed parameter small circle on the sphere surface is \[\gamma = \{r \cos(t/r),r\sin(t/r), \sqrt{a^2 - r^2}\}\] The small circle expressed in term of surface coordinates is \[x^i(t)=\{\theta=t/r,\phi=\arcsin(r/a)\}\]

We use the curve tangent vector as vector field \(X^i\) along the curve, so that we are doing a covariant derivative of a surface curve, otherwise it is called geodesic equations. So we have \[X^i=\frac{dx^i}{dt}=\{\frac{1}{r},0\}\] \[X'^i =\frac{d^2x^i}{dt^2}=\{0,0\}\]

Gammaphi=c("-cos(phi)  * sin(phi)","0","0","0")
Gammaphi=matrix(Gammaphi,nrow=2)
Gammatheta=c("0","cos(phi)/sin(phi)","cos(phi)/sin(phi)","0")
Gammatheta=matrix(Gammatheta,nrow=2)

\[\frac{DX^1}{dt}=\frac{d^2x^1}{dt^2}+\Gamma^{\phi}_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}=0+\left[ {\begin{array}{cc} \frac{1}{r} \\ 0 \\ \end{array} } \right]^T \left[ {\begin{array}{cc} -\cos(\phi) \sin(\phi) & 0 \\ 0 & 0 \\ \end{array} } \right] \left[ {\begin{array}{cc} \frac{1}{r} \\ 0 \\ \end{array} } \right]=-\cos(\phi) \sin(\phi)/r^2\] \[\frac{DX^2}{dt}=\frac{d^2x^2}{dt^2}+\Gamma^{\theta}_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}=0+\left[ {\begin{array}{cc} \frac{1}{r} \\ 0 \\ \end{array} } \right]^T \left[ {\begin{array}{cc} 0 & \cos(\phi)/\sin(\phi) \\ \cos(\phi)/\sin(\phi) & 0 \\ \end{array} } \right] \left[ {\begin{array}{cc} \frac{1}{r} \\ 0 \\ \end{array} } \right] =0\]

There must be a way to convert geodesic equations to curvature. I’m guessing it is called Riemann curvature. So to be continued.